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3 years ago

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Calculate the volume occupied by 25.2 g of co2 at 0.84 atm and 25°c. r = 0.08206 latm/kmol.

1 answer:

Nitella [24]3 years ago

5 0

Answer:

16.68 L

Explanation:

First of all, we need to calculate the number of moles corresponding to 25.2 g of CO2.

The molar mass of CO2 is 44.01 g/mol, therefore the number of moles is:

$n=\frac{m}{M_m}=\frac{25.2 g}{44.01 g/mol}=0.573 mol$

So now we can use the ideal gas equation:

$pV=nRT$

where:

p = 0.84 atm is the gas pressure

V = ? is the volume

n = 0.573 mol is the number of moles

R = 0.08206 L atm / K mol is the gas constant

T = 25°c = 298 K is the gas temperature

Substituting into the equation and re-arranging it, we find the volume:

$V=\frac{nRT}{p}=\frac{(0.573 mol)(0.08206 Latm/Kmol)(298 K)}{0.84 atm}=16.68 L$

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yanalaym [24]

Answer:

$V=17.9\ Volt$

Explanation:

<u>Joule's Law in Electricity</u>

The Joule's law allows us to calculate the power dissipated in a resistor of resistance R through which goes a current I.

$P=I^2R$

The relation between the voltage and the current is given by Ohm's law:

$V=RI$

Solving for I and replacing int the first equation

$\displaystyle P=\frac{V^2}{R}$

Solving for V

$V=\sqrt{PR}$

$V=\sqrt{0.8\cdot 400}=17.9$

$\boxed{V=17.9\ Volt}$

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3 years ago

How much work would a child do while puling a 12-kg wagon a distance of 3m with a 22 N force directed 30 degrees with respect to

Alika [10]

Answer:

The work done will be 57.15 J

Explanation:

Given that,

Mass = 12 kg

Distance = 3 m

Force = 22 N

Angle = 30°

We need to calculate the work done

The work done is defined as,

$W = Fd\cos\theta$

Where, F = force

d = displacement

Put the value into the formula

$W=22\times3\times\cos30^{\circ}$

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$W = 57.15\ J$

Hence, The work done will be 57.15 J

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3 years ago

Under certain conditions, it is possible to get more work out of a machine than you put into it.

Iteru [2.4K]

IF that were true, then you could take some of the work from the output,
put it back into the input to keep the machine running, and use the rest
to run the lights and the hot water heater in your house. That way, you'd
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You NEVER get more work out of a machine than you put into it.
Never never never.

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3 years ago

A 60kg60 kg board that is 6 m6 m long is placed at the edge of a platform, with 4 m4 m of its length extending over the edge. Th

scoundrel [369]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The minimum mass of $M_1 = 90\ kg$ correct option is E

Explanation:

Free body diagram of the set up in the question is shown on the third uploaded image

The mass of board is $M = 60kg$

The length of the board is $L = 6 \ m$

The length extending over the edge is $L_e = 4 \ m$

The second mass is $M_2 = 30kg$

Now to obtain $M_1$ we take moment about the edge of the platform

$M_1 g L_1 = Mg \frac{L}{2} + M_2 g L_2$

$M_1 L_1 = M \frac{L}{2} + M_2 L_2$

Substituting value

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8 0

4 years ago

Cyclist A is moving at 20.0 m/s whereas cyclist B is moving at 12.0 m/s in the same direction and is initially ahead ofA. When t

stealth61 [152]

Answer:

$v_{fA} = 28 \frac{m}{s}$

Explanation:

The kinematic parameters from the moment the two cyclists begin to accelerate until they meet are:

Initial parameters:

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$v_{oB} = 12 m/s$ : Initial speed of cyclist B

Final parameters:

$v_{fB} = 20 m/s$ : Final speed of cyclist B

$d_{A} = d_{B}$

distance of cyclist A = distance of cyclist B

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time of cyclist A = time of cyclist B

Cyclist B Kinematics

$v_{fB} = v_{oB} +a_{B} *t\\$

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$a_{B} = 2 \frac{m}{s^{2} }$

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$a_{A} = \frac{288-240}{72}$

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$v_{fA} = 20 + 0.67*12$

$v_{fA} = 28 \frac{m}{s}$

7 0

3 years ago