Answer:
7.5 × 10¹⁵ Hz
Explanation:
Given data
- Wavelength of the radio waves (λ): 40 nm = 40 × 10⁻⁹ m = 4.0 × 10⁻⁸ m
- Frequency of the radio waves (ν): ?
- Speed of light (c): 3.00 × 10⁸ m/s
We can determine the frequency of the radio waves using the following expression.
c = λ × ν
ν = c/λ
ν = (3.00 × 10⁸ m/s)/4.0 × 10⁻⁸ m
ν = 7.5 × 10¹⁵ s⁻¹ = 7.5 × 10¹⁵ Hz
Explanation:
We know that relation between and is as follows.
= 14
As it is given that is 8.18. Therefore, calculate the value of as follows.
= 14
= 14
= 14 - 8.18
= 5.82
Similarly, as value of pH is given as 7.18. Therefore, value of pOH will be as follows.
pH + pOH = 14
7.18 + pOH = 14
pOH = 6.82
Let us take that B represents the enzyme. Hence, its reaction with proton will be as follows.
(protonated active enzyme)
Hence, pOH =
6.82 = 5.82 +
= 10
Therefore, percentage of active enzyme = % =
% = 90.9%
Thus, we can conclude that 90.9% is the percentage of the enzyme which is active in a buffer at pH 7.18.
Answer:
62.3 moles of Ca
Explanation:
1 mole of a substance is defined as 6.022x10²³ particles of this substance. For example:
1mol of electrons = 6.022x10²³electrons
1mol of atoms = 6.022x10²³atoms
1 mol of molecules = 6.022x10²³ molecules
And 1mol of atoms of Ca = 6.022x10²³ atoms of Ca
That means 3.75x10²⁵ atoms of Ca are:
3.75x10²⁵ atoms of Ca * (1 mole of Ca / 6.022x10²³ atoms) =
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62.3 moles of Ca
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I'm not 100% sure on this one, but I think it's C.
Sorry, I'm not really good with Chemistry.
Hope this helps :)