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katrin2010 [14]
3 years ago
11

A certain satellite orbiting Earth has a speed of about 17,000 miles/hour. What is its approximate speed expressed to the correc

t number of significant figures in kilometers/second? One kilometer is about 0.62 mile, and there are 3,600 seconds in an hour.
Chemistry
1 answer:
algol [13]3 years ago
6 0

Answer:- The speed of the satellite is 7.6\frac{kilometer}{second} .

Solution:- It's a unit conversion problem that could be solved using dimensional analysis which is also known as train trail method.

The given speed of the satellite is 17000 miles per hour and we are asked to convert it to kilometers per second.

given:

1 Kilometer = 0.62 mile

1 hour = 3600 second

Let's use this given info and plug in the values to make the set up as:

17000\frac{mile}{hour}(\frac{1 kilometer}{0.62 mile})(\frac{1 hour}{3600 second})

= 7.616\frac{kilometer}{second}

We round the answer to two significant figures, since the given number(17000) has only two significant figures.

Hence, the speed of the satellite is 7.6\frac{kilometer}{second} .

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mass = 5.36 x 10²⁴ / 6.02 x 10²³

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The ideal gas heat capacity of nitrogen varies with temperature. It is given by:
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Answer:

A)  1059 J/mol

B)  17,920 J/mol

Explanation:

Given that:

Cp = 29.42 - (2.170*10^-3 ) T + (0.0582*10^-5 ) T2 + (1.305*10^-8 ) T3 – (0.823*10^-11) T4

R (constant) = 8.314

We know that:

C_p=C_v+R

We can determine C_v from above if we make C_v the subject of the formula as:

C_v=C_p-R

C_V = 29.42-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4-8.314

C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4

A).

The formula for calculating change in internal energy is given as:

dU=C_vdT

If we integrate above data into the equation; it implies that:

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4\,) du

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1= 1059J/mol

Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.

B).

If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.

then T = 273 K & T2 = 1073 K

∴

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})273/1+(5.82*10^{-7})1073/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1= 17,920 J/mol

3 0
3 years ago
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