<u>Answer:</u> The moles of hydroxide ions present in the sample is 0.0008 moles
<u>Explanation:</u>
To calculate the concentration of acid, we use the equation given by neutralization reaction:

where,
are the n-factor, molarity and volume of acid which is HCl.
are the n-factor, molarity and volume of base which is 
We are given:

Putting values in above equation, we get:

To calculate the moles of hydroxide ions, we use the equation used to calculate the molarity of solution:

Molarity of solution = 0.011 M
Volume of solution = 36.0 mL
Putting values in above equation, we get:

1 mole of calcium hydroxide produces 1 mole of calcium ions and 2 moles of hydroxide ions.
Moles of hydroxide ions = (0.0004 × 2) = 0.0008 moles
Hence, the moles of hydroxide ions present in the sample is 0.0008 moles
Answer:
Arsenic (As) anion has -‐3 charge, but also forms cations with +3 or +5 charge.
Explanation:
The answer is true!!!!!!!!!!!!!!!!!!!!!!!
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.