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skad [1K]
3 years ago
12

How do you solve this problem

Chemistry
1 answer:
Butoxors [25]3 years ago
4 0
Re-read and go back and look for A & B and then whatever A & B is solve for it and then get your answer

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3 years ago
Given the equilibrium reaction: 2 A (aq) + 3 B (aq) &lt;— —&gt; 2 C (aq) + D (aq) and equilibrium concentrations of [A] = 0.60M,
Alex787 [66]
Given the equilibrium reaction: 2 A (aq) + 3 B (aq) <— —> 2 C (aq) + D (aq) and equilibrium concentrations of [A] = 0.60M, [B] = 0.30 M, [C] = 0.10 M and [D] = 0.50 M. The Kc value will be:
a. 1.9 c. 2.4 b. 0.15 d. 0.51





Answer . A
5 0
3 years ago
I need help someone plsss answer!!<br><br>​
Papessa [141]

Answer:

c

Explanation:

4 0
3 years ago
Read 2 more answers
Foods labeled "fat free" can actually contain less than 0.5 grams of fat.
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5 0
2 years ago
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1. In this experiment, what property of NaCl is used to separate it from the other two components? Is this a chemical or physica
irinina [24]

Answer 1) In the mixture of sand, NaCl and CaCO_{3} first separation was done by dissolving the mixture in water. In the first step the NaCl will get dissolved whereas, CaCO_{3} will be undissolved and sand will settle at the bottom. Here, the physical property of solubility of NaCl in water is taken into consideration. The collected water is then filtered off and evaporated to get the NaCl back from the mixture with some loss. No chemical change occurs in case of NaCl extraction from water.


Answer 2) When CaCO_{3} was to be removed from the undissolved part. The chemical property of CaCO_{3} was used. Where it gets dissolved in acidic medium. And this way we can extract CaCO_{3} and remove sand from it. We change the chemical composition of CaCO_{3} by adding HCl to the mixture and dissolve CaCO_{3} to form CaCl_{2} which gets dissolved into the solution of HCl. Here, first decantation occurs and then extraction is done. Second, extraction is done using potassium carbonate in it which separates CaCl_{2} from sand.


Answer 3) Here, Unknown mixture of salt and sand weighed =7.52 g (before washing);


Unknown mixture of salt and sand weighed =3.45 g (after washing);


To calculate the amount of salt in it, we can simply subtract the values of before and after washing change in weights.


The property of NaCl being soluble in water it will go away with washing leaving behind the sand only, after washing.


So, Weight of salt was = 7.52g - 3.45 g = 4.07g


To find the percentage of sand that was mixed with salt =

(3.45 g sand / 7.52g of mixture) X 100 = 45.9% sand was mixed with salt.


To verify whether the correct percentage of salt and sand was calculated we can recheck the value for salt as well.

(4.07 g of salt / 7.52g of mixture) X 100 = 54.1% salt


On adding we get, 45.9% + 54.1% = 100%.


Which confirms that the calculations are correct.

4 0
3 years ago
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