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3 years ago

What is the volume occupied by 2.00 moles of Ar(g) at STP? Another one I am stuck on...please help!

1 answer:

3 0

Our only unknown is the volume of H2(g) . Our known variables are P,n,R, and T. At STP, the temperature is 273K and the pressure is 1 atm. Rearrange the equation to solve for V: n×R×TP. V=2.20mol ×0.0821L×atm mol×K ×273K 1atm. V=49.3L

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Calculate δs∘rxn for the reaction2no(g) o2(g)→2no2(g)express your answer to one decimal place and include the appropriate units

Anna35 [415]

The δs∘rxn for the reaction $2NO(g) + O_{2}$ → $2NO_{2} (g)$ will be -146 J/K.

Entropy would be a measurable physical characteristic and a scientific notion that is frequently connected to a condition of disorder, unpredictability, or uncertainty.

Entropy would be a measurement of the system's unpredictability or disorder. The entropy increases as randomness do. It has broad properties as well as a state function. It has the unit $JK^{-1} mol^{-1}$ .

Entropy of the reaction can be calculated by the reaction.

Δ $S^{0} rxn$ = 2 mol × $S^{0} (NO_{2} (g) - 2 mol$ × $S^{0} NO (g)$ - 1 mol × $S^{0} (O_{2} )$

Δ $S^{0} rxn$ = 2 mol × 240 J/mol.K - 2 mol × 210 J/mol.K-1 mol ×205.2 J/mol.K

Δ $S^{0} rxn$ = -146.8 J/K

Therefore, the δs∘rxn for the reaction $2NO(g) + O_{2}$ → $2NO_{2} (g)$ will be -146 J/K.

To know more about reaction

brainly.com/question/20305735

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4 0

1 year ago

CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7

padilas [110]

Answer: The $\Delta G$ for the reaction is 54.425 kJ/mol

Explanation:

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

$\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]$

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

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 = 46900 J

R = Gas constant = $8.314J/K mol$

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

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.425 kJ/mol

7 0

3 years ago

Calculate the volume of 1M NaOH required to neutralize 200 cc of 2M HCI. What mass of sodium

Nezavi [6.7K]

NaOH+HCl-> NaCl+H2O

1 mole of NaOH

1 mole of HCl.

To calculate volume of NaOH

CaVa/CbVb= Na/Nb

Where Ca=2M

Cb=1M

Va=200cm³

Vb=xcm³

Substitute into the equation.

2×200/1×Vb=1/1

400/Vb=1/1

Cross multiply

Vb×1=400×1

Vb=400cm³

To calculate the mass of sodium chloride, NaCl from the neutralization rxn.

Mole of NaCl=1

Molar mass of NaCl= 23+35.5=58.5

Mass=xgrammes.

Mass of NaCl=Number of moles × Molar mass.

Substitute

Mass of NaCl= 1×58.5

=58.5g

This is what I could come up with.

5 0

3 years ago

Cane sugar is being manufactured at the rate of 500 kg/hr from sugar canes that have the following composition (% by weight):

My name is Ann [436]

Answer:

a. 4166,67 b. 74553.0

Explanation:

6 0

3 years ago

Day and night are caused by what

Aleks04 [339]

Answer:

THE GOD ALMIGHTY!

Explanation:

6 0

3 years ago