What volume of 0.25 mol/L solution of lead(II)nitrate would be required to form 500mL of a 0.15 mol/L solution of lead(II)nitrat
1 answer:
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Answer:
d. To the left because Q > K_p
Explanation:
Hello,
In this case, for the given reaction:
The pressure-based equilibrium expression is:
In such a way, since Kp is given we rather compute the reaction quotient at the specificed pressure of carbon dioxide as shown below:
Therefore, since Q>Kp we can see that there are more products than reactants, which means that the reaction must shift leftwards towards the reactants in order to reestablish equilibrium, thus, answer is d. To the left because Q > Kp.
Regards.
0.85 moles formula units of lead nitrate will produce 0.57 moles formula units of chromium (III) nitrate.
<h3>Explanation</h3>Typically, the oxidation state of Pb in lead nitrate tend to be +2. In other words, Pb in lead nitrate tends to exist as ions. The formula for a nitrate ion is
. The charge on each of the nitrate ion is -1. The charge on the two ions should balance. As a result, each
ion in lead nitrate would pair up with two
ions. The formula for lead nitrate will be
. Each formula unit of lead nitrate will contain one
ion and two
ions.
The "III" in the name "chromium (III) nitrate" is a Roman Numeral. It indicates that the oxidation state of Cr in chromium (III) nitrate is +3. The Cr in that compound will exist as . Similarly, each
will pair up with three
ions. The formula for chromium (III) nitrate will be
. Each formula unit of chromium (III) nitrate will contain one
ion and three
ions.
0.85 moles formula units of lead nitrate will contain 0.85 × 2 = 1.7 moles of ions. Those nitrate ions will end up in 1.7 / 3 = 0.57 moles formula units of chromium (III) nitrate. As a result, the reaction will produce 0.57 moles formula units of chromium (III) nitrate.