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Natasha2012 [34]
3 years ago
5

A 250 mL sample of nitrogen(N2) has a pressure of 745 mm Hg at 30 C.What is the mass of nitrogen?

Physics
1 answer:
Lera25 [3.4K]3 years ago
3 0
We assume that nitrogen gas sample here is an ideal gas. So that, we can use the ideal gas equation which is expressed as follows:

PV = nRT

First, we calculate the number of moles from the equation above. Then, use molar mass to calculate mass.

n = PV/RT
n = (745/760) (0.250) / 0.08206 (30+273.15)
n = 0.01 mol 

m = 0.01 ( 28) = 0.28 g
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Answer:

Explanation:

Initial kinetic energy of M = 1/2 M vi²

let final velocity be vf

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vf² =  vi² + 2 (F / M) x D

Kinetic energy

= 1/2 Mvf²

= 1/2 M ( vi² + 2 (F / M) x D

1/2 M vi² + FD

Ratio with initial value

1/2 M  vi² + FD) / 1/2 M  vi²

RK = 1 + FD / 2 M  vi²

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Why silicon have large forward current as compared to germanium?​
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Answer:

The structure of Germanium crystals will be destroyed at higher temperature. However, Silicon crystals are not easily damaged by excess heat. Peak Inverse Voltage ratings of Silicon diodes are greater than Germanium diodes. Si is less expensive due to the greater abundance of element.

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George and Harriot walk with an average velocity of .95 m/s eastward. If it takes them 30
artcher [175]

Answer:

1.71 km

Explanation:

Convert 30 minutes to seconds:

30 min × (60 s / min) = 1800 s

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0.95 m/s × 1800 s = 1710 m

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After being struck by a bowling ball, a 1.3 kg bowling pin sliding to the right at 5.0 m/s collides head-on with another 1.3 kg
GuDViN [60]

Answer:

a) 4.2m/s

b) 5.0m/s

Explanation:

This problem is solved using the principle of conservation of linear momentum which states that in a closed system of colliding bodies, the sum of the total momenta before collision is equal to the sum of the total momenta after collision.

The problem is also an illustration of elastic collision where there is no loss in kinetic energy.

Equation (1) is a mathematical representation of the the principle of conservation of linear momentum for two colliding bodies of masses m_1 and m_2 whose respective velocities before collision are u_1 and u_2;

m_1u_1+m_2u_2=m_1v_1+m_2v_2..............(1)

where v_1 and v_2 are their respective velocities after collision.

Given;

m_1=1.3kg\\u_1=5m/s\\m_2=1.3kg\\u_2=0m/s

Note that u_2=0 because the second mass m_2 was at rest before the collision.

Also, since the two masses are equal, we can say that m_1=m_2=m so that equation (1) is reduced as follows;

mu_1+mu_2=mv_1+mv_2\\\\m(u_1+u_2)=m(v_1+v_2)..............(2)

m cancels out of both sides of equation (2), and we obtain the following;

u_1+u_2=v_1+v_2.............(3)

a) When v_1=0.8m/s, we obtain the following by equation(3)

5+0=0.8+v_2\\hence\\v_2=5-0.8\\v_2=4.2m/s

b) As m_1 stops moving v_1=0, therefore,

5+0=0+v_2\\v_2=5m/s

5 0
3 years ago
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