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2 years ago

What is the density of sand if the volume is 40.0 ml and the mass is 62.3 g?

Chemistry

2 answers:

4vir4ik [10]2 years ago

7 0

Density = mass/volume. 62.3g/40.0ml=1.5575g/ml which should be rounded to 1.6g/ml

Llana [10]2 years ago

3 0

Density: mass÷ volume
so 62.3 ÷ 40 = 1.5575

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2 years ago

A ball is rolling across the floor. Every 3 seconds it travels 12 meters. Its speed, therefore, is 36 m/s. Is this problem solve

kirill [66]

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2 years ago

Science, please help and thank you!!

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2 years ago

Which is the limiting reactant if we start with 30.0 g Al and 30.0 h O2

lorasvet [3.4K]

Explanation:

The limiting reactant will be Al:

4Al + 3O₂ → 2Al₂O₃

The limiting reactant is the reactant in short supply in a chemical reaction.

Given parameters:

Mass of Al = 30g Molar mass = 27g/mol

Number of moles = $\frac{mass}{molar mass}$ = $\frac{30}{27}$

Number of moles of Al = 1.111 mole

Mass of O₂ = 30g, molar mass = 32g/mol

Number of moles = $\frac{30}{32}$ = 0.94mol

In the reaction:

4 moles of Al reacted with 3 moles of O₂

1.11moles of Al will require $\frac{1.11 x3}{4}$ = 0.83mole to react

But we have been given 0.94mole of O₂. This is more than required.

Therefore O₂ is in excess and Al is the limiting reactant.

Learn more:

Limiting reagents brainly.com/question/6078553

#learnwithBrainly

4 0

2 years ago

An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the

Brrunno [24]

Answer:

Approximately $81.84\%$ .

Explanation:

Balanced equation for this reaction:

${\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_{3}}\, (s)$ .

Look up the relative atomic mass of elements in the limiting reactant, $\rm CaCl_{2}$ , as well as those in the product of interest, $\rm CaCO_{3}$ :

$\rm Ca$ : $40.078$ .
$\rm Cl$ : $35.45$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass for both the limiting reactant and the product of interest:

$\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}$ .

$\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the quantity of the limiting reactant ( $\rm CaCl_{2}$ ) available to this reaction:

$\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}$ .

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant ( $\rm CaCl_{2}$ ) and the product ( ${\rm CaCO_{3}}$ ) are both $1$ . Thus:

$\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1$ .

In other words, for every $1\; \rm mol$ of $\rm CaCl_{2}$ formula units that are consumed, $1\; \rm mol\!$ of $\rm CaCO_{3}$ formula units would (in theory) be produced. Thus, calculate the theoretical yield of $\rm CaCO_{3}\!$ in this experiment:

$\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}$ .

Calculate the theoretical yield of this experiment in terms of the mass of $\rm CaCO_{3}$ expected to be produced:

$\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}$ .

Given that the actual yield in this question (in terms of the mass of $\rm CaCO_{3}$ ) is $13.19\; \rm g$ , calculate the percentage yield of this experiment:

$\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}$ .

6 0

2 years ago