Answer:
2.74 M
Explanation:
Given data:
Mass of sodium chloride = 80.0 g
Volume of water = 500.0 mL
Molarity of solution = ?
Solution:
Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.
Formula:
Molarity = number of moles of solute / L of solution
Now we will convert the mL into L.
500.0 mL× 1 L /1000 mL = 0.5 L
In next step we will calculate the number of moles of sodium chloride.
Number of moles = mass/molar mass
Number of moles = 80.0 g/ 58.4 g/mol
Number of moles = 1.37 mol
Molarity:
M = 1.37 mol/ 0.5 L
M = 2.74 M
Answer:
solvent (such as water, oil or isopropyl alcohol) is allowed to absorb up the paper strip. ... Different molecules run up the paper at different rates. As a result, components of the solution separate and, in this case, become visible as strips of color on the chromatography paper.
Explanation:
Hope this helps leave a heart c:
Answer:
1. Hydrogen will diffuse faster.
2. The ratio of diffusion of hydrogen gas to that of the unknown gas is 4 : 1
Explanation:
Let the rate of diffusion of hydrogen gas, H2 be R1
Let the molar mass of H2 be M1
Let the rate of diffusion of the unknown gas be R2.
Let the molar mass of the unknown gas be M2.
Molar mass of H2 (M1) = 2x1 =2g/mol
Molar mass of unknown gas (M2) = 16 times that of H2
= 16 x 2 = 32g/mol
1. Determination of the gas that will diffuse faster. This is illustrated below:
R1/R2 = √(M2/M1)
R1/R2 = √(32/2)
R1/R2 = √16
R1/R2 = 4
Cross multiply
R1 = 4R2
From the above calculations, we can see that the rate of diffusion H2 (R1) is four times the rate of diffusion of the unknown gas (R2).
Therefore, hydrogen will diffuse faster.
2. Again, from the calculations made above, the ratio of diffusion of hydrogen (R1) to that of the unknown gas (R2) is given by;
R1/R2 = 4
Therefore, the ratio of diffusion of hydrogen (R1) to that of the unknown gas (R2) is:
4 : 1
Answer:
H30+ First H20 Second N03 Third HNO3 Fourth
Explanation:
