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Alecsey [184]
1 year ago
13

Balance each reaction and write its reaction quotient, Qc:(b) SF₆(g) + SO₃(g) → SO₂F₂(g)

Chemistry
1 answer:
IceJOKER [234]1 year ago
4 0

When we balance the given equation

      SF₆(g) + SO₃(g) → SO₂F₂(g)

We will get

      SF₆(g) + 2SO₃(g) → 3SO₂F₂(g)

Solution:

Balancing the given equaation

      SF₆(g) + SO₃(g) → SO₂F₂(g)

We have to balance the given number of O

      SF₆(g) + 2SO₃(g) → 3SO₂F₂(g)

We get balanced equation

      SF₆(g) + 2SO₃(g) → 3SO₂F₂(g)

The reaction quotient will be

     Qc = [product] / [reactant]

     Qc ​= [SO₂F₂(g)] / [SF₆(g) + SO₃(g)]

To learn more click the given link

brainly.com/question/16025432

#SPJ4

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His the answer I hope it helps
3 0
3 years ago
A 150.0 mL solution of 2.888 M strontium nitrate is mixed with 200.0 mL of a 3.076 M sodium fluoride solution. Calculate the mas
Lelechka [254]

Answer:

Mass SrF2 produced = 38.63 g SrF2 produced

[Na^+]:  = 1.758 M

[NO3^-]:  = 1.238 M

[Sr^2+] = 0.3589 M

[F^-] = 2.36*10^-5 M

Explanation:

Step 1: Data given

Volume of 2.888M strontium nitrate = 150.0 mL = 0.150 L

Volume of 3.076 M sodium fluoride = 200.0 mL = 0.200 L

Step 2 : The balanced equation

Sr(NO3)2(aq) + 2NaF(aq) → SrF2(s) + 2NaNO3(aq) → Sr2+ + 2F- + 2

Step 3: Calculate moles strontium nitrate

Moles Sr(NO3)2 = Molarity * volume  

Moles Sr(NO3)2 = 2.888 M * 0.150 L

Moles Sr(NO3)2 = 0.4332 moles

Step 4: Calculate moles NaF

Moles NaF = 3.076 M * 0.200 L

Moles NaF = 0.6152 moles

It takes 2 moles F^- to precipitate 1 mole Sr^2+, so F^- is limiting.

Step 5: Calculate limiting reactant

For 1 mol of Sr(NO3)2 we need 2 moles of NaF to produce 1 mol of SrF2 and 2 moles of NaNO3

NaF is the limiting reactant. It will completely be consumed (0.6152 moles).

Sr(NO3)2 is in excess. There will react 0.6152/2 = 0.3076 moles

Moles Sr^2+ precipitated by F^- = 0.3076

There will remain 0.4332 - 0.3076 = 0.1256 moles of Sr(NO3)2

Moles Sr^2+ no precipitated (left over) = 0.1256 moles

Step 6: Calculate moles SrF2  

For 1 mol of Sr(NO3)2 we need 2 moles of NaF to produce 1 mol of SrF2 and 2 moles of NaNO3

For 0.6152 moles NaF we have 0.6152/2 = 0.3076 moles of SrF2

Mass SrF2 produced:  0.3076 mol * 125.6 g/mol = 38.63 g SrF2 produced

Step 7: Calculate concentration of [Na+] and [NO3-]

Since both Na^+ and NO3^- are spectator ions, and the final volume is 150 ml + 200 ml = 350 ml (0.350 L), the concentrations of Na^+ and NO3^- can be calculated as follows:

[Na^+]:  (200 ml)(3.076 M) = (350 ml)(x M) and x = 1.758 M

[NO3^-]:  (150 ml)(2.888 M)(2) = (350 ml)(x M) = 1.238 M

Step 8: Calculate [Sr^2+] and [F^-]

[Sr^2+] = 0.1256 moles/0.350 L = 0.3589 M

To find [F^-], one needs the Ksp for SrF2.  There are several values listed in the literature. I am using a value of 2x10^-10.

SrF2(s) <==> Sr^2+(aq) + 2F^-(aq)

Ksp = [Sr^2+][F^-]²

2x10^-10 = (0.3589)(x)²

x² = 5.57*10^-10

x = [F^-] = 2.36*10^-5 M

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3 years ago
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An unsaturated solution is formed when 80 grams of a salt is dissolved in 100 grams of water at 40. This salt could be ?
LekaFEV [45]
A simple way to go about this is that we look at the solubility curve, on the x axis we first look at the temperature and then the corresponding value of solute/100g H2O on the y axis, from the 4 curves above only NaNO3 has a curve that can accommodate  80g of salt at 40 without being Saturated since at 40 degrees it can accommodate 105g of salt to become completely Saturated.
7 0
3 years ago
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