I don’t know what the question is asking
Answer:
The molarity is 0.56
Explanation:
In a mixture, the chemical present in the greatest amount is called a solvent, while the other components are called solutes. Then, the molarity or molar concentration is the number of moles of solute per liter of solution.
In other words, molarity is the number of moles of solute that are dissolved in a given volume.
The Molarity of a solution is determined by:
Molarity is expressed in units (
).
Then you must know the number of moles of Cu(NO₂)₂. For that it is necessary to know the molar mass. Being:
-
Cu: 63.54 g/mol
- N: 14 g/mol
- O: 16 g/mol
the molar mass of Cu(NO₂)₂ is:
Cu(NO₂)₂= 63.54 g/mol + 2*(14 g/mol + 2* 16 g/mol)= 155.54 g/mol
Now the following rule of three applies: if 155.54 g are in 1 mole of the compound, 225 g in how many moles are they?
moles= 1.45
So you know:
- number of moles of solute= 1.45 moles
- volume=2.59 L
Replacing in the definition of molarity:
Molarity= 0.56
<u><em>The molarity is 0.56</em></u><u><em></em></u>
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.
