Answer:
The pH of the sweater containing Hydrogen ion concentration
![[H^{+}]=1\times 10^{-8}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%3D1%5Ctimes%2010%5E%7B-8%7D)
is
<u>8</u>
<u></u>
Explanation:
pH = It is the negative logarithm of activity (concentration) of hydrogen ions.
pH = -log([H+])
Now, In the question the concentration of [H+] ions is :
use the relation:
pH = 8
Note : <em><u> 1 times 10 to the power of 8 must be" 1 times 10 to the power of -8"</u></em>
If the concentration is
![[H^{+}]=1\times 10^{8}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%3D1%5Ctimes%2010%5E%7B8%7D)
Then pH = -8 , which is not possible . So in that case the pH calculation is by other method
<span>Carrying capacity is the number of organisms an ecosystem can support. It is the maximum size of a population that can survive in the ecosystem. If the animals reach the carrying capacity, the population may crash. As the consequence, the number of animals will decrease due to predators or diseases.</span>
Answer:
Electronegativity in group 1 decreases as we go from Lithium to Francium.
Explanation:
Electronegativity is defined as the tendency of an element to attract an electron pair towards itself.
In a group generally this tendency decreases from top to bottom as the size of the atom increases and hence the positive nucleus get far from the outer orbital.
In the same way group 1 elements i.e. from Lithium to Francium electronegativity decreases.
Answer:
C. The half-life of C-14 is about 40,000 years.
Explanation:
The only false statement from the options is that the half-life of C-14 is 40,000yrs.
The half-life of an isotope is the time it takes for half of a radioactive material to decay to half of its original amount. C-14 has an half-life of 5730yrs. This implies that during every 5730yrs, C-14 will reduce to half of its initial amount.
- All living organisms contain both stable C-12 and the unstable isotope of C-14
- The lower the C-14 compared to the C-12 ratio in an organism, the older it is.
Answer:
Molecular formula => C₃H₈O₃
Explanation:
From the question given above, the following data were obtained:
Carbon (C) = 39.12%
Hydrogen (H) = 8.75%
Oxygen (O) = 51.12%
Molar mass of compound = 92.09 g/mol
Molecular formula =?
Next, we shall determine the empirical formula of the compound. This can be obtained as follow:
C = 39.12%
H = 8.75%
O = 51.12%
Divide by their molar mass
C = 39.12 / 12 = 3.26
H = 8.75 / 1 = 8.75
O = 51.12 / 16 = 3.195
Divide by the smallest
C = 3.26 / 3.195 = 1
H = 8.75 / 3.195 = 2.7
O = 3.195 / 3.195 = 1
Thus, the empirical formula is CH₂.₇O
Finally, we shall determine the molecular formula of the compound. This can be obtained as follow:
Empirical formula = CH₂.₇O
Molar mass of compound = 92.09 g/mol
Molecular formula =?
Molecular formula = Empirical formula × n
Molecular formula = [CH₂.₇O]ₙ
92.09 = [12 + (2.7×1) + 16] × n
92.09 = 30.7n
Divide both side by 30.7
n = 92.09 / 30.7
n = 3
Molecular formula = [CH₂.₇O]ₙ
Molecular formula = [CH₂.₇O]₃
Molecular formula = C₃H₈O₃
