Answer:
volume
Explanation:
The volume of the two samples of water will be different because volume is the amount of space occupied by a body. It is dependent on the amount of materials it contains.
- The 50g sample will have a higher volume compared to the 10g sample.
- The boiling point and density are intensive properties and do not depend on the amount of matter present.
- Since both samples are from the same source, they will have the same color.
Heat capacity of aluminium = 0.900 J/g°C
While heat capacity of water = 4.186 J/g°C
Heat = heat gained by water + heat gained by aluminium
Heat gained by water = 100 × 4.186 × 30.5
= 12767.3 Joules
Heat gained by aluminium = 15 × 0.9 × 30.5
= 411.75 Joules
Heat required = 13179.05 Joules or 13.179 kJoules
Answer:
Lithium has 5 elements, sulphur has 2 elements and oxygen has 4 elements
Here we have to calculate the amount of
ion present in the sample.
In the sample solution 0.122g of
ion is present.
The reaction happens on addition of excess BaCl₂ in a sample solution of potassium sulfate (K₂SO₄) and sodium sulfate [(Na)₂SO₄] can be written as-
K₂SO₄ = 2K⁺ + 
(Na)₂SO₄=2Na⁺ + 
Thus, BaCl₂+
= BaSO₄↓ + 2Cl⁻ .
(Na)₂SO₄ and K₂SO₄ is highly soluble in water and the precipitation or the filtrate is due to the BaSO₄ only. As a precipitation appears due to addition of excess BaCl₂ thus the total amount of
ion is precipitated in this reaction.
The precipitate i.e. barium sulfate (BaSO₄)is formed in the reaction which have the mass 0.298g.
Now the molecular weight of BaSO₄ is 233.3 g/mol.
We know the molecular weight of sulfate ion (
) is 96.06 g/mol. Thus in 1 mole of BaSO₄ 96.06 g of
ion is present.
Or. we may write in 233.3 g of BaSO₄ 96.06 g of
ion is present. So in 1 g of BaSO₄
g of
ion is present.
Or, in 0.298 g of the filtered mass (0.298×0.411)=0.122g of
ion is present.