Answer:
To prepare 50L of 32% solution you need: 11L of 30% solution, 22L of 50% solution and 17L of 10% solution.
Explanation:
A 32% solution of acid means 32L of acid per 100L of solution. As the chemist wants to make a solution using twice as much of the 50% solution as of the 30% solution it is possible to write:
2x*50% + x*30% + y*10% = 50L*32%
<em>130x + 10y = 1600 </em><em>(1)</em>
<em>-Where x are volume of 30% solution, 2x volume of 50% solution and y volume of 10% solution-</em>
Also, it is possible to write a formula using the total volume (50L), thus:
<em>2x + x +y = 50L</em>
<em>3x + y = 50L </em><em>(2)</em>
If you replace (2) in (1):
130x + 10(50-3x) = 1600
100x + 500 = 1600
100x = 1100
<em>x = 11L -Volume of 30% solution-</em>
2x = 22L -Volume of 50% solution-
50L - 22L - 11L = 17 L -Volume of 10% solution-
I hope it helps!
Hello!
Ok so for this problem we use the ideal gas law of PV=nRT and I take it that the scientist needs to store 0.400 moles of gas and not miles.
So if we have
n=0.400mol
V=0.200L
T= 23degC= 273k+23c=296k
R=ideal gas constant= 0.0821 L*atm/mol*k
So now we rearrange equation for pressure(P)
P=nRT/V
P=((0.400mol)*(0.0821 L*atm/mol*k)*(296k))/(0.200L) = 48.6 atm of pressure
Hope this helps you understand the concept and how to solve yourself in the future!! Any questions, please feel free to ask!! Thank you kindly!!!
The ccorrect answer is C a generator tubrine
Answer:
Vapour pressure of benzene over the solution is 253 torr
Explanation:
According to Raoult's law for a mixture of two liquid component A and B-
vapour pressure of a component (A) in solution = 
vapour pressure of a component (B) in solution = 
Where
are mole fraction of component A and B in solution respectively
are vapour pressure of pure A and pure B respectively
Here mole fraction of benzene in solution is 0.340 and vapour pressure of pure benzene is 745 torr
So, vapour pressure of benzene in solution = 
= 253 torr
Answer : The mass of sodium bromide added should be, 18.3 grams.
Explanation :
Molality : It is defined as the number of moles of solute present in kilograms of solvent.
Formula used :

Solute is, NaBr and solvent is, water.
Given:
Molality of NaBr = 0.565 mol/kg
Molar mass of NaBr = 103 g/mole
Mass of water = 315 g
Now put all the given values in the above formula, we get:


Thus, the mass of sodium bromide added should be, 18.3 grams.