The driving thrust of the car produced by the engine is the main forward force.
The main opposing forces are air resistance (from the wind) and friction (between the tyres and the road)
Since the air resistance + friction = driving force the car moves at a constant speed.
Mass of molecule (g) = Mr of substance over avarogado constant
Balance each one by adding electrons to make the charges on both sides the same:
Sn--> Sn2+ + 2 e-
Ag+ + 1 e- --> Ag
Now, you have to have the same number of electrons in the two half-reactions, so multiply the second one by 2 to get:
2 Ag+ + 2 e- --> 2 Ag
Now, just add the two half reactions together, cancelling anything that's the same on both sides:
2 Ag+ + Sn --> Sn2+ + 2 Ag
And you're done.
Answer:
<h2>464.85 mL</h2>
Explanation:
The new volume can be found by using the formula for Boyle's law which is

Since we're finding the new volume

100.7 kPa = 100,700 Pa
95.1 kPa = 95,100 Pa
We have

We have the final answer as
<h3>464.85 mL</h3>
Hope this helps you
Answer:
28.16 g/mol
Explanation:
From Graham's law;
Let the rate of diffusion of gas X be 1.25
Let the rate of diffusion of CO2 be 1
Molecular mass of gas X= M
Molecular mass of CO2 = 44g/mol
1.25/1=√44/M
(1.25/1)^2 = 44/M
1.5625 = 44/M
M= 44/1.5625
M= 28.16 g/mol