Answer:
In the original sample there's 72.78% KCl
Explanation:
KCl(aq) + AgNO₃(aq) → AgCl(s) + KNO₃(aq)
- To solve this problem we need to calculate the mass of <em>pure</em> potasssium chloride. To do that we calculate the moles of KCl that reacted into AgCl:
1.26 g AgCl *
- Now with the molecular weight of KCl, we can calculate the mass of KCl that reacted:
8.792 * 10⁻³ mol KCl * 74.55 g/mol = 0.655 g KCl
- Finally we divide the mass of <em>pure</em> KCl by the mass of the sample, to calculate the percentage KCl:
0.655 / 0.900 * 100% = 72.78%
seaweed having 50 g salt in 1 L water. The bucket contains 150 g of salt in 2 L of water
amount of water present in bucket is twice to amount of water in weed
At equilibrium, volume of water in weed is x and volume in bucket is y but concentration remain same as follows:
At equilibrium, weed loose z L from 1 L water to bucket containing 2 L as follows:
Thus, Weed will loose 0.25 L of water
Hi , well I think that you're referring to a pure substance , in other words a substance that contains only one kind of matter.
Answer:
1.09 x 10⁻⁴ M
Explanation:
The equation of the reaction in given by
Ni²⁺ (aq) + 6NH₃ (aq) ⇔ Ni(NH₃)₆
At the beginning o the reaction, we have 0.18M concentration of Ni and 1.2M concentration of aqueous NH₃ and zero concentration of the product
As the reaction proceeds towards equilibrium, the concentration of the reactants decrease as the concentration of the product starts to increase
From the equation,
1 mole of Ni²⁺ reacts with 6 moles of aqueous NH₃ to give 1mole of Ni(NH3)
therefore
0.18 M of Ni would react with 1.08M (6 x 0.18M) aqueous NH₃ to give 0.18M of Ni(NH₃)6
At equilibrium,
1.08M of NH3 would have reacted to form the product leaving
(1.2 - 1.08)M = 0.12M of aqueous NH₃ left as reactant.
Therefore, formation constant K which is the ratio of the concentration of the product to that of the reactant is given by
K = [Ni(NH₃)₆} / [Ni²⁺] 6[NH₃]
5.5 x 10⁸ = 0.18 M / [Ni²⁺] [0.12]⁶
[Ni²⁺]= 0.18 M / (5.5 x 10⁸) (2.986 x 10⁻⁶)
=0.18 M / 0.00001642
= 1.09 x 10⁻⁴ M
[Ni]²⁺ = 1.09 x 10⁻⁴ M
Hence the concentration of Ni²⁺ is 1.09 x 10⁻⁴ M
At temperature 332.13 °C
<h3>Further explanation</h3>
Charles's Law stated that :
When the gas pressure is kept constant, the gas volume is proportional to the temperature
T₁ = 89 °C=89+273=362 K
V₁ = 0.67 L
V₂ = 1.12 L