Uses of nonmetals in our daily life: Oxygen which is 21% by volume helps in the respiration process.
Nonmetals used in fertilizers: Fertilizers contain nitrogen.
Nonmetals used in crackers: Sulphur and phosphorus are used in fireworks.
<span>Avogadro's number
represents the number of units in one mole of any substance. This has the value
of 6.022 x 10^23 units / mole. This number can be used to convert the number of
atoms or molecules into number of moles. We do as follows:
</span>10 mol NH3 ( 6.022 x 10^23 molecules / 1 mol ) = 6.022x10^24 molecules NH3
Answer:
The molarity of the solution is 7.4 mol/L
Explanation:
From the question above
0.400 ml of water contains 1.00 g of hydrochloride form of cocaine
Therefore 1000 ml of water will contain x g of hydrochloride form of cocaine
x = 1000 / 0.400
x = 2500 g
2500g of hydrochloride form of cocaine is present in 1000 ml of water.
Mole of hydrochloride form of cocaine = mass /molar mass of hydrochloride
Mole of hydrochloride form of cocaine = 2500/339.8
= 7.4 mol
Molarity = mol/ volume in liter (L)
molarity = 7.4 /1
Molarity = 7.4 mol/L
Answer:
0.189 g.
Explanation:
- This problem is an application on <em>Henry's law.</em>
- Henry's law states that the solubility of a gas in a liquid is directly proportional to its partial pressure of the gas above the liquid.
- Solubility of the gas ∝ partial pressure
- If we have different solubility at different pressures, we can express Henry's law as:
<em>S₁/P₁ = S₂/P₂,</em>
S₁ = 0.0106/0.792 = 0.0134 g/L and P₁ = 0.321 atm
S₂ = ??? g/L and P₂ = 5.73 atm
- So, The solubility of the gas at 5.73 atm (S₂) = S₁.P₂/P₁ = (0.0134 g/L x 5.73 atm) / (0.321 atm) = 0.239 g/L.
<em>The quantity in (g) = S₂ x V = (0.239 g/L)(0.792 L) = 0.189 g.</em>
<em></em>
Answer : The molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.
Solution : Given,
Density of solution = 
Molar mass of sulfuric acid (solute) = 98.079 g/mole
98.0 % sulfuric acid by mass means that 98.0 gram of sulfuric acid is present in 100 g of solution.
Mass of sulfuric acid (solute) = 98.0 g
Mass of solution = 100 g
Mass of solvent = Mass of solution - Mass of solute = 100 - 98.0 = 2 g
First we have to calculate the volume of solution.
Now we have to calculate the molarity of solution.
Now we have to calculate the molality of the solution.
Therefore, the molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.
