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?
Answer:
Neutrons released during a fission reaction cause other nuclei to split
Answer:-
Solution:- As is clear from the given Ka value, Cinnamic acid is a weak acid. let's calculate the moles of acid and KOH added to it from their given molarities and mL.
For KOH, 
= 0.002 mol
For Cinnamic acid, 
= 0.003 mol
Acid and base react as:
The reaction takes place in 1:1 mol ratio. Since the moles of acid are in excess, the acid is still remaining when all the kOH is used.
0.002 moles of KOH react with 0.002 moles of Cinnamic acid to form 0.002 moles of potassium cinnamate. Excess moles of Cinnamic acid = 0.003 - 0.002 = 0.001
As the solution have weak acid and it's salt(or we could say conjugate base), it is a buffer solution and the pH of the buffer solution could easily be calculated using Handerson equation:
pKa could be caluted from given Ka value using the formula:
pKa = - log Ka
pKa = 4.44
let's plug in the values in Handerson equation and calculate the pH:
pH = 4.44+0.30
pH = 4.74
So, the first choice is correct, pH is 4.74.
The energy of the 434 nm emission line is 4.58×10¯¹⁹ J (Option A)
<h3>Data obtained from the question </h3>
The following data were obtained from the question:
- Wavelength (λ) = 434 nm = 434×10¯⁹ m
- Planck's constant (h) = 6.626×10¯³⁴ Js
- Speed of light (v) = 3×10⁸ m/s
- Energy (E) =?
<h3>How to determine the energy </h3>
The energy of the 434 nm emission line can be obtained as follow:
E = hv / λ
E = (6.626×10¯³⁴ × 3×10⁸) / 434×10¯⁹ 9.58×10¹⁴
E = 4.58×10¯¹⁹ J
Thus, the energy of the 434 nm emission line is 4.58×10¯¹⁹ J
Learn more about energy:
brainly.com/question/10703928
#SPJ1
