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IrinaVladis [17]
3 years ago
11

You are given an unknown sample that when treated with hydrochloric acid forms a precipitate. After decantation, you add boiling

water to the solid; some of the solid disappears. You decant the solution and add a few drops of K2CrO4; a yellow solid appears. To the solid left from the addition of boiling water, you add ammonium hydroxide and centrifuge the product. You then decant the liquid and add concentrated nitric acid to the solution until acidic; a white precipitate forms.
Which group 1 cations were present in your unknown?
Chemistry
1 answer:
MrRa [10]3 years ago
6 0

Answer:

Part B : Hg

Part C : Ag

Explanation:

When we added ammonium hydroxide to the solid remains after addition hot water , AgCl dissolve due to the formation of soluble complex [Ag(NH3)2]+ and Hg2Cl2 gives black precipitate with ammonium hydroxide due to formation of elemental mercury.

Black precipitate is not formed upon addition of amonium hydroxide to the precipitate remains after addition of hot water, so Hg22+ is shown not to be present.

The precepite remains after addition of hot water is dissolved by the addition of ammonium hydroxide, so Ag+ may be present. The presence of Ag+ is not confiremed by K2CrO4 test.

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kondor19780726 [428]

Answer:

B

Explanation:

It's the same substance but in different states.

HETEROGENEOUS mixtures contain substances that are

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4 0
3 years ago
A solution is prepared by adding 100 mL of 1.0 M HC₂H₃O₂(aq) to 100 mL of 1.0 M NaC₂H₃O₂(aq). The solution is stirred and its pH
anastassius [24]

Answer: Option (c) is the correct answer.

Explanation:

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Thus, we can conclude that equation H_{3}O^{+}(aq) + C_{2}H_{3}O^{-}_{2}(aq) \rightarrow HC_{2}H_{3}O_{2}(aq) + H_{2}O(l) represents the chemical reaction that accounts for the fact that acid was added but there was no detectable change in pH.

3 0
3 years ago
Need some help please?
VladimirAG [237]
26. B. 28. B. 20. A.
4 0
2 years ago
A 400 ml sample of gas is heated from -20 c to 60
Viktor [21]
Using p1v1/t1=p2v2/t2
p1=50
p2=225
v1=400ml
v2=?
t1=-20=253k
t2=60=333k
50x400/253=225xv2/333
7.9=0.7xv2
v2=7.9/0.7
v2=11.3ml
6 0
3 years ago
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2. What is the smallest unit of an organism that is classified as living? *
Leviafan [203]

Answer:

D

Explanation:

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3 years ago
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