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3 years ago

What is power? (2 points)

Chemistry

1 answer:

damaskus [11]3 years ago

3 0

Answer:

I believe

Explanation:

I believe that the answer is B, but I could be wrong. I think it's How much work can be done in a given time, because, In physics, power is the amount of energy transferred or converted per unit time.

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How to test for sugar in your gingerbread

N76 [4]

Answer:

first of all we have to put sugar in bread and we have to add some ginger paste over it

4 0

3 years ago

A single atom of an element has 21 neutrons, 20 electrons, and 20 protons. Which element is it?

jeyben [28]

Answer:

Calcium

Explanation:

Clacium has atomic number 20. The atomic number. This means that it has 20 protons and 20 electrons. it has 21 neutrons.

5 0

3 years ago

Water is produced from the reaction of hydrogen and oxygen gas, according to the equation below. What is the excess reactant in

liq [111]

Explanation:

Mole ratio of Oxygen to Hydrogen gas = 1 : 2.

If we use 3.0 moles of oxygen gas, we would need 3.0 * 2 = 6.0 mol of hydrogen gas.

However we only have 4.2 mol of hydrogen. Therefore hydrogen is limiting and oxygen is in excess. (B)

3 0

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:

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

3 0

3 years ago

Find the place of the digit 1 in the number 81389398.

dlinn [17]

Answer:

millions

Explanation:

the 1 is seven decimbal places to the left (seven places left of zero), which is the millions place

5 0

3 years ago

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