Answer:
The concentration of sodium chloride in an aqueous solution that is 2.23 M and that has a density of 1.01 g/mL is 12.90% by mass
Explanation:
2.23 M aqueous solution of NaCl means there are 2.23 moles of NaCl in 1000 mL of solution.
We know that density is equal to ratio of mass to volume.
Here density of solution is 1.01 g/mL.
So mass of 1000 mL solution = (
) g = 1010 g
molar mass of NaCl = 58.44 g/mol
So mass of 2.23 moles of NaCl = (
) g = 130.3 g
% by mass is ratio of mass of solute to mass of solution and then multiplied by 100.
Here solute is NaCl.
So % by mass of 2.23 M aqueous solution of NaCl = 
% = 12.90%
Answer:
The mass of water = 219.1 grams
Explanation:
Step 1: Data given
Mass of aluminium = 32.5 grams
specific heat capacity aluminium = 0.921 J/g°C
Temperature = 82.4 °C
Temperature of water = 22.3 °C
The final temperature = 24.2 °C
Step 2: Calculate the mass of water
Heat lost = heat gained
Qlost = -Qgained
Qaluminium = -Qwater
Q = m*c*ΔT
m(aluminium)*c(aluminium)*ΔT(aluminium) = -m(water)*c(water)*ΔT(water)
⇒with m(aluminium) = the mass of aluminium = 32.5 grams
⇒with c(aluminium) = the specific heat of aluminium = 0.921 J/g°C
⇒with ΔT(aluminium) = the change of temperature of aluminium = 24.2 °C - 82.4 °C = -58.2 °C
⇒with m(water) = the mass of water = TO BE DETERMINED
⇒with c(water) = 4.184 J/g°C
⇒with ΔT(water) = the change of temperature of water = 24.2 °C - 22.3 °C = 1.9 °C
32.5 * 0.921 * -58.2 = -m * 4.184 * 1.9
-1742.1 = -7.95m
m = 219.1 grams
The mass of water = 219.1 grams
Answer:
0.0011 mol/L.s
Explanation:
The average rate of disappearing of the reagent is the variation of the concentration of it divided by the time that this variation is being measured. The reaction rate, is proportional to the coefficient of the substance, so, for a generic reaction:
aA + bB --> cC + dD
rate = -(1/a)Δ[A]/Δt = -(1/b)Δ[B]/Δt = (1/c)Δ[C]/Δdt = (1/d)Δ[D]/dt
The minus sign is because of the reagent is desapering, so:
rate = -(1/2)*(0.0209 - 0.0300)/(10 - 6)
rate = 0.0011 mol/L.s
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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!
