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4 years ago

A sample of gold (Au) has a mass of 35.12 g.

Chemistry

1 answer:

rjkz [21]4 years ago

7 0

Answer:

a) 0.1783 mol

b) 1.07*10^23 atoms

Explanation:

Step 1: Data given

Mass of gold (Au) = 35.12 grams

Molar mass of gold (Au) = 196.97 g/mol

Number of Avogadro = 6.022 *10^23 /mol

Step 2: Calculate number of moles gold

Number of moles = Mass of gold / molar mass of golf

Number of moles = 35.12 grams / 196.97 g/mol

Number of moles = 0.1783 moles

Step 3: Calculate number of atoms of gold

Number of atoms = number of moles * Number of Avogadro

Number of atoms = 0.1783 * 6.022 *10^23/ mol

Number of atoms = 1.07*10^23 atoms

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Kr 5s2 is the correct noble gas configuration for strontium

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4 years ago

A compound is 7.74% hydrogen and 92.26% carbon by mass. At 100°C a 0.6883 g sample of the gas occupies 250 mL when the pressure

ycow [4]

Answer: The molecular formula for the compound is $C_6H_6$

Explanation:

We are given:

Percentage of C = 92.26 %

Percentage of H = 7.74 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 92.26 g

Mass of H = 7.74 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{92.26g}{12g/mole}=7.68moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{7.74g}{1g/mole}=7.74moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 7.68 moles.

For Carbon = $\frac{7.68}{7.68}=1$

For Hydrogen = $\frac{7.74}{7.68}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 1

The empirical formula for the given compound is $CH$

Calculating the molar mass of the compound:

To calculate the molecular mass, we use the equation given by ideal gas equation:

PV = nRT

Or,

$PV=\frac{m}{M}RT$

where,

P = pressure of the gas = 820 torr

V = Volume of gas = 250 mL = 0.250 L (Conversion factor: 1 L = 1000 mL )

m = mass of gas = 0.6883 g

M = Molar mass of gas = ?

R = Gas constant = $62.3637\text{ L. torr }mol^{-1}K^{-1}$

T = temperature of the gas = $100^oC=(100+273)K=373K$

Putting values in above equation, we get:

$820torr\times 0.250L=\frac{0.6883g}{M}\times 62.3637\text{ L torr }mol^{-1}K^{-1}\times 373K\\\\M=\frac{0.6883\times 62.3637\times 373}{820\times 0.250}=78.10g/mol$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 78.10 g/mol

Mass of empirical formula = 13 g/mol

Putting values in above equation, we get:

$n=\frac{78.10g/mol}{13g/mol}=6$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 6)}H_{(1\times 6)}=C_6H_6$

Hence, the molecular formula for the compound is $C_6H_6$

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An experiment produced 0.10 g CO2, with a volume of 0.056 L at STP. If the accepted density of CO2 at STP is 1.96 g/L, what is t

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The density of CO2 getting from experiment is 0.1/0.056 = 1.79 g/L. The percent error of this is (1.96 -1.79)/1.96*100%=8.67%. So the approximate percent error is 8.67%.

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Answer:

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Explanation:

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