Answer:
The amount of Chlorodecane in the unknown is 0.105nmols
Explanation:
a) Since the GC is in an isothermal state, Chlorohexane C6H13Cl (1.69 nmols) because of its lower boiling point will elute first and Chlorodecane C12H21Cl will elute second.
The area of the first peak corresponding to Chlorohexane is 32434 units.
The area of the second peak corresponding to chlorodecane is 2022 units.
Since the response factor of the compound is not given in question and considering the response factor is same for both the compounds, the answer will be as follow:
1.69 nmols of Chlorohexane gives 32434 units
How much of chlorodecane gives 2022 units
By cross multiplication;
Moles of Chlorodecane = 2022*1.69/32434
=0.105nmols
Answer: There are 6.9 mol of 
are required to react completely with 2.30 mol of S.
Explanation:
The given reaction equation is as follows.
Here, 1 mole of S is reaction with 3 moles of which means 1 mole of S requires 3 moles of .
Therefore, moles of required to react completely with 2.30 moles S are calculated as follows.
Thus, we can conclude that there are 6.9 mol of are required to react completely with 2.30 mol of S.
Answer:
b. potassium.
Explanation:
Potassium-sparing diuretics and salt substitutes are diuretics that eliminate salt and water but save potassium. They act by inhibiting the conducting sodium channels in the collecting tubule, such as amiloride and triamterene, or by blocking aldosterone, such as spironolactone.
Concomitant use of potassium-sparing diuretics together with salt substitutes may result in dangerously high blood levels of serum potassium. For this reason, it is important to consult a physician before taking these substances at the same time to avoid potential problems with potassium accumulation.
Answer:
No
Explanation:
One mole of P₄ react with six moles of I₂ and gives 4 moles of PI₃.
When one gram phosphorus and 6 gram of iodine react they gives 8.234 g
ram of PI₃ .
Given data:
Mass of phosphorus = 1 g
Mass of iodine = 6 g
Mass of PI₃ = ?
Solution:
Chemical equation:
P₄ + 6I₂ → 4PI₃
Number of moles of P₄:
Number of moles = Mass /molar mass
Number of mole = 1 g / 123.9 g/mol
Number of moles = 0.01 mol
Number of moles of I₂:
Number of moles = Mass /molar mass
Number of moles = 6 g / 253.8 g/mol
Number of moles = 0.024 mol
Now we will compare the moles of PI₃ with I₂ and P₄.
I₂ : PI₃
6 : 4
0.024 :
4/6×0.024 = 0.02
P₄ : PI₃
1 : 4
0.01 : 4 × 0.01 = 0.04 mol
The number of moles of PI₃ produced by I₂ are less it will be limiting reactant.
Mass of PI₃ = moles × molar mass
Mass of PI₃ = 0.02 mol × 411.7 g/mol
Mass of PI₃ = 8.234 g
6 Atoms!
Mg = 1 atom.
O = 4 atoms.
A = 1 atom.
