If solid NaCl is added to a saturated water solution of PbCl2 at 20o C, a precipitate is formed. How would this affect the value
2 answers:
Answer:
c) The Ksp remains the same.
Explanation:
The solubility product constant, Ksp is essentially an equilibrium constant which represents the extent to which a solute dissolves in a solvent.
In the given situation, the PbCl2 equilibrium can be represented as:
Addition of NaCl will introduce a common ion i.e. Cl-. This would increase its concentration and solubility. This will change the reaction quotient. However, the value of Ksp which is the 'equilibrium constant' will reamin unaffected since it is a constant at a given temperature.
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Answer:
964ug
Explanation:
The problem here involves converting from one unit to another.
We are to convert from ounces to micrograms.
1ug = 1 x 10⁻⁶g
1oz = 28.35g
So we first convert to grams from oz then take to ug:
Solving:
1oz = 28.35g
3.4 x 10⁻⁵oz will then give 3.4 x 10⁻⁵ x 28.35 = 9.64 x 10⁻⁴g
So;
1 x 10⁻⁶g = 1ug
9.64 x 10⁻⁴g will give = 9.64 x 10²ug or 964ug
Answer:
Explanation:
We usually approximate the density of water to about at room temperature. In terms of the precise density of water, this is not the case, however, as density is temperature-dependent.
The density of water decreases with an increase in temperature after the peak point of its density. The same trend might be spotted if the temperature of water is decreased from the peak point.
This peak point at which the density of water has the greatest value is usually approximated to about . For your information, I'm attaching the graph illustrating the function of the density of water against temperature where you could clearly indicate the maximum point.
To a higher precision, the density of water has a maximum value at , and the density at this point is exactly
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