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Nataly [62]
3 years ago
13

1. If we used 0.0100 moles of K2CO3, how many moles of SrCO3 can be expected to form?​

Chemistry
1 answer:
faltersainse [42]3 years ago
3 0

Answer:

0.01 moles of SrCO₃

Explanation:

In this excersise we need to propose the reaction:

K₂CO₃ + Sr(NO₃)₂  →  2KNO₃ + SrCO₃

As we only have data about the potassium carbonate  we assume the strontium nitrite as the excess reactant.

1 mol of K₂CO₃ react to 1 mol of Sr(NO₃)₂ in order to produce 2 moles of potassium nitrite and 1 mol of strontium carbonate.

Ratio is 1:1. In conclussion,

0.01 mol of K₂CO₃ must produce 0.01 moles of SrCO₃

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Convert each quantity to the indicated units. a. 3.01g to cg. b. 6200m to km
miss Akunina [59]

a. 301 cg

b. 6.2 km

Explanation:

a. knowing that 1 gram (g) is equal to 100 centigrams (cg) we devise the following reasoning:

if        1 g is equal to 100 cg

then  3.01 g are equal to X cg

X = (3.01 × 100) / 1 = 301 cg

b. knowing that 1 kilometer (km) is equal to 1000 meters (m) we devise the following reasoning:

if         1 km is equal to 1000 m

then   Y km are equal to 6200 m

Y = (6200 × 1) / 1000 = 6.2 km

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3 years ago
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A solution contains 0.036 M Cu2+ and 0.044 M Fe2+. A solution containing sulfide ions is added to selectively precipitate one of
Ratling [72]

Answer:

The precipitate is CuS.

Sulfide will precipitate at  [S2-]= 3.61*10^-35 M

Explanation:

<u>Step 1: </u>Data given

The solution contains 0.036 M Cu2+ and 0.044 M Fe2+

Ksp (CuS) = 1.3 × 10-36

Ksp (FeS) = 6.3 × 10-18

Step 2:  Calculate precipitate

CuS → Cu^2+ + S^2-         Ksp= 1.3*10^-36

FeS → Fe^2+ + S^2-      Ksp= 6.3*10^-18

Calculate the minimum of amount needed to form precipitates:

Q=Ksp

<u>For copper</u>  we have:  Ksp=[Cu2+]*[S2-]

Ksp (CuS) = 1.3*10^-36 = 0.036M *[S2-]

[S2-]= 3.61*10^-35 M

<u>For Iron</u>  we have: Ksp=[Fe2+]*[S2-]

Ksp(FeS) = 6.3*10^-18 = 0.044M*[S2-]

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CuS will form precipitates before FeS., because only 3.61*10^-35 M Sulfur Ions are needed for CuS. For FeS we need 1.43*10^-16 M Sulfur Ions which is much larger.

The precipitate is CuS.

Sulfide will precipitate at  [S2-]= 3.61*10^-35 M

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Answer:

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Explanation:

7 0
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