Answer: measure the mass (48.425g) of KCl
Explanation:
To prepare the solution 0.65M KCl we must measure the mass of KCl that would be dissolved in 1L of the solution. This can be achieved by:
Molar Mass of KCl = 39 + 35.5 = 74.5g/mol
Number of mole (n) = 0.65
Mass conc of KCl = n x molar Mass
Mass conc of KCl = 0.65 x 74.5 = 48.425g
Therefore, to make 0.65M KCl, we must measure 48.425g
Answer:
Carbon dioxide reacts with limewater (a solution of calcium hydroxide, Ca(OH) 2), to form a white precipitate (appears milky) of calcium carbonate, CaCO 3.
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Answer:
A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)
B - Increase(P), Increase(q), Decrease (R)
C - Triple (P) and reduce (q) to one third
Explanation:
<em>According to Le Chatelier principle, when a system is in equilibrium and one of the constraints that affect the rate of reaction is applied, the equilibrium will shift so as to annul the effects of the constraint.</em>
P and Q are reactants, an increase in either or both without an equally measurable increase in R (a product) will shift the equilibrium to the right. Also, any decrease in R without a corresponding decrease in either or both of P and Q will shift the equilibrium to the right. Hence, Increase(P), Increase(q), and Decrease (R) will shift the equilibrium to the right.
In the same vein, any increase in R without a corresponding increase in P and Q will shift the equilibrium to the left. The same goes for any decrease in either or both of P and Q without a counter-decrease in R will shift the equilibrium to the left. Hence, Increase (R), Decrease (P), Decrease(q), and Triple both (Q) and (R) will shift the equilibrium to the left.
Any increase or decrease in P with a commensurable decrease or increase in Q (or vice versa) with R remaining constant will create no shift in the equilibrium. Hence, Triple (P) and reduce (q) to one third will create no shift in the equilibrium.
I would be difficult to remove an electron from a Noble or Inert Gas (also known as the group 8 or 0 elements). This is because they all have filled outermost shells and as such the outermost shell would be held tightly to the nucleus and as such make it difficult to remove. Examples Helium, Neon, Argon, Xenon, Krypton and Radon
