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Dennis_Churaev [7]
2 years ago
11

During the analysis, 0.00905 mol H2O is formed. Calculate the amount (mol) H in 0.00905 mol H2O.

Chemistry
1 answer:
sergeinik [125]2 years ago
5 0

Answer:

0.0181 mol H

Explanation:

Step 1: Given data

Moles of water (H₂O) formed during the analysis: 0.00905 mol H₂O

Step 2: Calculate the amount (mol) H in 0.00905 mol H₂O

According to the chemical formula of water, the molar ratio of water to hydrogen is 1:2, that is, there are 2 moles of H per 1 mole of H₂O. We will use this conversion factor to calculate the moles of H in 0.00905 moles of H₂O.

0.00905 mol H₂O × (2 mol H/1 mol H₂O) = 0.0181 mol H

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A bird building a nest for its young is a reproductive strategy. Based on this example, what do you think the term “reproductive
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If a 60-g object has a volume of 30 cm3, what is its density? 2 g/cm3 0.5 cm3/g 1800 g * cm3 none of the above
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Solution : Given,

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Volume of an object = 30cm^3

Formula used :

\text{Density of an object}=\frac{\text{Mass of an object}}{\text{Volume of an object}}

Now put all the given values in this formula, we get the density of an object.

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if 5.50 mol of calcium carbide (CaC 2 ) reacts with an excess of water, how many moles of acetylene (C 2 H 2 ) , a gas used in w
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Answer:

5.5moles

Explanation:

CaC2 + 2H2O —> Ca(OH)2 + C2H2

From the equation, the following were observed:

1mole of CaC2 reacted to produced 1mol of C2H2.

Therefore, 5.5moles of CaC2 will also produce 5.5moles of C2H2

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The decomposition of copper(II) nitrate on heating is endothermic reaction. 2Cu(NO3)2(s) → 2C10(s) + 4NO2(g) + O2(g) Calculate t
Basile [38]

Answer:

The enthalpy change for the given reaction is 424 kJ.

Explanation:

2Cu(NO_3)_2(s)\rightarrow 2CuO(s) + 4NO_2(g) + O_2(g),\Delta H_{rxn}=?

We have :

Enthalpy changes of formation of following s:

\Delta H_{f,Cu(NO_3)_2}=-302.9 kJ/mol

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

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

\Delta H_{rxn} =

=(2 mol\times \Delta H_{f,CuO}+4\times \Delta H_{f,NO_2}+1 mol\times \Delta H_{f,O_2})-(2mol\times \Delta H_{f,Cu(NO_3)_2})

\Delta H_{rxn}=

(2mol\times (-157.3 kJ/mol)+4\times 33.2 kJ/mol=1 mol\times 0 kJ/mol)-(2 mol\times (-302.9 kJ/mol)

\Delta H_{rxn}=424 kJ

The enthalpy change for the given reaction is 424 kJ.

6 0
2 years ago
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