Answer:
Molecular formula = H2O
It's molecular formula and empirical formula are same
Explanation:
Answer:
1- Yes, we can calculate the solubility of mineral compound X.
2- 0.012 g/mL.
Explanation:
<em>1- Using only the information above, can you calculate the solubility of X in water at 15.0 °C? </em>
The information available is:
The volume of water sample = 25.0 mL.
Weight of the mineral compound X after evaporation, drying, and washing = 0.30 g.
∴ Yes, we can calculate the solubility of mineral compound X.
<u><em>2- If you said yes, calculate it.</em></u>
∵ 25.0 mL of water sample contains → 0.30 g of the mineral compound X.
∴ 1.0 mL of water sample contains → ??? g of the mineral compound X.
1.0 ml of water sample will contain (0.3 g/25.0 mL) 0.012 g.
<em>∴ The solubility of the mineral compound X in the water sample is</em> <u><em>0.012 g/mL.</em></u>
<u><em></em></u>
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Answer:
Cools to a temperature below freezing.
Explanation:
Hope this helps :D
This is a bit long right now.
(n l m s) these are the numbers.
n=2, second orbital level.
l is the type of orbit: l=1 elongated, l=0 spherical
m=0, magnetic number: 0
s=spin, both have spin positive
You need still to round it up. Srry!