Answer:
Your answer should be 15.68 grams.
Explanation:
Seeing as 1 mole has a mass of 56 g, 56*0.28 would get you 15.68 g.
The mass of 165.0 g sample that remains after 90.0 minutes is 5.16 grams
calculation
lambda㏑2/18= 0.0385
m(t)= 165 x e( 0.0385 x90) =5.16g
Yes this is true as in cold conditions our body doesn't feel the urge of water therefore we may become dehydrated not even knowing about it and of course in hot coditions we sweat therefore we loose water.
Hope this helps :).
Answer:
3,4–diethylheptane
Explanation:
To name the compound given in the question above, the following must be obtained:
1. Determine the functional group of the compound.
2. Determine the longest continuous carbon chain. This gives the parent name of the compound.
3. Identify the substituent groups attached to the compound.
4. Give the substituent groups the lowest possible count by naming them alphabetically.
5. Combine the above to obtain the name of the compound.
Now, we shall name the compound given in the question above as follow:
1. The compound belongs to the alkane family since it has only single bonds.
2. The longest continuous carbon chain is 7 i.e the parent name of the compound is heptane.
3. The substituent group attached to the compound is ethyl (–CH2CH3). There are two ethyl groups attached to the compound.
4. The two ethyl group is at carbon 3 and 4 (i.e number from the right to the upper branch chain).
5. The name of the compound is:
3,4–diethylheptane.
Answer:
a. Rate constant: 1.2118x10⁻⁴ yrs⁻¹
b. The age of the object is 20750 years
Explanation:
a. We can solve the rate constant in an isotope decay by using Half-Life, as follows:
K = Ln 2 / Half-life
K = ln 2 / 5720 years =
<h3>1.2118x10⁻⁴ yrs⁻¹</h3><h3 />
b. The general equation of isotope decay is:
Ln [A] = -kt + Ln [A]₀
<em>Where [A] is concentration of the isotope after time t, </em>
<em>k is rate constant</em>
<em>and [A]₀ initial concentration of the isotope.</em>
<em />
Computing the values of the problem:
Ln [0.89x10⁻¹⁴] = -1.2118x10⁻⁴ yrs⁻¹t + Ln [1.1x10⁻¹³]
-2.5144 = -1.2118x10⁻⁴ yrs⁻¹t
20750 years = t
The age of the object is 20750 years
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