Answer:
The water would act as a base and would produce an undesired product of ethanol (CH3OH) through a dissociation reaction. If doing a titration reaction, it will likely yield inaccurate results.
Which one of the following reactions must be carried out in an electrolytic cell rather than in an electrochemical cell?
a. Zn2+ + Ca → Zn + Ca2+
b. Al3+ + 3Br– → Al + (3/2)Br2
c. 2Al + 3Fe2+ → 2Al3+ + 3Fe
d. H2 + I2(s) → 2H+ + 2I–
Answer:
b. Al3+ + 3Br– → Al + (3/2)Br2
Explanation:
If we look at this reaction closely, aluminum was reduced while bromine was oxidized. The reduction potential of aluminum is -1.66 V while that of bromine is + 1.087. Recall that the more negative the redox potential of a chemical specie, the greater its tendency to function as a reducing agent donating electrons in an electrochemical reaction.
However, in this reaction, aluminium is found to accept rather than donate electrons. Therefore, the process is non spontaneous and can only occur in an electrolytic cell, hence the answer.
b. [tex]Al^{3+} + 3 Br^{-}[/tex] → [tex]Al + \frac{3}{2} Br_{2}[/tex]
Electrolytic Cell v/s Electrochemical Cell:
Electrochemical cells convert chemical energy into electrical energy or vice versa while an electrolytic cell is a type of electrochemical cell in which electrical energy is converted into chemical energy.Electrochemical cells consist of a cathode (+) and an anode(-) while Electrolytic cells consist of a positively charged anode and a negatively charged cathode.Out of the given reactions, [tex]Al^{3+} + 3 Br^{-}[/tex] → [tex]Al + \frac{3}{2} Br_{2}[/tex] is carried out in an electrolytic cell rather than in an electrochemical cell.
As we know, In electrolytic cells, like galvanic cells, are composed of two half-cells
one is a reduction half-cell, the other is an oxidation half-cell.In the given reaction, aluminum is being reduced while bromine gets oxidized and the value of reduction potential of aluminum is -1.66 V while that of bromine is + 1.087. Therefore, the process is non spontaneous and can only occur in an electrolytic cell.
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When 50.0 mL of 1.27 M of HCl(aq) is combined with 50.0 mL of 1.32 M of NaOH(aq) in a coffee-cup calorimeter, the temperature of the solution increases by 8.49°C. What is the change in enthalpy for this balanced reaction? HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) Assume that the solution density is 1.00 g/mL and the specific heat capacity of the solution is 4.18 J/g⋅°C. Hint: You need to determine the limiting reagent.
Answer:
-55.9kJ/mol is the change in enthalpy of the reaction
Explanation:
In the reaction:
HCl(aq) + NaOH(aq) → H₂O(l) + NaCl
Some heat is released per mole of reaction.
To know how many moles reacts we need to find limiting reactant:
Moles HCl = 0.050L ₓ (1.27mol / L) = 0.0635 moles HCl
Moles NaOH = 0.050L ₓ (1.32mol / L) = 0.066 moles NaOH
As there are more moles of NaOH than moles of HCl, HCl is limiting reactant and moles of reaction are moles of limiting reactant, 0.0635 moles
Using the coffee-cup calorimeter equation we can find how many heat was released thus:
Q = C×m×ΔT
Where Q is heat released, C is specific heat of the solution (4.18J/g°C), m is mass of solution (100g because there are 100mL of solution -50.0mL of HCl and 50.0mL of NaOH- and density is 1g/mL) and ΔT is change in temperature (8.49°C)
Replacing:
Q = 4.18J/g°C×100g×8.49°C
Q = 3548.8J of heat are released in the reaction
Now, change in enthalpy, ΔH, is equal to change in heat (As is released heat ΔH < 0) per mole of reaction, that is:
ΔH = Heat / mol of reaction
ΔH = -3548.8J / 0.0635 moles of reaction
Negative because is released heat.
ΔH = -55887J / mol
ΔH =
-55.9kJ/mol is the change in enthalpy of the reaction
The heat of reaction is -54.7 kJ/mol.
The equation of the reaction is;
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Number of moles of HCl = 50/1000 L × 1.27 M = 0.064 moles
Number of moles of NaOH = 50/1000 L × 1.32 M = 0.066 moles
The limiting reactant is HCl
Total volume of solution = 100mL
Total mass of solution = 100 g
Temperature rise = 8.49°C
Heat capacity of solution = 4.18 J/g⋅°C
Using;
H = mcdT
m = mass of solution
c = heat capacity of solution
dT = temperature rise
H = 100 g × 4.18 J/g⋅°C × 8.49°C = 3548.82 J
The heat of reaction = -ΔH/n = -(3.5kJ/0.064 moles)
= -54.7 kJ/mol
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When a solution is diluted with water, the ratio of the initial to final
volumes of solution is equal to the ratio of final to initial molarities
Select one:
True
-
When a solution is diluted with water, the ratio of the initial to final volumes of solution is equal to the ratio of final to initial molarities. The statement is True.
Concentration refers to the amount of a substance in a defined space. Another definition is that concentration is the ratio of solute in a solution to either solvent or total solution.
There are various methods of expressing the concentration of a solution.
Concentrations are usually expressed in terms of molarity, defined as the number of moles of solute in 1 L of solution.
Solutions of known concentration can be prepared either by dissolving a known mass of solute in a solvent and diluting to a desired final volume or by diluting the appropriate volume of a more concentrated solution (a stock solution) to the desired final volume.
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2) Os foguetes são utilizados para levar pessoas ao espaço (os astronautas), mas principalmente cargas como, por exemplo, os satélites artificiais, os telescópios espaciais, levar sondas a outros planetas etc. Escreva V(verdadeiro) ou F (falso) em cada afirmação.
( ) Foguetes só levam astronautas ao espaço.
( ) Satélites artificiais servem para ajudar na previsão do clima.
( ) Satélites artificiais "fotografam" o planeta para descobrir queimadas ilegais.
( ) Satélites artificiais permitem vermos jogos ao vivo até do Japão.
( ) Foguetes são movidos com pólvora e dinamite.
Answer:
F, V, V , V, F
Explanation:
1 - "Os foguetes são utilizados para levar pessoas ao espaço (os astronautas), mas principalmente cargas como, por exemplo, os satélites artificiais, os telescópios espaciais, levar sondas a outros planetas etc".
2 - Tipo Meteorologia: utilizados para monitorar o tempo e o clima no planeta Terra, por exemplo, os da série Meteosat.
3 - ...
4 - ...
5 - Usam combustivel solido, liquido, hibridos (solido e liquido), iônica:
Solido:
São sistemas simples que unem os dois propelentes envolvidos em uma massa sólida que, quando inflamada, não para de queimar até o esgotamento completo.
Liquido:
São muito mais complexos e envolvem o bombeamento de quantidades imensas de propelentes para as câmaras de combustão dos motores.
Hibridos:
O propelente sólido – normalmente o combustível – é distribuído ao longo do tanque de maneira homogênea. O propelente líquido ou gasoso "normalmente o oxidante" fica armazenado em tanques.
Podem ser desligados depois de sofrerem ignição, além de permitirem um controle de queima relativamente preciso.
Iônica:
Usando eletricidade (captada por painéis solares ou gerada por reatores atômicos) para ionizar átomos (normalmente gases nobres, como xenônio), e expulsá-los em velocidades altíssimas.
Osmosis is the process responsible for carrying nutrients and water from groundwater supplies to the upper parts of trees. The osmotic pressures required for this process can be as high as 19.1 atm . What would the molar concentration of the tree sap have to be to achieve this pressure on a day when the temperature is 32 ∘C ? Express your answer to three significant figures and include the appropriate units. View Available Hint(s)
Answer:
[tex]M=0.763\frac{mol}{L}=0.763M[/tex]
Explanation:
Hello,
In this case, as the osmotic pressure (π) is widely known as a colligative property, we can see that the solution in this case is formed by water and tree sap, that is mathematically defined by:
[tex]\pi =iMRT[/tex]
Thus, since tree sap is a covalent substance that is nonionizing, we can infer its van't Hoff factor to be 1, therefore, for the given osmotic pressure and temperature, we can compute the molar concentration (in molar units mol/L) as follows:
[tex]M=\frac{\pi }{RT} =\frac{19.1atm}{0.082\frac{atm*L}{mol*K}*(32+273.15)K} \\\\M=0.763\frac{mol}{L}=0.763M[/tex]
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Give the name of the following molecule
Answer:
[tex]\boxed{Heptene}[/tex]
Explanation:
Double Bond => An Alkene molecule
So, the suffix will be "-ene"
7 Carbons => So, we'll use the prefix "Hept-"
Combining the suffix and prefix, we get:
=> Heptene
Answer:
[tex]\boxed{\mathrm{Heptene}}[/tex]
Explanation:
Alkenes have double bonds. The molecule has one double bond.
Suffix ⇒ ene
The molecule has 7 carbon atoms and 14 hydrogen atoms.
Prefix ⇒ Hept (7 carbons)
The molecule is Heptene.
[tex]\mathrm{C_7H_{14}}[/tex]
Arrange the following oxides in order of increasing acidity.
Rank from least acidic to most acidic. To rank items as equivalent,overlap them.
CaO
P2O5
SO3
SiO2
Al2O3
CO2
Answer:
Based on the Modern Periodic table, there is an increase in the electropositivity of the atom down the group as well as increases across a period. On comparing the electropositivities of the mentioned oxides central atom, it is seen that Ca is most electropositive followed by Al, Si, C, P, and S is the least electropositive.
With the decrease in the electropositivity, there is an increase in the acidity of the oxides. Thus, the increasing order of the oxides from the least acidic to the most acidic is:
CaO > Al2O3 > SiO2 > CO2 > P2O5 > SO3. Hence, CaO is the least acidic and SO3 is the most acidic.
Since acidity increases across a period from left to right, and decreases down a group, the oxides can be ranked from the least acidic to the most acidic as follows:
[tex]\mathbf{CaO < Al_2O_3 < SiO_2 < CO_2 < P_2O_5 <SO_3}[/tex]
The least acidic is CaOThe most acidic is [tex]SO_3[/tex]Note the following:
Acidity of an oxide depends on its electronegativity.Non-metals are more electronegative, while metals are less electronegative.Acidity of oxides increases across a period as you move from left to the right side of a periodic table.Acidity of oxides decreases down a group (column) in a periodic table.Using the periodic table diagram given in the attachment below, we can rank the given oxides according to their increasing acidity.
CaO, is the least, because it is an oxide of the metal, Calcium, which is at the far left in group 2 in the periodic table.The next is, [tex]Al_2O_3[/tex]. Aluminum is a metal from group 3.[tex]SiO_2[/tex] is an oxide of Silicon, also in group 4 but below Carbon.[tex]CO_2[/tex] is an oxide of Carbon, from group 4.
[tex]P_2O_5[/tex] is an oxide of the non-metal, Phosphorus, a group 5 element
[tex]SO_3[/tex] is an oxide of the non-metal, Sulphur, a group 6 element.
Therefore, since acidity increases across a period from left to right, and decreases down a group, the oxides can be ranked from the least acidic to the most acidic as follows:
[tex]\mathbf{CaO < Al_2O_3 < SiO_2 < CO_2 < P_2O_5 <SO_3}[/tex]
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If powdered platinum metal is used to speed up the following reaction: Cl2(g) 3F2(g) --> 2ClF3(g), what would you classify the platinum as
Answer:
Catalyst
Explanation:
For the reaction:
[tex]Cl_2_(_g_)~+~3F_2_(_g_)->2ClF_3_(_g_)[/tex]
We have a main observation: When platinum is added the reaction goes faster. With this in mind, we have to remember the kinetic equilibrium theory. In figure 1, we have an energy diagram. In which we have an specific energy for the reagents and the products. When the reaction takes place, the reaction has to must go through an energy peak. This energy peak is called "activation energy". When platinum is added the activation energy decreases and the reaction can go faster. Therefore, platinum is a "catalyst", a substance with the ability to reduce the activation energy.
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Night vision glasses detect
energy emitted from cooling objects?
ultraviolet
infrared
X-ray
Answer:
I think the answer is " Night vision glasses detect Infrared" energy emitted from cooling objects.
Explanation:
Treatment of 1 mole of dimethyl sulfate with 2 moles of sodium acetylide results in the formation of propyne as the major product.
A) Draw a reasonable mechanism accounting for the formation of the byproduct 2-butyne.
B) 2-Butyne is observed as a minor product of this reaction. Draw a mechanism accounting for the formation of this minor product and explain how your proposed mechanism is consistent with the observation that acetylene is present among the reaction products.
C) Predict the major and minor products that are expected if diethyl sulfate is used in place of dimethyl sulfate.
Answer:
(a) appended underneath is the inorganic ion shaped in the reaction and the mechanism of its formation
(b) 2-butyne framed as a minor product is appeared in the connection. It is shaped when the monosodium subordinate of dimethylsulphoxide gets a hydrogen from the propyne and reacts again with monosodium methylsulphoxide.
(c) The major product framed when diethylsulphoxide is utilized, would be butyne and minor product would be 3-hexyne.
Explanation:
attached below is diagram
What class of organic product results when 1-heptyne is treated with a mixture of mercuric acetate (HgSO4) in aqueous sulfuric acid (H2O/H2SO4)
Answer:
heptan-2-one
Explanation:
In this case, the final product would be a ketone: heptan-2-one. To understand why this molecule is produced we have to check the reaction mechanism.
The first step is the protonation of the triple bond to produce the more stable carbocation (a secondary one) by the action of sulfuric acid [tex]H_2SO_4[/tex]. The next step is the attack of water to the carbocation to produce a new bond between C and the O, producing a positive charge in the oxygen. Then, a deprotonation step takes place to produce an enol. Finally, we will have a rearrangement (keto-enol tautomerism) to produce the final ketone.
See figure 1
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A 1.0 kg object absorbs 1,303 J of heat energy and experiences a temperature increase of 5.2∘C. What is the object’s specific heat, in joules per gram-degree celsius? Report your answer with the correct number of significant figures.
Answer:
c = 250.58 J/kg/[tex]^{0}C[/tex]
Explanation:
The specific heat of a substance is the required quantity of heat to increase or decrease the temperature of its unit mas by 1 kelvin.
Q = mcΔθ
where: Q is the quantity of heat absorbed or released, m is the mass of the substance, c is its specific heat and Δθ is the change in temperature of the substance.
Given that; m = 1.0 kg, Q = 1303 J and Δθ = 5.2 [tex]^{0}C[/tex], then;
c = Q ÷ (mΔθ)
= 1303 ÷ (1.0 × 5.2)
= 1303 ÷ 5.2
= 250.58 J/kg/[tex]^{0}C[/tex]
The specific heat of the object is 250.58 J/kg/[tex]^{0}C[/tex].
Answer:
0.25
Explanation:
What is an ideal gas?
Answer:
a hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the gas laws exactly.
Find the molecular formula of each compound CCl , 189.83 g/mol C3H2N , 156.23 g/mol
Answer:
Explanation:
The given formula is empirical formula
Let the molecular formula of first be
[tex]( CCl )_n[/tex]
molecular weight = n x ( 12 + 35.5 )
= 47.5 n
Given molecular weight = 189.83 so
47.5 n = 189.83
n = 3.99 or 4 approx
Molecular formula =
[tex]( CCl )_4[/tex]
= C₄ Cl₄
Let the molecular formula of second compound be
[tex]( C_3H_2N)_n[/tex]
molecular weight = n x ( 3 x 12 +2+14 )
= 52 n
Given molecular weight = 156.23 so
52 n = 156.23
n = 3.0044 or 3 approx
Molecular formula =
[tex]( C_3H_2N )_3[/tex]
= C₉H₆ N₃
Cyclohexane (C6H12) undergoes a molecular rearrangement in the presence of AlCl3 to form methylcyclopentane (CH3C5H9) according to the equation: C6H12 ⇌ CH3C5H9 If Kc = 0.143 at 25°C for this reaction predict the direction in which the system
Answer:
The reaction will shift leftwards, towards the formation of more cyclohexane at 25 °C
Explanation:
Hello,
In this case, for the given chemical reaction, we can write the law of mass action (equilibrium expression) as shown below:
[tex]Kc=\frac{[CH_3C_5H_9]}{[C_6H_{12} ]}[/tex]
Thus, since Kc < 1, we can conclude there are more moles of cyclohexane at equilibrium (denominator is greater than numerator), therefore, the reaction will shift leftwards, towards the formation of more cyclohexane at 25 °C.
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True or False
1. Density is considered a chemical (i.e., not a physical) property. TRUE FALSE
2. When naming an ionic compound containing a transition element such as iron (Fe), the name must include a Roman numeral to indicate the charge of the metal ion. TRUE FALSE
3. The neutron was discovered about 20 years after the electron and proton because it has no charge (in order for it to be detected). TRUE FALSE
4. When we balance a chemical equation, we are observing the law of conservation of mass as well as the part of Dalton’s theory that atoms are neither created or destroyed in a chemical reaction TRUE FALSE
5. When a gas is heated up in a closed container, the kinetic energy of the molecules or atoms of the gas increase, which leads to a decrease in the pressure of the gas. TRUE FALSE
6. The amount of enthalpy (heat energy) for a reaction is directly proportional to the amount (number of moles or grams) of the reactants. TRUE FALSE
7. The combined gas law works for any gas (i.e., you do not need to know the chemical formula). TRUE FALSE
8. A balloon with 10.0 g of CO2 gas will have more molecules than a 10.0 g sample of NO gas. TRUE FALSE
9. Unless a sample is at absolute zero (kelvins), the particles in the sample will have kinetic energy and have some kind of motion. TRUE FALSE
Answer:
1. False
2. True
3. True
4. True
5. True
6. True
7. True
8. False
9. True
Explanation:
Density is a physical property since its measurement does not involve any chemical process.
Since transition elements exhibit variable oxidation states, the actual oxidation state of the transition element must be specified in the compound.
Due to the fact that neutron has no charge, it was discovered by Chadwick long after the electron and proton were discovered.
The balancing of chemical reaction equations is a demonstration that atoms are neither created no destroyed. It also shows that mass is neither created nor destroyed in chemical reactions.
When a gas is heated, it expands. Its volume and its kinetic energy increases. Since volume and pressure are inversely proportional (Boyle's law) the pressure decreases.
Enthalpy is said to be an extensive property. This implies that the magnitude of change in enthalpy is known to depend on the amount of reactants that is actually reacted.
The combined gas law is applicable to all ideal gas systems irrespective of their individual chemical formulas.
10g of CO2 contains 0.227 moles of CO2 while 10g of NO contains 0.33 moles of NO hence 10.0 g of NO will contain more molecules than 10.0g of CO2.
If a sample is not at absolute zero, the particles are known to possess kinetic energy which decreases continuously until absolute zero is attained.
The table below shows the electronegativity values of various elements on the periodic table. Electronegativities A partial periodic table. Which pair of atoms would form a covalent bond ? calcium (Ca) and bromine (Br) rubidium (Rb) and sulfur (S) cesium (Cs) and nitrogen (N) oxygen (O) and chlorine (Cl)
Answer:
Oxygen and Chlorine
Explanation:
Covalent bonds involve the sharing of electrons between nonmetals.
Answer:
oxygen (O) and chlorine (Cl)
Explanation:
cuz i said so
An aqueous solution containing 5.06 g of lead(II) nitrate is added to an aqueous solution containing 6.03 g of potassium chloride.Enter the balanced chemical equation for this reaction. Be sure to include all physical states.balanced chemical equation:What is the limiting reactant?lead(II) nitratepotassium chlorideThe percent yield for the reaction is 82.9% . How many grams of the precipitate are formed?precipitate formed:gHow many grams of the excess reactant remain?excess reactant remaining:
Answer:
Pb(NO3)2(aq) + 2KCl(aq) ------> 2KNO3(aq) + PbCl2(s)
3.52 g of PbCl2
3.76 g of KCl
Explanation:
The equation of the reaction is;
Pb(NO3)2(aq) + 2KCl(aq) ------> 2KNO3(aq) + PbCl2(s)
Number of moles of Pb(NO3)2 =mass/molar mass 5.06g/331.2 g/mol = 0.0153 moles
Number of moles of KCl= mass/ molar mass= 6.03g/74.5513 g/mol= 0.081 moles
Next we obtain the limiting reactant; the limiting reactant yields the least number of moles of products.
For Pb(NO3)2;
1 mole of Pb(NO3)2 yields 1 mole of PbCl2
Therefore 0.0153 moles of Pb(NO3)2 yields 0.0153 moles of PbCl2
For KCl;
2 moles of KCl yields 1 mole of PbCl2
0.081 moles of KCl yields 0.081 moles ×1/2 = 0.041 moles of PbCl2
Therefore Pb(NO3)2 is the limiting reactant.
Theoretical Mass of precipitate obtained = 0.0153 moles of PbCl2 × 278.1 g/mol = 4.25 g of PbCl2
% yield = actual yield/theoretical yield ×100
Actual yield = % yield × theoretical yield /100
Actual yield= 82.9 ×4.25/100
Actual yield = 3.52 g of PbCl2
If 1 mole of Pb(NO3) reacts with 2 moles of KCl
0.0153 moles of Pb(NO3)2 reacts with 0.0153 moles × 2 = 0.0306 moles of KCl
Amount of excess KCl= 0.081 moles - 0.0306 moles = 0.0504 moles of KCl
Mass of excess KCl = 0.0504 moles of KCl × 74.5513 g/mol = 3.76 g of KCl
Calculate the amount of energy absorbed by 45.0 g sample of water to raise its temperature from 18.0C to 48.0 C. The specific heat of water is 4.18 J/g C. 1000 J= 1kj
Answer:
5.643 kJ
Explanation:
The quantity of heat released or absorbed by a substance (Q) is given by the equation:
Q = mcΔT
Where m is the mass of the substance, c is the specific heat of substance and ΔT is the difference between the final temperature and the initial temperature.
Given that:
m = 45 g, Final temperature = 48°C, Initial temperature = 18°C, c = specific heat of water = 4.18 J/g°C
ΔT = Final temperature - Initial temperature = 48°C - 18°C = 30°C
The quantity of heat is:
Q = mcΔT = 45 g × 4.18 J/g°C × 30°C = 5643 J
Q = 5.643 kJ
what is 1 +1 (a) 11 (b) 3 (c) 6 (d) 2
Answer:
(d) 2
Explanation:
Lets say that there are 2 apples or 1+1 if you count them you would do 1,2 so 2 would be the final answer
Suppose you titrate 25.00 mL of 0.200 M KOBr with 0.200M H2SO4. The pH at half-equivalence point is 7.75 a). What is the initial pH of the 25.00mL of 0.200M KOBr mentioned above
Answer:
Approximately [tex]10.88[/tex].
Explanation:
Equilibrium constant[tex]\rm OBr^{-}[/tex] can act as a weak Bronsted-Lowry base:
[tex]\rm OBr^{-}\; (aq) + H_2O\; (l) \rightleftharpoons HOBr\; (l) + OH^{-}\; (aq)[/tex].
(Side note: the state symbol of [tex]\rm HOBr[/tex] in this equation is [tex]\rm (l)[/tex] (meaning liquid) because [tex]\rm HOBr[/tex] is a weak acid.)
However, the equilibrium constant of this reaction, [tex]K_\text{eq}[/tex], isn't directly given. The idea is to find [tex]K_\text{eq}[/tex] using the [tex]\rm pH[/tex] value at the half-equivalence point. Keep in mind that this system is at equilibrium all the time during the titration. If temperature stays the same, then the same [tex]K_\text{eq}[/tex] value could also be used to find the [tex]\rm pH[/tex] of the solution before the acid was added.
At equilibrium:
[tex]\displaystyle K_\text{eq} = \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]}[/tex].
At the half-equivalence point of this titration, exactly half of the base, [tex]\rm OBr^{-}[/tex], has been converted to its conjugate acid, [tex]\rm HOBr[/tex]. Therefore, the half-equivalence concentration of [tex]\rm OBr^{-}[/tex] and [tex]\rm HOBr[/tex] should both be equal to one-half the initial concentration of [tex]\rm OBr^{-}[/tex].
As a result, the half-equivalence concentration of [tex]\rm OBr^{-}[/tex] and [tex]\rm HOBr[/tex] should be the same. The expression for [tex]K_\text{eq}[/tex] can thus be simplified:
[tex]\begin{aligned}& K_\text{eq} \\&= \frac{\left(\text{half-equivalence $[\rm HOBr\; (l)]$}\right)\cdot \left(\text{half-equivalence $[\rm OH^{-}\; (aq)]$}\right)}{\text{half-equivalence $[\rm OBr^{-}\; (l)]$}}\\ &=\text{half-equivalence $[\rm OH^{-}\; (aq)]$}\end{aligned}[/tex].
In other words, the [tex]K_\text{eq}[/tex] of this system is equal to the [tex]\rm OH^{-}[/tex] concentration at the half-equivalence point. Assume that [tex]\rm p\mathnormal{K}_\text{w}[/tex] the self-ionization constant of water, is [tex]14[/tex]. The concentration of [tex]\rm OH^{-}[/tex] can be found from the [tex]\rm pH[/tex] value:
[tex]\begin{aligned}& \text{half-equivalence $[\rm OH^{-}\; (aq)]$} \\ &= 10^{\rm pH - p\mathnormal{K}_\text{w}}\;\rm mol \cdot L^{-1} \\ &= 10^{7.75 - 14}\; \rm mol \cdot L^{-1}\\ &= 10^{-6.25}\; \rm mol \cdot L^{-1}\end{aligned}[/tex].
Therefore, [tex]\begin{aligned} K_\text{eq} &= 10^{-6.25}\end{aligned}[/tex].
Initial pH of the solutionAgain, since [tex]\rm KOBr[/tex] is a soluble salt, all that [tex]0.200\; \rm M[/tex] of [tex]\rm KOBr[/tex] in this solution will be in the form of [tex]\rm K^{+}[/tex] and [tex]\rm OBr^{-}[/tex] ions. Before any hydrolysis takes place, the concentration of [tex]\rm OBr^{-}[/tex] should be equal to that of [tex]\rm KOBr[/tex]. Therefore:
[tex]\text{$[\rm OBr^{-}\; (aq)]$ before hydrolysis} = 0.200\; \rm M[/tex].
Let the equilibrium concentration of [tex][\rm OH^{-}\; (aq)][/tex] be [tex]x\; \rm M[/tex]. Create a RICE table for this reversible reaction:
[tex]\begin{array}{c|ccccccc} & \rm OBr^{-}\; (aq) &+&\rm H_2O\; (l)& \rightleftharpoons & \rm HOBr\; (l)& + & \rm OH^{-}\; (aq) \\ \textbf{I}& 0.200\; \rm M & & & & 0 \; \rm M & & 0\; \rm M \\ \textbf{C} & -x\; \rm M & & & & +x \; \rm M & & +x\; \rm M \\ \textbf{E}& (0.200 + x)\; \rm M & & & & x \; \rm M & & x\; \rm M \end{array}[/tex].
Assume that external factors (such as temperature) stays the same. The [tex]K_\text{eq}[/tex] found at the half-equivalence point should apply here, as well.
[tex]\displaystyle K_\text{eq} = \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]}[/tex].
At equilibrium:
[tex]\displaystyle \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]} = \frac{x^2}{0.200 + x}[/tex].
Assume that [tex]x[/tex] is much smaller than [tex]0.200[/tex], such that the denominator is approximately the same as [tex]0.200[/tex]:
[tex]\displaystyle \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]} = \frac{x^2}{0.200 + x} \approx \frac{x^2}{0.200}[/tex].
That should be equal to the equilibrium constant, [tex]K_\text{eq}[/tex]. In other words:
[tex]\displaystyle \frac{x^2}{0.200} \approx K_\text{eq} = 10^{-6.25}[/tex].
Solve for [tex]x[/tex]:
[tex]x \approx 3.35\times 10^{-4}[/tex].
In other words, the [tex]\rm OH^{-}[/tex] before acid was added was approximately [tex]3.35\times 10^{-4}\; \rm M[/tex], which is the same as [tex]3.35\times 10^{-4}\; \rm mol \cdot L^{-1}[/tex]. Again, assume that [tex]\rm p\mathnormal{K}_\text{w} = 14[/tex]. Calculate the [tex]\rm pH[/tex] of that solution:
[tex]\begin{aligned}\rm pH &= \rm p\mathnormal{K}_\text{w} + \log [\mathrm{OH^{-}}] \approx 10.88\end{aligned}[/tex].
(Rounded to two decimal places.)
Predict the order of acid strengths in the following series of cationic
species: CH3CH2NH3
+, CH3CH=NH2
Answer:
CH3CH=NH2+>CH3CH2NH3+
Explanation:
If we look at the both species under review, we will realize that they are both amines hence they possess the polar N-H bond.
Electrons are ordinarily attracted towards the nitrogen atom hence making both compounds acidic. It is worthy of note that certain features of a compound may make it more acidic than another of close structural proximity. 'More acidic' simply means that the proton is more easily lost.
CH3CH=NH2+ contains an sp2 hybridized carbon atom which is highly electronegative and further withdraws electron density from the N-H bond thereby leading to a greater acidity of CH3CH=NH2+ compared to CH3CH2NH3+
Liquid octane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water . What is the theoretical yield of carbon dioxide formed from the reaction of of octane and of oxygen gas
Answer:
24.6g of CO₂ is theoretical yield
Explanation:
The reaction of 8.00g of octane with 38.9g of oxygen.
The reaction of octane with oxygen is:
C₈H₁₈(l) + 25/2O₂ → 9H₂O + 8CO₂
1 mole of octane reacts with 25/2 moles of oxygen to produce 8 moles of CO₂
Theoretical yield is the amount of carbon dioxide formed assuming a yield of 100%. To calculate theoretical yield, first, we need to find limiting reactant and, with the chemical reaction, we can obtain the theoretical moles of CO₂ produced and its mass to obtain theoretical yield.
Limiting reactant:
Moles octane (Molar mass: 114.23g/mol) in 8.00g:
8.00g × (1mol / 114.23g) = 0.0700 moles octane.
Moles oxygen (Molar mass: 32g/mol) in 38.9g:
38.9g × (1mol / 32g) = 1.2156 moles oxygen.
For a complete reaction of 1.2156 moles of O₂ there are necessaries:
1.2156 moles O₂ ₓ (1mol C₈H₁₈ / 25/2 moles O₂) = 0.0973 moles octane
As we have just 0.0700 moles,
octane is limiting reactant.Moles and mass of carbon dioxide:
As limiting reactant is octane, 0.0700 moles of C₈H₁₈ will produce:
0.0700mol C₈H₁₈ × (8 moles CO₂ / 1 mol C₈H₁₈) = 0.56 moles of CO₂ are theoretically produced. In mass (Molar mass CO₂ = 44.01g/mol):
0.56moles CO₂ × (44.01g / mol) =
24.6g of CO₂ is theoretical yield-Theoretical yield because we are assuming all octane is reacting. In real life, never happens like that-
A compound X has a molecular ion peak in its mass spectrum at m/z 136. What information does this tell us about X
Explanation:
The mass to charge ratio =136
The specific rotation of (S)-carvone (at 20°C) is +61. A chemist prepared a mixture of (R)-carvone and its enantiomer, and this mixture had an observed rotation of -55°.
A) What is the specific rotation of (R)-carvone at 20°C?
B) Calculate the % ee of this mixture.
C) What percentage of the mixture is (S)-carvone?
Answer:
a) Specific Rotation of (R)-carvone = -61°
b) The enantiomeric excess of (R)-carvone in the mixture = 90.2%
c) The percentage of (S)-carvone in the mixture = 4.9%
Explanation:
a) The specific rotation of the enantiomer of a substance is given simply as the negative of the specific rotation of that substance.
Hence, the specific rotation of (R)-carvone is simply the negative of the specific rotation of (S)-carvone.
Specific Rotation of (R)-carvone = -(61°) = -61°
b) Enantiometic excess is used to measure the optical purity of an enantiomeric mixture.
The enantiomeric excess is given mathematically as
ee% = (Observed rotation × 100)/(Specific rotation)
Hence, to calculate the enantiomeric excess of (R)-carvone,
Observed rotation of the mixture = -55°
Specific Rotation of (R)-carvone = -61°
ee% = (-55×100)/(-61) = 90.16% = 90.2%
c) An enantiomeric excess of 90.2% for (R)-carvone indicates that it's actual percentage is 90.2% more than the percentage of its enantiomeric partner, (S)-carvone, in the mixture.
Let the percentage of (R)-carvone in the mixture be x
Let the percentage of (S)-carvone in the mixture be y
x + y = 100
x - y = 90.2
2x = 190.2
x = (190.2/2) = 95.1%
y = 100 - x = 100 - 95.1 = 4.9%
Hence, the percentage of (R)-carvone in the mixture = 95.1%
The percentage of (S)-carvone in the mixture = 4.9%
Hope this Helps!!!
a) Specific Rotation of (R)-carvone = -61°
b) The enantiomeric excess of (R)-carvone in the mixture = 90.2%
c) The percentage of (S)-carvone in the mixture = 4.9%
a) Calculation of Specific Rotation:The specific rotation of the enantiomer of a substance is given simply as the negative of the specific rotation of that substance.
Hence, the specific rotation of (R)-carvone is simply the negative of the specific rotation of (S)-carvone.
Specific Rotation of (R)-carvone = -(61°) = -61°
b) Calculation for Enantiomeric excess:
The enantiomeric excess is given mathematically as
ee% = (Observed rotation × 100)/(Specific rotation)
Hence, to calculate the enantiomeric excess of (R)-carvone,
Observed rotation of the mixture = -55°
Specific Rotation of (R)-carvone = -61°
ee% = (-55×100)/(-61) = 90.16% = 90.2%
c) Calculation of percentage:
Let the percentage of (R)-carvone in the mixture be x
Let the percentage of (S)-carvone in the mixture be y
x + y = 100
x - y = 90.2
2x = 190.2
x = (190.2/2) = 95.1%
y = 100 - x = 100 - 95.1 = 4.9%
Hence, the percentage of (R)-carvone in the mixture = 95.1%
The percentage of (S)-carvone in the mixture = 4.9%
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How many grams are in 5.87 x 10^21 molecules of sulfur?
Answer:
0.312g
Explanation:
From Avogadro's hypothesis, 1mole of any substance contains 6.02x10^23 molecules. This means that 1mole of sulphur also contains 6.02x10^23 molecules
1mole of sulphur = 32g
If 1 mole(i.e 32g) of sulphur contains 6.02x10^23 molecules
Then, Xg of sulphur will contain 5.87x10^21 molecules i.e
Xg of sulphur = (32x5.87x10^21)/6.02x10^23 = 0.312g
just saying this is not my work, thank Eduard22sly
he answered it on a different page
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What do chemists use percent yield calculations for in the real world?
A. To balance the reaction equation.
B. To determine how much product they will need.
C. To determine how efficient reactions are.
D. To determine how much reactant they need.
Answer:
C. To determine how efficient reactions are.
D. To determine how much reactant they need.
Explanation:
When you are doing a reaction, you are hoping for a percent yield to close of 100%. You make the reaction and determine how many product you obtain. If you know the percent yield of a reaction you can calculate the amount of reactant you need to obtain a determined amount of product.
Having this in mind:
A. To balance the reaction equation. false. To calculate percent yield you need to balance the reaction before. You don't use percent yield to balance the reaction
B. To determine how much product they will need. false. You determine how much product you obtain after the reaction. How much product you need is independent of percent yield
C. To determine how efficient reactions are. true. A way to determine efficience of a reaction is with percent yield. An efficient reaction has a high percent yield.
D. To determine how much reactant they need. true. If you know percent yield of a reaction you can know how many reactant you must add to obtain the amount of product you want.
4 Al + 3O2 → 2Al2O3 If 14.6 grams Al are reacted, how many liters of O2 at STP would be required?
Answer: 9.08 L
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Al=\frac{14.6g}{27g/mol}=0.54moles[/tex]
[tex]4Al+3O_2\rightarrow 2Al_2O_3[/tex]
According to stoichiometry :
4 moles of [tex]Al[/tex] require = 3 moles of [tex]O_2[/tex]
Thus 0.54 moles of [tex]Al[/tex] will require=[tex]\frac{3}{4}\times 0.54=0.405moles[/tex] of [tex]O_2[/tex]
Standard condition of temperature (STP) is 273 K and atmospheric pressure is 1 atm respectively.
According to the ideal gas equation:
[tex]PV=nRT[/tex]
P = Pressure of the gas = 1 atm
V= Volume of the gas = ?
T= Temperature of the gas = 273 K
R= Gas constant = 0.0821 atmL/K mol
n= moles of gas= 0.405
[tex]V=\frac{nRT}{P}=\frac{0.405\times 0.0821\times 273}{1}=9.08L[/tex]
Thus 9.08 L of [tex]O_2[/tex] at STP would be required
Considering the reaction stoichiometry and STP conditions, 9.072 L of O₂ at STP would be required.
The balanced reaction is:
4 Al + 3 O₂ → 2 Al₂O₃
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
Al: 4 moles O₂: 3 moles Al₂O₃: 2 moles
Being 27 g/mole the molar mass of Al, this is the amount of mass that a substance contains in one mole, then if 14.6 grams Al are reacted, the number of moles of Al that react is calculated as:
[tex]14.6 gramsx\frac{1 mole}{27 grams}= 0.54 moles[/tex]
Then you can apply the following rule of three: if by stoichiometry 4 moles of Al react with 3 moles of O₂, 0.54 moles of Al react with how many moles of O₂?
[tex]amount of moles of O_{2} =\frac{0.54 moles of Alx3 moles of O_{2} }{4 moles of Al}[/tex]
amount of moles of O₂= 0.405 moles
On the other side, the STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.
Then you can apply the following rule of three: if by definition of STP 1 mole of O₂ occupies 22.4 L, 0.405 moles of O₂, how much volume does it occupy?
[tex]volume=\frac{0.405 moles of O_{2}x22.4 L }{1 mole of O_{2} }[/tex]
volume= 9.072 L
Finally, 9.072 L of O₂ at STP would be required.
Learn more:
brainly.com/question/16487206?referrer=searchResults brainly.com/question/14446695?referrer=searchResults brainly.com/question/11564309?referrer=searchResults brainly.com/question/4025026?referrer=searchResults brainly.com/question/18650135?referrer=searchResultsIf the heat of combustion for a specific compound is −1380.0 kJ/mol and its molar mass is 44.53 g/mol, how many grams of this compound must you burn to release 822.00 kJ of heat?mass:g
Answer:
Mass = 26.53 g
Explanation:
Heat of combustion = −1380.0 kJ/mol
This means 1 mol of the compound releases 1380 kJ
Molar mass = 44.53 g/mol
This means 1 mol of the compound has a mass of 44.53 g
How many grams would release 822kJ..?
First, we have to obtain the number of moles
1 mol = 1380
x = 822
x = 0.5957 moles
Moles = Mass / Molar mass
Mass = Molar mass * moles
Mass = 44.53 * 0.5957
Mass = 26.53 g
All of the following reactions can be described as displacement reactions except:____________.
a.) Zn(s) + FeCl2(aq) → ZnCl2(aq) + Fe(s).
b.) C6H6(l) + Cl2(g) → C6H5Cl(l) + HCl(g).
c.) 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g).
d.) Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2Ag(s).
e.) CuSO4(aq) + Fe(s) → Cu(s) + FeSO4(aq).
Answer:
b
Explanation:
The reaction that is not a displacement reaction from all the options is [tex]C_6H_6_{(l)} + Cl_{2(g)} --> C_6H_5Cl_{(l)} + HCl_{(g)}[/tex]
In a displacement reaction, a part of one of the reactants is replaced by another reactant. In single displacement reactions, one of the reactants completely displaces and replaces part of another reactant. In double displacement reaction, cations and anions in the reactants switch partners to form products.
Options a, c, d, and e involves the displacement of a part of one of the reactants by another reactant while option b does not.
Correct option = b.
The reaction given in Option A is not a displacement reaction. In Displacement reaction functional group of one reactant is replaced by the functional group of the another reactant.
Displacement reaction:
In this reaction functional group of one reactant is replaced by the functional group of the another reactant.
[tex]\bold { Zn(s) + FeCl_2(aq) \rightarrow ZnCl_2(aq) + Fe(s).}[/tex]
In the above reaction Zinc does not any functional group to exchange with iron chloride.
Therefore, the reaction given in Option A is not a displacement reaction.
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