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Nataly_w [17]
3 years ago
7

The net force on a car accelerating

Chemistry
1 answer:
mafiozo [28]3 years ago
7 0

Answer:

<h2>800 kg</h2>

Explanation:

The mass of the car can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m =  \frac{2000}{2.5}  \\

We have the final answer as

<h3>800 kg</h3>

Hope this helps you

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Which substance below will exhibit hydrogen bonding between the molecules of the substance?
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I think it might just might be e
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The formation of SO3 from SO2 and O2 is an intermediate step in the manufacture of sulfuric acid, and it is also responsible for
jeka57 [31]

Answer:

118.22 atm

Explanation:

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)      

KP = 0.13 = \frac{p(SO_{3})^{2}}{p(SO_{2})^{2}p(O_{2})}

Where p(SO₃) is the partial pressure of SO₃, p(SO₂) is the partial pressure of SO₂ and p(O₂) is the partial pressure of O₂.

  • With 2.00 mol SO₂ and 2.00 mol O₂ if there was a 100% yield of SO₃, then 2 moles of SO₃ would be produced and 1.00 mol of O₂ would remain.
  • With a 71.0% yield, there are only 2*0.71 = 1.42 mol SO₃, the moles of SO₂ that didn't react would be 2 - 1.42 = 0.58; and the moles of O₂ that didn't react would be 2 - 1.42/2 = 1.29.

The total number of moles is 1.42 + 0.58 + 1.29 = 3.29. With that value we can calculate the molar fraction (X) of each component:

  • XSO₂ = 0.58/3.29 = 0.176
  • XO₂ = 1.29/3.29 = 0.392
  • XSO₃ = 1.42/3.29 = 0.432

The partial pressure of each gas is equal to the total pressure (PT) multiplied by the molar fraction of each component.

  • p(SO₂) = 0.176 * PT
  • p(O₂) = 0.392 * PT
  • p(SO₃) = 0.432 * PT

Rewriting KP and solving for PT:

\frac{p(SO_{3})^{2}}{p(SO_{2})^{2}p(O_{2})}=0.13\\\frac{(0.432*P_{T})^{2}}{(0.176*P_{T})^{2}(0.392*P_{T})} =0.13\\\frac{0.1866*P_{T}^{2}}{0.0121*P_{T}^{3}} =0.13\\\frac{15.369}{P_{T}}=0.13\\P_{T}=118.22 atm

5 0
4 years ago
How many liters are in 2.75 ounces? Use the conversion factor: 1 liter = 33.814 ounces Rounded to the result to 3 significant fi
Serggg [28]
1L = 33.814 oz
xL = 2.75 oz

so it's a proportion

1L / 33.814 oz = xL / 2.75

solve for x

(1/33.814) * 2.75 = 0.0813272609 on your calculator, but it's not the answer.

the number in your problem, 2.75 oz, has 3 significant figures. so you can only round this number to 3 significant figures too.

your equipment isn't accurate enough to give a reading to 10 significant figures if that makes sense. you have to give the answer in terms of the term you use with the lowest significant figures.

so with 3 significant figures,
0.0813272609 rounds to
0.0813 L
4 0
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
3 years ago
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