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Crank
3 years ago
5

Determine the molar mass of Cao​

Chemistry
1 answer:
e-lub [12.9K]3 years ago
6 0

Answer:

Molar mass of CaO

(40 + 16)g

= 56g

hope it helps you

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no

Explanation:

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Calcuate the number of<br> grams of solute in 453.9mL<br> of 0.237 M calcium acetate
AysviL [449]

The number of  grams : 17.082 g

<h3>Further explanation</h3>

Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

\large {\boxed {\bold {M ~ = ~ \dfrac {n} {V}}}

Where

M = Molarity

n = Number of moles of solute

V = Volume of solution

453.9 mL  of 0.237 M calcium acetate

  • mol

\tt mol=M\times V=0.237\times 0.4539=0.108

  • mass

MW Ca(C₂H₃OO)₂ : 158,17 g/mol

\tt mass=mol\times MW\\\\mass=0.108\times 158.17=17.082~g

4 0
2 years ago
According to the kinetic-molecular theory, the average kinetic energy of gas
vladimir1956 [14]

Answer:

The average kinetic energy of a gas depends only on its temperature.

Explanation:

The average kinetic energy of particles in a gas can be found using the equation

\displaystyle \text{Average K.E.} = \frac{3}{2}k\cdot T,

where

  • k is the Stefan-Boltzmann constant, and
  • T is the absolute temperature of this gas (the one in degree Kelvins.)

As seen in this equation, the average kinetic energy of particles in a gas depends only on the temperature of the gas. Also, since the question is asking for the average not the total kinetic energy, the number of particles in this gas doesn't matter, either.

8 0
3 years ago
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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

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