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Travka [436]
2 years ago
8

PLEASE HELP ASAP!!!!!!!!!!!!

Chemistry
1 answer:
Aleks04 [339]2 years ago
7 0

In the scientific sense the two examples of people doing work is: A and D

<h3>Meaning of Work </h3>

Work can be defined as a result of applying force to a particular object that changes its displacement in the direction of the applied force.

Work can only be done when their is a force that is being applied and a displacement that occurs.

In conclusion, In the scientific sense the two examples of people doing work is: A and D

Lear more about Work: brainly.com/question/25573309

#SPJ1

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3 years ago
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An iron ore contains 39% Fe2O3 by mass. What is the maximum mass of iron that can be recovered from 13.2 kg of this ore?
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Answer:

3.6036

Explanation:

0.39 * 13.2 = 5.148

5.148 * 0.7 = 3.6036

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3 years ago
Acetic acid is a weak acid with a pKa of 4.76. What is the concentration of acetate in a buffer solution of 0.2M at pH 4.9. Give
Ghella [55]

Answer:

[base]=0.28M

Explanation:

Hello,

In this case, by using the Henderson-Hasselbach equation one can compute the concentration of acetate, which acts as the base, as shown below:

pH=pKa+log(\frac{[base]}{[acid]} )\\\\\frac{[base]}{[acid]}=10^{pH-pKa}\\\\\frac{[base]}{[acid]}=10^{4.9-4.76}\\\\\frac{[base]}{[acid]}=1.38\\\\

[base]=1.38[acid]=1.38*0.20M=0.28M

Regards.

6 0
3 years ago
Read “The Ozone Hole” and answer the question below. Write a short evaluation of the article’s reliability and scientific worth.
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6 0
3 years ago
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The maximum amount of nickel(II) cyanide that will dissolve in a 0.220 M nickel(II) nitrate solution is...?
sweet [91]

Answer : The maximum amount of nickel(II) cyanide is 5.84\times 10^{-12}M

Explanation :

The solubility equilibrium reaction will be:

                       Ni(CN)_2\rightleftharpoons Ni^{2+}+2CN^-

Initial conc.                        0.220       0

At eqm.                             (0.220+s)   2s

The expression for solubility constant for this reaction will be,

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Now put all the given values in this expression, we get:

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s=5.84\times 10^{-12}M

Therefore, the maximum amount of nickel(II) cyanide is 5.84\times 10^{-12}M

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