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sattari [20]
3 years ago
13

What is the molar mass of an ideal gas if a 0.800 g sample of this gas occupies a volume of 200. mL at 50.0 oC and 720. mm Hg?

Physics
1 answer:
Paladinen [302]3 years ago
7 0
PV=nRT
(720/760)(0.200)=(0.800/x)(0.08206)(323.15)
(0.1894736842)=(0.800/x)(0.08206)(323.15)
.0071451809=(0.800/x)
x=MM=111.9635758 g/mol
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Answer in detail? Describe an activity that gives an understanding of ‘Electrostatic force’.
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8 0
3 years ago
A 0.75 kg model car is moving west at a speed of 9.0 m/s when it collides head-on with a 2.00 kg model truck that is traveling e
Lyrx [107]

Answer:

  2.5 m/s east

Explanation:

Let east be the positive direction for velocity.

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  m1·v2 -m1·v1 = (0.75 kg)(11 m/s) -(0.75 kg)(-9 m/s)

  = (0.75 kg)(20 m/s) = 15 kg·m/s

The change in momentum of the 2.0 kg model car is the opposite of this, so the total change in momentum is zero.

  m2·v2 -m2·v1 = (2 kg)(v2 m/s) -(2 kg)(10 m/s) = 2(v2 -10) kg·m/s

The required relation is ...

  15 kg·m/s = -2(v2 -10) kg·m/s

  -7.5 = v2 -10 . . . . divide by -2

  2.5 = v2 . . . . . . . add 10

The velocity of the model truck after the collision is 2.5 m/s east.

3 0
2 years ago
How can you prove that acceleration is a derived unit​
Aliun [14]

a = (dx / dt)²

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7 0
3 years ago
Read 2 more answers
Help with these three
Elenna [48]
The first: alright, first: you draw the person in the elevator, then draw a red arrow, pointing downwards, beginning from his center of mass. This arrow is representing the gravitational force, Fg.
You can always calculate this right away, if you know his mass, by multiplying his weight in kg by the gravitational constant
g = 9.81 \frac{m}{s {}^{2} }
let's do it for this case:
f_{g}  = m \times g \\ f _{g}  = 65kg \times 9.81 \frac{m}{s {}^{2} }  = 637.65
the unit of your fg will be in Newton [N]
so, first step solved, Fg is 637.65N
Fg is a field force by the way, and at the same time, the elevator is pushing up on him with 637.65N, so you draw another arrow pointing upwards, ending at the tip of the downwards arrow.
now let's calculate the force of the elevator
f = m \times a \\ f = 65 \times 5 \frac{m}{s {}^{2} }  \\ f = 325n
so you draw another arrow which is pointing downwards on him, because the elevator is accelating him upwards, making him heavier
the elevator force in this case is a contact force, because it only comes to existence while the two are touching, while Fg is the same everywhere
8 0
3 years ago
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