Solubility of barium chloride at 30 degree Celsius is 38.2g /100 g water and solubility of barium chloride at 60 degree Celsius is 46.6 g / 100 g water.
The quantity of barium chloride that is dissolved in water at 30 degree Celsius = 38.2 * [150/100] = 57.30 g.
The quantity of barium chloride that will be dissolved in water at 60 degree Celsius = 46.6 * [150/100] = 69.90 g
The difference between these quantities is the amount of barium chloride that can be dissolved by heating the barium chloride to 60 degree Celsius.
69.90 - 57.30 = 12.60 g. Therefore, 12.60 g of barium chloride can still be dissolved in the water by heating the water to 60 degree Celsius.
PH= pKa + log( conjugate base/ conjugate acid)
Letter D ,because algae is food for fish (natural)
Answer:
ΔH vaporization of chloroform is 30.1kJ/mol
Explanation:
It is possible to find ΔH of vaporization of certain compound knowing vapor pressure under 2 different absolute temperatures (In Kelvin) by using Clausius-Clapeyron equation:
<em>Where P is vapor pressure. R is gas constant (8.314J/molK) and T absolute temperature of 1, first state and 2, final state.</em>
Absolute temperatures in the problem are:
T₁ = 24.1°C + 273.15 = 297.25K
T₂ = -6.3°C + 273.15 = 266.85K
Replacing:
30073J/mol = 30.1kJ/mol = ΔHVap
<h3>ΔH vaporization of chloroform is 30.1kJ/mol</h3>
Fe 3+ + SCN- --> FeSCN 2+
<span>.......Fe 3+ .......SCN-.........FeSCN 2+ </span>
<span>I.......0.04..........0.001.............. </span>
<span>C........-x...............-x............. </span>
<span>E.....0.04-x.....0.001-x...........x </span>
<span>Keq = 203.4 = x / (0.04-x)(0.001-x) </span>
<span>203.4 = x / (x^2 - 0.041x + 4x10^-5) </span>
<span>203.4x^2 - 8.34x + 0.00094 = x </span>
<span>203.4x^2 - 9.34x + 0.00094 = 0 </span>
<span>x = -0.0001M or 0.0458M </span>
<span>so, using your Keq, there would be no SCN- or Fe 3+ left.....all would be in the form of FeSCN 2+</span>