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
A solution is a homogeneous mixture of two or more substances. a solution doesn't allow light to scatter. the solute cannot be separated mechanically from the solution.
Answer:
a) 30 moles
Explanation:
2C6H6 + 1502 -------> 12CO2 + 6H20
from reaction 2 mol 15 mol
given 4.0 mol x mol
x = 4.0*15/2 = 30. mol
Trisulfur nonaiodide
Step-by-step explanation:
The name for a binary molecular compound has the form
Multiplying prefix+name of first element multiplying prefix+stem of second element element+ide (two words)
The multiplying prefixes for three and nine are tri and nona, respectively.
The stem of iodine is iod.
Put them together, and the name of S₃I₉ is
trisulfur nonaiodide.
I'm pretty sure it's D. Compounds
Salt water is an example of a solution, and a solution is a type of mixture, so by multiple choice logic none can be the right answer! Seriously though - mixture are combinations of two compounds that are physically combined but do not undergo a chemical reaction with one another to produce a new compound. For instance, when you dissolve salt in water, you don't produce any new products, and you can boil away the water to leave only the salt (a physical method of separation). You cannot separate a compound by physical means because by definition this would be a chemical reaction as you're breaking bonds.
