Answer: -
The hydrogen at 10 °C has slower-moving molecules than the sample at 350 K.
Explanation: -
Temperature of the hydrogen gas first sample = 10 °C.
Temperature in kelvin scale of the first sample = 10 + 273 = 283 K
For the second sample, the temperature is 350 K.
Thus we see the second sample of the hydrogen gas more temperature than the first sample.
We know from the kinetic theory of gases that
The kinetic energy of gas molecules increases with the increase in temperature of the gas. The speed of the movement of gas molecules also increase with the increase in kinetic energy.
So higher the temperature of a gas, more is the kinetic energy and more is the movement speed of the gas molecules.
Thus the hydrogen at 10 °C has slower-moving molecules than the sample at 350 K.
The answer you are looking for is A. If you need me to show you how I got the answer let me know. :)
Answer:
K = Ka/Kb
Explanation:
P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?
P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka
PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb
K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)
Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)
Kb = [PCl₅]/ ([PCl₃] [Cl₂])
Since [PCl₅] = [PCl₅]
From the Ka equation,
[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)
From the Kb equation
[PCl₅] = Kb ([PCl₃] [Cl₂])
Equating them
Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])
(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)
(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)
Comparing this with the equation for the overall equilibrium constant
K = Ka/Kb
I believe that sugar is a compound because there are elements that make up sugar
Answer:
Methanol and butanol are alcohols. Alcohols have the same_______________ as alkanes and the __________ identifies the compound as an alcohol.
Explanation:
Alcohols belong to a group of organic compounds which contain -OH group as the functional group.
So alcohols have the same carbon -hydrogen bonds as alkanes and the -OH functional group identifies the compound as an alcohol.
