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Anettt [7]
3 years ago
15

Calculate the root-mean-square velocity for a hydrogen molecule at

Chemistry
1 answer:
Y_Kistochka [10]3 years ago
3 0

Answer:

56.5 m/s

Explanation:

Step 1: Convert the temperature to Kelvin

We will use the following expression.

K = °C + 273.15

K = -15.0°C + 273.15 = 258.2 K

Step 2: Calculate the root-mean-square velocity

The root-mean-square velocity is the square root of the average square velocity. We can calculate it using the following expression.

v_{rms} = \sqrt{\frac{3RT}{M} }

where,

  • R: ideal gas constant
  • T: absolute temperature
  • M: molar mass of the gas

v_{rms} = \sqrt{\frac{3 \times \frac{8.314J}{mol.K} \times 258.2K }{2.02g/mol} } = 56.5m/s

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Calculate the density of helium if a balloon with a capacity of 5.00 L holds 0.890 g.
liubo4ka [24]

Answer:

Explanation:

Density is m/V. Also, 1 liter = 1000 cm^3. So, we get 0.890/(5*1000) = 1.78*10^{-4} g/cm^3. You can convert this to kg/m^3 as well by multiplying it by 10. Depends which one you want.

5 0
3 years ago
An alloy is a mixture of what?
Nastasia [14]

Answer:

An alloy is a mixture of chemical elements, which form an impure substance that retains the characteristics of a metal.

Explanation:

Have a nice day. Auf Wiedersehen.

5 0
4 years ago
Which of the following elements makes up the tissues and organs of animals? A.Carbon B.Hydrogen C. Nitrogen D.Oxygen
telo118 [61]
Hydrogen if im wrong by texting me 

5 0
4 years ago
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After a reaction, a new compound contains 0.73 g Mg and 0.28 g N. What is the empirical formula of this compound?
FromTheMoon [43]

Answer:

Mg₃N₂

Explanation:

The empirical formula of a chemical compound is defined as the simplest positive integer ratio of atoms present in a compound. Using molecular mass of Mg (24,305g/mol) and mass of nitrogen (14,006g/mol), moles of each element are:

0,73g × (1mol / 24,305g) = 0,03 moles of Mg

0,28g × (1mol / 14,006g) = 0,02 moles of N

Dividing each value in 0,01 to obtain natural numbers:

0,03 moles of Mg / 0,01 = 3

0,02 moles of N / 0,01 = 2.

Thus, empirical formula is: <em>Mg₃N₂</em>

<em></em>

I hope it helps!

5 0
3 years ago
If a single gold atom has a diameter of 2.9×x10−8 cm, how many atoms thick was rutherford's foil
ale4655 [162]

27,586

<h3>Further explanation</h3>

<u>Given:</u>

A single gold atom has a diameter of \boxed{ \ 2.9 \times 10^{-8} \ cm. \ }

From a reference, the Rutherford gold foil used in his scattering experiment had a thickness of approximately \boxed{ \ 8 \times 10^{-3} \ mm. \ }

<u>Question:</u>

How many atoms thick were Rutherford's foil?

<u>The Process:</u>

Convert thickness from mm to cm.

\boxed{ \ 8 \times 10^{-3} \ mm = 8 \times 10^{-3} \times 10{-1} \ cm \ } \rightarrow \boxed{ \ 8 \times 10^{-4} \ cm \}

The number of atoms is calculated from gold foil thickness divided by the atomic diameter.

\boxed{ \ = \frac{8 \times 10^{-4} \ cm}{2.9 \times 10^{-8} \ cm} \ }

\boxed{ \ =2.7586 \times 10^4 \ atoms \ }

Therefore, we get an atomic thickness of 27,586 atoms.

<u>Notes:</u>

  • In 1909-1910, Ernest Rutherford with two of his assistants, namely Hans Geiger and Ernest Marsden, conducted a series of experiments to find out more about the arrangement of atoms. They fired at a very thin gold plate with high-energy alpha particles.
  • One of their observations is that a small portion of alpha particles are reflected. This greatly surprised Rutherford. The reflected alpha particle must have hit something very dense in the atom. This fact is incompatible with the atomic model proposed by J.J. Thomson where the atoms are described as homogeneous in all parts with electrons and protons evenly distributed.
  • In 1911, Rutherford was able to explain the scattering of alpha rays by proposing ideas about atomic nuclei. According to him, most of the mass and positive charge of atoms are concentrated at the center of the atom, hereinafter referred to as the nucleus.
<h3>Learn more</h3>
  1. The energy density of the stored energy  brainly.com/question/9617400
  2. The theoretical density of platinum which has the FCC crystal structure. brainly.com/question/5048216
  3. Compound microscope brainly.com/question/4000241

Keywords: if a single gold atom, has a diameter of 2.9 x 10⁻⁸ cm, how many, atoms thick, Rutherford's foil, his scattering experiment

7 0
3 years ago
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