Answer:
(a) The final pressure of the sample becomes one-fourth of the original pressure.
(b) The pressure of the sample remains unchanged.
(c) The final pressure of the sample becomes four times of the original pressure.
Explanation:
(a)
The volume of sample doubled and kelvin temperature halved.
Therefore, the final pressure of the sample becomes one-fourth of the original pressure.
(b)
Volume and temperature of sample doubled.
Therefore, the pressure of the sample unchanged.
(c)
Volume of sample halved and temperature double.
Therefore, the pressure of the sample becomes four times of the original pressure.
Answer:
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Colligative properties calculations are used for this type of problem. Calculations are as follows:
ΔT(boiling point) = 101.02 °C - 100.0 °C= 1.02 °C
<span>ΔT(boiling point) = (Kb)m
</span>m = 1.02 °C / 0.512 °C kg / mol
<span>m = 1.99 mol / kg
</span><span>ΔT(freezing point) = (Kf)m
</span>ΔT(freezing point) = 1.86 °C kg / mol (<span>1.99 mol / kg)
</span>ΔT(freezing point) = 3.70 <span>°C
</span>Tf - T = 3.70 <span>°C
T = -3.70 </span><span>°C</span>
Answer: Groups and periods are two ways of categorizing elements in the periodic table. Periods are horizontal rows (across) the periodic table, while groups are vertical columns (down) the table. Atomic number increases as you move down a group or across a period.
Explanation:
<span>The two populations will show cospeciation.</span>
