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Natasha_Volkova [10]
3 years ago
5

Liquid water can be separated into hydrogen gas and oxygen gas through electrolysis. 1 mole of hydrogen gas and 0.5 moles of

Physics
1 answer:
den301095 [7]3 years ago
8 0

The temperature of the oxygen gas is  243.75 K.

Using ideal gas law to explain the answer, the absolute temperature of the gas will decrease if the number of moles of the gas increases and it will increase if the volume and/or pressure of the gas increases.

The reaction of the given elements;

H_2 \ + \ \frac{1}{2} O_2 \ --->\ \ H_2O

volume of the collected oxygen gas, V = 10 L

pressure of the gas, P = 1 atm

number of moles of the gas, n = 0.5

Using ideal law the temperature of the oxygen gas is calculated as follows;

PV = nRT\\\\T = \frac{PV}{nR} \\\\where;\\R \ is \ the \ ideal \ gas \ constant = 0.08205 \ L.atm/K.mol\\\\T = \frac{1 \times 10 }{0.5 \times 0.08205} \\\\T = 243.75 \ K

Thus, the temperature of the gas is 243.75 K.

Using ideal gas law to explain the answer. The absolute temperature of the oxygen gas is directly proportional to the product of its pressure and volume and inversely proportional to its number of moles. That is the absolute temperature of the gas will decrease if the number of moles of the gas increases and it will increase if the volume and/or pressure increases.

Learn more here: brainly.com/question/16617695

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Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

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R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0

where R_1,R_2 are the magnitudes of the tensions in ropes 1 and 2, respectively;

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R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0

where m=1300\,\rm kg and g=9.8\frac{\rm m}{\mathrm s^2}.

Eliminating R_2, we have

\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)

R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2

R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2

-R_1 \sin(50^\circ) = -\dfrac{mg}2

R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}

Solve for R_2.

\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0

\dfrac{R_2}2 = -mg\cot(110^\circ)

R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}

8 0
1 year ago
A 90 kg body is taken to a planet where the acceleration due to
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Answer:

2250N

Explanation:

W= mg,

where W= weight

m= mass

g= acceleration due to gravity

Given that the body is 90kg, m= 90kg.

Acceleration due to gravity of planet

= 2.5(10)

= 25 m/s²

Weight of body on planet

= 90(25)

= 2250N

*Mass is the amount of matter an object has and is constant (same on earth and the planet).

6 0
3 years ago
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0.5 kg air hockey puck is initially at rest. What will it’s kinetic energy be after a net force of .8 N acts on it for a distanc
weqwewe [10]

Answer:

1.6 J

Explanation:

Work = change in energy

W = ΔKE

Fd = KE

(0.8 N) (2 m) = KE

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3 years ago
A student sits at rest on a piano stool that can rotate without friction. The moment of inertia of the student-stool system is 4
irina [24]

Here We can use principle of angular momentum conservation

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Since there is no torque so we can say the angular momentum is conserved

mvL = (I + mL^2)\omega

now we know that

m = 2 kg

v = 2.5 m/s

L = 0.35 m

I = 4.5 kg-m^2

now plug in all values in above equation

2\times 2.5 \times 0.35 = (4.5 + (2\times 0.35^2))\omega

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1.75 = 4.745\omega

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so the final angular speed will be 0.37 rad/s

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Answer:

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