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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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Answer:

The answer is a. I learned that they can multiply by using your cells

6 0
2 years ago
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Consider the reaction between acetylene, c2h2, and oxygen in a welding torch: 2c2h2(g) + 5o2(g) → 4co2(g) + 2h2o(g) if 5.4 moles
fredd [130]
<span>                                      2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)
   from the reaction     2 mol                              4 mol
from the problem      5.4 mol                         10.8 mol

M(CO2) = 12.0 +2*16.0 = 44.0 g/mol
10.8 mol CO2 * 44.0 g CO2/1 mol CO2 = 475.2 g CO2 </span>≈480 = 4.8 * 10² g
Answer is  C. 4.8*10² g.
4 0
3 years ago
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What is the mass in grams of 358 mL of ethylene glycol?
frez [133]
The density of ethylene glycol is: D = 1.11 g/mL
D = m / V
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m = D * V
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6 0
3 years ago
Write the balanced chemical equation between H2SO4H2SO4 and KOHKOH in aqueous solution. This is called a neutralization reaction
const2013 [10]

Answer:

0.185M sulfuric acid

Explanation:

Based on the reaction:

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

<em>1 mole of sulfuric acid reacts with 2 moles of KOH</em>

Initial moles of H₂SO₄ and KOH are:

H₂SO₄: 0.750L ₓ (0.470mol / L) = <em>0.3525 moles of H₂SO₄</em>

KOH: 0.700L ₓ (0.240mol / L) = <em>0.168 moles of KOH</em>

The moles of sulfuric acis that react with KOH are:

0.168mol KOH ₓ (1 mole H₂SO₄ / 2 moles KOH) = 0.0840 moles of sulfuric acid.

Thus, moles that remain are:

0.3525moles - 0.0840 moles = <em>0.2685 moles of sulfuric acid remains</em>

As total volume is 0.700L + 0.750L = 1.450L, concentration is:

0.2685mol / 1.450L = <em>0.185M sulfuric acid</em>

8 0
2 years ago
What are the limitations of litmus paper and Phenolphthalein indicators? Name to other indicators that can be used that do not h
jenyasd209 [6]

Answer:

Here's what I find.

Explanation:

An indicator is usually is a weak acid in which the acid and base forms have different colours. Most indicators change colour over a narrow pH range.

(a) Litmus

Litmus is red in acid (< pH 5) and blue in base (> pH 8).

This is a rather wide pH range, so litmus is not much good in titrations.

However, the range is which it changes colour includes pH 7 (neutral), so it is good for distinguishing between acids and bases.

(b) Phenolphthalein

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This is a narrow pH range, so phenolphthalein is good for titrating acids with strong bases..

However, it can't distinguish between acids and weakly basic solutions.

It would be colourless in a strongly acid solution with pH =1 and in a basic solution with pH = 8.

(c) Other indicators  

Other acid-base indicators have the general limitations as phenolphthalein. Most of them have a small pH range, so they are useful in acid-base titrations.

The only one that could serve as a general acid-base indicator is bromothymol blue, which has a pH range of 6.0 to 7.6.

5 0
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