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>
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If it is cooled the motion of the particles decreases as they lose energy.
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Explanation:
The mass of one atom of calcium is the same as the molar mass of the calcium. Also, the atomic mass of calcium is 40.078 g/mol.
Answer:
CaF2 will not precipitate
Explanation:
Given
Volume of Ca(NO3)2
ml
Molar concentration of Ca(NO3)2 
Volume of NaF
ml
Molar concentration of NaF 
Ksp for CaF2 
CaF2 will precipitate if Q for the reaction is greater than ksp of CAF2
Moles of calcium ion

![[Ca2+] = \frac{0.01}{10 + 10} \\= \frac{0.01}{20} \\= 5 * 10^{-4}](https://tex.z-dn.net/?f=%5BCa2%2B%5D%20%3D%20%5Cfrac%7B0.01%7D%7B10%20%2B%2010%7D%20%5C%5C%3D%20%5Cfrac%7B0.01%7D%7B20%7D%20%5C%5C%3D%205%20%2A%2010%5E%7B-4%7D)
Moles of F- ion

![[F-] = \frac{0.001}{10 + 10} \\= \frac{0.001}{20} \\= 5 * 10^{-5}](https://tex.z-dn.net/?f=%5BF-%5D%20%3D%20%5Cfrac%7B0.001%7D%7B10%20%2B%2010%7D%20%5C%5C%3D%20%5Cfrac%7B0.001%7D%7B20%7D%20%5C%5C%3D%205%20%2A%2010%5E%7B-5%7D)
![Q = [Ca2+] [F-]^2\\= (5 * 10^{-4}) * (0.5* 10^-4)\\= 1.25 * 10^{-12}](https://tex.z-dn.net/?f=Q%20%3D%20%5BCa2%2B%5D%20%5BF-%5D%5E2%5C%5C%3D%20%285%20%2A%2010%5E%7B-4%7D%29%20%2A%20%280.5%2A%2010%5E-4%29%5C%5C%3D%201.25%20%2A%2010%5E%7B-12%7D)
Q is lesser than Ksp value of CaF2. Hence it will not precipitate
For the purpose we will here use t<span>he ideal gas law:
p</span>×V=n×R×<span>T
V= </span><span>5.0 L
T= </span><span>373K
p= </span><span>203kPa
</span><span>
R is </span> universal gas constant, and its value is 8.314 J/mol×<span>K
</span>
Now when we have all necessary date we can calculate the number of moles:
n=p×V/R×T
n= 203 x 5 / 8.314 x 373 = 0.33 mole