Answer: The answer is D. This has a Carboxylic Acid group, and is acetic acid, or Ethanoic Acid.
ALWAYS LOOK for the Functional Group in question.
A. Would likely not stay in water, or at least not be acidic, for it is butane gas.
B. Is 1-propanol, and alcohols are not acidic as a rule. Certainly not in water.
C. This is an Ether. It will not give up an H+, it it not an acid.
E. This functional group is an amine, which is more “base” like, since the lone pairs of the Nitrogen atom would tend to attract a H+.
Calcium is stored in bones in the body
Answer:
Mean Partial pressure of Nitrogen in Mars' atmosphere = 15.86 Pa
Explanation:
According to Dalton's law of Partial Pressure, the total pressure exerted by a mixture of ideal gases (that do not react together) is the sum of the partial pressures of the individual gases that make up the mixture. It goes further to explain that the partial pressure of a gas in a mixture of gases is equal to its mole fraction of that gas multipled by the total pressure exerted by the mixture of gases.
Total Pressure exerted by the mixture of gases in the atmosphere on Mars = Mean atmospheric pressure on Mars = 610 Pa
Partial pressure of Nitrogen = (mole fraction or mole percentage of Nitrogen in the atmosphere) × (total pressure exerted by all the gases in the atmosphere)
Mole percentage of Nitrogen in the atmosphere of Mars = 2.6%
Partial pressure of Nitrogen = 2.6% × 610 = 15.86 Pa
Mean Partial pressure of Nitrogen in Mars' atmosphere = 15.86 Pa
Hope this Helps!!!
The specific heat of the unknown metal is 0.859 J/g⁰C.
<h3>Specific heat of the unknown metal</h3>
Heat lost by the unknown metal = Heat gained by water
m₁c₁Δθ₁ = m₂c₂Δθ₂
where;
- c₁ is specific heat of the unknown metal
- c₂ is specific heat of water
(15)(c₁)(99 - 26) = (75)(4.18)(26 - 23)
1095c₁ = 940.5
c₁ = 940.5/1095
c₁ = 0.859 J/g⁰C
Learn more about specific heat here: brainly.com/question/16559442
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With reference to radioactive material, half-life is the time required to 50% depletion of initial amount of material.
Given: Initial amount of radioactive material = 40 g
Half life = 4 days.
Therefore, After 4 days, amount of compound left = 40/2 = 20 g
After 8 days, i.e 2 half-life, amount of compound left = 20/2 = 10 g
Finally after 12 days, i.e. 3 half-life, amount of compound left = 10/2 = 5 g
Thus, 5 <span>grams will remain after 12 days</span>
