The answer to the question is D.
Boiling-point elevation is a colligative property.
That means, the the boiling-point elevation depends on the molar content (fraction) of solute.
The dependency is ΔTb = Kb*m
Where ΔTb is the elevation in the boiling point, kb is the boiling constant, and m is the molality.
A solution of 6.00 g of Ca(NO3) in 30.0 g of water has 4 times the molal concentration of a solution of 3.00 g of Ca(NO3)2 in 60.0 g of water.:
(6.00g/molar mass) / 0.030kg = 200 /molar mass
(3.00g/molar mass) / 0.060kg = 50/molar mass
=> 200 / 50 = 4.
Then, given the direct proportion of the elevation of the boiling point with the molal concentration, the solution of 6.00 g of CaNO3 in 30 g of water will exhibit a greater boiling point elevation.
Or, what is the same, the solution with higher molality will have the higher boiling point.
Answer:
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Explanation:
If he was 30.8% too low, it means that he was at 69.2% of the boiling point needed. So 50o C is 69.2% of total.
In order to know what 100% is, you can divide the number by it's percentage and then multiply it by a hundred.
So: 50/30.8=1.623
1.623*100=162.3
So the correct boiling point of the liquid he was working with in the lab is 162.3 oC
Answer:
Nitrification is a biological process by which ammonia is oxidized to nitrites (
) and then to nitrates (
) by action of specialized bacteria in nature (such as <em>nitrosomonas</em> and <em>nitrobacter</em>).
Explanation:
This two-step process utilizes molecular atmospheric oxygen and ammonia and ammonium components of soil and nature.
<em>(a.) Ammonia and ammonium compounds to nitrite</em>
→ 
→ 
<em>(b.) Nitrite to nitrate</em>
→ 
<em>(c.) The reverse process involves reduction of nitrates to molecular nitrogen, called denitrification (with succinic acid in the equation below).</em>
→
(
and
is an intermediate product in this process.)