Answer:
Formula of EPSOM salt = MgSO4.7H2O
molecular mass of MgSO4.7H2O = atomic mass of Mg + atomic mass of S + 4 × atomic mass of O + 7 { 2 × atomic mass of H + atomic mass of O }
= 24 + 32 + 4× 16 + 7{ 2 × 1 + 16 } g/mol
= (24 + 32 + 64+ 126 ) g/mol
= 246 g/mol
molecular mass of total water = 7 × ( 2× atomic mass of H + atomic mass of O )
= 7 × 18 = 126 g/mol
now ,
% mass of H2O in EPSOM salt = {total molar mass of H2O/molar mass of Epsom salts }× 100
= {126/246 } × 100
= 12600/246
= 51.21 %
Explanation:
i have done it hope it helps
The specific heat capacity of this chunk of metal is equal to 0.32 J/g°C.
<u>Given the following data:</u>
- Quantity of energy = 400 Joules
- Initial temperature = 20°C
To determine the specific heat capacity of this chunk of metal:
<h3>
The formula for quantity of heat.</h3>
Mathematically, quantity of heat is given by the formula;

<u>Where:</u>
- Q represents the quantity of heat.
- m represents the mass of an object.
- c represents the specific heat capacity.
- ∅ represents the change in temperature.
Making c the subject of formula, we have:

Substituting the given parameters into the formula, we have;

Specific heat, c = 0.32 J/g°C.
Read more on specific heat here: brainly.com/question/2834175
Oxidation
iron+oxygen happened
We have to know the molarity of solution obtained when 5.71 g of Na₂CO₃.10 H₂O is dissolved in water and made up to 250 cm³ solution.
The molarity of solution obtained when 5.71 g of sodium carbonate-10-water (Na₂CO₃.10 H₂O) is dissolved in water and made up to 250.0 cm^3 solutionis: (A) 0.08 mol dm⁻³
The molarit y of solution means the number of moles of solute present in one litre of solution. Here solute is Na₂CO₃.10 H₂O and solvent is water. Volume of solution is 250 cm³.
Molar mass of Na₂CO₃.10 H₂O is 286 grams which means mass of one mole of Na₂CO₃.10 H₂O is 286 grams.
5.71 grams of Na₂CO₃.10 H₂O is equal to
= 0.0199 moles of Na₂CO₃.10 H₂O. So, 0.0199 moles of Na₂CO₃.10 H₂O present in 250 cm³ volume of solution.
Hence, number of moles of Na₂CO₃.10 H₂O present in one litre (equal to 1000 cm³) of solution is
= 0.0796 moles. So, the molarity of the solution is 0.0796 mol/dm³ ≅ 0.08 mol/dm³