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1 year ago

I need help with my homework.Calculate the number of formula units in: 41.5 grams CaBr2

Chemistry

1 answer:

lana66690 [7]1 year ago

5 0

Answer:

$\text{ }1.25\times10^{23}\text{ formula units}$

Explanation:

Here, we want to calculate the number of formula units in the given molecule

We start by getting the number of moles

To get the number of moles, we have to divide the mass given by the molar mass

The molar mass is the mass per mole

The molar mass of calcium bromide is 200 g/mol

Thus, we have the number of moles as follows:

$\frac{41.5}{200}\text{ = 0.2075 mol}$

The number of formula units in a mole is:

$1\text{ mole = 6.02 }\times10^{22}\text{ formula units}$

The number of formula units in 0.2075 mole will be:

$0.2075\text{ }\times6.02\times10^{23}\text{ = }1.25\times10^{23}\text{ formula units}$

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3 years ago

A gas system has an initial number of moles of 0.693 moles with the volume unknown. When the number of moles changes to 0.928 mo

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Answer:

The initial volume in mL is 5959.2 mL

Explanation:

As the number of moles of a gas increases, the volume also increases. Hence, number of moles and volumes are directly proportional i.e

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Where n is the number of moles and V is the volume

Then, n = cV

c is the proportionality constant

∴n/V = c

Hence n₁/V₁ = n₂/V₂

Where n₁ is the initial number of moles

V₁ is the initial volume

n₂ is the final number of moles

and V₂ is the final volume.

From the question,

n₁ = 0.693 moles

V₁ = ?

n₂ = 0.928 moles

V₂ = 7.98 L

Putting the values into the equation

n₁/V₁ = n₂/V₂

0.693 / V₁ = 0.928 / 7.98

Cross multiply

∴ 0.928V₁ = 0.693 × 7.98

0.928V₁ = 5.53014

V₁ = 5.53014/0.928

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2 years ago

The combustion of 0.570 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera

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Answer:

The temperature change from the combustion of the glucose is 6.097°C.

Explanation:

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Enthaply of combustion of benzoic acid = 3,228 kJ/mol

Mass of benzoic acid = 0.570 g

Moles of benzoic acid = $\frac{0.570 g}{122.12 g/mol}=0.004667 mol$

Energy released by 0.004667 moles of benzoic acid on combustion:

$Q=3,228 kJ/mol \times 0.004667 mol=15.0668 kJ=15,066.8 J$

Heat capacity of the calorimeter = C

Change in temperature of the calorimeter = ΔT = 2.053°C

$Q=C\times \Delta T$

$15,066.8 J=C\times 2.053^oC$

$C=7,338.92 J/^oC$

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Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=2.900 g

Moles of glucose = $\frac{2.900 g}{180.16 g/mol}=0.016097 mol$

Energy released by the 0.016097 moles of calorimeter combustion:

$Q'=2,780 kJ/mol \times 0.016097 mol=44.7491 kJ=44,749.1 J$

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

$Q'=C\times \Delta T'$

$44,749.1 J=7,338.92 J/^oC\times \Delta T'$

$\Delta T'=6.097^oC$

The temperature change from the combustion of the glucose is 6.097°C.

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2 years ago

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Answer: all elements in the periodic table is classified as elements

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