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sleet_krkn [62]
2 years ago
13

You have a meteorite sample and you decide to use the uranium-235/lead-207 system to date it. After analysis, you find that it h

as 22,500 atoms of 235U remaining, and 1,477,500 atoms of 207Pb that the 235U decayed to. What percentage of the original amount of 235U is still present?
Physics
1 answer:
storchak [24]2 years ago
5 0

Originally there must been

1,4775E6 + 2.25E4 = 147.75E4 + 2.25E4 = 150E4 present at start

% = 2.25 / 150 = 1.5 %      of 235 U left

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Observe yourself breathing and count the number of times you inhale per second. During each breath you probably inhale 0.66 L of
Pavel [41]

To solve this exercise it is necessary to apply the concepts related to Robert Boyle's law where:

PV=nRT

Where,

P = Pressure

V = Volume

T = Temperature

n = amount of substance

R = Ideal gas constant

We start by calculating the volume of inhaled O_2 for it:

V = 21\% * 0.66L

V = 0.1386L

Our values are given as

P = 1atm

T=293K R = 0.083145kJ*mol^{-1}K^{-1}

Using the equation to find n, we have:

PV=nRT

n = \frac{PV}{RT}

n = \frac{(1)(0.1386)}{(0.0821)(293)}

n = 5.761*10^{-3}mol

Number of molecules would be found through Avogadro number, then

\#Molecules = 5.761*10^{-3}*6.022*10^{23}

\#Molecules = 3.469*10^{21} molecules

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3 years ago
When you ask you friend for hoemwork answers and he answers like a year later hahahah
Hitman42 [59]
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5 0
3 years ago
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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)

<u>FOR HYDROGEN:</u>

v = √(3KT)/m = 2000 m/s  _____ eqn (2)

<u>FOR OXYGEN:</u>

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)

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

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

<u>K.E of Oxygen = 4000 J</u>

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)

<u>v = 2149.24 m/s</u>

<u></u>

8 0
3 years ago
What factor affects weight( what force causes weight to change)
Alex Ar [27]

Answer:

Gravitational force affects weight, weight changes with change in gravity

Hope it helped u,

mark as the brainliest

^_^

3 0
3 years ago
How would decreasing the volume of the reaction vessel affect each of the following equilibria?2NOBr(g)⇌2NO(g)+Br2(g)
mafiozo [28]

Answer:

The equilibrium position will shift towards the left hand side or reactants side

Explanation:

Decreasing the volume (increasing the pressure) of the system will shift the equilibrium position towards the lefthand side or reactants side. This is because, decreasing the volume (increasing the pressure) implies shifting the equilibrium position towards the side having the least number of moles.

There are two moles of reactants and a total of three moles of products(total). Hence decreasing the volume and increasing the pressure of the gas phase reaction will shift the equilibrium position towards the lefthand side.

5 0
3 years ago
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