QUESTION 36

A vessel containing 200 ml hydrogen gas at pressure 500torr at temperature 10 degree C.Find its volume when pressure increased t

Margaret [11]

Answer:

When the pressure and the temperature are increased the volume is 285.7 ml.

Explanation:

We can find the new volume by using the Ideal Gas Law:

$PV = nRT$

Where:

P: is the pressure

V: is the volume

n: is the number of moles

R: is the gas constant

T: is the temperature

Initially, when V₁ = 200 ml, P₁ = 500 torr and T₁ = 10 °C, we have:

$nR = \frac{P_{1}V_{1}}{T_{1}}$ (1)

And finally, when P₂ = 700 torr and T₂ = 20 °C, we have:

$nR = \frac{P_{2}V_{2}}{T_{2}}$ (2)

By equating (1) with (2):

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

$V_{2} = \frac{P_{1}V_{1}T_{2}}{T_{1}P_{2}} = \frac{500 torr*200 ml*20 ^{\circ} C}{10 ^{\circ} C*700 torr} = 285.7 ml$

Therefore, when the pressure and the temperature are increased the volume is 285.7 ml.

I hope it helps you!

3 0

3 years ago

Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).

Yuliya22 [10]

Answer: The mean of the samples is 32.325.

Proportion of the defective unit is = $\frac{2}{40}=\frac{1}{20}=0.05$

Explanation:

The point estimate of the mean of the sample:

First calculate the mean of the 40 samples:

$\frac{\text{Sum of all the samples}}{\text{Number of observations}}$

$\frac{1,293}{40}=32.325$

The mean of the samples is 32.325.

Defected pieces in the samples whose life span is less than the 26 days = 2

Proportion of the defective unit is = $\frac{2}{40}=\frac{1}{20}=0.05$

7 0

4 years ago

Read 2 more answers

Which is a sign of a chemical change?

Molodets [167]

Answer: When chemical change occurs you will see like, the change of color or bubbles will form, the temperature will change, and maybe the odor too.

5 0

3 years ago

Read 2 more answers

A person in the passenger basket of a hot-air balloon throws a ball horizontally outward from the basket with a speed of 10.0 m/

Aleksandr-060686 [28]

Answer:

Explanation:

Given

balloon is rising with a speed of $u_y=6\ m/s$

Person throws a ball out of basket with a horizontal velocity of $u_x=10\ m/s$

Considering upward direction to be positive

When ball is thrown it has two velocity i.e. in upward direction and in horizontal direction so net velocity is

$v_{net}=\sqrt{(u_x)^2+(u_y)^2}$

$v_{net}=\sqrt{(6)^2+(10)^2}$

$v_{net}=\sqrt{36+100}$

$v_{net}=\sqrt{136}$

$v_{net}=11.66\ m/s$

Direction of velocity

$\tan \theta =\dfrac{u_y}{u_x}$

$\tan \theta =\dfrac{6}{10}$

$\theta =30.96^{\circ}$

where $\theta$ is angle made by net velocity with horizontal .

7 0

3 years ago

a baseball is hit straight up into the air. If the initial velocity was 20 m/s, how high will the ball go?How long will it be un

AleksAgata [21]

1) The motion of the ball is an uniformly accelerated motion, with constant acceleration equal to $g=-10 m/s^2$ (the negative sign means it is directed towards the ground).

For an uniformly accelerated motion, we can use the following relationship:
$2gh=v_f^2-v_i^2$
where h is the maximum height reached by the ball, $v_i = 20 m/s$ is its initial velocity and $v_f$ the velocity of the ball when it reaches the maximum height. But $v_f=0$ (when the ball reaches the maximum height, it stops before going down, so its velocity at that moment is zero), so we can use the relationship to calculate h, the maximum height:
$h=- \frac{v_i^2}{2g} =- \frac{20 m/s)^2}{2 \cdot (-10 m/s^2)} =20 m$

2) We can find the time the ball takes to return to the ground by requiring that the space covered by the ball returns to zero:
$S(t)=0$
where for an uniformly accelerated motion,
$S(t)=v_i t + \frac{1}{2} gt^2 =0$
By solving this, we have two solutions: one is t=0, which corresponds to the moment the player hits the ball, the second one is
$t=- \frac{2 v_i}{g}=- \frac{2 \cdot 20 m/s}{-10 m/s^2}=4 s$
so, the ball returns to the ground after 4 s.

3) The velocity of the ball when it returns to the ground is given by:
$v(t) = v_i +gt=20 m/s + (-10 m/s^2)(4 s)=-20 m/s$
where the negative sign means the ball is going downwards.

8 0

3 years ago