The mass of hydrated salt - 2.123 g
mass of anhydrous salt - 1.861 g
mass that has been reduced is the mass of water that has been heated and lost from the compound thereby making the salt anhydrous.
therefore mass of water lost - 2.123 - 1.861 = 0.262 g
number of moles of water lost - 0.262 g / 18 g/mol = 0.0146 mol
number of moles of salt - 1.861 g / 380.6 g/mol = 0.00490 mol
molar ratio of moles of water to moles of salt
molar ratio = 0.146 mol / 0.00490 mol = 2.98 rounded off to 3
for every 1 mol of salt there are 3 moles of water
therefore empirical formula - Cu₃(PO₄)₂.3H₂O
Enthalpy is a thermodynamic quantity that describes the heat content of a system, that can not be measured directly. That's why we measure change in enthaply, measured in the units joules. The statement that e<span>nthalpy change depends on the rate at which a substance is heated or cooled is false. Enthalpy change depends only on the following factors:
-</span><span>physical state of reactants and products
- quantity of reactants</span><span>
- allotropic modifications
- temperature and pressure</span><span>
</span>
Answer:
34.3 g NH3
Explanation:
M(H2) = 2*1 = 2 g/mol
M(N2) = 2*14 = 28 g/mol
M(NH3) = 14 + 3*1 = 17 g/mol
23.6 g H2* 1 mol/2 g = 11.8 mol H2
28.3 g N2 * 1 mol/28 g = 1.01 mol N2
3H2 + N2 ------> 2NH3
from reaction 3 mol 1 mol
given 11.8 mol 1.01 mol
We can see that H2 is given in excess, N2 is limiting reactant.
3H2 + N2 ------> 2NH3
from reaction 1 mol 2 mol
given 1.01 mol x
x = 2*1.01/1= 2.02 mol NH3
2.02 mol * 17g/1 mol ≈ 34.3 g NH3
Answer:
Explanation:
Here, we want to describe the relationship between the volume and temperature of an ideal gas
This relationship is defined by Charles' law
From this law, we know that the volume of a given mass of gas is directly proportional to its temperature at a fixed pressure
What this means is that as long as the pressure remains unchanged, when the volume increases, the temperature increases, and when the volume decreases, the temperature decreases
These can be represented by the mathematical formula below:
Dissolving sugar in water
