A 613 ml sample of hydrogen gas at 968 torr is held at a constant temperature while it is compressed to 157 ml. what is the new
1 answer:
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The question is incomplete, complete question is;
A solution of is added dropwise to a solution that contains
of
and
and
.
What concentration of is need to initiate precipitation? Neglect any volume changes during the addition.
value
value
What concentration of is need to initiate precipitation of the first ion.
Answer:
Cadmium carbonate will precipitate out first.
Concentration of is need to initiate precipitation of the cadmium (II) ion is
.
Explanation:
1)
The expression of an solubility product of iron(II) carbonate :
2)
The expression of an solubility product of cadmium(II) carbonate :
On comparing the concentrations of carbonate ions for both metallic ions, we can see that concentration to precipitate out the cadmium (II) carbonate from the solution is less than concentration to precipitate out the iron (II) carbonate from the solution.
So, cadmium carbonate will precipitate out first.
And the concentration of carbonate ions to start the precipitation of cadmium carbonate we will need concentration of carbonate ions greater than the concentration.
The answer for the following question is mentioned below.
<u><em>Therefore no of moles present in the gas are 1.12 moles</em></u>
Explanation:
Given:
Pressure of gas (P) = 1.2 atm
Volume of a gas (V) = 50.0 liters
Temperature (T) =650 K
To calculate:
no of moles present in the gas (n)
We know;
According to the ideal gas equation;
We know;
<u>P × V = n × R × T </u>
where,
P represents pressure of the gas
V represents volume of the gas
n represents no of the moles of a gas
R represents the universal gas constant
where the value of R is 0.0821 L atm mole^{-1} K^-1
T represents the temperature of the gas
As we have to calculate the no of moles of the gas;
n =
n = \frac{1.2*50.0}{0.0821*650}
n = \frac{60}{53.365}
n = 1.12 moles
<u><em>Therefore no of moles present in the gas are 1.12 moles</em></u>