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Veronika [31]
2 years ago
6

Which statement best explains why heating a liquid affects its viscosity?

Chemistry
2 answers:
mylen [45]2 years ago
4 0
The statement that best explains why heating a liquid affects its viscosity is that "The molecules move faster at higher temperatures and overcome attractions more easily."<span>. Remember that viscosity is a physical property of the fluids that measure the resistance (opposition) to flow and it, generally decreases, as the temperature increases and the intermolecular force decrease.</span>
AleksAgata [21]2 years ago
4 0

Answer:

aka: A

Explanation:

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You are out on a hike, and you notice that the line of telephone wires you have been following along a slope is crooked. Two of
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73.5 mol of P4O10 contains how many moles of P ? moles of P:
Sholpan [36]

Answer:

294 moles of P

Explanation:

For every 1 mol of P4O10 contains 4 mol of P

so;

73.5 mol P4O10 × <u> </u><u> </u><u> </u><u> </u><u> </u><u>4 mol P</u><u> </u><u> </u><u> </u><u> </u><u> </u>

1 mol P4O10

= 73.5 × 4

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7 0
2 years ago
During an experiment, 95 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. If the percent yield of
almond37 [142]

Answer:

Actual yield: 86.5 grams.

Explanation:

How many moles of formula units in 95 grams of calcium carbonate \rm CaCO_3?

Refer to a modern periodic table for relative atomic mass data:

  • Ca: 40.078;
  • C: 12.011;
  • O: 15.999.

Formula mass of \rm CaCO_3:

M(\mathrm{CaCO_3})  = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{1\times 12.011}_{\rm C} + \underbrace{3\times 15.999}_{\rm O} = \rm 100.086\;g\cdot mol^{-1}.

\displaystyle n(\mathrm{CaCO_3}) = \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} = \rm \frac{95\;g}{100.086\;g\cdot mol^{-1}} = 0.949184\;mol.

How many moles of \rm CaCl_2 will be produced?

The coefficient in front of \rm CaCO_3 in the chemical equation is the same as that in front of \rm CaCl_2. That is:

\displaystyle \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = 1.

\displaystyle n(\mathrm{CaCl_2}) = n(\mathrm{CaCO_3})\cdot \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = n(\mathrm{CaCO_3}) = \rm 0.949184\;mol.

What's the theoretical yield of calcium chloride? In other words, what's the mass of \rm 0.949184\;mol of \rm CaCl_2?

Again, refer to a periodic table for relative atomic data:

  • Ca: 40.078;
  • Cl: 35.45.

M(\mathrm{CaCl_2}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{2\times 35.45}_{\rm Cl} = \rm 110.978\;g\cdot mol^{-1}.

\begin{aligned}m(\mathrm{CaCl_2}) &= n(\mathrm{CaCl_2})\cdot M(\mathrm{CaCl_2})\\ &= \rm 0.949184\;mol\times 110.978\;g\cdot mol^{-1}\\ &= \rm 105.339\; g\end{aligned}.

What's the actual yield of calcium chloride?

\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%.

\displaystyle \begin{aligned}\text{Actual Yield} &= \text{Theoretical Yield}\cdot \frac{\text{Percentage Yield}}{100\%}\\ &=\rm 105.339\; g \times \frac{82.15\%}{100\%}\\&= \rm 86.5\;g \end{aligned}.

8 0
3 years ago
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