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loris [4]
3 years ago
9

Mercury (Hg) poisoning is a debilitating disease that is often fatal. In the human body, mercury reacts with essential enzymes l

eading to irreversible inactivity of these enzymes. If the amount of mercury in a polluted lake is 0.4 \muμg Hg/mL, what is the total mass in kilograms of mercury in the lake? (The lake has a surface area of 100 mi2 and an average depth of 20 ft.) (1 mile = 5280 ft; 1 ft = 12 in; 2.54 cm = 1 in) (Use scientific notation)
Chemistry
1 answer:
alisha [4.7K]3 years ago
3 0

Answer:

6 x 10⁵ kg Hg

Explanation:

The mass of mercury in the entire lake is found by multiplying the concentration of the mercury by the volume of the lake.

The volume of the lake is calculated in cubic feet:

V = (SA)x(depth) = (100mi²)(5280ft/mi)² x (20ft) = 5.57568 x 10¹⁰ ft³

Cubic feet are then converted to mL (1cm³=1mL)

(5.57568 x 10¹⁰ ft³) x (12in/ft)³ x (2.54cm/in)³ = 1.578856752 x 10¹⁵ mL

The mass of mercury is then found:

m = CV = (0.4μg/mL)(1g/10⁶μg)(1kg/1000g) x (1.578856752 x 10¹⁵ mL) = 6 x 10⁵ kg Hg

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emiprical formula is 55

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The decomposition of hydrogen peroxide was studied, and the following data were obtained at a particular temperature: Time (s) (
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Answer:

Part a: The rate of the equation for 1st order reaction is given as  Rate=k[H_2O_2]

Part b: The integrated Rate Law is given as [H_2O_2]=[H_2O_2]_0 e^{-kt}

Part c: The value of rate constant is 7.8592 \times 10^{-4} s^{-1}

Part d: Concentration after 4000 s is 0.043 M.

Explanation:

By plotting the relation between the natural log of concentration of H_2O_2, the graph forms a straight line as indicated in the figure attached. This indicates that the reaction is of 1st order.

Part a

Rate Law

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Rate=k[H_2O_2]

Part b

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The integrated Rate Law is given as

[H_2O_2]=[H_2O_2]_0 e^{-kt}

Part c

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Value of the rate constant is given by using the relation between 1st two observations i.e.

t1=0, M1=1.00

t2=120 s , M2=0.91

So k is calculated as

-k(t_2-t_1)=ln{\frac{M_2}{M_1}}\\-k(120-0)=ln{\frac{0.91}{1.00}}\\k=\frac{-0.09431}{-120}\\k=7.8592 \times 10^{-4} s^{-1}

The value of rate constant is 7.8592 \times 10^{-4} s^{-1}

Part d

Concentration after 4000 s is given as

-k(t_2-t_1)=ln{\frac{M_2}{1.0}}\\-7.8592 \times 10^{-4}(4000-0)=ln{\frac{M_2}{1.00}}\\-3.1437=ln{\frac{M_2}{1.00}}\\M_2=e^{-3.1437}\\M_2=0.043 M

Concentration after 4000 s is 0.043 M.

7 0
3 years ago
Potassium superoxide is a yellow paramagnetic solid that reacts with water according to the following balanced equation. Calcula
babunello [35]

<u>Answer:</u> The mass of potassium superoxide required is 142.2 grams

<u>Explanation:</u>

The chemical equation for the reaction of potassium superoxide with water follows:

2KO_2+H_2O\rightarrow KOH+KHO_2+O_2;\Delta H=2405J

Number of moles of potassium superoxide reacted = 2 moles for given amount of heat released

To calculate the mass for given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of potassium superoxide = 71.1 g/mol

Moles of potassium superoxide = 2 moles

Putting values in above equation, we get:

2mol=\frac{\text{Mass of potassium superoxide}}{71.1g/mol}\\\\\text{Mass of potassium superoxide}=(2mol\times 71.1g/mol)=142.2g

Hence, the mass of potassium superoxide required is 142.2 grams

4 0
3 years ago
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