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3 years ago

Why is it important to keep good records during a scientific investigation?

Chemistry

1 answer:

love history [14]3 years ago

5 0

Answer:

everything needs to be documented in case you need to come back to it and to have evidence for yourself and others. as they say if it is not documented it wasn't done

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Explanation:

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3 years ago

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3 years ago

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.

6 0

2 years ago