The heat required to raise the temperature of a certain mass of sample to a specific temperature change, we use the formula mCpΔT where m is mass, Cp is the specific heat of the substance and ΔT is the temperature change. In this case, we substitute and form 1.25 g x 0.057 cal/g C *20 C equal to 1.425 calories.
<span>Pass the mixture through filter paper. The large particles in the suspension will filter out. to tell the difference between a solution and a colloid, shine a beam of light through the mixture, if it reflects then it is a colloid, if it doesn't then it is a solution</span>
Answer: The mass of the gas is 18.3 g/mol.
Explanation:
To calculate the rate of diffusion of gas, we use Graham's Law.
This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:
Squaring both sides and solving for 
Hence, the molar mas of unknown gas is 18.3 g/mol.
Answer:
BaI₂.5H₂O
Explanation:
Given Data:
Mass of Hydrated BaI₂ = 10.222 g
Mass of dried BaI₂ = 9.520 g
Mass of Water removed = 10.222-9.520 = 0.702 g
M.Mass of BaI₂ = 391.136 g/mol
M.Mass of Water = 18.02 g/mol
Now,
Calculate moles of dried BaI₂ as,
Moles = Mass / M.Mass
Moles = 9.520 g / 391.136 g/mol
Moles = 0.02434 moles
Calculate moles of Water as,
Moles = Mass / M.Mass
Moles = 0.702 g / 18.02 g/mol
Moles = 0.0389 moles
Then,
Calculate Mole ratio of BaI₂ and water as,
= 0.02434 moles BaI₂ / 0.0389 moles Water
= 0.625
Now,
We will convert this mole ratio to a whole number by multiplying it with a nearest integer,
= 0.625 × 8
= 5
Hence, this means for every one mole of BaI there are 5 moles of Water.
Result:
BaI₂.5H₂O
