Answer:
Trial 1: Moles acetic acid = 0.00686 moles;
Moles of Mg(OH)₂ = 0.00343 moles
Neutralization capacity = 0.00137 mol/g
Trial 2: Moles acetic acid = 0.00722 moles
Moles of Mg(OH)₂ = 0.00361 moles
Neutralization capacity = 0.00144 mol/g
Explanation:
Equation of the reaction: 2CH₃COOH + Mg(OH)₂ ---> Mg(CH₃COO)₂ + 2H₂O
Trial 1:
Moles of acetic acid = concentration * volume in litres
concentration of acetic acid = 0.88 M
volume of acid used = 7.8 mL = (7.8/1000) Litres = 0.0078 L
Moles acetic acid = 0.88 M * 0.0078 L
Moles acetic acid = 0.00686 moles
Moles of Mg(OH)₂:
From the equation of reaction, 2 moles of acetic acid reacts with 1 mole of Mg(OH)₂
Therefore, 0.00686 moles of acetic acid will react with 0.00686/2 moles of Mg(OH)₂ = 0.00343 moles of Mg(OH)₂
Moles of Mg(OH)₂ = 0.00343 moles
Neutralization capacity = moles of Mg(OH)₂/mass of milk of magnesia
Neutralization capacity = 0.00343 mole /2.5 g
Neutralization capacity = 0.00137 mol/g
Trial 2.
Moles of acetic acid = concentration * volume in litres
concentration of acetic acid = 0.88 M
volume of acid used = 8.2 mL = (8.2/1000) Litres = 0.0082 L
Moles acetic acid = 0.88 * 0.0082
Moles acetic acid = 0.00722 moles
Moles of Mg(OH)₂:
From the equation of reaction, 2 moles of acetic acid reacts with 1 mole of Mg(OH)₂
Therefore, 0.00722 moles of acetic acid will react with 0.00722/2 moles of Mg(OH)₂ = 0.00361 moles of Mg(OH)₂
Moles of Mg(OH)₂ = 0.00361 moles
Neutralization capacity = moles of Mg(OH)₂/mass of milk of magnesia
Neutralization capacity = 0.00361 mole /2.5 g
Neutralization capacity = 0.00144 mol/g
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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1. The following thermochemical equation is for the reaction of water(l) to form hydrogen(g) and oxygen(g). 2H2O(l)2H2(g) + O2(g) H = 572 kJ How many grams of H2O(l) would be made to react if 110 kJ of energy were provided? _____ grams
2. The following thermochemical equation is for the reaction of carbon monoxide(g) with hydrogen(g) to form methane(g) and water(g). CO(g) + 3H2(g) CH4(g) + H2O(g) H = -206 kJ When 6.27 grams of carbon monoxide(g) react with excess hydrogen(g),_____ kJ of energy are ____ a.evolved b.absorbed
Answer:
1. 6.92 g of H2O
2i. - 46 KJ of energy.
ii. Option A. Evolved.
Explanation:
1. Determination of the mass of H2O that would be made to react if 110 kJ of energy were provided.
This can be obtained as follow:
The equation for the reaction is given below
2H2O(l) —> 2H2(g) + O2(g) H = 572 kJ
Next, we shall determine the mass of H2O required to produce 572 kJ from the balanced equation.
Molar mass of H2O = (2x1) + 16 = 18 g/mol
Mass of H2O from the balanced equation = 2 x 18 = 36 g
From the balanced equation above, 36 g of H2O reacted to produce 572 kJ of energy.
Finally, we shall determine the mass of water (H2O) needed to produce 110 kJ of energy.
This is illustrated below:
From the balanced equation above, 36 g of H2O reacted to produce 572 kJ of energy.
Therefore, Xg of H2O will react to 110 kJ of energy i.e
Xg of H2O = (36 x 110)/572
Xg of H2O = 6.92 g
Therefore, 6.92 g of H2O is needed to react in order to produce 110 KJ of energy.
2i. Determination of the energy.
The balanced equation for the reaction is given below:
CO(g) + 3H2(g) —> CH4(g) + H2O(g) H = -206 kJ
Next, we shall determine the mass of CO that reacted to produce -206 kJ of energy from the balanced equation.
This is illustrated below:
Molar mass of CO = 12 + 16 = 28 g/mol
Mass of CO from the balanced equation = 1 x 28 = 28 g
From the balanced equation above,
28 g of CO reacted to produce -206 kJ of energy.
Finally, we shall determine the amount of energy produced by reacting 6.27 g of CO. This is illustrated below:
From the balanced equation above,
28 g of CO reacted to produce -206 kJ of energy.
Therefore, 6.27 g of CO will react to produce = (6.27 x -206)/28 = - 46 KJ of energy.
Therefore, - 46 KJ of energy were produced from the reaction.
2ii. Since the energy obtained is negative, it means heat has been given off to the surroundings.
Therefore, the heat is evolved.
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+
If 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
Air contains nitrogen, oxygen, argon, and trace gases. Ifthe partial pressure of nitrogen is 592 mm Hg, oxygen is160 mm Hg, argon is 7 mm Hg, and trace gas is 1 mm Hg,what is the atmospheric pressure
Answer:
760 mmHg
Explanation:
Step 1: Given data
Partial pressure of nitrogen (pN₂): 592 mmHgPartial pressure of oxygen (pO₂): 160 mmHgPartial pressure of argon (pAr): 7 mmHgPartial pressure of the trace gas (pt): 1 mmHgStep 2: Calculate the atmospheric pressure
Since air is a gaseous mixture, the atmospheric pressure is equal to the sum of the gases that compose it.
P = pN₂ + pO₂ + pAr + pt = 592 mmHg + 160 mmHg + 7 mmHg + 1 mmHg = 760 mmHg
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
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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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.)
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:
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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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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Help please.☺️☺️☺️☺️☺️
Answer:
A.∆s>0contribute to spontaneity.
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
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.
Which resulted from the study of chemistry?
A) Alchemy to turn base metals into noble metals
B) The understanding of earth, air, fire, and water as the basic components of matter.
C) A supernatural, mystical view of the world.
D) Discovering the role of oxygen in combustion
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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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:
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
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What is the name of the molecule below?
A) 2-pentene
B) pentane
C) 2-pentyne
D) 2-pentane
The name of the molecule which is given below is 2-pentene.
What are alkene?Alkenes are the organic compounds which are composed of carbon and hydrogen atoms, in which double bond is present.
In the given diagram:
Each corner and joints shows the carbon atoms and number of carbon atoms in it is 5.One double bond is present in the 2nd position.So the compound is 2 pentene.
Hence, 2 pentene is the name of the compound.
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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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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-
Carbon-14 has a half-life of 5720 years and this is a fast-order reaction. If a piece of wood has converted 75 % of the carbon-14, then how old is it?
Answer:
11445.8years
Explanation:
Half-life of carbon-14 = 5720 years
First we have to calculate the rate constant, we use the formula :
Drag each image to the correct location on the model. Each image can be used more than once. Apply the rules and principles of electron configuration to draw the orbital diagram of aluminum. Use the periodic table to help you.
Answer:
The answer to your question is given below.
Explanation:
Aluminium has atomic number of 13. Thus, the electronic configuration of aluminium can be written as:
Al (13) —› 1s² 2s²2p⁶ 3s²3p¹
The orbital diagram is shown on the attached photo.
Answer: screen shot
Explanation:
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]
The least acidic is CaOThe most acidic is [tex]SO_3[/tex]Learn more here:
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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=searchResultsWhy do you think sodium bicarbonate is included to neutralize an acidic spill rather than sodium hydroxide?
Imagine a hypothetical situation in which 250 mL of diethyl ether (SDS) has spilled inside of a chemical fume hood onto a stir plate that is plugged in and stirring. Discuss the risks associated with this situation (location, size, compound spilled, and external hazards), and then explain how this spill should be managed.
Answer:
Acid spills should be neutralized with sodium bicarbonate and then cleaned up with a paper towel or sponge.
Explanation:
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
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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