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Flura [38]
3 years ago
15

You have just completed assaying the change in absorbance for an enzyme. The initial absorbance was 0.022; the absorbance after

5 minutes was 0.444. Calculate the rate of enzyme activity for this sample (as increase in A per 10 minutes). Show your work. What is a RATE and name some RATES that you commonly encounter.
Chemistry
1 answer:
Hitman42 [59]3 years ago
4 0

Answer:

The answer is 0.844/10 minutes

Explanation:

You have an enzyme that catalizes a reaction which gives a product that can be quantified by an absorbance measurement. The more reaction time, the more product quantity and higher absorbance.

The rate of the reaction is the change in products quantity per time unit. As you are using the absorbance as a measure of the product quantity, you can calculate the rate as the change in absorbance (ΔA) per time (in minutes) as follows:

rate= ΔA/time

rate= (final absorbance - initial absorbance) /minutes

rate= (0.444-0.022)/5 min

rate= 0.422/5 min

In 10 minutes will be :

rate= 0.844/10 min

Commonly, a rate is the relation between two quantities measured in different units. For example, the speed of a car is the change in meters (traveled distance) per time (m/s or km/h). For an enzyme, is the same (quantity of product/time).

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A) Combustion analysis of toluene, a common organic solvent, gives 5.86 mg of CO2 and 1.37 mg of H2O. If the compound contains o
Mumz [18]

<u>Answer:</u>

<u>For a:</u> The empirical formula for the given compound is CH

<u>For b:</u> The empirical and molecular formula for the given organic compound are C_{10}H_{20}O

<u>Explanation:</u>

  • <u>For a:</u>

The chemical equation for the combustion of hydrocarbon follows:

C_xH_y+O_2\rightarrow CO_2+H_2O

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

We are given:

Conversion factor used:  1 g = 1000 mg

Mass of CO_2=5.86mg=5.86\times 10^{-3}g

Mass of H_2O=1.37mg=1.37\times 10^{-3}g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

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

So, in 5.86\times 10^{-3}g  of carbon dioxide, \frac{12}{44}\times 5.86\times 10^{-3}=1.60\times 10^{-3}g of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

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

So, in 1.37\times 10^{-3}g of water, \frac{2}{18}\times 1.37\times 10^{-3}=0.152\times 10^{-3}g of hydrogen will be contained.

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = \frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.60\times 10^{-3}g}{12g/mole}=0.133\times 10^{-3}moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.152\times 10^{-3}g}{1g/mole}=0.152\times 10^{-3}moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.133\times 10^{-3} moles.

For Carbon = \frac{0.133\times 10^{-3}}{0.133\times 10^{-3}}=1

For Hydrogen = \frac{0.152\times 10^{-3}}{0.133\times 10^{-3}}=1.14\approx 1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 1

Hence, the empirical formula for the given compound is CH

  • <u>For b:</u>

The chemical equation for the combustion of menthol 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  = 0.2829 g

Mass of H_2O = 0.1159 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

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

So, in 0.2829  g of carbon dioxide, \frac{12}{44}\times 0.2829=0.077g of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

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

So, in 0.1159 g of water, \frac{2}{18}\times 0.1159=0.013g of hydrogen will be contained.

Mass of oxygen in the compound = (0.1005) - (0.077 + 0.013) = 0.105 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = \frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.077g}{12g/mole}=0.0064moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.013g}{1g/mole}=0.013moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.0105g}{16g/mole}=0.00065moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00065 moles.

For Carbon = \frac{0.0064}{0.00065}=9.84\approx 10

For Hydrogen = \frac{0.013}{0.00065}=20

For Oxygen = \frac{0.00065}{0.00065}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 10 : 20 : 1

The empirical formula for the given compound is C_{10}H_{20}O

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 156 g/mol

Mass of empirical formula = 156 g/mol

Putting values in above equation, we get:

n=\frac{156g/mol}{156g/mol}=1

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(1\times 10)}H_{(1\times 20)}O_{(1\times 1)}=C_{10}H_{20}O

Hence, the empirical and molecular formula for the given organic compound are C_{10}H_{20}O

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