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2 years ago

A breeder reactor can be used to produce electricity by fission.

1 answer:

8 0

es, such reactors are possible. The main feature is that a blanket of uranium surrounds the active core, and by absorbing neutrons some of this transmutes to plutonium, which can then be used to make the active core of a further reactor and so on, hence the term 'breeder

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The density of benzene at 15 ∘C is 0.8787 g/mL. Calculate the mass of 0.1200 L of benzene at this temperature.

Pavel [41]

Use the density to convert volume into mass.

since the density is in g/ml and the volume was given in Liters, we need to first convert the Liters into mililiters. just multiply by 1000 or move the decimal three times.

0.1200 Liters= 120.0 mL

120.0 mL (0.8787 grams/ 1 mL)= 105 grams

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2 years ago

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Which is the formula mass of Na₂SO₄? * 1. 119 amu 2. 125 amu 3. 142 amu 4. 174 amu

ANTONII [103]

Answer:

the answer is option 3.

4 0

2 years ago

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One hundred students were surveyed about who they would vote for. What is

GuDViN [60]

Hey hope this helps ⬇️

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2 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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2 years ago

Which of the following is the correct angle between the atoms in a tetrahedral molecule? A. 90o B. 109o C. 120o D. 270o

sineoko [7]

The correct angle between the atones in a tetrahedral molecule would be B. 109 degrees.

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3 years ago