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3 years ago

How many mL are 2.3 mol of CO2 at STP?

Chemistry

2 answers:

kupik [55]3 years ago

8 0

Answer:

0.052mL

Explanation:

1mole of a gas occupy 22.4L.

Therefore, 1 mole of CO2 will also occupy 22.4L.

If 1mole of CO2 occupies 22.4L,

Then 2.3moles of CO2 will occupy = 2.3 x 22.4 = 51.52L

coverting this volume to mL, we simply divide by 1000 as shown below:

51.52/1000 = 0.05152mL = 0.052mL

Dovator [93]3 years ago

7 0

Answer:

The volume of 2.3 moles CO2 is 51.5 L = 51500 mL

Explanation:

Step 1: Data given

Number of moles CO2 = 2.3 moles

Step 2: Calculate volume CO2

p*V= n*R*T

⇒with p = the pressure = 1 atm

⇒with V = the volume = TO BE DETERMINED

⇒with n = the number of moles = 2.3 moles

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 273 K

V = (nRT)/p

V = (2.3*0.08206*273)/1

V = 51.5 L

The volume of 2.3 moles CO2 is 51.5 L = 51500 mL

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4 years ago

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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4 years ago

At 350°c, keq = 1.67 × 10-2 for the reversible reaction 2hi (g) ⇌ h2 (g) + i2 (g). what is the concentration of hi at equilibriu

mariarad [96]

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