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3 years ago

How many water molecules are in a block of ice containing 1.75 mol of water (H2O)?

2 answers:

6 0

1.75 mol H2O x (6.022x10^23 molecules H2O / 1 mol H2O) = 1.05x10^24 molecules H2O if you need a further example let me know

Nostrana [21]3 years ago

3 0

Answer: The number of molecules present in the given moles of water are $1.054\times 10^{24}$

Explanation:

We are given:

Moles of water = 1.75 moles

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules.

So, 1.75 moles of water will contain = $1.75\times 6.022\times 10^{23}=1.054\times 10^{24}$ number of molecules.

Hence, the number of molecules present in the given moles of water are $1.054\times 10^{24}$

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3 years ago

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When calcium carbonate is heated, it decomposes to produce calcium oxide and carbon dioxide, as shown in the diagram below. How

Vikki [24]

Explanation:

Before we start with any calculation, we need to determine the stoichiometric relationship between the reactants consumed and the products formed. In other words, we need to establish a balanced chemical equation that describes the decomposition of calcium carbonate as shown below.

$\text{CaCO}_{3} \ \ \text{(s)} \ \overset{\Delta}{\longrightarrow} \ \text{CaO} \ \text{(s)} \ + \ \text{CO}_{2} \ \text{(g)}$ ,

where the uppercase delta symbol, $\Delta$ , above the chemical reaction arrow denotes that heat is applied to make the reaction proceed in the direction of the arrow.

To calculate how many moles are there in 4 grams of calcium carbonate, we use the relation

$n \ = \ \displaystyle\frac{m}{M_{r}}$ ,

where $n$ is the number of moles, $m$ is the mass of the chemical substance and $M_{r}$ is the molar mass of the chemical substance. We first need to identify the molar mass of calcium carbonate,

$M_{r} \, [\text{CaCO}_{3}] \ = \ A_{r} \, [\text{Ca}] \ + \ A_{r} \, [\text{C}] \ + \ 3 \, \times \, A_{r} \, [\text{O}] \\ \\ M_{r} \, [\text{CaCO}_{3}] \ = \ 40.078 \ + \ 12.011 \ + \ 3 \, \times \, 15.999 \\ \\ M_{r} \, [\text{CaCO}_{3}] \ = \ 100.086 \ \text{g mol}^{-1}$

Hence,

$n \ = \ \displaystyle\frac{4 \ \text{g}}{100.086 \ \text{g mol}^{-1}} \\ \\ \\ n \ = \ 0.04 \ \text{mol}$ .

We know that, from the balanced chemical equation above, 1 mole of calcium carbonate following decomposition, will theoretically lead to the production of 1 mole of carbon dioxide. Then, if 0.04 moles of calcium carbonate is decomposed, then, 0.04 moles of carbon dioxide is produced.

To convert the number of moles into the mass of carbon dioxide produced, we need to rearrange the formula as follows.

$n \ = \ \displaystyle\frac{m}{M_{r}} \\ \\ \\m \ = \ n \ \times \ {M_{r}}$ .

Thus,

$M_{r} \, [\text{CO}_{2}] \ = \ A_{r} \, [\text{C}] \ + \ 2 \, \times \, A_{r} \, [\text{O}] \\ \\ M_{r} \, [\text{CO}_{2}] \ = \ (12.011 \ + \ 2 \, \times \, 15.999) \ \text{g mol}^{-1} \\ \\ M_{r} \, [\text{CO}_{2}] \ = \ 44.009 \ \text{g mol}^{-1}$

$\rule{12cm}{0.01cm}$

$m \ = \ 0.04 \ \text{mol} \ \times \ 44.009 \ \text{g mol}^{-1} \\ \\ m \ = \ 1.76 \ \text{g}$

Therefore, 4 grams of calcium carbonate yields 1.76 grams of carbon dioxide following decomposition.

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