Answer:
The correct answer is -2878 kJ/mol.
Explanation:
The reaction that takes place at the time of the oxidation of glucose is,
C₆H₁₂O₆ (s) + 6O₂ (g) ⇒ 6CO₂ (g) + 6H₂O (l)
The standard free energy change for the oxidation of glucose can be determined by using the formula,
ΔG°rxn = ∑nΔG°f (products) - ∑nΔG°f (reactants)
The ΔG°f for glucose is -910.56 kJ/mol, for oxygen is 0 kJ/mol, for H2O -237.14 kJ/mol and for CO2 is -394.39 kJ/mol.
Therefore, ΔG°rxn = 6 (-237.14) + 6 (-394.39) - (-910.56)
ΔG°rxn = -2878 kJ/mol
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
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.
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
I hope it helps!
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 does the state symbol (aq) mean when written after a chemical
compound in a chemical equation?
A. It means the compound is in the liquid phase.
B. It means the compound is dissolved in water.
C. It means the compound is in the gas phase.
D. It means the compound is in the solid phase.
B. it means the compound is dissolved in water
Answer:
b
Explanation:
a p e x :)
Calculate the mass of feso4 that would be produced by 0.5mole of Fe
Answer:76 grams
Explanation:
Fe+H₂SO₄-->FeSO₄+H₂
For one mole of Fe we get 1 mole of feso4, therefore for 0.5 moles of Fe we get 0.5 moles of feso4.
The molar mass of feso4 is AFe+AS+4AO(A is atomic mass)
56+32+4*16=152grams/mole
Now, we need to multiply the number of moles by the molar mass to get the mass that reacts
152*0.5=76 grams
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.
The following balanced equation describes the reduction of iron(III) oxide to molten iron within a blast furnace: Fe2O3(s) + 3CO(g) ---> 2Fe(l) + 3CO2(g) Steve inserts 450. g of iron(III) oxide and 260. g of carbon monoxide into the blast furnace. After cooling the pure liquid iron, Steve determines that he has produced 288g of iron ingots. What is the theoretical yield of liquid iron, in grams? Just enter a numerical value. Do not enter units.
Answer: 313.6
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Fe_2O_3=\frac{450g}{160g/mol}=2.8moles[/tex]
[tex]\text{Moles of} CO=\frac{260g}{28g/mol}=9.3moles[/tex]
[tex]Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(l)+3CO_2(g)[/tex]
According to stoichiometry :
1 mole of [tex]Fe_2O_3[/tex] require 3 moles of [tex]CO[/tex]
Thus 2.8 moles of [tex]Fe_2O_3[/tex] will require=[tex]\frac{3}{1}\times 2.8=8.4moles[/tex] of [tex]CO[/tex]
Thus [tex]Fe_2O_3[/tex] is the limiting reagent as it limits the formation of product and [tex]CO[/tex] is the excess reagent.
As 1 mole of [tex]Fe_2O_3[/tex] give = 2 moles of [tex]Fe[/tex]
Thus 2.8 moles of [tex]Fe_2O_3[/tex] give =[tex]\frac{2}{1}\times 2.8=5.6moles[/tex] of [tex]Fe[/tex]
Mass of [tex]Fe=moles\times {\text {Molar mass}}=2.6moles\times 56g/mol=313.6g[/tex]
Theoretical yield of liquid iron is 313.6 g
16. A metal element and a non-metal element are brought near each other and allowed to react. What's the most likely type of compound
that will form between these two elements?
A. lonic and covalent
B. lonic
C. Covalent
D. Neither, metals and non-metals don't react.
Answer:
B) Ionic
Explanation:
Assuming an efficiency of 34.90%, calculate the actual yield of magnesium nitrate formed from 139.6 g of magnesium and excess copper(II) nitrate.Mg+Cu(NO3)2⟶Mg(NO3)2+Cu
Answer:
300.44 g
Explanation:
The balanced equation for the reaction is given below:
Mg + Cu(NO3)2 —> Mg(NO3)2 + Cu
Next, we shall determine the mass of Mg that reacted and the mass of Mg(NO3)2 produced from the balanced equation.
This is illustrated below:
Molar mass of Mg = 24 g/mol
Mass of Mg from the balanced equation = 1 x 24 = 24 g
Molar mass of Mg(NO3)2 = 24 + 2[14 + (16x3)]
= 24 + 2[ 14 + 48]
= 24 + 124 = 148 g/mol
Mass of Mg(NO3)2 from the balanced equation =
1 x 148 = 148 g
From the balanced equation above,
24 g of Mg reacted to produce 148 g of Mg(NO3)2.
Next, we shall determine the theoretical yield of Mg(NO3)2.
This can be obtained as follow:
From the balanced equation above,
24 g of Mg reacted to produce 148 g of Mg(NO3)2.
Therefore, 139.6 g of Mg will react to = (139.6 x 148)/24 = 860.87 g of Mg(NO3)2
Therefore, the theoretical yield of Mg(NO3)2 is 860.87 g
Finally, we shall determine the actual yield of Mg(NO3)2 as follow:
Theoretical of Mg(NO3)2 = 860.87 g
Percentage yield = 34.90%
Actual yield of Mg(NO3)2 =?
Percentage yield = Actual yield /Theoretical yield x 100
34.90% = Actual yield /860.87
Cross multiply
Actual yield = 34.90% x 860.87
Actual yield = 34.9/100 x 860.87
Actual yield = 300.44 g
Therefore, the actual yield of Mg(NO3)2 is 300.44 g
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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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
Which of the following properties should carbon (C) have based on its position on
the periodic table?
A. Shiny
B. Dense
C. Malleable
D. Poor conductor
Answer:
D- poor conductor
Explanation:
metallic properties decrease as we go on the right of the periodic table. Carbon is a non metal hence it is dull and a poor conductor.
it has a low density and is ductile.
Answer: Poor conductor
Explanation:
What is the molar mass of a protein if a solution of 0.020 g of the protein in 25.0 mL of solution has an osmotic pressure of 0.56 torr at 25 ∘ C
Answer:
26.5 kD
Explanation:
Here we can apply the formula ∏ = iMRT, where ∏ = osmotic pressure = 0.56 - ( given ). This is only one part of the information we are given / can conclude in this case ....
i = van’t Hoff factor = 1 for a protein molecule,
R = gas constant = 62.36 L torr / K-mol,
T ( temperature in Kelvin ) = 25 + 273 - conversion factor C° + 273 = 298K
( Known initially ) ∏ = osmotic pressure = 0.56 torr
..... besides the part " M " in the formula, which we have no information on whatsoever, as we have to determine it's value.
_____
Substitute derived / known values to solve for M ( moles / liter ) -
∏ = iMRT
⇒ 0.56 = ( 1 )( M )( 62.36 )( 298 )
⇒ 0.56 = M( 18583.28 )
⇒ M = 0.56 / 18583.28 ≈ 0.00003013461 ....
_____
We know that M = moles / liter, so we can use this to solve for moles, and hence calculate the molar mass by the formula molar mass = g / mol -
M = mol / l
⇒ 0.00003013461 = 0.020 / 25 mL ( 0.025 L ),
0.020 / 0.025 = 0.8 g / L
⇒ 0.8 g = 0.00003013461 moles,
molar mass = 0.8 g / 0.00003013461 moles = 26,548 g / mol = 26.5 kD
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=searchResultsThe 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
Which of these substances has the highest pOH? 0.10 M HCl, pH = 1 0.001 M HNO3, pH = 3 0.01 M NaOH, pH = 12 The answer is 0.10 M HCI, pH=1
Answer:On these combined scales of pH and pH it can be shown that because for water when pH = pH = 7 that pH + pH = 14. This relationship is useful in the inter conversion of values. For example, the pH at a 0.01 M solution of sodium hydroxide is 2, the pH of the same solution must be 14-2 = 12.
Explanation:
The 0.10M HCI, pH = 1 solution has the highest pOH. Therefore, option (1) is correct.
What is the pOH?pOH of a solution can be determined from the negative logarithm of the hydroxide ions concentration in the solution.
The mathematically pOH of the solution can be expressed as:
pOH = -log [OH⁻] ..............(1)
Where [OH⁻] represents the concentration of hydroxide ions in an aqueous solution.
Given, the pH = 1 of HCl
pH + pOH = 14
1 + pOH = 14
pOH = 14 - 1
pOH = 13
Given, the pH = 3 of HNO₃
pH + pOH = 14
3 + pOH = 14
pOH = 14 - 3
pOH = 11
Given, the pH = 12 of NaOH = 0.01 M
pH + pOH = 14
12 + pOH = 14
pOH = 14 - 12
pOH = 2
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Complete ionic equation K2CO3(aq)+2CuF(aq) → Cu2CO3(s)+2KF(aq) Examine each of the chemical species involved to determine the ions that would be present in solution. Be sure to consider both the coefficients and subscripts of the molecular equation, and then write this precipitation reaction in the form of a balanced complete ionic equation. Express your answer as a chemical equation including phases.
Answer:
2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)
Explanation:
K2CO3(aq) + 2CuF(aq) → Cu2CO3(s) + 2KF(aq)
The complete ionic equation for the above equation can be written as follow:
In solution, K2CO3 and CuF will dissociate as follow:
K2CO3(aq) —› 2K+(aq) + CO3²¯(aq)
CuF(aq) —› Ca^2+(aq) + 2F¯(aq)
Thus, we can write the complete ionic equation for the reaction as shown below:
K2CO3(aq) + 2CuF(aq) —›
2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)
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:
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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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:
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.)
How many mL of calcium hydroxide are required to neutralize 25.0 mL of 0.50 M
nitric acid?
Answer:
6.5 mL
Explanation:
Step 1: Write the balanced reaction
Ca(OH)₂ + 2 HNO₃ ⇒ Ca(NO₃)₂ + 2 H₂O
Step 2: Calculate the reacting moles of nitric acid
25.0 mL of 0.50 M nitric acid react.
[tex]0.0250L \times \frac{0.50mol}{L} = 0.013 mol[/tex]
Step 3: Calculate the reacting moles of calcium hydroxide
The molar ratio of Ca(OH)₂ to HNO₃ is 1:2. The reacting moles of Ca(OH)₂ are 1/2 × 0.013 mol = 6.5 × 10⁻³ mol
Step 4: Calculate the volume of calcium hydroxide
To answer this, we need the concentration of calcium hydroxide. Since the data is missing, let's suppose it is 1.0 M.
[tex]6.5 \times 10^{-3} mol \times \frac{1,000mL}{1.0mol} = 6.5 mL[/tex]
Current is described as
A. moles of electrons.
B. the flow of electrons through a substance.
C. electricity.
D. the flow of ions through a substance.
Answer:
B!
Explanation:
I got it right in class!
If enough experimental data supports a hypothesis, then it
Answer:
Then the hypothesis is proved and becomes a theory.
If not, then another hypothesis should be proposed and tested.
The specific heat of a certain type of metal is 0.128 J/(g⋅∘C). What is the final temperature if 305 J of heat is added to 52.4 g of this metal, initially at 20.0 ∘C?
Answer:
65.47∘C
Explanation:
Specific heat capacity, c = 0.128 J/(g⋅∘C)
Initial temperature = 20.0 ∘C
Final temperature = ?
Mass = 52.4 g
Heat = 305 J
All these variables are related by the following equation;
H = m c ΔT
ΔT = H / mc
ΔT = 305 / (52.4 * 0.128)
ΔT = 45.47∘C
ΔT = Final Temperature - Initial Temperature
Final temperature = ΔT + Initial temperature
Final temperature = 45.47∘C + 20.0 ∘C = 65.47∘C
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
Learn more: https://brainly.com/question/6284546
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+
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.
Draw the curved arrow mechanism for the reaction between (2R,3R)-3,5-dimethylhexan-2-ol and PCl3.
Answer:
Sn2 mechanism
Explanation:
In this case, our nucleophile is the "OH" on (2R,3R)-3,5-dimethylhexan-2-ol. The alcohol group will attack the [tex]PCl_3[/tex] to produce a new bond between O and P with a positive charge in the oxygen. Additionally, when the OH attacks a Br atom leaves the molecule producing a bromide ion.
In the next step, the bromide ion produced will attack the carbon bonded to the OH that now is bonded to [tex]PCl_2[/tex]. An Sn2 reaction takes place and the substitution would be made in only one step. Due to this, we will have an inversion in the stereochemistry and the absolute configuration on carbon 2 will change from "R" to "S" to produce (2S,3R)-2-bromo-3,5-dimethylhexane.
I hope it helps!