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marshall27 [118]
2 years ago
10

Please help me thank u !​

Chemistry
1 answer:
konstantin123 [22]2 years ago
8 0
You should just try your best out what u think it will be okay ! I don’t know the answer I’m sorry
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Be sure to answer all parts. The equilibrium constant (Kp) for the reaction below is 4.40 at 2000. K. H2(g) + CO2(g) ⇌ H2O(g) +
nikklg [1K]

<u>Answer:</u>

<u>For 1:</u> The value of \Delta G for the chemical equation is -24.636 kJ/mol

<u>For 2:</u> The value of \Delta G for the chemical equation is -20.925 kJ/mol

<u>Explanation:</u>

For the given chemical equation:

H_2(g)+CO_2(g)\rightleftharpoons H_2O(g)+CO(g)

  • <u>For 1:</u>

To calculate the \Delta G for given value of equilibrium constant, we use the relation:

\Delta G=-RT\ln K_p      .....(1)

where,

\Delta G = ? kJ/mol

R = Gas constant = 8.314J/K mol

T = temperature = 2000 K

K_p = equilibrium constant in terms of partial pressure = 4.40

Putting values in above equation, we get:

\Delta G=-(8.314J/Kmol)\times 2000K\times \ln (4.40)\\\\\Delta G=-24636.12J/mol

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, -24636.12 J/mol = -24.636 kJ/mol

Hence, the value of \Delta G for the chemical equation is -24.636 kJ/mol

  • <u>For 2:</u>

The expression of K_p for the given chemical equation is:

K_p=\frac{p_{CO}p_{H_2O}}{p_{H_2}p_{CO_2}}

We are given:

p_{CO}=1.18atm\\p_{H_2O}=0.66atm\\p_{CO_2}=0.82atm\\p_{H_2}=0.27atm

Putting values in above equation, we get:

K_p=\frac{1.18\times 0.66}{0.27\times 0.82}\\\\K_p=3.52

Now, calculating the value of \Delta G by using equation 1:

R = Gas constant = 8.314J/K mol

T = temperature = 2000 K

K_p = equilibrium constant in terms of partial pressure = 3.52

Putting values in equation 1, we get:

\Delta G=-(8.314J/Kmol)\times 2000K\times \ln (3.52)\\\\\Delta G=-20925.68J/mol

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, -20925.68 J/mol = -20.925 kJ/mol

Hence, the value of \Delta G for the chemical equation is -20.925 kJ/mol

3 0
3 years ago
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1.....What are the 2 major divisions of Biochemistry?
kicyunya [14]

Answer:

Organic and inorganic

Explanation:

6 0
3 years ago
Hii pls help me to balance the equation thanksss​
tiny-mole [99]

▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓

\boxed{\pmb{\color{gold}{\sf{2SO_{2}(g) + O_{2}(g)\dashrightarrow 2SO_{3}(g)}}}}

▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓

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2 years ago
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Please help! Help me solve problems about naming structures with IUPAC rules
lianna [129]
A. The longest carbon chain is eight, and it has two methyl groups attached to carbon three, and a special group attached to carbon five. Its two names could be:

3-dimethyl-5-(1-methylethyl)octane
3-dimethyl-5-isopropyloctane

Both of these are correct. This is an alkane, because it has all single bonds.

B. This has a triple bond contained between carbons 2 and 3, and has a methyl group off carbon 4. The longest chain is 5. It’s name is:

4-methyl-2-pentyne

This is an alkene, because of the double bond.

C. This has a double bond contained between carbons 2 and 3, and has a methyl off of four and an methyl off of six. The longest chain is eight (follow the longest chain of carbons).

4,6-dimethyl-2-octene

This is an alkene, because of the double bond.

D. This has an ethyl group at 1 and a methyl group at 2 (rotate the compound to make it as clean as possible, in this case, the ring is flipped and rotated to make it alphabetical with the smallest numbers possible). The two names are:

1-ethyl-2-methylbenzene
ortho-ethylmethylbenzene

Both are correct, the ortho prefix telling the location of the ethyl and methyl groups. This is an aromatic structure because of its double bonded ring.

E. The longest chain is nine, and has methyls at three, five, and seven, along with a propyl at five. The name is:

3,5,7-trimethyl-5-propylnonane

This is an alkane, due to the single bonds.

Hope this helps!
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3 years ago
Physical activity and exercise typically lead to __________.
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<span>Physical activity and exercise typically lead to higher energy levels. As your body gets used to working hard consistently, you tend to be more prepared and in better physical condition. Your body begins to learn that you should burn up calories quickly to prepare to be active, and this results in increased energy levels even when you're not currently exercising.</span>
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3 years ago
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