<u>Answer:</u> The quantity of every prefix is written below as a power of ten.
<u>Explanation:</u>
In the metric system of measurement, the name of multiples and subdivision of any unit is done by combining the name of the unit with the prefixes.
<u>For Example:</u> deka, hecto and kilo means 10, 100 and 1000 respectively. Deci, centi and milli means one-tenth, one-hundredth, and one-thousandth respectively.
The quantity of these prefixes are written as the power of 10.
For the given prefixes:
<u>Nano:</u> The quantity will be 
<u>Kilo:</u> The quantity will be 
<u>Centi:</u> The quantity will be 
<u>Micro:</u> The quantity will be 
<u>Milli:</u> The quantity will be 
<u>Mega:</u> The quantity will be 
Hence, the quantity of every prefix is written above as a power of ten.
The answer for the following problem is mentioned below.
<u><em>Therefore volume occupied by methane gas is 184.78 × 10^-3 liters </em></u>
Explanation:
Given:
mass of methane(
) = 272 grams
pressure (P) = 250 k Pa =250×10^3 Pa
temperature(t) = 54°C =54 + 273 = 327 K
Also given:
R = 8.31JK-1 mol-1 ,
Molar mass of methane(
) = 16.0 grams
We know;
According to the ideal gas equation,
<u><em>P × V = n × R × T</em></u>
here,
n = m÷M
n =272 ÷ 16
<u><em>n = 17 moles</em></u>
Therefore,
250×10^3 × V = 17 × 8.31 × 327
V = ( 17 × 8.31 × 327 ) ÷ ( 250×10^3 )
V = 184.78 × 10^-3 liters
<u><em>Therefore volume occupied by methane gas is 184.78 × 10^-3 liters </em></u>
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Answer:
true im pretttyyy sure
Explanation:
bcuz the stronger the intermolecular forces the higher the boiling point :3
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