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3 years ago

What is the pressure exerted by 0.801 mol of co2 in a 12 l container?

1 answer:

8 0

We assume that the gas is an ideal gas so we can use the relation PV=nRT. Assuming that the temperature of the system is at ambient temperature, T = 298 K. We can calculate as follows:

PV = nRT
P = nRT / V
P = (0.801 mol ) (0.08205 L-atm / mol-K) (298.15 K) / 12 L
P = 1.633 atm

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A bowler who always left the same three pins standing could be considered a(n) ____ bowler.

RideAnS [48]

A bowler who always left the same 3 pins standing could be considered a C. Precise bowler as from bowling countless number of times he has observed the same amount of pins knocked down each time.

4 0

4 years ago

Peg P is driven by the forked link OA along the path described by r = eu, where r is in meters. When u = p4 rad, the link has an

8_murik_8 [283]

Answer:

The transverse component of acceleration is 26.32 $m/s^2$ where as radial the component of acceleration is 8.77 $m/s^2$

Explanation:

As per the given data

u=π/4 rad

ω=u'=2 rad/s

α=u''=4 rad/s

$r=e^u$

So the transverse component of acceleration are given as

$a_{\theta}=(ru''+2r'u')\\$

Here

$r=e^u\\r=e^{\pi/4}\\r=2.1932 m$

$r'=e^u.u'\\r'=2.1932 \times 2\\r'=4.3864 m$

$a_{\theta}=(ru''+2r'u')\\a_{\theta}=(2.1932\times 4+2\times 4.3864 \times 2)\\a_{\theta}=26.32 m/s\\$

The transverse component of acceleration is 26.32 $m/s^2$

The radial component is given as

$a_r=r''-r\theta'^2$

Here

$r''=e^u.u'^2+e^u u''\\r''=2.1932 \times (2)^2+2.1932\times 4\\r''=17.5456 m$

$a_r=r''-ru'^2\\a_r=17.5456-2.1932\times (2)^2\\a_r=8.7728 m/s^2$

The radial component of acceleration is 8.77 $m/s^2$

6 0

4 years ago

Which factor affects the angle of sunlight on Earth? The distance between Earth and the sun Earth's tilt from its axis The path

Rainbow [258]

Earth’s tilt from its axis.
For explanation:
The angle in which Earth is at is 23.5°. This causes its tilt which affects how the Sun’s light hits Earth

6 0

1 year ago

A 2 m tall, 0.5 m inside diameter tank is filled with water. A 10 cm hole is opened 0.75 m from the bottom of the tank. What is

valina [46]

Answer:

4.75 m/s

Explanation:

The computation of the velocity of the existing water is shown below:

Data provided in the question

Tall = 2 m

Inside diameter tank = 2m

Hole opened = 10 cm

Bottom of the tank = 0.75 m

Based on the above information, first we have to determine the height which is

= 2 - 0.75 - 0.10

= 2 - 0.85

= 1.15 m

We assume the following things

1. Compressible flow

2. Stream line followed

Now applied the Bernoulli equation to section 1 and 2

So we get

$\frac{P_1}{p_g} + \frac{v_1^2}{2g} + z_1 = \frac{P_2}{p_g} + \frac{v_2^2}{2g} + z_2$

where,

P_1 = P_2 = hydrostatic

z_1 = 0

z_2 = h

Now

$\frac{v_1^2}{2g} + 0 = \frac{v_2^2}{2g} + h\\\\V_2 < < V_1 or V_2 = 0\\\\Therefore\ \frac{v_1^2}{2g} = h\\\\v_1^2 = 2gh\\\\ v_1 = \sqrt{2gh} \\\\v_1 = \sqrt{2\times 9.8\times 1.15}$

= 4.7476 m/sec

= 4.75 m/s

6 0

3 years ago

Why are we not able to see immediate after we enter a dark hall from a lighter place.

Kipish [7]

Technically this is a Biology question;

The 'amount' we can see depends on how much light can get through our pupil to hit our retina.
When there is a lot of light the pupil is small; it doesn't need to be big to let a lot of light in.
When we move to a dark space there is much less light, so the pupil 'dilates' to let enough light so we can see properly.
The period in which one cant see is simply when the pupil hasn't had time to change shape yet so doesn't let in enough light.<span />

3 0

3 years ago