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Tom [10]
2 years ago
11

If an object has a 30 kg mass on Earth, what would be it's approximate weight on Earth?

Chemistry
1 answer:
Degger [83]2 years ago
3 0

Answer:

If something were to weigh 30 kgs on Earth, they would also weigh around 66 pounds on Earth.

Explanation:

1 pound = 2.20462 kilograms.

Next, multiply 2.20462 and 30 and you should get 66.1386, or 66 pounds.

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9900000

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Cells in different tissues or organs (skin, eye, bone, etc.) contain different DNA. true or fasle,?
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6 0
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How does black and coloured objects absorb light and convert it into heat?
Nezavi [6.7K]

Answer:

Light energy can be converted into heat energy. A black object absorbs all wavelengths of light and converts them into heat, so the object gets warm. A white object reflects all wavelengths of light, so the light is not converted into heat and the temperature of the object does not increase noticeably.

5 0
2 years ago
How many moles of oxygen are required to react completely with 5 mol C8H18?
Nata [24]

Answer:

62.5 moles of O₂.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2C₈H₁₈ + 25O₂ —> 16CO₂ + 18H₂O

From the balanced equation above,

2 moles of C₈H₁₈ reacted with 25 moles of O₂.

Finally, we shall determine the number of mole of O₂ needed to react with 5 moles of C₈H₁₈. This can be obtained as shown below:

From the balanced equation above,

2 moles of C₈H₁₈ reacted with 25 moles of O₂.

Therefore, 5 moles of C₈H₁₈ will react with = (5 × 25) / 2 = 62.5 moles of O₂.

Thus, 62.5 moles of O₂ is needed for the reaction.

6 0
2 years ago
Calculate the ph of a dilute solution that contains a molar ratio of potassium acetate to acetic acid (pka ???? 4.76) of (a) 2:1
givi [52]

According to Hasselbach-Henderson equation:

pH=pK_{a}+log\frac{[A^{-}]}{[HA]}

Here, [A^{-}] is concentration of conjugate base and [HA] is concentration of acid.

In the given problem, conjugate base is CH_{3}COOK and acid is CH_{3}COOH thus, Hasselbach-Henderson equation will be as follows:

pH=pK_{a}+log\frac{[CH_{3}COOK]}{[CH_{3}COOH]}...... (1)

(a) Ratio of concentration of potassium acetate and acetic acid is 2:1 thus,

\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=2

Also, pK_{a}=4.76

Putting the values in equation (1),

pH=4.76+log\frac{2}{1}=5.06

Therefore, pH of solution is 5.06.

(b) Ratio of concentration of potassium acetate and acetic acid is 1:3 thus,

\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{3}

Also, pK_{a}=4.76

Putting the values in equation (1),

pH=4.76+log\frac{1}{3}=4.28

Therefore, pH of solution is 4.28.

(c)Ratio of concentration of potassium acetate and acetic acid is 5:1 thus,

\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{5}{1}

Also, pK_{a}=4.76

Putting the values in equation (1),

pH=4.76+log\frac{5}{1}=5.45

Therefore, pH of solution is 5.45.

(d) Ratio of concentration of potassium acetate and acetic acid is 1:1 thus,

\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=1

Also, pK_{a}=4.76

Putting the values in equation (1),

pH=4.76+log\frac{1}{1}=4.76

Therefore, pH of solution is 4.76.

(e) Ratio of concentration of potassium acetate and acetic acid is 1:10 thus,

\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{10}

Also, pK_{a}=4.76

Putting the values in equation (1),

pH=4.76+log\frac{1}{10}=3.76

Therefore, pH of solution is 3.76.

4 0
2 years ago
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