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lyudmila [28]
2 years ago
11

What is the name of this moleculee?

Chemistry
1 answer:
o-na [289]2 years ago
6 0

Answer:

The chemical name of C6H6 is Benzene

Explanation:

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How many moles are in 4.5 g of<br> Sodium Chloride, NaCl?
Sergeu [11.5K]

Answer:

The answer to the question is 0.07 moles

6 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
What continent has no deserts
nevsk [136]

Answer:

Antarctica.

Explanation:

It’s all ice.

4 0
3 years ago
If element X has 65 protons, how many electrons does it have
Tomtit [17]
In an atom there are the same number of protons as electrons to start with. The answer would be 65 aswell.

8 0
3 years ago
Read 2 more answers
A chemist determines by measurements that 0.0550 moles of nitrogen gas participate in a chemical reaction. Calculate the mass of
shtirl [24]

The mass of nitrogen gas that participated in the chemical reaction is 1.54g

HOW TO CALCULATE MASS OF AN ELEMENT:

  • Mass of a substance can be calculated by multiplying the number of moles in mol of the substance by its molecular mass in g/mol. That is;

  • mass (M) = molar mass (MM) × number of moles (n)

According to this question, a chemist determines by measurements that 0.0550 moles of nitrogen gas (N2) participate in a chemical reaction.

  • The molecular mass of nitrogen gas (N2) = 14.01(2)

= 28.02g/mol

Hence, the mass of the nitrogen gas that participated in the chemical reaction is calculated as follows:

  • Mass (g) = 0.0550 mol × 28.02 g/mol

  • Mass = 1.5411

Therefore, the mass of nitrogen gas that participated in the chemical reaction is 1.54g

Learn more: brainly.com/question/18269198

6 0
2 years ago
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