Answer:
mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
Explanation:
The partition coefficient of X between ethoxy ethane (ether) and water, K is given by the formula
K = concentration of X in ether/concentration of X in water
Partition coefficient, K(X) between ethoxy ethane and water = 40
Concentration of X in ether = mass(g)/volume(dm³)
Mass of X in ether = m g
Volume of ether = 50/1000 dm³ = 0.05 dm³
Concentration of X in ether = (m/0.05) g/dm³
Concentration of X in water = mass(g)/volume(dm³)
Mass of X in water left after extraction with ether = (5 - m) g
Volume of water = 1 dm³
Concentration of X in water = (5 - m/1) g/dm³
Using K = concentration of X in ether/concentration of X in water;
40 = (m/0.05)/(5 - m)
(m/0.05) = 40 × (5 - m)
(m/0.05) = 200 - 40m
m = 0.05 × (200 - 40m)
m = 10 - 2m
3m = 10
m = 10/3
m = 3.33 g of X
Therefore, mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
Answer:
-411 kj
Explanation:
We solve by using this formula
∆U = ∆Q + ∆W
This formula is the first law of thermodynamics
Change in internal energy U = +241
Heat gained by system Q = 652
Putting the value into the equation
+241 = 652 + W
Workdone = 241 - 652
Workdone = -411 kj
Since work done is negative it means that work was done by the system
Answer is: line be long 3,011·10¹³ kilometers.
diametar of virus = 5·10⁻⁶ cm ÷ 100000 = 5·10⁻¹¹ km.
line lenght = 5·10⁻¹¹ km · 6,023·10²³.
line lenght = 3,011·10¹³ km.
Avogadro number = 6,023·10²³.
1 cm = 10⁻² m = 10⁻⁵ km.
Hey there!
Na + H₂O → NaOH + H₂
First, balance O.
One on the left, one on the right. Already balanced.
Next, balance H.
Two on the left, three on the right. Let's add a coefficient of 2 in front of NaOH and a coefficient of 2 in front of H₂O, so we have 4 on each side.
Na + 2H₂O → 2NaOH + H₂
Lastly, balance Na.
One on the left, two on the right. Add a coefficient of 2 in front of Na.
2Na + 2H₂O → 2NaOH + H₂
This is our final balanced equation.
Hope this helps!
