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3 years ago

What is the density of an 84.7 g sample of an unknown substance if the sample occupies 49.6cm cubed

Chemistry

1 answer:

Ahat [919]3 years ago

7 0

Answer:

1.70 g.cm⁻³

Solution:

Data Given;

Mass = 84.7 g

Volume = 49.6 cm³

Density = ?

Formula Used;

Density = Mass ÷ Volume

Putting values,

Density = 84.7 g ÷ 49.6 cm³

Density = 1.70 g.cm⁻³

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3 years ago

1.) Which of the following is an example of using creativity while doing background research?

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2 years ago

A) Find the gas speed of sulfur dioxide at 100.0 degrees Celsius? ______________

gtnhenbr [62]

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

Average velocities of gases can be expressed as root-mean-square averages. (V rms)

$\large {\boxed {\bold {v_ {rms} = \sqrt {\dfrac {3RT} {Mm}}}}$

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

$\tt v=\sqrt{\dfrac{3\times 8.314\times 373}{0.064} }\\\\v=381.27~m/s$

b. the effusion rates of two gases = the square root of the inverse of their molar masses:

$\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }$

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M₂ = molar mass nitrogen triodide = 395

$\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{395}{64} }=\dfrac{20}{8}=2.5$

the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide

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2 years ago