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2 years ago

How much water must be added to 6.0 M silver nitrate in order to make 500 mL of 1.2 M solution?

Chemistry

1 answer:

algol [13]2 years ago

7 0

The amount of water that must be added to 6.0 M silver nitrate to make 500mL of 1.2 M solution is : 2000 mL

<u>Given data :</u>

Concentration of siilver nitrate ( M₁ ) = 6.0 M

volume of solution ( V₁ ) = 500 mL

Conc of solution ( M₂ )= 1.2 M

<h3>Determine the amount of water that must be added</h3>

we will apply the equation below

M₁V₁ = M₂V₂ ---- ( 1 )

where : V₂ = V₁ + water added ---- ( 2 )

V₂ ( Final volume ) = ( M₁V₁ ) / M₂

= ( 6 * 500 ) / 1.2

= 2500 mL

Back to eqaution ( 2 )

2500 mL = 500 mL + added water

therefore ; added water = 2500 - 500

= 2000 mL

Hence we can conclude that The amount of water that must be added to 6.0 M silver nitrate to make 500mL of 1.2 M solution is : 2000 mL.

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3 years ago

I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 mol

galina1969 [7]

Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

Enthalpy of formation of iodine gas = $\Delta H_f_{(I_2)}=62.438 kJ/mol$

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The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]$

$=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol$

Enthaply change when 1.62 moles of iodine gas recast:

$\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ$

Entropy of the surrounding = $\Delta S^o_{surr}=\frac{\Delta H}{T}$

$=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K$

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

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3 years ago

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ycow [4]

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<u>Explanation:</u>

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The equation used to calculate enthalpy change is of a reaction is:

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

For the given chemical reaction:

$3CO(g)+2Fe_2O_3(s)\rightarrow Fe(s)+3CO_2(g)$

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