Answer:
V = 2.32 Liters
Explanation:
PV = nRT => V = nRT/P
n = 25.8g/122g/mole = 0.21 mole
R = 0.08206 L·atm/mol·K
T = 25.44°C + 273 = 298.44K
P = 2.22 atm (given in problem)
V = (0.21mol)(0.08206 L·atm/mol·K)(298.44K)/(2.22atm) = 2.32 Liters at 25.44°C & 2.22atm
It's difficult to write it down, but I'll attach you a good example of hydroboration of indene. I hope you'll find it helpful.
Answer:
Cp = 0.237 J.g⁻¹.°C⁻¹
Explanation:
Amount of energy required by known amount of a substance to raise its temperature by one degree is called specific heat capacity.
The equation used for this problem is as follow,
Q = m Cp ΔT ----- (1)
Where;
Q = Heat = 640 J
m = mass = 125 g
Cp = Specific Heat Capacity = <u>??</u>
ΔT = Change in Temperature = 43.6 °C - 22 °C = 21.6 °C
Solving eq. 1 for Cp,
Cp = Q / m ΔT
Putting values,
Cp = 640 J / (125 g × 21.6 °C)
Cp = 0.237 J.g⁻¹.°C⁻¹
<span>Answer: option D) Conditions over the oceans change slowly because water takes more time than land to gain or lose heat.
This is because water has a high heat capacity meaning that, with the same amount of heat, it will change its temperature less than what substances with lower heat capacities do.
</span>
