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IRISSAK [1]
2 years ago
11

Part a write an equation for the formation of nh3(g) from its elements in its standard states. express your answer as a chemical

equation. identify all of the phases in your answer. submitmy answersgive up part b find δh∘f for nh3(g) from appendix iib in the textbook. express your answer using three significant figures. δh∘f = kj/mol submitmy answersgive up part c write an equation for the formation of co2(g) from its elements in its standard states. express your answer as a chemical equation. identify all of the phases in your answer. submitmy answersgive up part d find δh∘f for co2(g) from appendix iib in the textbook. express your answer using four significant figures. δh∘f = kj/mol submitmy answersgive up part e write an equation for the formation of fe2o3(s) from its elements in its standard states. express your answer as a chemical equation. identify all of the phases in your answer. submitmy answersgive up part f find δh∘f for fe2o3(s) from appendix iib in the textbook. express your answer using four significant figures. δh∘f = kj/mol submitmy answersgive up part g write an equation for the formation of ch4(g) from its elements in its standard states. express your answer as a chemical equation. identify all of the phases in your answer. submitmy answersgive up part h find δh∘f for ch4(g) from appendix iib in the textbook. express your answer using three significant figures.
Chemistry
2 answers:
pochemuha2 years ago
8 0
Sourse 

https://quizlet.com/19584916/chemistry-chapter-6-and-7-flash-cards/
andrey2020 [161]2 years ago
3 0
A. N₂ (g) + 3 H₂ (g) --> 2 NH₃ (g)
B. The value for standard enthalpy of formation is empirical given that the reactants involved were pure elements. So, you can search this on the internet or in any textbook. The Hf for NH₃ is -46.0 kJ/mol.
C. C (s) + O₂ (g) --> CO₂ (g)
D. The Hf for CO₂ is <span>-393.5 kJ/mol
E. 4 Fe (s) + 3 O</span>₂ (g) --> 2 Fe₂O₃ (s)
F. The Hf for solid Fe₂O₃ is -826.0 kJ/mol.
G. C (s) + 2 H₂ (g) --> CH₄ (g)
H. The Hf for methane gas is -74.9 kJ/mol.
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Match the specific enzyme to its class. A. oxidoreductase B. hydrolase C. transferase D. ligase E. lyase aminase
lilavasa [31]

Answer: Every enzyme has a specific name that can give us insight into the specific reaction that that enzyme can catalyze. We divide them into six different categories.

1) Oxidoreductase - includes two different types of reactions by transferring electrons from either molecule A to B or vice versa. It is involved in oxidizing electrons away from a molecule.

2) Hydrolase - uses water to divide a molecule into two other molecules.

3) Transferase - you move some functional group X from molecule B to molecule A

4) Ligase - catalyzes reactions between two molecules, A and B, that are combining to form a complex between the two. (example: DNA replication)

5) Lyase - divides a molecule into two other molecules without using water and without reducing or oxidation

3 0
2 years ago
Imagine you needed to identify if an object has undergone a physical change or a chemical change. What information would you nee
pishuonlain [190]
How it looks. basically the thing that tells you how it change. for example if an ice cube was melted (heat), it only changed physically not chemically as the h20 molecules are still there. however lets say you burn woos— you cant get that would back. its ash now and it has changed chemically.
7 0
2 years ago
Consider a general reaction A ( aq ) enzyme ⇌ B ( aq ) A(aq)⇌enzymeB(aq) The Δ G ° ′ ΔG°′ of the reaction is − 5.980 kJ ⋅ mol −
Grace [21]

Answer : The value of K_{eq} is, 11.2

The value of \Delta G_{rxn} is -9.04 kJ/mol

Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

\Delta G^o=-RT\times \ln K_{eq}

where,

\Delta G^o = standard Gibbs free energy  = -5.980 kJ/mol = -5980 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = 25^oC=273+25=298K

K_{eq}  = equilibrium constant  = ?

Now put all the given values in the above formula, we get:

\Delta G^o=-RT\times \ln K_{eq}

-5980J/mol=-(8.314J/K.mol)\times (298K)\times \ln K_{eq}

K_{eq}=11.2

Thus, the value of K_{eq} is, 11.2

Now we have to calculate the \Delta G_{rxn}.

The formula used for \Delta G_{rxn} is:

The given reaction is:

A(aq)\rightleftharpoons B(aq)

\Delta G_{rxn}=\Delta G^o+RT\ln Q

\Delta G_{rxn}=\Delta G^o+RT\ln \frac{[B]}{[A]}    ............(1)

where,

\Delta G_{rxn} = Gibbs free energy for the reaction  = ?

\Delta G_^o =  standard Gibbs free energy  = -30.5 kJ/mol

R = gas constant = 8.314\times 10^{-3}kJ/mole.K

T = temperature = 37.0^oC=273+37.0=310K

Q = reaction quotient

[A] = concentration of A = 1.8 M

[B] = concentration of B = 0.55 M

Now put all the given values in the above formula 1, we get:

\Delta G_{rxn}=(-5980J/mol)+[(8.314J/mole.K)\times (310K)\times \ln (\frac{0.55}{1.8})

\Delta G_{rxn}=-9035.75J/mol=-9.04kJ/mol

Therefore, the value of \Delta G_{rxn} is -9.04 kJ/mol

3 0
2 years ago
How many moles of BCl3 are needed to produce 10.0 g of HCl(aq) in the following reaction? (HCl molar mass is 36.46 g/mol).
Scilla [17]

Answer:

Moles of BCl₃ needed = 0.089 mol

Explanation:

Given data:

Moles of BCl₃ needed = ?

Mass of HCl produced = 10.0 g

Solution:

Chemical equation:

BCl₃ + 3H₂O  →     3HCl + B(OH)₃

Number of moles of HCl:

Number of moles = mass/molar mass

Number of moles = 10.0 g/ 36.46 g/mol

Number of moles = 0.27 mol

Now we will compare the moles of HCl with BCl₃.

               HCl             :           BCl₃

                 3               :             1

             0.27             :            1/3×0.27 = 0.089 mol

6 0
2 years ago
Write a balanced equation for the reaction of NaCH3COO (also written as NaC2H3O2) and HCl.
hammer [34]
The balance chemical equation is:

NaCH₃COO + HCl → NaCl + HCH₃COO

Make 
4 0
2 years ago
Read 2 more answers
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