Answer:
The total kg from the mixer: 
Evaporated water: 
Jam produced: 
Explanation:
Step by step:
1) Kg of mixture from the mixer
It says that crushed fruit is added to the mixer with the sugar and pectin.
The fruit added is 1000 kg
The sugar added is 1.22 kg sugar/1 kg crushed fruit:
The pectin added is 0.0025 kg pectin/1 kg crushed fruit:
The total kg from the mixer:
2) Evaporated water
To calculate the evaporated water it's important to have in mind that the mix goes from a concentration of 14 wt% to 67 wt%. This difference is because of the evaporated water. So:
Initial: 1000 kg of fruit with 14 wt% of solids
<em>This amount of solids is constant </em>
Final: The mass of solids is the same but now it represents the 67 wt%
and
3) Jam produced
The correct answer to this question is a physical change.
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
Answer:
Explanation:The final homogenous solution, after cooling it to 40°C, will contain 47 g of potassium sulfate disolved in 150 g of water, so you can calculate the amount disolved per 100 g of water in this way:
[47 g of solute / 150 g of water] * 100 g of g of water = 31.33 grams of solute in 100 g of water.
So, when you compare with the solutiblity, 15 g of solute / 100 g of water, you realize that the solution has more solute dissolved with means that it is supersaturated.
To make a saturated solution, 15 grams of potassium sulfate would dissolve in 100 g of water.
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