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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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The amount of heat lost by the copper is 2402.4 J.

To Calculate the amount of heat lost, we use the formula below.

<h3>Formula:</h3>

Q = cm(t₂-t₁)................. Equation 1

<h3>Where:</h3>

Q = Amount of heat lost
c = specific heat capacity of copper
m = mass of copper
t₂ = Final temperature
t₁ = Initial temperature

From the question,

<h3>Given:</h3>

m = 78 g
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Substitute these values into equation 1.

Q = 78(0.385)(120-40)
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Learn more about heat here: brainly.com/question/13439286

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3 years ago

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AysviL [449]

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<u>Explanation:</u>

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

For the given chemical reaction:

$2NH_3(g)+3O_2(g)+2CH_4(g)\rightarrow 2HCN(g)+6H_2O(g)$

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

$\Delta H_{rxn}=[(2\times \Delta H_f_{(HCN)})+(6\times \Delta H_f_{(H_2O)})]-[(2\times \Delta H_f_{(NH_3)})+(3\times \Delta H_f_{(O_2)})+(2\times \Delta H_f_{(CH_4)})]$

We are given:

$\Delta H_f_{(H_2O)}=-241.8kJ/mol\\\Delta H_f_{(NH_3)}=-80.3kJ/mol\\\Delta H_f_{(CH_4)}=-74.6kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol\\\Delta H_{rxn}=-870.8kJ$

Putting values in above equation, we get:

$-870.8=[(2\times \Delta H_f_{(HCN)})+(6\times (-241.8))]-[(2\times (-80.3))+(3\times (0))+(2\times (-74.6))]\\\\\Delta H_f_{(HCN)}=135.1kJ$

Hence, the $\Delta H_f$ for HCN (g) in the reaction is 135.1 kJ/mol.

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densk [106]

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