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Alja [10]
3 years ago
14

Oxygen is denoted as 16O8. How many protons and neutrons does an oxygen atom have?

Chemistry
2 answers:
hichkok12 [17]3 years ago
6 0

Answer: The given element has 8 protons and 8 neutrons in it.

Explanation:

General representation in writing an element is given by: _{Z}^{A}\textrm{X}

where, X denotes the element

Z denotes the atomic number

A denotes the atomic mass

We know that:

Atomic number = Number of electrons = Number of protons

And, Atomic mass = Number of protons + Number of neutrons   ....(1)

In the given element: _8^{16}\textrm{O}

Z = 8 which means number of protons = 8

A = 16

Putting A = 16 in Equation 1, we get:

16 = 8 + Number of neutrons

Number of neutrons = 16 - 8 = 8

Hence, the given element has 8 protons and 8 neutrons in it.

MrRissso [65]3 years ago
4 0
<span>Oxygen-16 atomic number is 8 so it has 8 protons. Its atomic weight is 16 so 16-8 = 8 neutrons.

Hope I helped. :) </span>
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A sample of argon gas (molar mass 40 g) is at four times the absolute temperature of a sample of hydrogen gas (molar mass 2 g).
katen-ka-za [31]

Answer:

Ratio of Vrms of argon to Vrms of hydrogen = 0.316 : 1

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The root-mean-square speed measures the average speed of particles in a gas, and is given by the following formula:  

Vrms = \sqrt{3RT/M}

where R is molar gas constant = 8.3145 J/K.mol, T is temperature in kelvin, M is molar mass of gas in Kg/mol

For argon, M = 40/1000 Kg/mol = 0.04 Kg/mol, T = 4T , R = R

Vrms = √(3 * R *4T)/0.04 = √300RT

For hydrogen; M = 1/1000 Kg/mol = 0.001 Kg/mol, T = T, R = R

Vrms = √(3 * R *T)/0.001 = √3000RT

Ratio of Vrms of argon to that of hydrogen = √300RT / √3000RT = 0.316

Ratio of Vrms of argon to that of hydrogen = 0.316 : 1

4 0
4 years ago
What should be the mole fraction of O2 in the gas mixture the diver breathes in order to have the same partial pressure of oxyge
Nutka1998 [239]

The question is incomplete, here is the complete question:

A scuba diver is at a depth of 355 m, where the pressure is 36.5 atm.

What should be the mole fraction of O_2 in the gas mixture the diver breathes in order to have the same partial pressure of oxygen in his lungs as he would at sea level? Note that the mole fraction of oxygen at sea level is 0.209.

<u>Answer:</u> The mole fraction of oxygen in the gas mixture is 0.00573

<u>Explanation:</u>

To calculate the partial pressure, we use the equation given by Raoult's law, which is:

p_{A}=p_T\times \chi_{A}      ........(1)

where,

p_A = partial pressure of oxygen at sea level = ?

p_T = total pressure at sea level = 1.00 atm

\chi_A = mole fraction of oxygen at sea level = 0.209

Putting values in equation 1, we get:

p_{O_2}=1.00atm\times 0.209\\\\p_{O_2}=0.209atm

As, partial pressure of the oxygen in the diver's lungs is equal to the partial pressure of oxygen at sea level

We are given:

p_T=36.5atm\\p_{O_2}=0.209atm

Putting values in equation 1, we get:

0.209atm=36.5atm\times \chi_{O_2}\\\\\chi_{O_2}=\frac{0.209}{36.5}=0.00573

Hence, the mole fraction of oxygen in the gas mixture is 0.00573

3 0
3 years ago
A sample of water vapor at 105 C is cooled to 95 C.
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I beleive its A but not very sure though. If im wrong sorry. But if im right I hope it helps (:

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