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2 years ago

Assume that the variables x and y are inversely related. If k = 18, what is the value of y for each of the following points?

Chemistry

2 answers:

charle [14.2K]2 years ago

8 0

Answer:

The values of Y are: 18, 9, 6, 3, 2, 1

Explanation:

It is given that X and Y are inversely related. This can be depicted mathematically as:

$X \alpha \frac{1}{Y}$

i.e. as X increases the value of Y is expected to decrease.

(or) $X =k* \frac{1}{Y}$

where k = proportionality constant which is given as 18

In terms of Y, the above equation becomes:

$Y =k* \frac{1}{X}$

Based on the above equation the table can be filled as :

X Y K

1 18 18

2 9 18

3 6 18

6 3 18

9 2 18

18 1 18

kogti [31]2 years ago

4 0

6_____18 record your data to be used in the following problem.

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3 years ago

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Applied forces/or unbalanced:i hope that helps you

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3 years ago

Read 2 more answers

A plate moves 200 m in 10,000 years. What is its rate in cm/year?

Varvara68 [4.7K]

Answer:

The answer is

Explanation:

To find the rate in cm/year we must first convert 200 m into cm

1 m = 100 cm

if 1 m = 100 cm

Then 200 m = 200 × 100 = 20 ,000 cm

So the rate is

<u>Reduce the fraction with 10,000</u>

We have the final answer as

Hope this helps you

3 0

2 years ago

5. What are the relative rates of diffusion for methane, CH, and oxygen, O2? If O2 la travels 1.00 m in a certain amount of time

11Alexandr11 [23.1K]

Answer:

The relative rates of diffusion for methane and oxygen is 1.4142.

Methane gas will be able to travel 1.4142 meter in the same conditions.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. Mathematically written as:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

We are given:

Molar mass of methane gas, m = 16 g/mol

Molar mass of oxygen gas,m' = 32 g/mol

By taking their ratio, we get:

$\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{m'}{m}}$

$\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{32}{16}}=1.4142$

The relative rates of diffusion for methane and oxygen is 1.4142.

If oxygen gas travels 1 meters in time t.

Rate of diffusion of oxygen = $d_{O_2}=\frac{1 m}{t}$

If methane gas travels travels in y meters in time t.

Rate of diffusion of methane= $d_{CH_4}=\frac{y }{t}$

$\frac{d_{CH_4}}{d_{O_2}}=\frac{\frac{y }{t}}{\frac{1 m}{t}}=1.4142$

y = 1.4142 m

Methane gas will be able to travel 1.4142 meter in the same conditions.

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3 years ago

A 2.912 g sample of a compounds containing only C, H, and O was completely oxidized in a reaction that yielded 3.123 g of water

Taya2010 [7]

Answer:

Explanation:

18 gram of water contains 2 g of hydrogen

3.123 gram of water will contain 2 x 3.123 / 18 = .347 g of hydrogen .

44 gram of carbon dioxide contains 12 g of carbon

7.691 gram of carbon dioxide will contain 12 x 7.691 / 44 = 2.1 g of carbon .

So the sample will contain 2.912 - ( .347 + 2.1 ) g of oxygen .

= .465 g of oxygen .

moles of Carbon = 2.1 / 12 = .175

moles of hydrogen = .347 / 1 = .347

moles of oxygen = .465 / 16 = .029

Ratio of moles of carbon , oxygen and hydrogen ( C,O,H )

= 0.175 : 0.029 : 0.347

= .175/ .029 : 1 : .347 / .029

= 6 : 1 : 12

So empirical formula = C₆H₁₂O

Let the molecular formula be $(C_6H_{12}O)_n$

molecular weight = n ( 6 x 12 + 12x 1 + 16)

= 100 n

Given 100 n = 100.1

n = 1

Molecular formula = C₆H₁₂O.

3 0

2 years ago