Answer:
The molar mass of the gas is 44 g/mol
Explanation:
It is possible to solve this problem using Graham's law that says: Rates of effusion are inversely dependent on the square of the mass of each gas. That is:
If rate of effusion of nitrogen is Xdistance / 48s and for the unknown gas is X distance / 60s and mass of nitrogen gas is 28g/mol (N₂):
6,61 = √M₂
44g/mol = M₂
<em>The molar mass of the gas is 44 g/mol</em>
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I hope it helps!
Answer:
0.677 moles
Explanation:
Take the atomic mass of K = 39.1, O =16.0, P = 31.0
no. of moles = mass / molar mass
no. of moles of K3PO4 used = 4.79 / (39.1x3 + 31 + 16x4)
= 0.02256 mol
From the equation, the mole ratio of KOH : K3PO4 = 3 :1,
meaning every 3 moles of KOH used, produces 1 mole of K3PO4.
So, using this ratio, let the no. of moles of KOH required to be y.
y = 0.02256 x3
y = 0.0677 mol
If you don't find exactly 0.677 moles as one of the options, go for the closest one. A very slight error may occur because of taking different significant figures of atomic masses when calculating.
Answer:
(2R,3S)-2-ethoxy-3-methylpentane
and
(2S,3S)-2-ethoxy-3-methylpentane
Explanation:
For this case, we will have as nucleophile. Also, this compound is also in excess. So, we will have as solvent
a protic solvent. Therefore the Sn1 reaction would be favored.
The first step would be the carbocation formation followed by the attack of the nucleophile. In this case both isomers would be produced: R and S (see figure).
Answer: The empirical formula for the given compound is
Explanation:
The chemical equation for the combustion of compound having carbon and hydrogen follows:
where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.
We are given:
Mass of
We know that:
Molar mass of carbon dioxide = 44 g/mol
For calculating the mass of carbon:
In 44 g of carbon dioxide, 12 g of carbon is contained.
So, in 22.0 g of carbon dioxide, of carbon will be contained.
For calculating the mass of hydrogen:
Mass of hydrogen = Mass of sample - Mass of carbon
Mass of hydrogen = 7.0 g - 6 g
Mass of hydrogen = 1.0 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Carbon =
Moles of Hydrogen =
Step 2: 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 0.5 moles.
For Carbon =
For Hydrogen =
Step 3: Taking the mole ratio as their subscripts.
The ratio of Fe : C : H = 1 : 2
Hence, the empirical formula for the given compound is