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2 years ago

Which example best demonstrates stewardship of the atmosphere

Chemistry

1 answer:

Margarita [4]2 years ago

3 0

Incomplete question. However, I provided a specific example of stewardship of the atmosphere.

Explanation:

First, note that the term stewardship refers to the belief that humans are obligated to take care and look after our environment (which includes the atmosphere.

An example of this is: deciding to change our mode of transportation: What this entails is that we may decide to switch to riding a bicycle to places rather than going by car. By so doing you will be reducing the number of greenhouse gases emitted into the environment as bicycles do not emit harmful gases when used.

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Calculate the pH of each of following buffered solutions.?a. 0.10 M acetic acid/0.25 M sodium acetate b. 0.25 M acetic acid/0.10

Masteriza [31]

Answer:

a. 5.10.

b. 4.35.

c. 5.10.

d. 4.35.

Explanation:

a. 0.10 M acetic acid/0.25 M sodium acetate

For acidic buffer:

∵ pH = pKa + log [salt]/[Acid]

∴ pH = - log(Ka) + log [salt]/[Acid]

Ka for acetic acid = 1.8 x 10⁻⁵.

∴ pH = - log(1.8 x 10⁻⁵) + log(0.25)/(0.10)

∴ pH = 4.744 + 0.34 = 5.084 ≅ 5.10.

b. 0.25 M acetic acid/0.10 M sodium acetate

For acidic buffer:

∵ pH = pKa + log [salt]/[Acid]

∴ pH = - log(Ka) + log [salt]/[Acid]

Ka for acetic acid = 1.8 x 10⁻⁵.

∴ pH = - log(1.8 x 10⁻⁵) + log(0.10)/(0.25)

∴ pH = 4.744 - 0.34 = 4.346 ≅ 4.35.

c. 0.080 M acetic acid/0.20 M sodium acetate

For acidic buffer:

∵ pH = pKa + log [salt]/[Acid]

∴ pH = - log(Ka) + log [salt]/[Acid]

Ka for acetic acid = 1.8 x 10⁻⁵.

∴ pH = - log(1.8 x 10⁻⁵) + log(0.20)/(0.08)

∴ pH = 4.744 + 0.34 = 5.084 ≅ 5.10.

d. 0.20 M acetic acid/0.080 M sodium acetate

For acidic buffer:

∵ pH = pKa + log [salt]/[Acid]

∴ pH = - log(Ka) + log [salt]/[Acid]

Ka for acetic acid = 1.8 x 10⁻⁵.

∴ pH = - log(1.8 x 10⁻⁵) + log(0.08)/(0.20)

∴ pH = 4.744 - 0.34 = 4.346 ≅ 4.35.

6 0

2 years ago

Earth’s early atmosphere was very different from our current atmosphere. Explain the presence or absence of gases in our current

Ivanshal [37]

Answer:

heavier gases - sank to the surface

lighter gases - escaped into space

oxygen - produced through photosynthesis

carbon dioxide - released by volcanoes

Explanation:

Since the formation of the atmosphere on Earth, there has been constant changes in it, gradually leading to the atmosphere that we have nowadays. In the distant past, the atmosphere was much different, and it took lot of time for it to become what it is. The volcanoes with their activity contributed to constant release of carbon dioxide, resulting in the greenhouse effect. The organisms that are able to perform photosynthesis started to release lot of oxygen into the atmosphere, thus setting up better living conditions. The gravity of our planet contributed to the vertical distribution of the gasses, with the heavier being lower, and the lighter ones being further up or released into space.

4 0

2 years ago

In acidic solution, the breakdown of sucrose into glucose and fructose has this rate law: rate = k[H+][sucrose].

Karo-lina-s [1.5K]

Answer:

a)If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d) If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

Explanation:

Sucrose + $H^+\rightarrow$ fructose+ glucose

The rate law of the reaction is given as:

$R=k[H^+][sucrose]$

$[H^+]=0.01M$

[sucrose]= 1.0 M

$R=k[0.01M][1.0 M]$ ..[1]

The rate of the reaction when [Sucrose] is changed to 2.5 M = R'

$R'=[0.01 M][2.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][2.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

The rate of the reaction when [Sucrose] is changed to 0.5 M = R'

$R'=[0.01 M][0.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][0.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

The rate of the reaction when $[H^+]$ is changed to 0.001 M = R'

$R'=[0.0001 M][1.0 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.0001 M][1.0M]}{k[0.01M][1.0 M]}$

$R'=0.01\times R$

If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

The rate of the reaction when [sucrose] and $[H^+]$ both are changed to 0.1 M = R'

$R'=[0.1M][0.1M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.1M][0.1M]}{k[0.01M][1.0 M]}$

$R'=1\times R$

If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

5 0

2 years ago

Calculate the average time it took the car to travel 0.25 What is the average time it took the car to travel 0.25

DedPeter [7]

Answer:

Explanation:

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A teacher cut an apple into three wedges of the same size. She dipped one wedge in lemon juice, dipped another in water, and lef

aleksandrvk [35]

Answer:

The responding variable of this experement is the outcome and that would be that the one in lemon juice responded and the one in water didn't (the other one is the control). Thus the responding varible is the one in lemon juice.

Explanation:

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2 years ago

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