Answer : The vapor pressure of solution is 23.67 mmHg.
Solution:
As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.
The formula for relative lowering of vapor pressure will be,
where,
= vapor pressure of pure solvent (water) = 23.76 mmHg
= vapor pressure of solution= ?
= mass of solute (sucrose) = 12.25 g
= mass of solvent (water) = 176.3 g
= molar mass of solvent (water) = 18.02 g/mole
= molar mass of solute (sucrose) = 342.3 g/mole
Now put all the given values in this formula ,we get the vapor pressure of the solution.
Therefore, the vapor pressure of solution is 23.67 mmHg.
Answer:
30%
Explanation:
<em>This is the chemical formula for zinc bromate: Zn(BrO₃)₂. Calculate the mass percent of oxygen in zinc bromate. Round your answer to the nearest percentage.</em>
Step 1: Determine the mass of 1 mole of Zn(BrO₃)₂
M(Zn(BrO₃)₂) = 1 × M(Zn) + 2 × M(Br) + 6 × M(O)
M(Zn(BrO₃)₂) = 1 × 65.38 g/mol + 2 × 79.90 g/mol + 6 × 16.00 g/mol
M(Zn(BrO₃)₂) = 321.18 g/mol
Step 2: Determine the mass of oxygen in 1 mole of Zn(BrO₃)₂
There are 6 moles of atoms of oxygen in 1 mole of Zn(BrO₃)₂.
6 × m(O) = 6 × 16.00 g = 96.00 g
Step 3: Calculate the mass percent of oxygen in Zn(BrO₃)₂
%O = mO/mZn(BrO₃)₂ × 100%
%O = 96.00 g/321.18 g × 100% ≈ 30%
Answer:
24.525 g of sulfuric acid.
Explanation:
Hello,
Normality (units of eq/L) is defined as:
Since the sulfuric acid is the solute, and we already have the volume of the solution (500 mL) but we need it in liters (0.5 L, just divide into 1000), the equivalent grams of solute are given by:
Now, since the sulfuric acid is diprotic (2 hydrogen atoms in its formula) 1 mole of sulfuric acid has 2 equivalent grams of sulfuric acid, so the mole-mass relationship is developed to find its required mass as follows:
Best regards.
Answer:
Correct option is
B
5 liters of CH
4
(g)NO
2
at STP
No. of molecules=
22.4
5
mol=
22.4
5
×N
A
molecules
A) 5ℊ of H
2
(g)
No. of moles=
2
5
mol=
2
5
×N
A
molecules
B) 5l of CH
4
(g)
No. of moles of CH
4
=
22.4
5
mol=
22.4
5
N
A
molecules
C) 5 mol of O
2
=5N
A
O
2
molecules
D) 5×10
23
molecules of CO
2
(g)
Molecules of 5l NO
2
(g) at STP=5l of CH
4
(g) molecules at STP
Therefore, option B is correct.
Answer:
It's well Explained below.
Explanation:
First of Excess product of CaCO_3 would be produced due to the fact that there would not be enough CaCl_2 to react with Na_2•CO_3. The main purpose of having stoichiometric quantities is for us to know the correct amount or near the correct amount of each reactant in order to create a product that will be close to the theoretical amount and thus have a higher percent yield.
