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3 years ago

CH, +20, → CO2 + 2H,0 CH, +20, are the of the above chemical reaction.

Physics

1 answer:

rjkz [21]3 years ago

7 0

These are the reactants because they are on the left side of the equation.

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A gas expands at a constant pressure of 3 atm from a volume of 0.02 cubic meters to 0.10 cubic meters. In the process it experie

Nonamiya [84]

a) $2.4\cdot 10^4 J$

For a gas transformation occuring at a constant pressure, the work done by the gas is given by

$W=p(V_f -V_i)$

where

p is the gas pressure

V_f is the final volume of the gas

V_i is the initial volume

For the gas in the problem,

$p=3 atm = 3\cdot 1.013\cdot 10^5 Pa = 3.039\cdot 10^5 Pa$ is the pressure

$V_i = 0.02 m^2$ is the initial volume

$V_f = 0.10 m^3$ is the final volume

Substituting,

$W=(3.039\cdot 10^5 Pa)(0.10 m^3-0.02m^2)=24312 J = 2.4\cdot 10^4 J$

b) $3.24\cdot 10^5 J$

The heat absorbed by the gas can be found by using the 1st law of thermodynamics:

$\Delta U = Q-W$

where

$\Delta U$ is the change in internal energy of the gas

Q is the heat absorbed

W is the work done

Here we have

$\Delta U = 3.0\cdot 10^5 J$

$W=2.4\cdot 10^4 J$

So we can solve the equation to find Q:

$Q=\Delta U + W = 3.0\cdot 10^5 J +2.4\cdot 10^4 J = 3.24\cdot 10^5 J$

And this process is an isobaric process (=at constant pressure).

8 0

3 years ago

A 62.0 kg sprinter starts a race with an acceleration of 1.44 m/s2. If the sprinter accelerates at that rate for 30 m, and then

Dominik [7]

Answer:

The time taken for the race is 17.20 s.

Explanation:

It is given in the problem that a 62.0 kg sprinter starts a race with an acceleration of 1.44 meter per second square.The initial speed of the sprinter is zero as it starts from the rest.

Calculate the final speed of the sprinter.

The expression for the equation of the motion is as follows;

$v^{2}-u^{2}= 2as$

Here, u is the initial speed, v is the final speed, a is the acceleration and s is the distance.

Put u= 0, s=30 m and $a= 1.44 ms^{-2}$ .

$v^{2}-(0)^{2}= 2(1.44)(30)$

$v= 9.3 m^{-1}$

Calculate time taken to cover 30 m distance.

The expression for the equation of motion is as follows;

$s=ut+\frac{(1)}{2}at^2$

Put u= 0, s=30 m and $a= 1.44 ms^{-2}$ .

$30=(0)t+\frac{(1)}{2}(1.44)t^2$

t=6.45 s

Calculate the time taken to complete his race.

T= t+t'

Here, t is the time taken to cover 30 m distance and t' is the time taken to cover 100 m distance.

$T=6.45+\frac{s'}{v}$

Put s= 30 m, $v= 9.3 m^{-1}$ and s'= 100 m.

$T=6.45+\frac{(100)}{9.3}$

T= 17.20 s

Therefore, the time taken for the race is 17.20 s.

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Should i make my pet shark bark like a dog woof woof

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Answer: YES !!

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Elements with high atomic numbers tend to have____.

loris [4]

More protons than electron

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Rahul and Sonia were playing with blocks and each of them made a train out of them. Both of them thought of measuring the length

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Answer:

They measured their train using their handspan

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