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sp2606 [1]
3 years ago
11

Question 2 (2 points)

Chemistry
1 answer:
inysia [295]3 years ago
5 0

Answer:

v = 46.5 m/s

Explanation:

Given data:

Mass of car = 1210 kg

Momentum of car = 56250 kg m/s

Velocity of car = ?

Solution:

Formula:

p = mv

p = momentum

m = mass

v = velocity

Now we will put values in formula:

56250 kg m/s  = 1210 kg × v

v = 56250 kg m/s / 1210 kg

v = 46.5 m/s

So a car having mass of 1210 kg with momentum 56250 kg m/s having 46.5 m/s velocity.

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A mixture of gases with a pressure of 800.0 mm hg contains 60% nitrogen and 40% oxygen by volume. what is the partial pressure o
klasskru [66]
Hello!

<span>We have the following statement data:
</span>
Data:
P_{Total} = 800 mmHg
P\% N_{2} = 60\%
P\% O_{2} = 40\%
P_{partial} = ? (mmHg)

<span>As the percentage is the mole fraction multiplied by 100:

</span>P =  X_{ O_{2} }*100

<span>The mole fraction will be the percentage divided by 100, thus:
</span><span>What is the partial pressure of oxygen in this mixture? 
</span>
X_{ O_{2} }  =  \frac{P}{100}
X_{ O_{2}} =  \frac{40}{100}
\boxed{X_{ O_{2}} = 0.4}


<span>To calculate the partial pressure of the oxygen gas, it is enough to use the formula that involves the pressures (total and partial) and the fraction in quantity of matter:
</span>
In relation to O_{2} :

\frac{P O_{2} }{P_{total}} = X_O_{2}
\frac{P O_{2} }{800} = 0.4
P_O_{2} = 0.4*800
\boxed{\boxed{P_O_{2} = 320\:mmHg}}\end{array}}\qquad\quad\checkmark
<span>
Answer:
</span><span>b. 320.0 mm hg </span>
7 0
3 years ago
What is the volume of 1.9 moles of chlorine gas (Cl2) at standard temperature and pressure (STP)?
masya89 [10]
Considering ideal gas behavior, the volume of 1 mol of gas at STP is 22.4 L; then the volume occupied by 1.9 moles is 1.9mol*22.4L/mol = 42. 6 L.

Answer: 43 L
5 0
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A car traveling at a speed of 30.0 m/s encounters an emergency and comes to a complete stop in 7.5 seconds? What is the cars acc
Marianna [84]

Answer:

The acceleration is: 4m/s^2

Explanation:

Given

u = 30.0m/s --- The initial velocity

t = 7.5s --- time

v = 0m/s -- The final velocity

Required

Determine the acceleration

To do this, we make use of the first equation of motion

v = u - at

We used negative because the car was coming to stop.

This gives:

0 = 30 - 7.5 * a

0 = 30 - 7.5a

Collect like terms

7.5a = 30

Solve for a

a = \frac{30}{7.5}

a = 4m/s^2

6 0
3 years ago
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pshichka [43]

Answer:

use the rule of speed

Explanation:

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Which of the five type of equilibrium problems best applies to this question: Consider the following reaction at equilibrium. Wh
andreev551 [17]

Answer:

d

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