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Anni [7]
1 year ago
15

If 10.0 moles of sulfur dioxide react with excess chlorine gas, what is the theoretical yield (in grams) of CI2O produced?

Chemistry
1 answer:
xz_007 [3.2K]1 year ago
6 0

Answer:

869 g Cl₂O

Explanation:

To find the theoretical yield of Cl₂O, you need to (1) convert moles SO₂ to moles Cl₂O (via mole-to-mole ratio from reaction coefficients) and then (2) convert moles Cl₂O to grams Cl₂O (via molar mass). It is important to arrange the conversions/ratios in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 3 sig figs to reflect the sig figs of the given amount (10.0 moles).

1 SO₂ (g) + 2 Cl₂ (g) ----> 1 SOCl₂ (g) + 1 Cl₂O (g)

Molar Mass (Cl₂O): 2(35.453 g/mol) + 15.998 g/mol

Molar Mass (Cl₂O): 86.904 g/mol

10.0 moles SO₂         1 mole Cl₂O            86.904 g
------------------------  x  ----------------------  x  ------------------  = 869 g Cl₂O
                                    1 mole SO₂              1 mole

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C  The number and types of bonds within the molecule.

Explanation:

In a molecule, the number and types of bonds present determines the amount of available energy therein.

When bonds are broken or formed, energy is usually released.

  • Elements combine with one another in order to attain stability in this state.
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Learn more:

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4 0
3 years ago
When people walked on the Moon, they found that they could jump higher than they could back on Earth. Why is this true?
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8 0
3 years ago
when a solution of sodium chloride is added to a solution of copper(ii) nitrate, no precipitate is observed. Write the molexular
zhuklara [117]

Explanation:

1.

Cu(NO3)2 + 2NaCl(aq) --> CuCl2(aq) + 2NaNO3(aq)

2.

Cu(NO3)2 + 2NaOH(aq) --> Cu(OH)2(s) + 2NaNO3(aq)

A light blue precipitate of Cu(OH)2 is formed and NaNO3 in solution.

3.

Cu(NO3)2(aq) --> Cu2+(aq) + 2NO3^-2(aq)

2NaOH(aq) --> 2Na+(aq) + 2OH-(aq)

Cu2+(aq) + 2OH-(aq) --> Cu(OH)2(aq)

2Na+(aq) + 2NO3^-2(aq) --> 2NaNO3(aq)

4.

The reaction in both Questions 1 and 2 is called Double displacement reaction. A double-replacement reaction exchanges the cations and/or or the anions of two ionic compounds. A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate (precipitated) while the other in solution.

Since the cation and anions in Qustion 1 were exchanged, the same was done for Question 2, hence the identity of the precipitate in Question 2 was got.

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3 years ago
Which segment represents melting?"<br> Temperature (°C)<br> Time (minutes)
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3 years ago
A buffer solution contains 0.345 M acetic acid and 0.377 M sodium acetate . If 0.0613 moles of potassium hydroxide are added to
melamori03 [73]

Answer:

pH = 5.54

Explanation:

The pH of a buffer solution is given by the <em>Henderson-Hasselbach (H-H) equation</em>:

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For acetic acid, pKa = 4.75.

We <u>calculate the original number of moles for acetic acid and acetate</u>, using the <em>given concentrations and volume</em>:

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The number of CH₃COO⁻ moles will increase with the added moles of KOH while the number of CH₃COOH moles will decrease by the same amount.

Now we use the H-H equation to <u>calculate the new pH</u>, by using the <em>new concentrations</em>:

  • pH = 4.75 + log\frac{(0.0942+0.0613)mol/0.250L}{(0.0862-0.0613)mol/0.250L} = 5.54
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