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GaryK [48]
3 years ago
13

When mountains are young, like the Rocky Mountains, they are very tall, with sharp jagged shapes. When mountains are old, they

Chemistry
1 answer:
marta [7]3 years ago
6 0
I think A.weathering, is the best possible answer
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Activity 2 Directions: Select among the choices thbest type of material to be used in making the objects at the left and explain
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A Walk in the City

Make a list of the number of cars, jeeps, tricycle, and even trucks you've seen on your way to school. On a Decision Making Chart, answer the question, "Does the volume of traffic affect the air quality in my local community?" Write your reasons for saying YES on the Reasons for column, and the reasons for saying NO on the Reasons Against column. At the bottom of a chart, make a position by writing your decision on the same question.

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2 years ago
Isomers have the same physical properties because the chemical formulas are identical. True False
Luda [366]
The statement is False.
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3 years ago
Amy is in a car moving quickly toward a stationary siren. Just as Amy moves
sineoko [7]

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3 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

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
2 years ago
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