Answer:- 3. 
and 
Explanations:- An empirical formula is the simplest whole number ratio of atoms of each element present in the molecule/compound.
For example, the molecular formula of benzene is 
. The ratio of C to H in it is 6:6 that could be simplified to 1:1. So, an empirical formula of benzene is CH.
In the first pair, the ratio of C to H in first molecule is 2:4 that could be simplified to 1:2 and the empirical formula is 
. In second molecule the ratio of C to H is 6:6 and it could be simplified to 1:1. and the empirical formula is CH. Empirical formulas are different for both the molecules of first pair and so it is not the right choice.
In second pair, C to H ratio in first molecule is 1:2, so the empirical formula is . The C to H ratio for second molecule is 1:4, so the empirical formula is 
. Here also, the empirical formulas are not same and hence it is also not the right choice.
In third pair, C to H ratio in first molecule is 1:3, so the empirical formula is . In second molecule the C to H ratio is 2:6 and it is simplified to 1:3. So, the empirical formula for this one is also . Hence. this is the correct choice.
In fourth pair, first molecule empirical formula is CH. Second molecule has 2:4 that is 1:2 mole ratio of C to H and so its empirical formula is . As the empirical formulas are different, it is not the right choice.
So, the only and only correct pair is the third one. 3. and
According to Raoult's law the relative lowering of vapour pressure of a solution made by dissolving non volatile solute is equal to the mole fraction of the non volatile solute dissolved.
the relative lowering of vapour pressure is the ratio of lowering of vapour pressure and vapour pressure of pure solvent
Where
xB = mole fraction of solute=?
p = 22.8 torr
mole fraction is ratio of moles of solute and total moles of solute and solvent
moles of solvent = mass / molar mass = 500 /18 = 27.78 moles
putting the values
mass of glucose = moles X molar mass = 1.218 X 180 = 219.24 grams
<h2>Answer:</h2>
<h2>Explanations</h2>
The complete balanced equation for the given reaction is expressed as;
Given the following parameters
Mass of CH4 = 5.90×10^−3 g = 0.0059grams
Determine the moles of methane
According to stoichimetry, 1 mole of methane produces 2 moles of water, hence the moles of water required will be:
Determine the mass of water produced
Therefore the mass of water produced from the complete combustion of 5.90×10−3 g of methane is 1.33 * 10^-2grams
Explanation:
The scientific notation:
where
and k is integer.
We have the example:
You can write the numbers in a "normal" form:
Make the sum:
And next write it in the scientific notation:
<h3>Other method:</h3>
You can add numbers in scientific notation if the power of tens in both number is the same.
Therefore you must convert the first number:
Now, you can make the sum:
When mixture of NaCl and Al₂(SO₄)₃ is allowed to react with excess NaOH, only Al₂(SO₄)₃ reacts with it and NaCl does not react with NaOH due to presence of common ion (Na⁺). On reaction gelatinous precipitate of aluminium hydroxide [Al(OH)₃] is produced. The balanced chemical reaction is represented as-
Al₂(SO₄)₃ + 6NaOH → 2Al(OH)₃ + 3Na₂SO₄
On this reaction, 0.495 g = 0.495/78 moles =6.346 X 10⁻³ moles of Al(OH)₃.
As per balanced reaction, two moles of Al(OH)₃ is produced from one mole Al₂(SO₄)₃. So, 6.346 X 10⁻³ moles of Al(OH)₃ is produced from (6.346 X 10⁻³)/2 moles=3.173 X 10⁻³ moles of Al₂(SO₄)₃= 3.173 X 10⁻³ X 342 g of Al₂(SO₄)₃=1.085 g of Al₂(SO₄)₃.
So, mass percentage of Al₂(SO₄)₃ is= (amount of Al₂(SO₄)₃/total amount of mixture)X100 = 
=74.8 %.
