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3 years ago

The amount of stored chemical energy is what determines the temperature of a substance

Chemistry

2 answers:

Setler79 [48]3 years ago

3 0

Answer:

False

Explanation:

Temperature is the degree of hotness or coldness of a body. Temperature is as a result of how much heat a body has. Heat is a form of energy called thermal energy. Heat can be transferred from one body to another as a result of differences in temperature between two bodies.

Chemical energy does not determine the temperature of a body. It is the amount of heat it contains.

tigry1 [53]3 years ago

3 0

Answer:

False / Not True

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What is a chemical equation for the second ionization for lithium

Orlov [11]

The 2nd ionization energy is removing a 2nd electron from that resulting cation:
<span>Li+ --> Li2+ + 1e- </span>

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3 years ago

How many grams of water will form if 10.54 g H2 react with 95.10 g O2?

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Explanation:

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3 years ago

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:

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

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Answer:

A. Sympathetic

and

C. Parasympathetic

Explanation:

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What volume of a 3.50 M Na3PO4 solution would you use to make 1.50 L of a 2.55 M Na3PO4 solution?

kati45 [8]

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