New answers tagged concentration
1
Unfortunately, your calculations are a little bit wrong because of the poor choice of the total initial mass. It is $200 g$, and not $220~ g$. So here are the right calculations.
As $100 g$ water dissolves $109g$ $\ce{KNO3}$ at $60$°C, producing a total mass of $209g$, you may deduce that $200g$ of your initial solution contains $(200/209)·109g = 104.3 g$ $\...
0
When water is evaporated from a solution, the water molecules leave the solution so there are more molecules if the substance left in the solution, so therefore overall the amount of water decreases from the solution and hence there are more substance molecules dissolved than water molecules in comparison to the solution prior to hearing and as the amount of ...
0
Salts do not precipitate from unsaturated solutions.
When saturated salt solution is being evaporated, water content decreases. The solution keeps being saturated, as while water is evaporating, the excessive amount of salt is precipitating. There is less salt in the solution, because there is less of saturated solution.
10 L of saturated solution does not ...
5
The distribution of molecules in solution is not uniform but has a distribution. The calculation of nearest neighbour distances from one molecule to another has been worked out a long time ago see, S. Chandrasekhar, Rev. Mod. Phys. 1943, v15, p1.
Let $w(r)$ be the probability that the nearest neighbour occurs between distance $r$ and $r+dr$. This must be ...
3
An easy way to estimate the intermolecular distance is to break up the container into a regular cell array, each cell containing one solute molecule. First invert the concentration to arrive at the volume per molecule. For instance, if you have a 1M concentration, then the molar volume is 1 L/mol. If you divide the molar volume by Avogadro's number you ...
0
"In this case, the equilibrium constant suggests the products are favoured but if you look at how much product we have compared to reactants there's so much more reactants than products." This is what I take as the heart of your question. To that point:
I believe you need to compare the concentration of all the reactantS, not just one of the ...
1
You have calculated correctly for parts $\bf{a}$ and $\bf{b}$. However, to answer part $\bf{c}$, you need to use the given plot, the interception and slope of which can be calculated as follows:
The slope is $\tan \alpha = - \frac{1}{K_d}$:
$$\tan \alpha = \frac{1-0.2}{0.1-0.8} = -1.14 \ \Rightarrow \ \therefore \ \frac{1}{K_d} = 1.14$$
According to the ...
0
You have 1/84 + 2/106 moles (twice as many sodium ions in a mole of carbonate as in bicarbonate) of sodium ions from the carbonates. Add that to the (0.45 x 0.154) moles of sodium ions from the NaCl. This is your total moles of sodium ions. The total volume of solution is (0.05 + 0.45) L so divide your total moles of sodium ions by 0.5 to get the ...
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