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Tcecarenko [31]

3 years ago

15

The number of kilograms of water in a human body varies directly as the mass of the body. upper a 93-kg person contains 62 kg o

f water. how many kilograms of water are in an 84-kg person

1 answer:

Bad White [126]3 years ago

4 0

Answer;

56 kg of water

Solution;

Weight = 62 kg of water in 93 kg of a person

W=km

62=93k;

k=2/3

W=(2/3)84

=56 kg

Therefore; a 84-kg person will have 56 kg of water.

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Uranium-235 and uranium-238 are considered isotopes of one another

Bogdan [553]

If you are asking which is the most abundant, Uranium-238 is

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This compound, an eight-carbon alkane, serves as the 100 octane standard for gasoline. It has five methyl groups, one quaternary

Reika [66]

Answer:

2,2,4-Trimethyl-pentane

Explanation:

Structural characteristics of the compound is as follows:

Has five methyl group
Has one quaternary carbon
No. double bond
Gives four monochloro substitution products

Compound must have straight chain of 5 carbons.

Three methyl substituents are attached to 2 and 4 carbons.

Therefore, IUPAC name of the compound will be 2,2,4-Trimethyl-pentane.

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3 years ago

Arizona was the site of a 400,000-acre wildfire in June 2002. How much carbon dioxide (CO2) was produced into the atmosphere by

kodGreya [7K]

Answer:

2.97 × 10¹³ g

Explanation:

First, we have to calculate the biomass the is burned. We can establish the following relations:

2.47 acre = 10,000 m²
10 kg of C occupy an area of 1 m²
50% of the biomass is burned

The biomass burned in the site of 400,000 acre is:

$400,000acre\times\frac{10,000m^{2} }{2.47acre} \times \frac{10kgC}{m^{2} } \times 50\% = 8.10 \times 10^{9} kgC$

Let's consider the combustion of carbon.

C(s) + O₂(g) ⇒ CO₂(g)

We can establish the following relations:

The molar mass of C is 12.01 g/mol
1 mole of C produces 1 mole of CO₂
The molar mass of CO₂ is 44.01 g/mol

The mass of produced is CO₂:

$8.10 \times 10^{12}gC \times \frac{1molC}{12.01gC} \times \frac{1molCO_{2}}{1molC} \times \frac{44.01gCO_{2}}{1molCO_{2}} =2.97 \times 10^{13} gCO_{2}$

4 0

3 years ago

10. Which would not be a good solution to counter overpopulation?

Vikki [24]

Answer:

A. Encourage an increase in pollution in these countries

4 0

2 years ago

Problem PageQuestion Sulfuric acid is essential to dozens of important industries from steelmaking to plastics and pharmaceutica

Feliz [49]

The question is incomplete, here is the complete question:

Sulfuric acid is essential to dozens of important industries from steel making to plastics and pharmaceuticals. More sulfuric acid is made than any other industrial chemical, and world production exceeds 2.0×10¹¹ kg per year.

The first step in the synthesis of sulfuric acid is usually burning solid sulfur to make sulfur dioxide gas. Suppose an engineer studying this reaction introduces 1.8 kg of solid sulfur and 10.0 atm of oxygen gas at 650°C into an evacuated 50.0 L tank. The engineer believes Kp = 0.099 for the reaction at this temperature.

Calculate the mass of solid sulfur he expects to be consumed when the reaction reaches equilibrium. Round your answer to 2 significant digits.

<u>Answer:</u> The mass of solid sulfur that will be consumed is 19. grams

<u>Explanation:</u>

The chemical equation for the formation of sulfur dioxide gas follows:

$S(s)+O_2\rightarrow SO_2(g)$

<u>Initial:</u> 10.0

<u>At eqllm:</u> 10-x x

The expression of $K_p$ for above equation follows:

$K_p=\frac{p_{SO_2}}{p_{O_2}}$

We are given:

$K_p=0.099$

Putting values in above expression, we get:

$0.099=\frac{x}{10-x}\\\\x=0.901atm$

Partial pressure of sulfur dioxide = x = 0.901 atm

To calculate the number of moles, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the sulfur dioxide gas = 0.901 atm

V = Volume of the gas = 50.0 L

T = Temperature of the gas = $650^oC=[650+273]K=923K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of sulfur dioxide gas = ?

Putting values in above equation, we get:

$0.901atm\times 50.0L=n\times 0.0821\text{ L. atm}mol^{-1}K^{-1}\times 923K\\\\n=\frac{0.901\times 50.0}{0.0821\times 923}=0.594mol$

By stoichiometry of the reaction:

1 mole of sulfur dioxide gas is produced from 1 mole of sulfur

So, 0.594 moles of sulfur dioxide gas will be produced from = $\frac{1}{1}\times 0.594=0.594mol$ of sulfur

To calculate the number of moles, we use the equation:

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

Moles of sulfur = 0.594 moles

Molar mass of sulfur = 32 g/mol

Putting values in above equation, we get:

$0.594mol=\frac{\text{Mass of sulfur}}{32g/mol}\\\\\text{Mass of sulfur}=(0.594mol\times 32g/mol)=19.008g$

Hence, the mass of solid sulfur that will be consumed is 19. grams

8 0

3 years ago