Answer:
0.0005mol/kg
Explanation:
Molality = mole of solute/mass of solvent (kg)
From the question, we obtained the following data:
Mole of solute = 0.05mole
Mass of solvent = 100kg
Molality = 0.05/100
Molality = 0.0005mol/kg
The molality of the solution is 0.0005mol/kg
<h3>
Answer:</h3>
13 g CO₂
<h3>
General Formulas and Concepts:</h3>
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
<u>Stoichiometry</u>
- Using Dimensional Analysis
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
<em>Identify variables</em>
[Given] 6.7 L O₂
[Solve] g O₂
<u>Step 2: Identify Conversions</u>
[STP] 22.4 L = 1 mol
[PT] Molar Mass of O: 16.00 g/mol
[PT] Molar Mass of C: 12.01 g/mol
Molar Mass of CO₂: 12.01 + 2(16.00) = 44.01 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:

- [DA] Divide/Multiply [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
13.1637 g CO₂ ≈ 13 g CO₂
It would protect best against C) Alpha radiation, as Beta radiation is stopped by lighter metals such as aluminium, and Gamma radiation can only be stopped by heavier metals such as lead.
Answer:- molar mass of the unknown gas is 71.5 gram per mol.
Solution:- From Graham's law of effusion rates, the rate of effusion of a gas is inversely proportional to the square root of it's molar mass.
When we compare the effusion rates of two gases then the formula for Graham's law is:

In this formula, V stands for volume and M stands for molar mass
Rate is volume effused per unit time. Since, the volumes are same, the formula could be written as:

let's say in formula, subscript 1 is for hydrogen gas and 2 is for the unknown gas.
Molar mass of hydrogen is 2.02 grams per mol and the time taken to effuse it is 2.42 min. The time taken to effuse the unknown gas is 14.4 min and we are asked to calculate it's molar mass. let's plug in the values in the formula:


doing squares to both sides:



So, the molar mass of the unknown gas is 71.5 grams per mol.