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3 years ago

What is the molar mass of a gas if 1.30g of the gas has a volume of 245mL at STP? ...?

1 answer:

5 0

First, we assume this as an ideal gas so we use the equation PV=nRT. Then, we use the conditions at STP that would be 1 atm and 273.15 K. We calculate as follows:

PV= nRT
PV= mRT/MM

1 atm (.245 L) =1.30(0.08206)(273.15) / MM
MM = 118.94 g/mol <--- ANSWER

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3 years ago

Which of the following prefixes represents the biggest number?

Komok [63]

<span>kilo
hecto
deco
deci
centi
milli

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4 years ago

suppose that the man pictured on the front side is orbiting the earth (mass=5.98 x 10^24kg at a distance of 310 miles above the

wlad13 [49]

Answer:

a = 9.81[m/s^2]; v = 18683.5[m/s]

Explanation:

The stament of the problem is:

Suppose that the man pictured on the front side is orbiting the earth (mass = 5.98 x 1024kg) at a distance of 310 miles (1600 meters = 1 mile) above the surface of the earth (radius = 4000 miles).

a. What acceleration does he experience due to the earth's pull?

b. What tangential velocity must he possess in order that he orbit safely (in m/s)?

First we need to convert all the initial data to units of the SI

Rs = 310 [mil] = 498897 [m]

RT= 400 [mil] = 643738 [m]

r = Rs + RT = 1142635 [m] "distance from the center of the earth to the man"

$F=G*\frac{M*m}{r^{2} } \\where:\\$

G = universal gravitational constant

M = mass of the earth [kg]

m = mass of the man [kg]

r = distance from the center of the earth to the man [m]

The acceleration he is experimenting is the same acceleration given by the gravity, therefore:

a = g = 9.81[m/s^2]

To find the tangential velocity, we must determinate the force exerted by the earth.

Now we will find the force exerted by the gravity when the man is orbiting the earth at distance r.

$G = 6.67*10^{-11} [\frac{N*m^{2} }{kg^{2} } ]\\M=5.98*10^{24}[kg]\\ m=100 [kg]\\Replacing:\\F = G*\frac{M*m}{r^{2} }$

$F = 6.67*10^{-11}*\frac{5.98*10^{24}*100 }{1142635^{2} } \\ F= 30550 [N]$

And this force will be equal to the following expression:

$F = m*\frac{v^{2} }{r} \\where:\\v= tangential velocity [m/s]\\v=\sqrt{\frac{F*r}{m} } \\v=\sqrt{\frac{30550*1142635}{100} } \\v=18683.5[m/s]$

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3 years ago

When you approach a light source that in turn is moving towards you, your speed relative to the emitted light waves _____.?

Natalka [10]

This is a trick question:

The Doppler effect states that as you move closer to the source, the frequency of light(or sound/waves in general) increases, but technically the speed of light is always the same speed, even if you are moving at the speed of light.

Thus, the answer would be something along the lines of <u>don't change</u>.

7 0

3 years ago

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A helicopter’s speed increases from 30 m/s to 40 m/s in 5 seconds.

TEA [102]

Acceleration is defined as the rate of change of velocity, which, simply put, is a mouthful to describe how fast something speeds up, slows down, or turns. The equation for acceleration is

a = Δv / Δt,

or your final velocity - your starting velocity, then divided by the amount of time. It can also be expressed as

a = (Vf - Vi) / t,

Where Vf is your final velocity, Vi is your initial velocity, and t is the time traveled.

The question gives us that the helicopter moves from a starting velocity of 30 m/s to a final velocity of 40 m/s in the span of 5 seconds. This means we can fill in the variables to the equation, where

Vf = 40,
Vi = 30, and
t = 5.

Plug these known variables into the original equation, and we get

a = (Vf - Vi) / t = (40 - 30) / 5.

From here, the answer comes down to 10 / 5, or 2 m/s^2.

Hope this helps! If you have any questions, don't hesitate to ask :D

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4 years ago

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