Hey there!
MgCl₂
Find molar mass of magnesium chloride.
Mg: 1 x 24.305
Cl: 2 x 35.453
--------------------
95.211 grams
One mole of magnesium chloride has a mass of 95.211 grams.
We have 2.40 moles.
2.40 x 95.211 = 228.5
To 3 sig figs this is 229.
The mass of 2.40 moles of magnesium chloride is 229 grams.
Hope this helps!
D the chemical energy in a batery changes to electrical when its used
The given question is incomplete. The complete question is as follows.
Sodium sulfate is slowly added to a solution containing 0.0500 M
and 0.0390 M
. What will be the concentration of
(aq) when
begins to precipitate? What percentage of the
can be separated from the Ag(aq) by selective precipitation?
Explanation:
The given reaction is as follows.

= 0.0390 M
When
precipitates then expression for
will be as follows.
![K_{sp} = [Ag^{+}]^{2}[SO^{2-}_{4}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BAg%5E%7B%2B%7D%5D%5E%7B2%7D%5BSO%5E%7B2-%7D_%7B4%7D%5D)
![1.20 \times 10^{-5} = (0.0390)^{2} \times [SO^{2-}_{4}]](https://tex.z-dn.net/?f=1.20%20%5Ctimes%2010%5E%7B-5%7D%20%3D%20%280.0390%29%5E%7B2%7D%20%5Ctimes%20%5BSO%5E%7B2-%7D_%7B4%7D%5D)
= 0.00788 M
Now, equation for dissociation of calcium sulfate is as follows.

![K_{sp} = [Ca^{2+}][SO^{2-}_{4}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BCa%5E%7B2%2B%7D%5D%5BSO%5E%7B2-%7D_%7B4%7D%5D)
![4.93 \times 10^{-5} = [Ca^{2+}] \times 0.00788](https://tex.z-dn.net/?f=4.93%20%5Ctimes%2010%5E%7B-5%7D%20%3D%20%5BCa%5E%7B2%2B%7D%5D%20%5Ctimes%200.00788)
= 0.00625 M
Now, we will calculate the percentage of
remaining in the solution as follows.

= 12.5%
And, the percentage of
that can be separated is as follows.
100 - 12.5
= 87.5%
Thus, we can conclude that 87.5% will be the concentration of
when
begins to precipitate.
Answer:
97% of earth's water is in the ocean. The rest could be found underground, in glaciers and ice, and in rivers and lakes
To find the acceleration, use the formula: force= mass x acceleration.
12.5 N = 10 x Acceleration
Acceleration = 12.5 / 10
Acceleration = 1.25
Hope this helps :)