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2 years ago

Zn + AgCI - ZnCl2 + - Ag HELPPP?

Chemistry

1 answer:

garik1379 [7]2 years ago

8 0

Answer:

Explanation: Zn (s) + 2 AgCl (s) ⇒ ZnCl2 (s) + 2 Ag(s)

Less noble Zinc reduces more noble Silver

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Which molecule has a dipole moment of 0 d?

fiasKO [112]

The answer is

a symmetrical molecule.

5 0

3 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]

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

A drinking water plant adds 500 grams of fluoride to a water tank containing 500,000 liters of drinking water. what is the conce

nlexa [21]

Answer is: concentration of fluoride in the water in parts-per-million is 1 ppm.
Parts-per-million (10⁻⁶) is present at one-millionth of a gram per gram of sample solution, for example mg/kg.
m(fluoride) = 500 g · 1000 mg/g = 500000 mg.
m(water) = d(water) · V(water).
m(water) = 1 kg/L · 500000 L.
m(water) = 500000 kg.
arts-per-million = 500000 mg ÷ 500000 kg = 1 mg/kg = 1 ppm.

5 0

2 years ago

How many sub-energy levels does the 2nd MEL have?

DedPeter [7]

Answer:

n= | Shell | Maximum Number of Electrons

1 | 1st Shell | 2

2 | 2nd Shell | 8

3 | 3rd Shell | 18

4 | 4th Shell | 32

Explanation: cause :)

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2 years ago

What force pulls electrons and protons together?

ruslelena [56]

Electrostatic force ..

7 0

2 years ago