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.

You might be interested in

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!

We have the following statement data:

Data:
$P_{Total} = 800 mmHg$
$P\% N_{2} = 60\%$
$P\% O_{2} = 40\%$
$P_{partial} = ? (mmHg)$

As the percentage is the mole fraction multiplied by 100:

 $P = X_{ O_{2} }*100$

The mole fraction will be the percentage divided by 100, thus:
What is the partial pressure of oxygen in this mixture?

$X_{ O_{2} } = \frac{P}{100}$
$X_{ O_{2}} = \frac{40}{100}$
$\boxed{X_{ O_{2}} = 0.4}$

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:

In relation to $O_{2}$ :

$\frac{P O_{2} }{P_{total}} = X_O_{2}$
$\frac{P O_{2} }{800} = 0.4$
$P_O_{2} = 0.4*800$
$\boxed{\boxed{P_O_{2} = 320\:mmHg}}\end{array}}\qquad\quad\checkmark$

Answer:
b. 320.0 mm hg

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

3 years ago

Read 2 more answers

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: $4m/s^2$