Answer:
So ²³⁵UF₆ effuses 1.0043 times faster than ²³⁸UF₆
Explanation:
The rate of effusion of two gases A and B can be expressed by <em>Graham's law</em>:

Where M is the molar mass, and in this case A is ²³⁵UF₆ while B is ²³⁸UF₆.
So now we <u>calculate the molar mass of each mass</u>:
²³⁵UF₆ ⇒235 + 6*19 = 349 g/mol
²³⁸UF₆ ⇒238 + 6*19 = 352 g/mol
Putting the data in Graham's law:
= 1.0043
So ²³⁵UF₆ effuses 1.0043 times faster than ²³⁸UF₆.
True, because a liquid can be taken and added, but a solid stay the same never losses and never gains
Answer:
a. Raise the pH slightly
Explanation:
We know that
Pka of HNO2/KNO2 =3.39
Moles of HNO2 in the buffer=0.247 mol/L×1L=0.247 moles
Moles of NO2-=0.329mol/L×1L=0.329 moles
If 0.271 moles of Ca(OH)2 is added it will neutralise 0.136 moles of acid ,HNO2,remaining HNO2=0.247-0.136=0.111 moles
Moles of NO2- will increase as 0.0333 moles Ca(NO)2 will be formed =0.0333+0.036=0.0693 moles
pH=pka+log [base]/[acid] {henderson -hasselbach equation}
=3.39+log (0.0693/0.0317)=3.39+0.34=3.73
pH=3.73
Answer:
See figure 1
Explanation:
On this case we have a <u>base</u> (methylamine) and an <u>acid</u> (2-methyl propanoic acid). When we have an acid and a base an <u>acid-base reaction </u>will take place, on this specific case we will produce an <u>ammonium carboxylate salt.</u>
Now the question is: <u>¿These compounds can react by a nucleophile acyl substitution reaction?</u> in other words <u>¿These compounds can produce an amide? </u>
Due to the nature of the compounds (base and acid), <u>the nucleophile</u> (methylamine) <u>doesn't have the ability to attack the carbon</u> of the carbonyl group due to his basicity. The methylamine will react with the acid-<u>producing a positive charge</u> on the nitrogen and with this charge, the methylamine <u>loses all his nucleophilicity.</u>
I hope it helps!
Answer : 17.12 g
Explanation:
= elevation in boiling point
= boiling point elevation constant
m= molality

given 
Molar mass of solute = 46.0 
Weight of the solvent = 150.0 g = 0.15 kg
Putting in the values


x = 17.12 g