Answer:
THE MOLAR MASS OF XCL2 IS 400 g/mol
THE MOLAR MASS OF YCL2 IS 250 g/mol.
Explanation:
We derive the molar mass of XCl2 and YCl2 by recalling the molar mass formula when both mass and the number of moles are known.
Number of moles = mass / molar mass
Molar mass = mass / number of moles.
For XCl2,
mass = 100 g
number of moles = 0.25 mol
Thus, molar mass = mass / number of moles
Molar mass = 100 g / 0.25 mol
Molar mass = 400 g/mol.
For YCl2,
mass = 125 g
number of moles = 0.50 mol
Molar mass = 125 g / 0.50 mol
Molar mass = 250 g/mol.
Accordingly, the molar masses for XCl2 and YCl2 are 400 g/mol and 250 g/mol, respectively.
Response:
4.5 m³
Resolution:
The statement indicates the presence of two blocks on a lid of a container with a volume of 9 m³. The lid's weight is equal to that of the two blocks. Thus, there were initially four blocks (or 4 atm pressure) acting on a volume of 9 m³.
After adding four additional blocks on the lid, the pressure rises from 4 atm to 8 atm (2 atm from the lid, 2 atm from the original blocks, and 4 atm from the new blocks).
Hence, The data established is,
P₁ = 4 atm
V₁ = 9 m³
P₂ = 8 atm
V₂ =?
Using Boyle's Law,
P₁ V₁ = P₂ V₂
Resolving for V₂,
V₂ = P₁ V₁ / P₂
Substituting values yields:
V₂ = (4 atm × 9 m³) ÷ 8 atm
V₂ = 4.5 m³
The mass is 150,000 grams. Multiply 100 by 50 by 30 to determine the container's volume, which equals 150,000 cm^3. Since a milliliter is equivalent to one cubic centimeter, and given that the density of water is one gram per milliliter, it follows that the mass of water is 150,000 grams.
Response:
H₂SO₄
Clarification:
Given a compound consisting of 0.475 g H, 7.557 g S, and 15.107 g O, we must compute the empirical formula by following specific steps.
Step 1: Compute the total mass of the compound
Total mass = mass H + mass S + mass O = 0.475 g + 7.557 g + 15.107 g
Total mass = 23.139 g
Step 2: Determine the percentage composition.
H: (0.475g/23.139g) × 100% = 2.05%
S: (7.557g/23.139g) × 100% = 32.66%
O: (15.107g/23.139g) × 100% = 65.29%
Step 3: Divide each percentage by the element's atomic mass
H: 2.05/1.01 = 2.03
S: 32.66/32.07 = 1.018
O: 65.29/16.00 = 4.081
Step 4: Normalize all values by the smallest one
H: 2.03/1.018 ≈ 2
S: 1.018/1.018 = 1
O: 4.081/1.018 ≈ 4
Thus, the empirical formula for the compound is H₂SO₄.
Answer:
0.605 molal
Explanation:
Molality indicates the solute amount in a specific solvent mass.
Let’s find the amount of benzene solute.
Mass of benzene = 13.3g
Molar mass of C6H6 = 12*6 +1*6 =72+7=78g/mol
Amount of benzene = mass/molar mass
=13.3/78
=0.1705mol
Molality = amount of solute/mass of solvent in kg
Mass of solvent = 282g = 0.282kg
Molality = 0.1705/0.282
=0.605 molal
