Answer:
Molecular solid
Explanation:
A molecular solid has a low melting point, they are soft and do not conduct electricity.
We have been told in the question that the solid does not really dissolve in water and it's solution does not improve the electrical conductivity of water. Hence, it must be a molecular solid.
Answer:
[OH⁻] = 4.3 x 10⁻¹¹M in OH⁻ ions.
Explanation:
Assuming the source of the carbonate ion is from a Group IA carbonate salt (e.g.; Na₂CO₃), then 0.115M Na₂CO₃(aq) => 2(0.115)M Na⁺(aq) + 0.115M CO₃²⁻(aq). The 0.115M CO₃²⁻ then reacts with water to give 0.115M carbonic acid; H₂CO₃(aq) in equilibrium with H⁺(aq) and HCO₃⁻(aq) as the 1st ionization step.
Analysis:
H₂CO₃(aq) ⇄ H⁺(aq) + HCO₃⁻(aq); Ka(1) = 4.3 x 10⁻⁷
C(i) 0.115M 0 0
ΔC -x +x +x
C(eq) 0.115M - x x x
≅ 0.115M
Ka(1) = [H⁺(aq)][HCO₃⁻(aq)]/[H₂CO₃(aq)] = [(x)(x)/(0.115)]M = [x²/0.115]M
= 4.3 x 10⁻⁷ => x = [H⁺(aq)]₁ = SqrRt(4.3 x 10⁻⁷ · 0.115)M = 2.32 x 10⁻⁴M in H⁺ ions.
In general, it is assumed that all of the hydronium ion comes from the 1st ionization step as adding 10⁻¹¹ to 10⁻⁷ would be an insignificant change in H⁺ ion concentration. Therefore, using 2.32 x 10⁻⁴M in H⁺ ion concentration, the hydroxide ion concentration is then calculated from
[H⁺][OH⁻] = Kw => [OH⁻] = (1 x 10⁻¹⁴/2.32 x 10⁻⁴)M = 4.3 x 10⁻¹¹M in OH⁻ ions.
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NOTE: The 2.32 x 10⁻⁴M value for [H⁺] is reasonable for carbonic acid solution with pH ≅ 3.5 - 4.0.
Answer:
Relative humidity is low .
Explanation:
The wet bulb reads low temperature because due to low humidity of atmosphere , evaporation of water takes place from the wet bulb which makes the bulb cool and therefore it reads lower temperature . In the process of evaporation , heat equal to latent heat of vaporization is taken from the bulb and it loses temperature.
Answer:
What that means is that when pressure and number of moles are kept constant, increasing the temperature will result in an increase in volume. Likewise, a decrease in temperature will result in a decrease in volume. In your case, the volume of the gas decreased by a factor of about 3, from "140.0 mL" to "50.0 mL".
