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Veronika [31]
3 years ago
10

Think about a large pitcher of ice water and a small cup of ice water with a similar temperature. Which statement about these ob

jects is true? A. Each has the same thermal energy because each has the same temperature. B. The molecules in the cup are moving faster than the molecules in the pitcher. C. The cup of water has more thermal energy than the pitcher of water. D. The pitcher of water has more thermal energy than the cup of water.
Chemistry
1 answer:
Bezzdna [24]3 years ago
3 0
B because ice water is water which molocules move fast like air
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Vinil7 [7]

Answer:

3 Handsprings

Explanation:

18 divided by 6 is 3

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The compound trimethylamine, (CH3)3N, is a weak base when dissolved in water. Write the Kb expression for the weak base equilibr
Zina [86]

Answer:

Kb = [OH⁻] . [C₃H₉NH⁺] / [ C₃H₉N ]

Explanation:

The equation for the reaction of trimethylamine when it is dissolved in water is:

C₃H₉N  +  H₂O ⇄ C₃H₉NH⁺ + OH⁻   Kb

1 mol of trimethylamine catches a proton from the water in order to produce trimethylamonium.

It is a base, because it give OH⁻ to the medium

Expression for Kb (Molar concentration)

Kb = [OH⁻] . [C₃H₉NH⁺] / [ C₃H₉N ]

7 0
3 years ago
When molten rock undergoes crystallization above ground, what type of rock results?
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If you change the state of something, say, liquid to solid, is it a physical or chemical change?
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3 years ago
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A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c
natka813 [3]

Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

  • Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

  • The mass of one mole of CaCO₃ is the same as the molar mass of this compound: \rm 100\; g.
  • The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: \rm 40.078\; g.

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}.

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be \rm 30 \% \times 100\; g = 30\; g of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio \displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}:

\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}.

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%.

3 0
3 years ago
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