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3 years ago

The person to introduce the idea of an "atom" was _____.

Chemistry

2 answers:

Alexus [3.1K]3 years ago

8 0

Answer:

The correct answer is Democritus.

Explanation:

Democritus, a Greek philosopher, and scientist in 440 B. C. proposed the theory of the atom. Democritus made his theory of atoms depend on the work of previous philosophers according to different scientists.

His explanation of the atom begins a stone cut in half gives two halves. If stone again and again continuously cut, at some point the stone becomes so small that it could no longer be cut, that atoms present separately from each other.

Thus, the correct answer is Democritus.

nadezda [96]3 years ago

5 0

The answer is Democritus. He first started to propose the idea in the 5th century B.C.

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A 25.0 liter rigid container has a mixture of 32.00 grams of oxygen gas and 1 point

Ksivusya [100]

Answer:

$P_{T}$ = 2.94 atm

Explanation:

The total pressure ( $P_{T}$ ) in the container is given by:

$P_{T} = P_{O_{2}} + P_{He}$

The pressure of the oxygen ( $P_{O_{2}}$ ) and the pressure of the helium ( $P_{He}$ ) can be calculated using the ideal gas law:

$PV = nRT$

<u>Where</u>:

V: is the volume = 25.0 L

n: is the number of moles of the gases

R: is the gas constant = 0.082 Latm/(Kmol)

T: is the temperature = 298 K

First, we need to find the number of moles of the oxygen and the helium:

$n_{O_{2}} = \frac{m}{M}$

Where m is the mass of the gas and M is the molar mass

$n_{O_{2}} = \frac{32.00 g}{31.998 g/mol} = 1.00 moles$

And the number of moles of helium is:

$n_{He} = \frac{8.00 g}{4.0026 g/mol} = 2.00 moles$

Now, we can find the pressure of the oxygen and the pressure of the helium:

$P_{O_{2}} = \frac{nRT}{V} = \frac{1.00 moles*0.082 Latm/(Kmol)*298 K}{25.0 L} = 0.98 atm$

$P_{He} = \frac{nRT}{V} = \frac{2.00 moles*0.082 Latm/(Kmol)*298 K}{25.0 L} = 1.96 atm$

Finally, the total pressure in the container is:

$P_{T} = P_{O_{2}} + P_{He} = 0.98 atm + 1.96 atm = 2.94 atm$

Therefore, the total pressure in the container is 2.94 atm.

I hope it helps you!

6 0

4 years ago

The reaction below has an equilibrium constant kp=2.2×106 at 298 k. 2cof2(g)⇌co2(g)+cf4(g) you may want to reference (pages 680

OLEGan [10]

<span>Kp is the equilibrium pressure constant calculated from the partial pressures of a reaction equation.
Kp =[pCF4]*[p CO2] / [p COF2]^2 = 2.2 x 10^6
When the mole fraction gets doubled we have
Kp = [pCO2]^2*[pCF4]^2 / [pCOF2]^4
Kp = [[pCF4]*[p CO2] / [p COF2]^2] * 2
Kp = (2.2 * 10^6) * 2
Kp = 4.8 * 10^12</span>

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3 years ago