Which of the following statements is true? *

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at

denis-greek [22]

Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

Explanation:

A)

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

T = Absolute Temperature

but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m _____ eqn (1)

FOR HYDROGEN:

v = √(3KT)/m = 2000 m/s _____ eqn (2)

FOR OXYGEN:

velocity of oxygen = √(3KT)/(mass of oxygen)

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

velocity of oxygen = (1/4)(2000 m/s) = 500 m/s

B)

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

K.E of Oxygen = 4000 J

C)

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

v = 2149.24 m/s

8 0

3 years ago

Suppose the Sun moved twice the distance from Earth. Describe immediate and future changes to our planet. Justify your reasoning

Reptile [31]

Our year would now be 2.8 times longer, we would also be receiving only 1/4 of the energy from the sun that we currently do. This means that we’d now be out beyond the orbit of Mars and right at the edge of the asteroid belt, and things would rapidly get very cold with temperatures expected to drop by around 50 degrees Celsius on average, and that’s with our current atmospheric composition which would not be stable in the new conditions. And also, any living thing on earth would die.

5 0

2 years ago

An Na ion has a charge of +1 and an O ion has a charge of –2. How many Na ions does one O ion need to balance charges

Aleonysh [2.5K]

Because Na⁺ ion lacks an electron and O²⁻ has two extra electrons extra, to balance the charge, we need 2 Na⁺ ions.

All ions, atoms and molecules want to get to the minimum energy state, and that state is when the ion, atom, or molecule is neutral, that's why all of them want to balance their charges.

7 0

3 years ago

Read 2 more answers

A 6.8 kg bowling ball and 7.4 kg bowling ball rest on a rack 0.74 m apart. What is the force of gravity pulling each ball toward

zimovet [89]

The gravitational force between the two balls is $6.13\cdot 10^{-9} N$

Explanation:

The magnitude of the gravitational force between two objects is given by:

$F=G\frac{m_1 m_2}{r^2}$

where :

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between the objects

For the balls in this problem, we have

$m_1 = 6.8 kg$

$m_2 = 7.4 kg$

r = 0.74 m

Substituting into the equation, we find the gravitational force between the two balls:

$F=(6.67\cdot 10^{-11})\frac{(6.8)(7.4)}{(0.74)^2}=6.13\cdot 10^{-9}N$

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

4 0

3 years ago

When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa

MA_775_DIABLO [31]

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

$\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2$

In parallel

$\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}$

$\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0$

Solving the above quadratic equation

$\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}$

$\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega$

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

6 0

3 years ago