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brilliants [131]
3 years ago
7

What statement is incorrect about this oxidation-reduction reaction? 2 SO2(g) + O2(g) → 2 SO3(g) What statement is incorrect abo

ut this oxidation-reduction reaction? 2 SO2(g) + O2(g) → 2 SO3(g) SO2 gains electrons. O2 is reduced. SO2 is the reducing agent. O2 is the oxidizing agent.
Chemistry
1 answer:
Leno4ka [110]3 years ago
6 0

Answer:

The incorrect statement is: SO₂ gains electrons                      

Explanation:

A chemical reaction that involves the simultaneous transfer of electrons between two chemical species, is known as the redox reaction.

Given chemical reaction: 2SO₂(g) + O₂(g) → 2SO₃(g)

In this redox reaction, S is present in +4 oxidation state in SO₂ and +6 oxidation state SO₃. Whereas, O is present in 0 oxidation state in O₂ and -2 oxidation state in SO₃.

<u>Therefore, SO₂ loses electrons and thus gets oxidized. Whereas, O₂ gains electrons and thus gets reduced. </u>

<u>In this reaction, SO₂ is the reducing agent and O₂ is the oxidizing agent.</u>

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Answer:

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Explanation:

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3 0
3 years ago
Read 2 more answers
What is the total pressure exerted by a mixture of 48.0 grams of CH4 and 56.0 grams of
Oksi-84 [34.3K]

The pressure of the gas is obtained as 48 atm.

<h3>What is the total pressure?</h3>

Now we know that;

Number of moles of CH4 = 48.0 grams /16 g/mol = 3 moles

Number of moles of H2 =  56.0 grams/2 g/mol = 28 moles

Total number of moles present = 3 moles + 28 moles = 31 moles

Using;

PV =nRT

P = total pressure

V = total volume

n = total number of moles

R = gas constant

T = temperature

P = nRT/V

P = 31 * 0.082 * 286/15

P = 48 atm

Learn more about pressure of a gas:brainly.com/question/18124975

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4 0
1 year ago
What causes pressure inside a helium balloon?
kupik [55]

Explanation:

So the gas pressure of a helium balloon arises from the impact of the collisions of the helium atoms between themselves and with the inside surface of the balloon. Of course, the outside atmosphere similarly exerts a pressure on the outside of the balloon.

7 0
3 years ago
Iodine-131 has a half-life of 8 days. If the amount of iodine-131 in the orginial sample is 8 g, how much iodine-131 will remain
hram777 [196]

1/32

Explanation it stated with half on 8 days then that means you divide 24 and 8 so its 3 and you have to multiply ✖ ½

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3 years ago
A gas sample has a temperature of 22c with an unknown volume. The same gas has a volume of 456 mL when the temperature is 86c wi
Mkey [24]

The initial volume is 116.65 mL

<u>Explanation:</u>

<u />

Given:

Temperature, T₁ = 22°C

T₂ = 86°C

Volume, V₂ = 456 m

V₁ = ?

According to Charle's law:

\frac{V_1}{T_1} = \frac{V_2}{T_2}

Substituting the values:

\frac{V_1}{22} =\frac{456}{86} \\\\V_1 = \frac{456 X 22}{86} \\\\V_1 = 116.65 mL

Therefore, the initial volume is 116.65 mL

3 0
3 years ago
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