Help me please:

How much he is required to raise the temperature of 464g of lead from 120°C to 219°C?

24805.44 J

Explanation:

Step 1:

Data obtained from the question.

Mass (M) = 464g

Initial temperature (T1) = 120°C

Final temperature (T2) = 219°C

Change in temperature (ΔT) = T2 - T1 = 219°C - 120°C = 99°C

Specific heat capacity of lead (C) = 0.129cal/g°C = 4.184 x 0.129 = 0.54J/g°C

Heat (Q) =?

Step 2:

Determination of the heat Q, required the temperature of lead. This is illustrated below

Q = MCΔT

Q = 464 x 0.54 x 99

Q = 24805.44 J

Therefore, 24805.44 J of heat is required to raise the temperature of lead.

answer: a chemical reaction

explanation:

im not wrong

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%