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GenaCL600 [577]

2 years ago

8

Explain how the models you developed show that when methane combines with oxygen to form carbon dioxide and water, no atoms are

left over from the reaction.

1 answer:

Xelga [282]2 years ago

3 0

Answer: the bonds in the methane and oxygen come apart, the atoms rearrange and then re-bond to form water and carbon dioxide

Explanation:^

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Gaseous cyclobutene undergoes a first-order reaction to form gaseous butadiene. At a particular temperature, the partial pressur

Gnoma [55]

Answer:

41.3 minutes

Explanation:

Since the reaction is a first order reaction, therefore, half life is independent of the initial concentration, or in this case, pressure.

$t_{1/2}= \frac{0.693}{K}$

So, fraction of original pressure = $\frac{1}{2}^2$

n here is number of half life

therefore, $\frac{1}{8}= \frac{1}{2}^3$

⇒ n= 3

it took 124 minutes to drop pressure to 1/8 of original value, half life = 124/3= 41.3 minutes.

8 0

2 years ago

Read 2 more answers

2KOH+H2SO4=k2SO4+2H2O Is a balanced equation, displaying the combination of potassium hydroxide with sulfuric acid increase pota

Lerok [7]

Answer:

The number of moles of potassium hydroxide, KOH required to make 4 moles of K₂SO₄ is 8 moles of KOH

Explanation:

2KOH + H₂SO₄ → K₂SO₄ + 2H₂O

From the above reaction, we have 2 moles of KOH combining with 1 mole of H₂SO₄ to produce 1 mole of K₂SO₄ and 2 moles of H₂O.

Therefore the number of moles of potassium hydroxide that will be needed to make 4 moles of K₂SO₄ is;

8KOH + 4H₂SO₄ → 4K₂SO₄ + 8H₂O

8 moles of KOH is required to make 4 moles of K₂SO₄.

6 0

1 year ago

Sort the descriptions below to indicate whether they describe β oxidation of stearoyl-coa, oleoyl-coa, or both. items (6 items)

Oksi-84 [34.3K]

Answer:

The answer is given below

Explanation:

β oxidation of stearoyl-CoA------------ 8 FAD reduced

β oxidation of oleoyl-CoA -------------- 1- 7 FAD reduced and 2- reduced requires δ3,δ2-enoyl-CoA isomerase required for completion of round 4

in Both ----------- 1- 9 acetyl-CoA produced, 2- 8 NAD+ reduced and 3- β oxidation completed in 8 rounds

4 0

2 years ago

A solution of 20.0 g of which hydrated salt dissolved in 200 g H2O will have the lowest freezing point? (A) CuSO4 • 5 H2O (M = 2

Andrews [41]

Answer:

(D) Na₂SO₄•10H₂O (M = 286).

Explanation:

The depression in freezing point of water by adding a solute is determined using the relation:

ΔTf = i.Kf.m,

Where, <em>ΔTf </em>is the depression in freezing point of water.

<em>i</em> is van't Hoff factor.

<em>Kf </em>is the molal depression constant.

<em>m</em> is the molality of the solute.

Since, Kf and m is constant for all the mentioned salts. So, the depression in freezing point depends strongly on the van't Hoff factor (i).

van't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass.

(A) CuSO₄•5H₂O:

CuSO₄ is dissociated to Cu⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(B) NiSO₄•6H₂O:

NiSO₄ is dissociated to Ni⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(C) MgSO₄•7H₂O:

MgSO₄ is dissociated to Mg⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(D) Na₂SO₄•10H₂O:

Na₂SO₄ is dissociated to 2 Na⁺ and SO₄²⁻.

So, i = dissociated ions/no. of particles = 3/1 = 3.

∴ The salt with the high (i) value is Na₂SO₄•10H₂O.

So, the highest ΔTf resulted by adding Na₂SO₄•10H₂O salt.

4 0

2 years ago

Cu2O reacts with Cu2S to produce Cu and SO2 gas. How many moles of SO2 is produced when 0.48 mol Cu is produced in this reaction

nika2105 [10]

Answer:

2.88moles

Explanation:

To solve this question adequately, we need to have a balanced chemical equation

2Cu2O + Cu2S → 6Cu + SO2

From the chemical equation, we can see that the mole ratio of the sulfur oxide to the Cu is 6 to 1.

This means that if 0.48 mole is sulfur oxide is produced, then 6 of this number of moles would have been produced in the case of Cu.

Hence, to get the number of moles of copper produced, we simply multiply 0.48 by 6 to give 2.88 moles

3 0

2 years ago