A 5.00 L flask contains 7.94 g of a gas at STP. What is the molar mass of the gas?
1 answer:
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Answer:
E.F= OH
M.F=
Explanation:
Empirical Formula
Step 1: Calculate mols of each element
0.44gH(1 mol H/ 1.008g H)= 0.4365 mol H
*note leave extra sig figs for calculations
6.92gO(1 mol O/ 16 g O)=0.4325 mol O
Step 2: Identify which is the smallest mol
*in this case 0.4365 mol H> 0.4325 mol O so we will use 0.4325 mol O
Step 3: Divide above calculations by the smallest mol
0.4325 mol O/0.4325= 1
0.4365 mol H/0.4325= 1.009 *rounds to 1
Step 4: use calculations as subscripts
Oxygen = 1 so the subscript will 1 (O)
Hydrogen = 1 so the subscript will 1 (H)
making E.F= OH
Molecular Formula:
Step 1: identify molecular mass and mass from the E.F
the molecular mass is given 34.00g/mol
the mass of the E.F is
(oxygen mass from periodic table)+(hydrogen mass from periodic table)
16+1.008= 17.008
Step 2:Divide the molecular mass by the mass given by the emipirical formula.
round to 2
Step 3:Multiply the empirical formula (the subscripts) by this number to get the molecular formula. ANSWER: M.F=2(OH)-
Explanation:
(A) As we know that carbonic acid () and Sodium bicarbonate (
) forms an acidic buffer.
Therefore, pH of an acidic buffer is given by Hendeerson-Hasselbalch equation as follows.
pH = ........... (1)
So mathematically, if [Salt] = [Acid] then = 1
.
And, = 0
Therefore, equation (1) gives us the following.
pH = (when acid and salt are equal in concentration)
Hence, of
(carbonic acid) is 6.35.
And, with this we have following results.
In (A) and (D) we have the case \frac{[NaHCO_{3}]}{[H_{2}CO_{3}]}[/tex] i.e. [Salt] = [Acid].
Hence, for the cases pH = = 6.35.
(B) = 0.045 M and,
= 0.45 M
Hence, pH =
=
= 6.35 + (-1)
= 5.35
Therefore, it means that this buffer will be most suitable buffer as it has pH on acidic side and addition of slight excess base will not affect much of its pH value.
(C) = 0.45 M
= 0.045 M
So, pH =
=
= 6.35 + (+1)
= 7.35
This means that pH on Basic side makes it no more acidic buffer.