Answer:
The mass of NaCl is 0.029 grams
Explanation:
Step 1: Data given
Molecular weight of NaCl = 58.44 g/mol
Volume of solution = 100 mL = 0.100 L
Molarity = 0.0050 M
Step 2: Calculate moles NaCl
Moles NaCl = molarity * volume
Moles NaCl = 0.0050 M * 0.100 L
Moles NaCl = 0.00050 moles
Step 3: Calculate mass NaCl
Mass NaCl = moles NaCl * molar mass NaCl
Mass NaCl = 0.00050 moles * 58.44 g/mol
Mass NaCl = 0.029 grams
The mass of NaCl is 0.029 grams
The empirical formula for 200 grams of an organic sample which contains carbon,hydrogen, and oxygen is C3H6O
Good ventilation as a product of it is pure Cl2 gas
Answer:
A) 14. 25 × 10²³ Carbon atoms
B) 34.72 grams
Explanation:
1 molecule of Propane has 3 atoms of Carbon and 8 atoms of Hydrogen.
The sample has 3.84 × 10²⁴ H atoms.
If 8 atoms of Hydrogrn are present in 1 molecule of propane.
3.84 × 10²⁴ H atoms are present in

<u>= 4.75 × 10²³ molecules of Propane</u>.
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
No. of Carbon atoms in 1 molecule of propane = 3
=> C atoms in 4.75× 10²³ molecules of Propane = 3 × 4.75 × 10²³
<u>= 14.25 × 10²³ </u>
<u>________________________________________</u>
<u>Gram</u><u> </u><u>Molecular</u><u> </u><u>Mass</u><u> </u><u>of</u><u> </u><u>Propane</u><u>(</u><u>C3H8</u><u>)</u>
= 3 × 12 + 8 × 1
= 36 + 8
= 44 g
1 mole of propane weighs 44g and has 6.02× 10²³ molecules of Propane.
=> 6.02 × 10²³ molecules of Propane weigh = 44 g
=> 4. 75 × 10²³ molecules of Propane weigh =



<u>= 34.72 g</u>
The balanced chemical
reaction will be:
CH4 + 2O2 → CO2 + 2H2O
We are given the amount of carbon dioxide to produce from the reaction.
This will be our starting point.
560 L CH4 ( 1 mol CH4/ 22.4 L CH4 ) (2 mol O2/ 1 mol CH4 ) (
22.4 L O2 / 1 mol <span>O2</span><span>) = 1120 L O2</span>