Answer:
The empirical formula is same as the molecular formula = C6H10S2O
Explanation:
we start by dividing the percentage of each of the elements by their atomic mass
Carbon = 44.4/12 = 3.7
Hydrogen = 6.21/1 = 6.21
Sulphur = 39.5/32 = 1.234375
oxygen = 9.89/16 = 0.618125
That of oxygen is smallest so we divide all by that of oxygen
C = 3.7 / 0.618125 = 6
H = 6.21/ 0.618125 = 10
S = 1.234375/ 0.618125 = 2
Automatically, oxygen is 1
So the empirical formula is;
C6H10S2O
Let’s get its molecular formula. We multiply each of the subscript by the number;
(72 + 10 + 64 + 16)n= 162
162n = 162
n = 1
So the molecular formula is same as the empirical formula
2 mole Na needs 1 mole Cl2
4 mole Na = 2 mole Cl2
The answer is 0.5010
Number of moles (n) is equal to the quotient of mass (m) and molar mass of a sample (Mr):
n = m/Mr
We have:
n = ?
m = 40.10 g
Mr = 80.0432 g/mol
n = 40.10 g : 80.0432 g/mol = 0.5010
<span>40.10 has 4 significant digits,
</span><span>80.0432 has 6 significant digits.
Since 4 is less than 6, we choose 4 </span>significant digits
Answer:
15.75 grams of HNO3 was used and dissolved in 2.5 liters of solvent, to make a 0.10 M solution
Explanation:
Step 1: Data given
Nitric acid = HNO3
Molar mass of H = 1.01 g/mol
Molar mass of N = 14.0 g/mol
Molar mass O = 16.0 g/mol
Number of moles nitric acid (HNO3) = 0.25 moles
Molairty = 0.10 M
Step 2: Calculate molar mass of nitric acid
Molar mass HNO3 = Molar mass H + molar mass N + molar mass (3*O)
Molar mass HNO3 = 1.01 + 14.0 + 3*16.0
Molar mass HNO3 = 63.01 g/mol
Step 3: Calculate mass of solute use
Mass HNO3 = moles HNO3 * molar mass HNO3
Mass HNO3 = 0.25 moles * 63.01 g/mol
Mass HNO3 = 15.75 grams
15.75 grams of HNO3 was used and dissolved in 2.5 liters of solvent, to make a 0.10 M solution
Cl2 is the right answer since the molecule is composed of two chlorines.
