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kakasveta [241]
3 years ago
8

What are the approximate boiling points for the c2, c4, c6, and c8 alkanes?

Physics
1 answer:
lakkis [162]3 years ago
5 0

Alkanes are hydrocarbons with general formula C_nH_{2n+2}.

So alkane with two carbon atoms will be C_2H_6, alkane with four carbon atoms will be C_4H_{10}, alkane with six carbon atoms will be C_6H_{14}, alkane with eight carbon atoms will be C_8H_{18}.

As the number of carbon atoms increase, the surface area will increase and thus the vanderwaal forces will also increase, and the boiling point will also increase.

Thus the approximate boiling point order is: C_8>C_6>C_4>C_2

The approximate boiling points will be: 125° C >68° C >  -1 °C > -89° C

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A 5.0-kg block of wood is placed on a 2.0-kg aluminum frying pan. How much heat is required to raise the temperature of both the
Shalnov [3]

Heat required to raise the temperature of a given system is

Q = ms\Delta T

here we know that

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s = specific heat capacity

\Delta T = change in temperature

now as we know that

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4 0
3 years ago
An astronaut lands on an alien planet. He places a pendulum (L = 0.200 m) on the surface and sets it in simple harmonic motion,
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5 0
3 years ago
Suppose a police officer is 1/2 mile south of an intersection, driving north towards the intersection at 40 mph. At the same tim
blagie [28]

Answer:

75.36 mph

Explanation:

The distance between the other car and the intersection is,

x=x_{0}+V t \\ x=\frac{1}{2}+V t

The distance between the police car and the intersection is,

y=y_{0}+V t

y=\frac{1}{2}-40 t

(Negative sign indicates that he is moving towards the intersection)

Therefore the distance between them is given by,

z^{2}=x^{2}+y^{2}(\text { Using Phythogorous theorem })

z^{2}=\left(\frac{1}{2}+V t\right)^{2}+\left(\frac{1}{2}-40 t\right)^{2} \ldots \ldots \ldots(1)

The rate of change is,

2 z \frac{d z}{d t}=2\left(\frac{1}{2}+V t\right) V+2\left(\frac{1}{2}-40 t\right)(-40)

2 z \frac{d z}{d t}=V+2 V^{2} t-40+3200 t \ldots \ldots \ldots

Now finding z when t=0, from (1) we have

z^{2}=\left(\frac{1}{2}+V(0)\right)^{2}+\left(\frac{1}{2}-40(0)\right)^{2}

z^{2}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} \\ z=\sqrt{\frac{1}{2}} \approx 0.7071

The officer's radar gun indicates 25 mph pointed at the other car then, \frac{d z}{d t}=25 when t=0, from

From (2) we get

2(0.7071)(25)=V+2 V^{2}(0)-40+3200(0)

2(0.7071)(25)=V+2 V^{2}(0)-40

35.36=V-40

V=35.36+40=75.36

Hence the speed of the car is 75.36 mph

7 0
3 years ago
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