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2 years ago

9

Exactly 1 mol Hg2(NO3)2 contains how many moles of Hg , N , and O ?

1 answer:

marin [14]2 years ago

6 0

In exactly 1 mol Hg₂(NO₃)₂ , there are 2 mol Hg, 2 mol N and 6 mol O.

Since the molecular formula of Hg₂(NO₃)₂ shows that for every mole of the substance, we have 2 moles of Hg, 2 moles of N and 6 moles of O.

So, in exactly 1 mol Hg₂(NO₃)₂ , there are 2 mol Hg, 2 mol N and 6 mol O.

Learn more about number of moles here:

brainly.com/question/3935424

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C.) That all matter was composed of earth, fire, water and air

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Explain intermolecular forces cause water molecules to stick to a material like a glass pitcher

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2 years ago

Match the sentences with the steps of the scientific method.

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3 years ago

In a particular experiment, 2.50-g samples of each reagent are reacted. The theoretical yield of lithium nitride is ________ g.

Neporo4naja [7]

Answer:

4.18 g

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Given: For Li

Given mass = 2.50 g

Molar mass of Li = 6.94 g/mol

<u>Moles of Li = 2.50 g / 6.94 g/mol = 0.3602 moles</u>

Given: For $N_2$

Given mass = 2.50 g

Molar mass of $N_2$ = 28.02 g/mol

<u>Moles of $N_2$ = 2.50 g / 28.02 g/mol = 0.08924 moles</u>

According to the given reaction:

$6Li+N_2\rightarrow 2Li_3N$

6 moles of Li react with 1 mole of $N_2$

1 mole of Li react with 1/6 mole of $N_2$

0.3602 mole of Li react with $\frac {1}{6}\times 0.3602$ mole of $N_2$

Moles of $N_2$ that will react = 0.06 moles

Available moles of $N_2$ = 0.08924 moles

$N_2$ is in large excess. (0.08924 > 0.06)

Limiting reagent is the one which is present in small amount. Thus,

Li is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

6 moles of Li gives 2 mole of $Li_3N$

1 mole of Li gives 2/6 mole of $Li_3N$

0.3602 mole of Li react with $\frac {2}{6}\times 0.3602$ mole of $Li_3N$

Moles of $Li_3N$ = 0.12

Molar mass of $Li_3N$ = 34.83 g/mol

Mass of $Li_3N$ = Moles × Molar mass = 0.12 × 34.83 g = 4.18 g

<u>Theoretical yield = 4.18 g</u>

5 0

3 years ago

Butane C4H10 (g),(Hf = –125.7), combusts in the presence of oxygen to form CO2 (g) (Delta.Hf = –393.5 kJ/mol), and H2O(g) (Delta

shusha [124]

Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ

Explanation:

The balanced chemical reaction is,

$2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The expression for enthalpy change is,

$\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]$

Putting the values we get :

$\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]$

$\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)]$

$\Delta H=-5314.8kJ$

2 moles of butane releases heat = 5314.8 kJ

1 mole of butane release heat = $\frac{5314.8}{2}\times 1=2657.4kJ$

Thus enthalpy of combustion per mole of butane is -2657.4 kJ

3 0

3 years ago