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jok3333 [9.3K]
3 years ago
8

Helppp VVV URGENTT GENIUSES HELP MEEE

Chemistry
1 answer:
marta [7]3 years ago
6 0

Answer: Hydrogen

Explanation: Im pretty sure its Hydrogen since P is the cathode and it has a - charge meaning positively charged ions will be attracted to it and Hydrogen is the only gas with a positive charge in the answers.

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Consider the following reaction: CH3OH(g)⇌CO(g)+2H2(g) Part A Calculate ΔG for this reaction at 25 ∘C under the following condit
kati45 [8]

<u>Answer:</u> The \Delta G of the reaction at given temperature is -12.964 kJ/mol.

<u>Explanation:</u>

For the given chemical reaction:

CH_3OH(g)\rightleftharpoons CO(g)+2H_2(g)

The expression of K_p for the given reaction:

K_p=\frac{(p_{CO})\times (p_{H_2}^2)}{p_{CH_3OH}}

We are given:

p_{CO}=0.140atm\\p_{H_2}=0.180atm\\p_{CH_3OH}=0.850atm

Putting values in above equation, we get:

K_p=\frac{(0.140)\times (0.180)^2}{0.850}\\\\K_p=5.34\times 10^{-3}

To calculate the Gibbs free energy of the reaction, we use the equation:

\Delta G=\Delta G^o+RT\ln K_p

where,

\Delta G = Gibbs' free energy of the reaction = ?

\Delta G^o = Standard gibbs' free energy change of the reaction = 0 J (at equilibrium)

R = Gas constant = 8.314J/K mol

T = Temperature = 25^oC=[25+273]K=298K

K_p = equilibrium constant in terms of partial pressure = 5.34\times 10^{-3}

Putting values in above equation, we get:

\Delta G=0+(8.314J/K.mol\times 298K\times \ln(5.34\times 10^{-3}))\\\\\Delta G=-12963.96J/mol=-12.964kJ/mol

Hence, the \Delta G of the reaction at given temperature is -12.964 kJ/mol.

5 0
3 years ago
Which of the following is the correct molecule for dinitrogen trisulfide?
Setler79 [48]

Answer:

A. N2S3

yep u r right

Explanation:

Dinitrogen Trisulfide N2S3 Molecular Weight -- EndMemo.

4 0
3 years ago
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A compound distributes between benzene (solvent 1) and water (solvent 2) with a distribution coefficient, K = 2.7. If 1.0g of th
mote1985 [20]

Explanation:

The given data is as follows.

Solvent 1 = benzene,          Solvent 2 = water

 K_{p} = 2.7,         V_{S_{2}} = 100 mL

V_{S_{1}} = 10 mL,       weight of compound = 1 g

       Extract = 3

Therefore, calculate the fraction remaining as follows.

                  f_{n} = [1 + K_{p}(\frac{V_{S_{2}}}{V_{S_{1}}})]^{-n}

                                  = [1 + 2.7(\frac{100}{10})]^{-3}

                                  = (28)^{-3}

                                  = 4.55 \times 10^{-5}

Hence, weight of compound to be extracted = weight of compound - fraction remaining

                                  = 1 - 4.55 \times 10^{-5}

                                  = 0.00001

or,                               = 1 \times 10^{-5}

Thus, we can conclude that weight of compound that could be extracted is 1 \times 10^{-5}.

7 0
3 years ago
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We can preserve natural resources by
DedPeter [7]
The answer is c, relying on renewable energy sources
3 0
3 years ago
Low tide____.
Fantom [35]

Answer:

i think the answer is A

Explanation:

because in some areas, a regular pattern occurs of one high tide and one low tide each day,

8 0
3 years ago
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