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3 years ago

Consider the rate law below.

Chemistry

1 answer:

Serga [27]3 years ago

5 0

C. quadruples the rate

<h3>Further explanation</h3>

Given

The rate law :

R=k[A]²

Required

The rate

Solution

There are several factors that influence reaction kinetics :

1. Concentration
2. Surface area
3. Temperature
4. Catalyst
5. Pressure
6. Stirring

The rate is proportional to the concentration.

If the concentration increased, the reaction rate will increase

The reaction is second-order overall(The exponent is 2)

The concentration of A is doubled, the reaction rate will increase :

r = k[A]² ⇒ r= k[2A]²⇒r=4k[A]²

<em>The reaction rate will quadruple.</em>

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Answer:

Option C

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Explanation:

Carbonxylic acids are compounds which has the general formula

R–COOH where R is an alkyl group.

Considering the options given in the question above,

For A:

CH₃CH₂OCH₂CH₃ is an ether compound with general formula ROR' where R and R' are both alkyl group.

For B:

CH₃CH₂CH₂CH₂OH is an alcohol with general formula ROH where R is an alkyl group.

For C:

CH₃CH₂CH₂COOH is a carbonxylic acid with general formula R–COOH where R is an alkyl group.

For D:

CH₃CH₂C=OCH₂CH₃ is a ketone compound with general formula RC=OR' where R and R' are both alkyl group

For E:

ClCH₂CH₂CH₂CH₂CH₂CH₂Br is simply an Alkyl halide with general formula XRX where X is an halogen (i.e F, Cl, Br or I) and R is an alkyl group.

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Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2$ = 12.24 g

Mass of $H_2O$ = 2.505 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 12.24 g of carbon dioxide, = $\frac{12}{44}\times 12.24=3.338g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 2.505 g of water, = $\frac{2}{18}\times 2.505=0.278g$ of hydrogen will be contained.

Mass of oxygen in the compound = (5.287) - (3.338+0.278) = 1.671 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{3.338g}{12g/mole}=0.278moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.278g}{1g/mole}=0.278moles$

Moles of O= $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{1.671g}{16g/mole}=0.104moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.278}{0.104}=3$

For H = $\frac{0.278}{0.104}=3$

For O = $\frac{0.104}{0.104}=1$

The ratio of C : H : O = 3: 3: 1

Hence the empirical formula is $C_3H_3O$ .

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