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?
a. 381.27 m/s
b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide
<h3>Further explanation</h3>
Given
T = 100 + 273 = 373 K
Required
a. the gas speedi
b. The rate of effusion comparison
Solution
a.
Average velocities of gases can be expressed as root-mean-square averages. (V rms)
R = gas constant, T = temperature, Mm = molar mass of the gas particles
From the question
R = 8,314 J / mol K
T = temperature
Mm = molar mass, kg / mol
Molar mass of Sulfur dioxide = 64 g/mol = 0.064 kg/mol
b. the effusion rates of two gases = the square root of the inverse of their molar masses:
M₁ = molar mass sulfur dioxide = 64
M₂ = molar mass nitrogen triodide = 395
the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide
Answer:
Objective
Optics
For an optical instrument, the aperture is the diameter of the objective lens (refracting telescope) or the primary mirror (reflecting telescope). The larger the aperture, the more light the telescope can gather, and the fainter the limiting magnitude of the instrument.
Explanation:
<h2>♨ANSWER♥</h2>
pl (25*C)
Arginine -----> 10.76
Glutamic -----> 3.08
Asparagine -----> 5.43
Tyrosine -----> 5.63
<u>☆</u><u>.</u><u>.</u><u>.</u><u>hope this helps</u><u>.</u><u>.</u><u>.</u><u>☆</u>
_♡_<em>mashi</em>_♡_
