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shutvik [7]
3 years ago
13

72 Points Please Answer! An organism has the scientific name Canis lupus. Using complete sentences, write the information that y

ou can obtain about the organism from its scientific name. Also give one reason that scientists use scientific names.
If you spam I will report you and get my points refunded
Chemistry
1 answer:
Vanyuwa [196]3 years ago
8 0

Answer:

From the name <em>Canis lupus</em>, one can find that the organism is a wolf. <em>Canis</em> is a genus meaning "dog". <em>Lupus</em> is the species meaning "wolf". So in short, <em>Canis Lupus</em> means "dog wolf".

One reason scientist use scientific names is to avoid confusion. For example, there are many species of bass, a fish, but there's only one kind of <em>Micropterus salmoides</em>, or a Largemouth bass for short.

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

\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} }

M₁ = molar mass sulfur dioxide = 64

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

4 0
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