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LekaFEV [45]
3 years ago
13

Which of the following statements describes a physical property?

Chemistry
2 answers:
castortr0y [4]3 years ago
8 0


The substance has a higher density than water


enot [183]3 years ago
3 0

its The substance that has the higher density in water.


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Why does the lack of snow in the mountains make drought cycles worse
expeople1 [14]

Answer:

lack of snow in the mountains makes drought cycles worse because mountains are a natural part of the water cycle.

Explanation:

Mountains play a critical part In the water cycle by capturing moisture in the air and it precipitates as snow and when it does this the snow will melt in the warmer seasons and flow down stream then following the rest of the water cycle if we didn't have this then the streams would run less and it would greatly disrupt the water cycle.

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2 years ago
How can the rate constant be determined from the rate law?
Vinil7 [7]

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B. The rate constant is the reaction rate divided by the concentration

terms.

Explanation:

The rate constant can be determined from the rate law because it is the reaction rate divided by the concentration terms. I hope I could help! :)

7 0
2 years ago
In order to prepare very dilute solutions, a lab technician chooses to perform a series of dilutions instead of measuring a very
laiz [17]

Answerit is a

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3 years ago
PLEASE HELP <br> C. Balance these fossil-fuel combustion reactions. (1 point)
Damm [24]

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CH4(g) + _2___O2(g) -> ____CO2(g) + _2___H2O(g) + heat

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I hope it helps!

7 0
3 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
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