<u>Answer:</u> The pH of the buffer is 4.61
<u>Explanation:</u>
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
![pH=pK_a+\log(\frac{[\text{conjuagate base}]}{[\text{acid}]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5B%5Ctext%7Bconjuagate%20base%7D%5D%7D%7B%5B%5Ctext%7Bacid%7D%5D%7D%29)
We are given:
= negative logarithm of acid dissociation constant of weak acid = 4.70
= moles of conjugate base = 3.25 moles
= Moles of acid = 4.00 moles
pH = ?
Putting values in above equation, we get:

Hence, the pH of the buffer is 4.61
The given equation from the problem above is already balance,
N2O5 ---> 2NO2 + 0.5O2
Since, in every mole of N2O5 consumed, 2 moles of NO2 are formed, we can answer the problem by multiplying the given rate, 7.81 mol/L.s with the ratio.
(7.81 mol/L.s) x (2 moles NO2 formed/ 1 mole of N2O5 consumed)
= 15.62 mol/L.s
The answer is the rate of formation of NO2 is approximately 15.62 mol/L.s.
Answer:
aldehyde
carbon-1
ketone
carbon-2
Explanation:
Monosaccharides are colorless crystalline solids that are very soluble in water. Moat have a swwet taste. D-Fructose is the sweetest monosaccharide.
In the open chain form, monosaaccharides have a carbonuyl group in one of their chains. If the carbonyl group is in the form of an aldehyde group, the monosaccharide is an aldose; if the carbonyl group is in the form of a ketone group, the monosaccharide is known as a ketose. glucose is an aldose while fructose is a ketose.
In D-glucose, there is an aldehyde functional group, and the carbonyl group is at carbon-1 when looking at the Fischer projection.
In D-fructose, there is a ketone functional group, and the carbonyl group is at carbon-2 when looking at the Fischer projection.
Answer:
Collision theory states that when suitable particles of the reactant hit each other, only a certain fraction of the collisions cause any noticeable or significant chemical change; these successful changes are called successful collisions.
Explanation:
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