Answer:
They are 1.204×10^24 atoms of hydrogen present in 18 grams of water. In order to calculate this,it is necessary to compute the number of hydrogen moles present in the sample.
To get the number of gold atoms, you have to divide the mass of the gold by the mass of the gold atom. It follows this simple equation

.
Let x be the number of gold atoms. Plug in the values to a calculator.
x =

Both have the same units so the unit gram(g) can be cancelled.
x then would be equal to 1.53x10^22. So there are 1.53x10^22 atoms of gold in 5 g of gold
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
Answer:
I don't really get the options but it favoures the reactant side.
Explanation:
Increasing pressure favours the side with fewer moles of gas while decreasing pressure favours the side with the more moles of gas. E.g
If there is 0 moles of gas particles in the reactant side and 1 mole of gas particle in the product side, increasing pressure favours the reactants while decreasing pressure favours the product side.
With the explanations I have made, I hope the question is now clear to you.