Glucose and fructose are monosaccharides that have the same chemical formula, C6H12O6. However, they do not react the same chemi
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Answer:
0.92 kg
Explanation:
The volume occupied by the air is:
The moles of air are:
The heat required to heat the air by 10.0 °C (or 10.0 K) is:
Methane's heat of combustion is 55.5 MJ/kg. The mass of methane required to heat the air is:
Answer: Option (C) is the correct answer.
Explanation:
A water molecule is made up of two hydrogen atoms and one oxygen atom. Its molecular formula is .
It is known that hydrogen is electropositive in nature whereas oxygen is electronegative in nature. Therefore, oxygen will pull the electron of hydrogen towards itself as a result, there will be formation of partial positive charge on hydrogen and partial negative charge on oxygen atom.
Thus, the water molecule becomes polar. So, we can conclude that water molecule is a polar molecule because water molecules have a positive charge on one end and a negative charge on the other.
Answer:
a. 478.69 K
b. 939.43
c. 19.30 J
d. 64.5J
Explanation:
From the question, we can identify the following;
= 785
= 0.000785
= 400K
= 125 Kpa = 125 000 Pa
Using the ideal gas equation,
PV = nRT
where R is the molar gas constant = 8.314 ⋅Pa⋅
⋅
Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol
a. Steam temperature in K
To calculate this, we use the constant pressure process;
q = nΔH
Where q is 83.8J according to the question
Thus;
83.8 = 0.03 × [34980 + 35.5 - (34980 + 35.5
)]
83.8 = (0.03 × 35.5) ( - 400K)
83.8 = 1.065 ( - 400)
78.69 = ( - 400)
= 400 + 78.69
= 478.69 K
b. Final cylinder volume
To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)
/
=
/
=
/
= (785 × 478.69)/400
= 939.43
c. Work done by the system
Mathematically, the work done by the system is calculated as follows;
w = P(-
) = 125 KPa ( 939.43 - 785) = 19.30 J
d. Change in internal energy of the steam in J
ΔU = q - w = 83.8 - 19.3 = 64.5J