The percent by mass sugar of a solution : 11.07%
<h3>Further explanation</h3>
Given
mass of sugar = 12.45 g
mass of water = 100 g
Required
The percent by mass
Solution
Mass of solution :

Percent mass of sugar :

Answer:
0.641 moles of ethane
Explanation:
Based on the equation:
C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H2O(l)
We can determine ΔH of reaction using Hess's law. For this equation:
<em>Hess's law: ΔH products - ΔH reactants</em>
ΔH = {2ΔHCO2 + 3ΔHH2O} - {ΔHC2H6}
<em>Pure monoatomic substances have a ΔH = 0kJ/mol; ΔHO2 = 0kJ/mol</em>
<em />
ΔH = {2*-393.5kJ/mol + 3*-285.8kJ/mol} - {-84.7kJ/mol}
ΔH = -1559.7kJ/mol
That means when 1 mole of ethane is in combustion there are released 1559.7kJ of heat. To produce 1.00x10³kJ there are needed:
1.00x10³kJ * (1mole ethane / 1559.7kJ) =
<h3>0.641 moles of ethane</h3>
Explanation:
12)
a) Sodium + oxygen = ?
When sodium reacts oxygen its forms sodium oxide as a product.

b) magnesium + fluorine = ?
When magnesium reacts with fluorine its forms magnesium fluoride as a product.

13)
a) 
Decomposition of mercury(II) oxide on heating gives out mercury and oxygen gas.
b)
The electrolytic decomposition of water gives out hydrogen gas amd oxygen gas.
Pressure of the gas P1 = 30.7 kpa
When it doubled P2 = 61.4 kpa
Temperature T1 = 0 => T1 =. 0 + 273 =273
Temperature T2 =?
We have pressure temperature equation P1T1 = P2T2
=> T2 = P1T1 / P2 = (30.7 x 273) / 61.4 = 136.5
So the temperature for doubling the pressure is 136.5.