I believe that answer is D
The heat from the Bunsen burner transfers to the water and the pot, then the heat from the pot transfers to the person’s hand.
We can calculate the final temperature from this formula :
when Tf = (V1* T1) +(V2* T2) / (V1+ V2)
when V1 is the first volume of water = 5 L
and V2 is the second volume of water = 60 L
and T1 is the first temperature of water in Kelvin = 80 °C +273 = 353 K
and T2 is the second temperature of water in Kelvin = 30°C + 273= 303 K
and Tf is the final temperature of water in Kelvin
so, by substitution:
Tf = (5 L * 353 K ) + ( 60 L * 303 K) / ( 5 L + 60 L)
= 1765 + 18180 / 65 L
= 306 K
= 306 -273 = 33° C
Answer: Seastars prey on mussels and shellfish which would otherwise have no other natural predators. Herbivorous fish like the butterflyfish pictured to the left prey on marine algae. Without this crucial predator-prey balance, the algae would over-grow, which would then kill coral, as they compete for the same resources.
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.
