Answer: 600 mL
Explanation:
Given that;
M₁ = 5.85 m
M₂ = 1.95 m
V₁ = 200 mL
V₂ = ?
Now from the dilution law;
M₁V₁ = M₂V₂
so we substitute
5.85 × 200 = 1.95 × V₂
1170 = 1.95V₂
V₂ = 1170 / 1.95
V₂ = 600 mL
Therefore final volume is 600 mL
Answer:
Your question is complex, because I think you wrote it wrong.
Although in front of this what I can help you is that the carbons are associated between a single, double or triple union.
This depends on whether they are attached to more or less carbons or hydrogens, the carbons have the possibility of joining 4 radicals, both other carbons and hydrogens.
Simple junctions talks about compound organisms called ALKANS.
The double unions, in organic these compounds are called as ALQUENOS.
And as for the tertiary unions, the organic chemistry names them as ALQUINOS.
These compounds that we write, a simple union, the less energy, the less this union, that is why the triple bond is the one that contains the most energy when breaking or destroying it in a reaction.
Explanation:
In a chemical compound the change of these unions if we modified them we would generate changes even in the classifications naming them as well as different compounds and not only that until they change their properties
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
