All categories

3 years ago

Which of the following ions came from an atom that gained 2 electrons?

1 answer:

4 0

Group 6 elements like Oxygen like to take on the extra two electrons to become a 2- anion and take on the noble gas configuration. so the O^2-, which looks like O=O, will have two oxygens sharing two electrons.... answer is O^2-

You might be interested in

HELP ME ASAP

gtnhenbr [62]

Answer:D

Explanation:

7 0

2 years ago

How Earth and the sun move in space relative to each other

liberstina [14]

The two theories were:

1.Geocentric theory stating that earth is the centre of the universe and sun revolves around it.

2.Heliocentric theory stating that the earth revolves around the sun.

6 0

2 years ago

⚠️⚠️ ⚠️⚠️

Anastaziya [24]

Answer/Explanation: A weight lifted vertically to acquire gravitational potential energy. A weight lifted vertically to acquire gravitational potential energy. If the force were to be removed, the object would fall back down to the ground and the gravitational potential energy would be transferred to kinetic energy of the falling object.

3 0

2 years ago

Read 2 more answers

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

List 10 ways to reduce air pollution at home

Natasha_Volkova [10]

Don't smoke, drive less, don't let toast burn, use an electrical fire instead of a wood one....
I can't think of anything else.

8 0

3 years ago