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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:

\Delta v= 4.66\frac{m}{s}

Explanation:

In this case we have to use the Principle of conservation of Momentum:

<em>This principle says that in a system  the total momentum is constant if no external forces act in the system. The formula is:</em>

m_1v_1+m_2v_2=m_1u_1+m_2u_2

<em>Where:</em>

m_1: Mass of the first object.

m_2: Mass of the second object.

v_1: Initial velocity of the first object.

v_2: Initial velocity of the second object.

u_1: Final velocity of the first object.

u_2: Final velocity of the second object.

In <u>this problem</u> we have:

m_1=81kg\\m_2=40kg\\v_1_2=2.3\frac{m}{s}

u_1=0\frac{m}{s}

Observation: v_1_2: Is because the system has the same initial velocity.

First we have to find u_2,

m_1v_1+m_2v_2=m_1u_1+m_2u_2

We can rewrite it as:

(m_1+m_2)v_1_2=m_1u_1+m_2u_2

Replacing with the data:

(m_1+m_2)v_1_2=m_1u_1+m_2u_2\\\\(81kg+40kg)2.3\frac{m}{s}=81kg(0\frac{m}{s})+40kg(u_2)\\\\(121kg)2.3\frac{m}{s}=40kg(u_2)\\\\\frac{(121kg)2.3\frac{m}{s}}{40kg}=u_2\\\\\frac{278.3}{40}\frac{m}{s}=u_2\\\\6.96\frac{m}{s}=u_2

We found the final velocity of the cart, but the problem asks for the resulting change in the cart speed, this means:

\Delta v=u_2-v_2\\\Delta v=6.96\frac{m}{s}-2.3\frac{m}{s}\\\Delta v= 4.66\frac{m}{s}

Then, the resulting change in the cart speed is:

\Delta v= 4.66\frac{m}{s}

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