Answer:

Explanation:
The oxidation state of S is (-2) and that of Mg is (+2) since Mg loses two electrons to S
Hi there! Let's solve this problem shall we!
⠀Volume = 10g
Mass = 2 mL
In this specific problem, they are asking us to find the <u><em>density </em></u>of the object. So,<u><em> using the information given to us</em></u> (volume and mass), let's solve the problem!
Now, if you remember, D = M ÷ V
So, let's fill in the blanks!
D = Our unknown value
M = 2mL
V = 10g
Here is the filled out formula:
D = M ÷ V
D = 2mL ÷ 10g
D = 5 g/mL
*Make sure you put the units for your final solution!*
It Emeterio near the North Pole of the worlddddddd
The substance that releases the greatest amount of ions will have the greatest attractive forces within its solution, resulting in a reduced freezing point.
K₂SO₄ yields 3 ions
NH₄I yields 2 ions
CoCl₃ yields 4 ions
Freezing points:
CoCl₃ < K₂SO₄ < NH₄I
Answer:
1.58×10E18
Explanation:
Since we have the reduction potentials we could make decisions regarding which one will be the anode or cathode. Evidently, bromine having the more positive reduction potential will be the cathode while the iodine will be the anode.
E°cell= 1.07- 0.53= 0.54 V
E°cell= 0.0592/n logK
0.54 = 0.0592/2 logK
logK= 0.54/0.0296
logK= 18.2
K= Antilog (18.2)
K= 1.58×10^18