Photon has no mass at all
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>
Answer:
When a front passes over an area, it means a change in the weather. Many fronts cause weather events such as rain, thunderstorms, gusty winds, and tornadoes. At a cold front, there may be dramatic thunderstorms. At a warm front, there may be low stratus clouds. Usually, the skies clear once the front has passed.
Explanation:
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Answer:
Explanation:
Molal freezing point depression constant of butanol Kf = 8.37⁰C /m
ΔTf = Kf x m , m is no of moles of solute per kg of solvent .
mol weight of butanol = 70 g
235.1 g of butanol = 235.1 / 70 = 3.3585 moles
3.3585 moles of butanol dissolved in 4.14 kg of water .
ΔTf = 8.37 x 3.3585 / 4.14
= 6.79⁰C
Depression in freezing point = 6.79
freezing point of solution = - 6.79⁰C .
Answer: Physical change
Explanation: Condensation is the phase change of a gas to the liquid state. There is no formation of new products so it does not undergoes chemical change. Any phase change is considered a physical change.
