Answer:
Explanation:
Hello!
In this case, since the ionization of ammonia, which is a weak base, is written as:
We can see that the ammonium ion is the conjugate acid whereas the hydroxide ions the conjugate base; that is why we use the Henderson-Hasselbach equation to compute the pH, given the pKb of ammonia 4.75:
![pH=pKb+log(\frac{[conj\ acid]}{[base]} )](https://tex.z-dn.net/?f=pH%3DpKb%2Blog%28%5Cfrac%7B%5Bconj%5C%20acid%5D%7D%7B%5Bbase%5D%7D%20%29)
In such a way, for the given moles of ammonia, base, and those of ammonium chloride, conjugate acid form, we obtain:
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The cell model/theory is a big one
the atomic model/theory
P₁ = 0.90 atm
V₁ = 50.0 mL
T₁ = 298 K
P₂ = 1 atm
T₂ = 273 K
V₂ = P₁ x V₁ x T₂ / T₁ x P₂
V₂ = 0.90 x 50.0 x 273 / 298 x 1
V₂ = 12285 / 298
V₂ = 41 mL
Answer (1)
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Answer:
b) Linear molecule with two domains
Explanation:
The image represents a linear molecule with two domains.
The compound shown here is carbon dioxide in which a central carbon atom is surrounded by two oxygen atoms.
- This bond geometry is of the type AX₂
- In this linear molecule the bond angle is a perfect 180°
- The lone pairs are perfectly balanced out and will not cause distortion of the central carbon.
Answer:- Frequency is 
.
Solution:- frequency and wavelength are inversely proportional to each other and the equation used is:
where, 
is frequency, c is speed of light and 
is the wavelength.
Speed of light is 
.
We need to convert the wavelength from nm to m.
( 
)
= 
Now, let's plug in the values in the equation to calculate the frequency:
= 
or
since, 
So, the frequency of the green light photon is .
