Answer:
It helps the body remove heat through sweating.
Explanation:
When the weather is hot, the body tries to keep cool by sweating. The high specific heat capacity means that the body doesn't have to lose much water to stay cool.
The high specific heat capacity of water doesn’t heat the body, but it slows down the rate of heat loss when the weather is cool.
B is wrong. The body uses glucose, not water, as an energy source.
C is wrong. The high specific heat capacity of water is not connected with the body's ability to store it.
D is wrong. The high specific heat capacity of water doesn't heat the body, but it slows the rate at which it cools.
Answer:
V = 240.79 L
Explanation:
Given data:
Volume of butane = ?
Temperature = 293°C
Pressure = 10.934 Kpa
Mass of butane = 33.25 g
Solution:
Number of moles of butane:
Number of moles = mass/ molar mass
Number of moles = 33.25 g/ 58.12 g/mol
Number of mole s= 0.57 mol
Now we will convert the temperature and pressure units.
293 +273 = 566 K
Pressure = 10.934/101 = 0.11 atm
Volume of butane:
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
V = nRT/P
V = 0.57 mol × 0.0821 atm.L/ mol.K ×566 K / 0.11 atm
V = 26.49 L/0.11
V = 240.79 L
Answer:
ΔE = -2661 KJ/mole
ΔH = -2658 KJ/mole
Explanation:
ΔH = q - PΔV
ΔE = q + w
<u>First, to find ΔE:</u>
The reaction PRODUCES 2658 kJ of h (q), and does 3 kJ of work (w).
2658 kJ(q) + 3 kJ(w) = 2661 kJ, BUT the reaction <u><em>PRODUCES</em></u> heat, which means ΔE is negative.
ΔE = -2661 KJ/mole
<u>Second, to find ΔH:</u>
ΔH = q - PΔV
ΔH = 2658 kJ(q) - PΔV
Now, the question states that butane burns at a constant pressure; that just translates to the pressure of the reaction is equal to 0.
ΔH = 2658 KJ(q) - (0)ΔV
ΔH = 2658 KJ - 0
ΔH = 2658 kJ, BUT, like before, the reaction PRODUCES heat, which also mean ΔH is negative.
ΔH = -2658 KJ/mole
I hope this helped! Have a nice week.
