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2 years ago

Why is the atomic radius of sodium much smaller than the atomic radius of potassium

Chemistry

1 answer:

Bezzdna [24]2 years ago

8 0

As you go down a group the atomic radius increases and so does the number of electron shells added to an atom.

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3. You are asked to prepare an aqueous solution. What would be a better choice for a solute potassium nitrate or

liubo4ka [24]

Answer:

i do not knowwww

Explanation:

8 0

2 years ago

What advice would you give a person who handles hydrogen and oxygen in their workplace?

Murljashka [212]

Regard the principle of utilization of two gas.

Make a consistent control of hardware containing gas.

Make a consistent control of weight diminishing valves giving gas.

No smoking zone.

3 0

2 years 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]

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

<u>Explanation:</u>

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

2 years ago

Convert 1176 mmHg to kPa

OLEGan [10]

Answer:

156.8kPa

Explanation:

The problem here is to convert mmHg to kPa;

We have been given:

1176mmHg and the problem is to convert to kPa;

1000Pa = 1kPa

1 mmHg = 133.322Pa

1176mmHg will give 1176 x 133.322 = 156787.1Pa

To kPa;

156.8kPa

8 0

2 years ago

What is the normal boiling point of etanoic acid?

Nezavi [6.7K]

Answer: 244.6°F

Explanation:

3 0

2 years ago