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love history [14]
3 years ago
8

Will mark the brainiest later for correct answers! Please show work.

Chemistry
1 answer:
Elodia [21]3 years ago
7 0

Answer:

According to avogadro's law, 1 mole of every substance contains avogadro's number 6.023\times 10^{23} of particles and weighs equal to its molecular mass.

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

\text{Number of moles}=\frac{\text{Given molecules}}{\text {Avogadros number}}

a. moles in 14.08 g of C_{12}H_{22}O_{11} = \frac{14.08g}{342.3g/mol}=0.04113moles

molecules in 14.08 g of C_{12}H_{22}O_{11} = 0.04113\times 6.023\times 10^{23}=0.2477\times 10^{23}

b. moles in 17.75 g of NaCl = \frac{17.75g}{58.5g/mol}=0.3034moles

molecules in 17.75 g of NaCl = 0.3034\times 6.023\times 10^{23}=1.827\times 10^{23}

formula units 17.75 g of NaCl = 0.3034\times 6.023\times 10^{23}=1.827\times 10^{23}

c. moles in 20.06 g of  CuSO_4.5H_2O= \frac{20.06g}{249.68g/mol}=0.08034moles

formula units in 20.06 g of  CuSO_4.5H_2O= 0.08034\times 6.023\times 10^{23}=0.4839\times 10^{23}

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The enthalpy change in a reaction is given by-

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ΔH°rxn = [ (C-C) + 2(C-O) + 4(C-H) + 2(O-H) ] - [ (C-C) + 2(C-O) + 4(C-H) + 2(O-H)  = 0 kJ/mol

<h3>What is enthalpy change?</h3>

Enthalpy change is a measure of the energy emitted or consumed in a reaction. This can be determined using the following equation which involves standard enthalpy of reactant and product formation:

ΔH°rxn = ∑nΔH°f,products - ∑nΔH°f,reactants

<h3>What is bond energy?</h3>

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An unknown compound with a molar mass of 223.94 g/mol consists of 32.18% c, 4.50% h, and 63.32% cl. find the molecular formula f
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The actual number of atoms of each element present in the molecule of the compound is represented by the formula known as molecular formula.

Molar mass of the unknown compound = 223.94 g/mol (given)

Mass of each element present in the unknown compound is determined as:

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\frac{32.18}{100}\times 223.94 = 72.06 g

  • Mass of hydrogen, H:

\frac{4.5}{100}\times 223.94 = 10.08 g

  • Mass of chlorine, Cl:

\frac{63.32}{100}\times 223.94 = 141.79 g

Now, the number of each element in the unknown compound is determined by the formula:

number of moles = \frac{given mass}{molar mass}

  • Number of moles of C:

number of moles = \frac{72.06}{12} = 6.005 mole\simeq 6 mole

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number of moles = \frac{10.08}{1} = 10.08 mole\simeq 10 mole

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number of moles = \frac{141.79}{35.5} = 3.99 mole\simeq 4 mole

Dividing each mole with the smallest number of mole, to determine the empirical formula:

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Multiplying with 2 to convert the numbers in formula into a whole number:

So, the empirical formula is C_{3}H_{5}Cl_{2}.

Empirical mass = 12\times 3+1\times 5+2\times 35.5 = 112 g/mol

In order to determine the molecular formula:

n = \frac{molar mass}{empirical mass}

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So, the molecular formula is:

2\times C_{3}H_{5}Cl_{2} =  C_{6}H_{10}Cl_{4}

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