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3 years ago

Which equations represent precipitation reactions?

Chemistry

1 answer:

viktelen [127]3 years ago

3 0

In a precipitation reaction, one of the product formed is an insoluble precipitate.

The given reactions can be written with phases included as:

A. $Na_{2}S(aq)+FeBr_{2}(aq)-->2NaBr(aq)+FeS(s)$ : In this reaction one of the product, FeS is insoluble. Therefore, this is a precipitation reaction.

B. $MgSO_{4}(aq)+CaCl_{2}(aq)-->MgCl_{2}(aq)+CaSO_{4}(s)$ : In this reaction, the product $CaSO_{4}$ is a solid(insoluble). So, this is a precipitation reaction too.

C. $LiOH(aq)+NH_{4}I(aq)-->LiI(aq)+NH_{4}OH(aq)$ : In this reaction, both the products are soluble. So this is not a precipitation reaction.

D. $2NaCl(aq)+K_{2}S(aq)-->Na_{2}S(aq)+2KCl(aq)$ : In this reaction, both the products are soluble. So this is not a precipitation reaction.

E. $AgNO_{3}(aq)+NaCl(aq)-->AgCl(s)+NaNO_{3}(aq)$ : In this reaction, the product AgCl is a precipitate. So, it is a precipitation reaction.

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3 years ago

A 250-mL aqueous solution contains 1.56 10–5 g of methanol and has a density of 1.03 g/mL. What is the concentration in ppm?

Hoochie [10]

First, calculate for the mass of the aqueous solution by multiplying the given volume (in mL) by the density (in g/mL). In mathematical equation, that is,

m = ρV

where m is mass, ρ is density, and V is volume. Substituting the known values,

m = (1.03 g/mL)(250 mL) = 257.5 g

To get the concentration in ppm, divide the given mass of methanol by the mass of the solution. Note that the parts-per million (ppm) is equal to mass of solute in milligram(mg) divided by the mass of solution in kilogram (kg)

C (in ppm) = (1.56 x 10^-6 g)(1000 mg/1 g) / (257.5 g)(1 kg/1000 g)

Simplifying,
C (in ppm) = (1.56 x 10^-3 mg)/ 0.2575 kg

C (in ppm) = 0.00606 ppm

Answer: 0.00606 ppm

7 0

3 years ago

Read 2 more answers

What volume will 3.0 moles of O2 gas occupy at S.T.P?

REY [17]

Answer:

3 moles of oxygen at STP will occupy a volume of 3×22. 4=67. 2 L.

Explanation:

6 0

2 years ago

What's what's the Lewis structure for 2H2O

Umnica [9.8K]

Lewis Structure:

H -- O -- H (bent, Oxygen has 2 lone pairs)

Percent Composition

Hydrogen percent composition = [ 2 * (Hydrogen mass) ] / [Total mass of H2O]

Oxygen percent composition = [Oxygen mass] / [Total mass of H2O]

dose this make any sense...??

5 0

3 years ago

Consider the following balanced equation. SiO2(s)+3C(s)→SiC(s)+2CO(g) Complete the following table, showing the appropriate numb

vlada-n [284]

Answer:

mol(SiO₂) mol(C) mol(SiC) mol(CO)

3 9 3 6

1 3 1 2

13 39 13 26

2.5 7.5 2.5 5.0

1.4 4.2 1.4 2.8

Explanation:

From the balanced equation:

SiO₂(s) + 3C(s) → SiC(s) + 2CO(g),

It is clear that 1.0 mole of SiO₂ reacts with 3.0 moles of C to produce 1.0 mole of SiC and 2.0 moles of CO.

We can complete the table of no. of moles of each component:

1. 9.0 moles of C:

We use the triple amount of C, so we multiply the others by 3.0.

So, it will be 3.0 moles of SiO₂ with 9.0 moles of C that produce 3.0 moles of SiC and 6.0 moles of CO.

2. 1.0 mole of SiO₂:

We use the same amount of SiO₂ as in the balnced equation, so the no. of moles of other components will be the same as in the balanced equation.

So, it will be 1.0 moles of SiO₂ with 3.0 moles of C that produce 1.0 moles of SiC and 2.0 moles of CO.

3. 26.0 moles of CO:

We use the amount of CO higher by 13 times than that in the balanced equation, so we multiply the others by 13.0.

So, it will be 13.0 moles of SiO₂ with 39.0 moles of C that produce 13.0 moles of SiC and 26.0 moles of CO.

4. 7.5 moles of C:

We use the amount of C higher by 2.5 times than that in the balanced equation, so we multiply the others by 2.5.

So, it will be 2.5 moles of SiO₂ with 7.5 moles of C that produce 2.5 moles of SiC and 5.0 moles of CO.

5. 1.4 moles of SiO₂:

We use the amount of SiO₂ higher by 1.4 times than that in the balanced equation, so we multiply the others by 1.4.

So, it will be 1.4 moles of SiO₂ with 4.2 moles of C that produce 1.4 moles of SiC and 2.8 moles of CO.

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3 years ago