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:
Molality = 7.5 mol/kg
Explanation:
Given data:
Mass of NH₄Cl = 6.30 g
Mass of water = 15.7 g (15.7/1000 =0.016 kg)
Molality = ?
Solution:
Formula of molality:
Molality = Moles of solute / mass of solvent in gram
Now we will first calculate the number of moles of solute( NH₄Cl )
Number of moles = mass/ molar mass
Molar mass of NH₄Cl = 53.491 g/mol
Number of moles = 6.30 g/ 53.491 g/mol
Number of moles = 0.12 mol
Now we will calculate the molality.
Molality = Moles of solute / mass of solvent in gram
Molality = 0.12 mol / 0.016 kg
Molality = 7.5 m
or (m=mol/kg)
Molality = 7.5 mol/kg
Answer:
A substance that can conduct electricity in solution
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