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3 years ago

If 41.5 mol of an ideal gas occupies 86.5 L at 27.00 °C, what is the pressure of the gas?

Physics

2 answers:

Dmitry [639]3 years ago

7 0

PV=nRT
(P)(86.5)=(41.5)(.08206)(300.15)
(P)(86.5)=(1022.157824)
P=11.81685345 atm

AlexFokin [52]3 years ago

5 0

YUP!!and I think its important to note that when using this value of R

<em>It is crucial to match your units of Pressure, Volume, number of mole, and Temperature with the units of R. If you use the first value of </em>R,<em> which is </em>0.082057 L <u><em>your unit for Pressure must be atm, Volume must be Liter, & Temperature must be Kelvin.</em></u>

Clarity in Chemistry is crucial.

*_^

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Jody has a mass of 50 kilograms what is his weight on earth

WARRIOR [948]

Answer:

Weight of body = 490 Newton

Explanation:

Mass of the body is the total amount of matter contained in the body but Weight of the body is the force exerted by gravity on the body. Weight of the body depends upon acceleration due to gravity of the planet.

Thus,

Weight = mass × acceleration due to gravity

W = mg

W = weight of the body

m = mass of the body

g = acceleration due to gravity

W = 50 × 9.8

W = 490 $\frac{kg m}{s^{2}}$

W = 490 Newton

Hence, weight of the body is 490 Newton.

5 0

3 years ago

Read 2 more answers

A ball thrown vertically upward is caught by the thrower after 2.00 s. Find (a) the initial velocity of the ball and (b) the max

ankoles [38]

Answer:

a) 9.8 m/s

b) 4.9 m

Explanation:

This problem is a good example of Vertical motion, where the main equations for this situation are:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

$V^{2}={V_{o}}^{2}-2gy$ (2)

Where:

$y$ is the height of the ball at a given time

$y_{o}=0m$ is the initial height of the ball (assuming the hand of the thrower the origin of the system)

$V_{o}$ is the initial velocity of the ball

$V$ is the final velocity of the ball

$t=2s$ is the time it takes for the ball to make the complete movement (from the moment it is thrown until it falls back into the pitcher's hands)

$g=9.8 m/s^{2}$ is the acceleration due to gravity

Knowing this, let's begin with the answers:

<h3>a) Initial velocity </h3>

In order to find the initial velocity $V_{o}$ of the ball, we will use equation (1) and $t=2s$ , taking into account that $y=0 m$ and $y_{o}=0m$ at this given time:

$0=0+V_{o}t-\frac{1}{2}gt^{2}$ (3)

Isolating $V_{o}$ :

$V_{o}=\frac{1}{2}gt$ (4)

$V_{o}=\frac{1}{2}(9.8 m/s^{2})(2 s)$ (5)

Then:

$V_{o}=9.8 m/s$ (6)

<h3>b) Maximum height </h3>

In this part, we will use equation (2), knowing the value of the height is maximum when $V=0$ . So, we will name this height as $y_{max}$ :

$0={V_{o}}^{2}-2gy_{max}$ (7)

Isolating $y_{max}$ :

$y_{max}=\frac{{V_{o}}^{2}}{2g}$ (8)

$y_{max}=\frac{{(9.8 m/s)}^{2}}{2(9.8 m/s^{2})}$ (9)

Finally:

$y_{max}=4.9 m$

4 0

3 years ago

4. A driver travels 135 km, east in 1.5 h, stops for 45 minutes for lunch,

stira [4]

Answer:

The driver's average velocity is 82.35 km/h.

Explanation:

Given:

The motion of the driver can be divided into 3 parts:

i. Displacement of the driver in 1.5 hours = 135 km

ii. Rest for 45 minutes.

iii. Displacement in next 2 hours = 215 km

The direction of motion remains same (east).

Now, total displacement of the driver is, $D_{Total}=135+215=350$ km.

Rest time is 45 minutes. Converting it to hours, we need to use the conversion factor $1\textrm{ min} = \frac{1}{60}$ hour.

So, 45 minutes in hours is equal to $\frac{45}{60}=0.75$ hours.

Now, total time taken for the complete journey is, $\Delta t=1.5+\frac{45}{60}+2=1.5+0.75+2=4.25\textrm{ h}$

Average velocity is given as:

$v_{avg}=\frac{\textrm{Total displacement}}{Total time}=\frac{350}{4.25}=82.35\textrm{ km/h}$

Therefore, the driver's average velocity is 82.35 km/h

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3 years ago

Your teacher's SUV can go from rest to 25 m/s (roughly 55mph) in 10 seconds, The car's velocity changes at a uniform rate.

maw [93]

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2 years ago

10.

aliina [53]

The normal force is the supporting force that is exerted on an object that is in contact with another stable object.

Answer: Option C

<u>Explanation: </u>

Normal force is forward or upward pushing force acting on an object. Mostly the normal force acts as supporting force exerted on the object by the neighbouring stable object with which the object in question is in contact. So normal force falls under the category of contact forces.

Generally, normal force will be acting to support the weight of any object placed on another object. The best examples of normal forces are the weight of the book supported by table or by the pushing force of the wall on the person leaning on the wall.

5 0

2 years ago

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