Given the percentage composition of HC as C → 81.82 % and H → 18.18 %
So the ratio of number if atoms of C and H in its molecule can will be:
C : H = 81.82 12 : 18.18 1 C : H = 6.82 : 18.18 = 6.82 6.82 : 18.18 6.82 = 1 : 2.66 ≈ 3 : 8
So the Empirical Formula of hydrocarbon is:
C 3 H 8
As the mass of one litre of hydrocarbon is same as that of C O 2 The molar mass of the HC will be same as that of C O 2 i.e 44 g mol
Now let Molecular formula of the HC be ( C 3 H 8 ) n
Using molar mass of C and H the molar mass of the HC from its molecular formula is:
( 3 × 12 + 8 × 1 ) n = 44 n So 44 n = 44 ⇒ n = 1
Hence the molecular formula of HC is C 3 H 8
Does that help?
Balance the reaction of Fe2(SO4)3 + KOH = K2SO4 + Fe(OH)3 using this chemical equation balancer! ... Fe2(SO4)3 + 6KOH → 3K2SO4 + 2Fe(OH)3 ...
Missing: _K[ | Must include: _K[
Answer:
0.047 %
Explanation:
Step 1: Given data
- Partial pressure of ozone (pO₃): 0.33 torr
- Total pressure of air (P): 695 torr
Step 2: Calculate the %v/v of ozone in the air
Air is a mixture of gases. We can find the %v/v of ozone (a component) in the air (mixture) using the following expression.
<em>%v/v = pO₃/P × 100%</em>
%v/v = 0.33 torr/695 torr × 100%
%v/v = 0.047 %
Explanation:
The number of moles of solute present in liter of solution is defined as molarity.
Mathematically, Molarity = 
Also, when number of moles are equal in a solution then the formula will be as follows.
It is given that 
is 8.00 M, 
is 7.00 mL, and 
is 0.80 M.
Hence, calculate the value of 
using above formula as follows.
= 70 ml
Thus, we can conclude that the volume after dilution is 70 ml.
Explanation:
what you have to do is to multiply the given grams with the ratio of grams of that certain element in it's full compound in order to isolate the compound and get the element.
because the question did not specify how many grams of NO2 is formed, we can assume that the mass is neglegible, thus 1 mole was assigned to Nitrogen.
