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alexdok [17]
3 years ago
5

Calculate the number of SO2 molecules formed when 14 S8 molecules and 184 O2 molecules react

Chemistry
1 answer:
Scilla [17]3 years ago
3 0

Answer:

112 molecules are formed from the reaction

Explanation:

Let's make the reaction:

S₈ + 8O₂  → 8SO₂

1 mol of S₈ reacts with 8 moles of O₂ to produce 8 moles of sulfur dioxide

Let's determine the moles ( amount of molecules / NA)

14 / 6.02x10²³ = 2.32x10⁻²³ mol of S

184 / 6.02x10²³ = 3.05x10⁻²² mol of O₂

As ratio is 1:8, the limiting reactant would be the sulfur. It's obviouly I do not have enough the amount of molecules.

8 mol of O₂ need 1 mol of S₈

3.05x10⁻²² mol of O₂ would need (3.05x10⁻²² / 8) = 3.81ₓ10²¹

Now, that we get the limiting, we can work with the reaction

1 mol of S₈ produce 8 mol of dioxide

Then, 2.32x10⁻²³ mol of S will produce (2.32x10⁻²³ . 8) = 1.85x10⁻²² moles of SO₂

Let's convert the moles in, amount of molecules (mol . NA)

1.85x10⁻²² mol . 6.02x10²³ = 112 molecules

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A gas is collected at 20.0 °C and 725.0 mm Hg. When the temperature is
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Answer:

676mmHg

Explanation:

Using the formula;

P1/T1 = P2/T2

Where;

P1 = initial pressure (mmHg)

P2 = final pressure (mmHg)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question;

P1 = 725.0mmHg

P2 = ?

T1 = 20°C = 20 + 273 = 293K

T2 = 0°C = 0 + 273 = 273K

Using P1/T1 = P2/T2

725/293 = P2/273

Cross multiply

725 × 273 = 293 × P2

197925 = 293P2

P2 = 197925 ÷ 293

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3 years ago
The higher the pH, the less acidic the solution
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Answer:

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Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find
FromTheMoon [43]

<u>Answer:</u> The concentration of SO_4^{2-} at equilibrium is 0.00608 M

<u>Explanation:</u>

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

       H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)

            0.025          0.025       0.025

Equation for the second dissociation of sulfuric acid:

                    HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)

<u>Initial:</u>            0.025            0.025      

<u>At eqllm:</u>      0.025-x          0.025+x        x

The expression of second equilibrium constant equation follows:

Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}

We know that:

Ka_2\text{ for }H_2SO_4=0.01

Putting values in above equation, we get:

0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of SO_4^{2-} at equilibrium is 0.00608 M

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