If 0.5 L of solution contains 4 mol
then let 1 L of solution contain x mol
⇒ (0.5 L) x = (4 mol) (1 L)
x = (4 mol · L) ÷ (0.5 L)
x = 8 mol
Thus the molarity of the Sodium Chloride solution is 8 mol / L OR 8 mol/dm³.
Answer:
70.0 %
Explanation:
Step 1: Given data
- Mass of nitrogen (mN): 74.66 g
- Mass of the compound (mNxOy): 250 g
Step 2: Calculate the mass of oxygen (mO) in the compound
The mass of the compound is equal to the sum of the masses of the elements that form it.
mNxOy = mN + mO
mO = mNxOy - mN
mO = 250 g - 74.66 g = 175 g
Step 3: Determine the percent composition of oxygen in the sample
We will use the following expression.
%O = mO / mNxOy × 100%
%O = 175 g / 250 g × 100% = 70.0 %
Answer:
Qsp > Ksp, BaCO3 will precipitate
Explanation:
The equation of the reaction is;
Na2CO3 + BaBr2 -------> 2NaBr + BaCO3
Since BaCO3 may form a precipitate we can determine the Qsp of the system.
Number of moles of Na2CO3 = 0.96g/106 g/mol = 9.1 * 10^-3 moles
concentration of NaCO3 = number of moles/volume of solution = 9.1 * 10^-3 moles/10 L = 9.1 * 10^-4 M
Number of moles of BaBr2 = 0.20g/297 g/mol = 6.7 * 10^-4 moles
concentration of BaBr2 = 6.7 * 10^-4 moles/10 L = 6.7 * 10^-5 M
Hence;
[Ba^2+] = 6.7 * 10^-5 M
[CO3^2-] = 9.1 * 10^-4 M
Qsp = [6.7 * 10^-5] [9.1 * 10^-4]
Qsp = 6.1 * 10^-8
But, Ksp for BaCO3 is 5.1*10^-9.
Since Qsp > Ksp, BaCO3 will precipitate
There are 6.022*10^23 molecules in 1 mole of carbon
So how many will moles will be 7.87*20^7?
Let the required number of moles be ‘x’.
1 mole ———6.022*10^23
x moles———7.87*10^7
(Cross multiplication)
x=7.87*10^7/6.022*10^23
Therefore x=1.3*10^-16
Answer:
No, a mole of oxygen is about 16 grams
Explanation:
