Answer:
700 calories
Explanation:
Using the formula below:
Q = m × c × ∆T
Where;
Q = amount of heat required (calories)
m = mass of substance (g)
c = specific heat of substance (cal/g°C)
∆T = change in temperature (°C)
According to this question, the following information was provided;
Q = ?
m = 20g
c = 1.0 cal/g °C
∆T = 40°C - 5°C = 35°C
Using the formula; Q = m × c × ∆T
Q = 20 × 1 × 35
Q = 700 calories
Hence, 700 cal of heat energy is needed to raise 20 g of H2O from 5°C to 40°C.
The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of 
) =-1275.0
enthalpy of combustion of oxygen(Δ of 
) = zero
enthalpy of combustion of carbon dioxide(Δ of 
) = -393.5
enthalpy of combustion of water(Δ of 
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ (products) - ∈Δ (reactants)
(s) +6 (g) → 6 (g)+ 6 (l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
If Robert has 4 grams of a substance and Jill has 10 grams of the same substance <span>Jill's sample will weigh more than Robert's sample.</span>
Answer:
more
Explanation:
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