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Vanyuwa [196]
3 years ago
11

Explain how it is possible for 10 grams of methane fuel to burn and emit 27 grams of carbon dioxide. Discuss whether or not this

reaction obeys the law of conservation of mass.
Chemistry
1 answer:
Rufina [12.5K]3 years ago
7 0

This reaction obeys the law of conservation of mass.

Explanation:

In the burning reaction, methane (CH₄) react with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O):

CH₄ + 2 O₂ → CO₂ + 2 H₂O

Now we calculate the number of moles of methane and carbon dioxide:

number of moles = mass / molar weight

number of moles of methane = 10 / 16.04 = 0.62 moles

number of moles of carbon dioxide = 27 / 44.01 = 0.61 moles

From the chemical reaction we see that 1 mole of methane produces 1 moles of carbon dioxide so 0.6 moles of methane produces 0.6 mole of carbon dioxide. This reaction obeys the law of conservation of mass because the mass of reactants is equal to the mass of products.

Learn more about:

combustion reaction

brainly.com/question/14117102

#learnwithBrainly

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What is the pH if the [H+] concentration is 3 x10^-13​
Shkiper50 [21]

Answer:

pH = 12.52

Explanation:

Given that,

The [H+] concentration is 3\times 10^{-13}.

We need to find its pH.

We know that, the definition of pH is as follows :

pH=-log[H^+]

Put all the values,

pH=-log[3\times 10^{-13}]\\\\pH=12.52

So, the pH is 12.52.

6 0
3 years ago
A) Compute the repeat unit molecular weight of polystyrene. B) Compute the number-average molecular weight for a polystyrene for
Andrej [43]

Answer:

Explanation:

In Polystrene, the molecular formula for the repeat unit = C_8H_8;

and the atomic weights of Carbon C = 12.01 g/mol

For Hydrogen, it is 1.01 g/mol

Hence, the repeat unit molecular weight is:

m = 8 (12.01 g/mol)+8(1.01 g/mol)

m = 96.08 g/mol + 8.08 g/mol

m = 104.16 g/mol

The degree of polymerization = no-average molecular weight/repeat unit molecular weight.

Mathematically;

DP = \dfrac{\overline M_n}{m}

\overline M_n= DP \times m

\overline M_n= 25000 \times 104.16 \ g/mol

\overline M_n= 2604000  \ g/mol

7 0
2 years ago
When methyloxirane is treated with HBr, the bromide ion attacks the less substituted position. However, when phenyloxirane is tr
konstantin123 [22]

Answer:

See explanation and picture below

Explanation:

First, in the case of methyloxirane (Also known as propilene oxide) the mechanism that is taking place there is something similar to a Sn2 mechanism. Although a Sn2 mechanism is a bimolecular substitution taking place in only step, the mechanism followed here is pretty similar after the first step.

In both cases, the H atom of the HBr goes to the oxygen in the molecule. You'll have a OH⁺ in both. However, in the case of methyloxirane the next step is a Sn2 mechanism step, the bromide ion will go to the less substitued carbon, because the methyl group is exerting a steric hindrance. Not a big one but it has a little effect there, that's why the bromide will rather go to the carbon with more hydrogens. and the final product is formed.

In the case of phenyloxirane, once the OH⁺ is formed, the next step is a Sn1 mechanism. In this case, the bond C - OH⁺ is opened on the side of the phenyl to stabilize the OH. This is because that carbon is more stable than the carbon with no phenyl. (A 3° carbon is more stable than a 2° carbon). Therefore, when this bond opens, the bromide will go there in the next step, and the final product is formed. See picture below for mechanism and products.

4 0
3 years ago
The component that dissolves the other component is called the
zheka24 [161]

Answer:

The component that dissolves the other component is called the solvent. Solute – The component that is dissolved in the solvent is called solute

6 0
2 years ago
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50 POINTS
Trava [24]
My guess is b for the question
3 0
2 years ago
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