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!
Answer: 3 moles solute x 1 dm^3/0.60 moles solute = 5 dm^3
Explanation:
Answer:
The maximum mass of water produced is 
Explanation:
From the question we are told that
The mass of sucrose is 
The chemical formula for sucrose is 
The chemical equation for the dissociation of sucrose is
The number of moles of sucrose can be evaluated as
Where Z is the molar mass of sucrose which has a constant value of
So
From the chemical equation one mole of sucrose produces 11 moles of water so 0.585 moles of sucrose will produce x moles of water
Therefore
Now the mass of water produced is mathematically represented as
Where 
is the molar mass of water with a constant values of 
So
Your going to want to round 8.01 to 8.01 and change 3.127 rounded too 3.1
then u want to take 8.0 and 3.1 and times it and u will get 24.8 or 25
Answer:
Explanation:
we know that
ΔH=m C ΔT
where ΔH is the change in enthalpy (j)
m is the mass of the given substance which is water in this case
ΔT IS the change in temperature and c is the specific heat constant
we know that given mass=2.9 g
ΔT=T2-T1 =98.9 °C-23.9°C=75°C
specific heat constant for water is 4.18 j/g°C
therefore ΔH=2.9 g*4.18 j/g°C*75°C
ΔH=909.15 j
