C) volume
.........
The volume of gas depends on e.g. temperature and pressure.
The difference in elements??
Answer:
Neutral nucleophile are: H2O, CH3OH, NH3, RNH2, R2NH, R3N, RCOOH, RSH and PR3. The products by nucleophilic substitution are diverse depending on the different nucleophiles, obtaining alcohol, eter, amines, ester and tioeter considering only the nucleophiles with a hydrogen available.
Explanation:
Please see the images attached.
Nucleophilic subtitution with water occurs under Sn1 mechanism. That's it because water as nucleophile is so weak. With the other neutral nucleophiles, the reaction occur under Sn2 mechanism.
RSH + CH3I -----> RSCH3 + HI
<u>Answer:</u>
<u>For a:</u> The empirical formula of the compound is 
<u>For b:</u> The empirical formula of the compound is 
<u>Explanation:</u>
We are given:
Percentage of P = 43.6 %
Percentage of O = 56.4 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of P = 43.6 g
Mass of O = 56.4 g
To formulate the empirical formula, we need to follow some steps:
- <u>Step 1:</u> Converting the given masses into moles.
Moles of Phosphorus =
Moles of Oxygen = 
- <u>Step 2:</u> Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.406 moles.
For Phosphorus = 
For Oxygen = 
Converting the moles in whole number ratio by multiplying it by '2', we get:
For Phosphorus = 
For Oxygen = 
- <u>Step 3:</u> Taking the mole ratio as their subscripts.
The ratio of P : O = 2 : 5
Hence, the empirical formula for the given compound is 
We are given:
Percentage of K = 28.7 %
Percentage of H = 1.5 %
Percentage of P = 22.8 %
Percentage of O = 56.4 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of K = 28.7 g
Mass of H = 1.5 g
Mass of P = 43.6 g
Mass of O = 56.4 g
To formulate the empirical formula, we need to follow some steps:
- <u>Step 1:</u> Converting the given masses into moles.
Moles of Potassium =
Moles of Hydrogen =
Moles of Phosphorus =
Moles of Oxygen = 
- <u>Step 2:</u> Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.735 moles.
For Potassium = 
For Hydrogen = 
For Phosphorus = 
For Oxygen = 
- <u>Step 3:</u> Taking the mole ratio as their subscripts.
The ratio of K : H : P : O = 1 : 2 : 1 : 4
Hence, the empirical formula for the given compound is 