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ElenaW [278]
3 years ago
15

A social movement dedicated to protecting the natural world and people from undesirable changes caused by human activity. *

Chemistry
2 answers:
Marta_Voda [28]3 years ago
4 0
B environmentalism.
viktelen [127]3 years ago
3 0

I think its b accauly no im positive its b

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Consider the following reaction: CO(g)+2H2(g)⇌CH3OH(g) Kp=2.26×104 at 25 ∘C. Calculate ΔGrxn for the reaction at 25 ∘C under eac
valentinak56 [21]

Answer : The value of \Delta G_{rxn} is, 8.867kJ/mole

Explanation :

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

\Delta G_{rxn}=\Delta G^o+RT\ln Q   ............(1)

where,

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

\Delta G_^o =  standard Gibbs free energy

R = gas constant = 8.314 J/mole.K

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

Q = reaction quotient

First we have to calculate the \Delta G_^o.

Formula used :

\Delta G^o=-RT\times \ln K_p

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

\Delta G^o=-(8.314J/mole.K)\times (298K)\times \ln (2.26\times 10^{4})

\Delta G^o=-24839.406J/mole=-24.83\times 10^3J/mole=-24.83kJ/mole

Now we have to calculate the value of 'Q'.

The given balanced chemical reaction is,

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

The expression for reaction quotient will be :

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

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

Q=\frac{(1.4)}{(1.2\times 10^{-2})\times (1.2\times 10^{-2})^2}

Q=8.1\times 10^{5}

Now we have to calculate the value of \Delta G_{rxn} by using relation (1).

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

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

\Delta G_{rxn}=-24.83kJ/mole+(8.314\times 10^{-3}kJ/mole.K)\times (298K)\ln (8.1\times 10^{5})

\Delta G_{rxn}=8.867\times 10^3J/mole=8.867kJ/mole

Therefore, the value of \Delta G_{rxn} is, 8.867kJ/mole

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3 years ago
Why must we do the a lot of quantity urine​
serg [7]

Answer:

because

ExplanatioN:

<em>BeCaUsE</em>

7 0
3 years ago
While doing a lab a student found the density of a piece of pure aluminum to be 2.85 g/cm3 the accepted value for the density of
____ [38]

Answer:

THE PERCENT ERROR IS 5.55 %

Explanation:

To calculate the percent error, we use the formula:

Percent error = Found value - accepted value / accepted value * 100

Found value = 2.85 g/cm3

Accepted value = 2.70 g/cm3

Solving for the percent error, we have:

Percent error = 2.85 g/cm3 - 2.70 g/cm3 / 2.70 g/cm3 * 100

Percent error = 0.15 / 2.70 * 100

Percent error = 0.05555 * 100

Percent error = 5.55 %

In conclusion, the percent error is 5.55 %

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3 years ago
Determine the empirical formula of a compound containing 24.74% potassium, 34.74% manganese, and 40.50% oxygen. What is the name
AlexFokin [52]
The answer would be KMnO4, please let me know if you would like me to explain how i got this
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List the 3 mole equalities.
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12.000 g of 12c is one

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