For i: 33mL
For ii: 87-88mL
For iii:22.3mL
C + O2 --> CO2
18g C = 18/12.01 = 1.49 moles (Limiting Reagent)
62.3-14.3 = 48g oxygen that reacted
48g O = 3 moles
Answer:
0.1 M
<h3>
Explanation:</h3>
- Molarity refers to the concentration of a solution in moles per liter.
- It is calculated by dividing the number of moles of solute by the volume of solvent;
- Molarity = Moles of the solute ÷ Volume of the solvent
<u>In this case, we are given;</u>
- Number of moles of the solute, NH₄Cl as 0.42 moles
- Volume of the solvent, water as 4200 mL or 4.2 L
Therefore;
Molarity = 0.42 moles ÷ 4.2 L
= 0.1 mol/L or 0.1 M
Thus, the molarity of the solution will be 0.1 M
1) Use the fact that 1 mol of gas at STP occupies 22.4 liter
=> 1 mol / 22.4 l = x / 0.125 l => x = 0.125 l * 1 mol / 22.4 l = 0.00558 mol
2) Now use the molar mass of the gas
molar mass of CO2 ≈ 44 g / mol
Formula: molar mass = mass in grams / number of moles =>
mass in grams = molar mass * number of moles = 44 g/mol * 0.00558 moles
mass = 0.246 g
Answer: 0.246 g
Technically it's grp 1 elements...the answer is ribidium Rb
