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2 years ago

Calculate the mass of silver nitrate (in grams) in a 145 mL solution of 5.88 M AgNO3?

Chemistry

2 answers:

Alika [10]2 years ago

7 0

Answer: The mass of silver nitrate is 137.5 grams.

Explanation:

Molarity is defined as the number of moles present in one liter of solution.

Mathematically,

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Or,

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = 5.88 M

Molar mass of silver nitrate = 170 g/mol

Volume of solution = 145 mL

Mass of silver nitrate = ? g

Putting values in above equation, we get:

$5.58mol/L=\frac{\text{Mass of silver nitrate}\times 1000}{170g/mol\times 145mL}\\\\\text{Mass of silver nitrate}=\frac{5.58\times 170\times 145}{1000}=137.5g$

Hence, the mass of silver nitrate is 137.5 grams.

oee [108]2 years ago

6 0

1 mole of AGNO3 weighs 170g
so 0.6322 miles will weigh 170 X 0.6322=107.474 g

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The mass percent of potassium chloride is 1.386%

calculation

mass percent = actual mass/ Theoretical mass x 100

Actual mass = 9.35 g

Theoretical mass is calculated as below

Step 1 : write the equation for reaction

KCl + H₂O → KOH + HCl

Step 2: find the moles of H₂O

moles = mass÷ molar mass

The molar mass of H₂O = (2 x1 ) +(16) = 18 g/mol

moles is therefore = 162.98 g÷ 18 g/mol =9.054 moles

Step 3: use the mole ratio to determine the moles of KCl

KCl: H₂O is 1:1 therefore the moles of KCl is also = 9.054 moles

Step 4: find the theoretical mass of KCl

mass = moles x molar mass

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2 years ago

What is the bond energy per mole for breaking all the bonds of oxygen, O 2 ?

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2 years ago

Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O,

WITCHER [35]

Answer: Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

$2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

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

$\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]$

We are given:

$\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?$

Putting values in above equation, we get:

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Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

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