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storchak [24]
3 years ago
12

How did the ancient greek understanding of atomic particles likley influence dalton research?

Chemistry
2 answers:
GalinKa [24]3 years ago
7 0

Answer:

The Greeks called these particles atomos, meaning indivisible, and the modern word “atom” is derived from this term. Democritus proposed that different types and combinations of these particles were responsible for the various forms of matter.

Explanation:

hope it helps

Inessa [10]3 years ago
6 0

How did the ancient Greek view the atom?

The Greeks called these particles atomos, meaning indivisible, and the modern word “atom” is derived from this term

I don't know if this is right and not so don't get mad at me I hope it's right though

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How would you prepare 175 mL of a 0.350kmol/m^3 calcium nitrate solution
SVEN [57.7K]
0,35 kmol/m³ = 0,35 mol/dm³ = 0,35 mol/L
175 mL = 0,175 L
*-*-*-*-*-*-*-*-*-*-*-*
C = n/V
n = 0,35×0,175
n = 0,06125 mol

mCa(NO₃)₂: 40+(14×2)+(16×6) = 164 g/mol

1 mol --------- 164g
0,06125 ---- X
X = 10,045g

To prepare 175 mL of 0,35M solution, add 10,045g of calcium nitrate and add water to a volume of 175ml.
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Based on the three formulas shown, use one of them to solve for the purple yellow and red box and explain how you did it.
zysi [14]

P = 11.133 atm (purple)

T = -236.733 °C(yellow)

n = 0.174 mol(red)

<h3>Further explanation  </h3>

Some of the laws regarding gas, can apply to ideal gas (volume expansion does not occur when the gas is heated),:  

  • Boyle's law at constant T, P = 1 / V  
  • Charles's law, at constant P, V = T  
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So that the three laws can be combined into a single gas equation, the ideal gas equation  

In general, the gas equation can be written  

\large {\boxed {\bold {PV = nRT}}}

where  

P = pressure, atm  

V = volume, liter  

n = number of moles  

R = gas constant = 0.08206 L.atm / mol K  

T = temperature, Kelvin  

To choose the formula used, we refer to the data provided

Because the data provided are temperature, pressure, volume and moles, than we use the formula PV = nRT

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T= 10 +273.15 = 373.15 K

V=5.5 L

n=2 mol

\tt P=\dfrac{nRT}{V}\\\\P=\dfrac{2\times 0.08205\times 373.15}{5.5}\\\\P=11.133~atm

  • Yellow box

V=8.3 L

P=1.8 atm

n=5 mol

\tt T=\dfrac{PV}{nR}\\\\T=\dfrac{1.8\times 8.3}{5\times 0.08205}\\\\T=36.42~K=-236.733^oC

  • Red box

T = 12 + 273.15 = 285.15 K

V=3.4 L

P=1.2 atm

\tt n=\dfrac{PV}{RT}\\\\n=\dfrac{1.2\times 3.4}{0.08205\times 285.15}\\\\n=0.174~mol

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