Answer:
Option C. 4.03 g
Explanation:
Firstly we analyse data.
12 % by mass, is a sort of concentration. It indicates that in 100 g of SOLUTION, we have 12 g of SOLUTE.
Density is the data that indicates grams of solution in volume of solution.
We need to determine, the volume of solution for the concentration
Density = mass / volume
1.05 g/mL = 100 g / volume
Volume = 100 g / 1.05 g/mL → 95.24 mL
Therefore our 12 g of solute are contained in 95.24 mL
Let's finish this by a rule of three.
95.24 mL contain 12 g of sucrose
Our sample of 32 mL may contain ( 32 . 12) / 95.24 = 4.03 g
Answer:
2 moles of KCl will be produced
Explanation:
Given parameters:
Number of moles of K = 2 moles
Unknown:
Number of moles of KCl produced = ?
Solution:
To solve this problem;
Obtain a balanced chemical equation:
2K + Cl₂ → 2KCl ;
Since K is the limiting reactant, its amount will determine the extent of this reaction.
From the balanced equation;
2 moles of K will produce 2 moles of KCl
Given that 2 moles of K reacted, 2 moles of KCl will be produced
The answer is NO. During combustion of hydrogen, water is produced as an end product. The chemical formulae for combustion of hydrogen are:
2H2 + O2 -->2 H2O
another advantage is that the whole of the fuel load is combusted hence little wastage. This has made the fuel advantageous in its use as rocket fuel since there is no baggage fuel.
Here we apply the Clausius-Clapeyron equation:
ln(P₁/P₂) = ΔH/R x (1/T₂ - 1/T₁)
The normal vapor pressure is 4.24 kPa (P₁)
The boiling point at this pressure is 293 K (P₂)
The heat of vaporization is 39.9 kJ/mol (ΔH)
We need to find the vapor pressure (P₂) at the given temperature 355.3 K (T₂)
ln(4.24/P₂) = 39.9/0.008314 x (1/355.3 - 1/293)
P₂ = 101.2 kPa
<span>D the temperature decreases and the pressure increases</span>
