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Vaselesa [24]
3 years ago
5

Which scientist was able to quantify the electron's charge and estimate its specific mass? A. J. J. Thomson B. Robert Millikan C

. Eugen Goldstein D. Ernest Rutherford
Chemistry
2 answers:
Katarina [22]3 years ago
8 0

Answer:

B. Robert Millikan

Explanation:

J.J. Thompson gave the plum pudding model of the atom.  

Millikan determined the charge and mass of an electron. Millikan’s oil drop experiment was conducted to measure the elementary charge ( or a unit charge).  

Eugen Goldstein discovered the anode rays or protons.  

Rutherford conducted the gold foil experiment and proved that the center of the nucleus has a positive charge and most of the atom’s mass lies in a small volume.  

Sidana [21]3 years ago
6 0

Millikan C

Hope this helps you out!

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The metallic pan iis most likely going to be used on a stove.

The stove is heating something, and the conductive metallic pan will, well, conduct that heat throughout the entire body of the pan. Doing this will spread the heat to the handle, burning your hands.

Both wood and plastic are insulators, and they do not conduct heat or electricity. They will insulate your hands and protect them from the heat.

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3 years ago
Find the pH of the equivalence point and the volume (mL) of 0.0372 M NaOH needed to reach the equivalence point in the titration
elixir [45]

Answer:

8.54

Explanation:

At equivalence point :  

42.2 X 0.052 = Vol. NaOH X 0.0372

Vol of NaOH = 2.1944/0.0372 = 58.99 ml

So volume of NaOH recquired to reach equivalence point = 58.99 ml

Number of miliimoles of CH3COOH = molarity X volume in ml = 42.2 X 0.052             = 2.1944 millimoles

Number of millimoles of NaOH = 58.99 X 0.0372 = 2.1944

Now CH₃COOH and NaOH reacts to give CH₃COONa according to the reaction :

CH₃COOH + NaOH ------> CH₃COONa + H₂O

1 mole of CH₃COOH reacts with 1 mole of NaOH to give 1 mole of CH₃COONa  

So 2.1944 millimoles of CH₃COOH will react with 2.1944 millimoles of NaOH to give 2.1944 millimoles of CH₃COONa

So all the acid (CH₃COOH) and base (NaOH) has been converted into salt (CH₃COONa) so there is no acid or base left.

Now molarity of CH₃COONa = number of millimoles of CH₃COONa/total volume in ml = 2.1944/(58.99 + 42.2) = 2.1944/101.19 = 0.02169 M

So using the hydrolysis equation :  

pH = 1/2 [ pKw + pKa + log c ]  

Ka for acetic acid = 1.75 X 10⁻⁵  

so pKa = -log (1.75 X 10⁻⁵) = 4.74  

Kw = 10⁻¹⁴

so pKw = -log 10⁻¹⁴ = 14

c = 0.02169  

so log c = log 0.02169 = -1.66  

putting the values....  

pH = 1/2 [14 + 4.74 - 1.66 ]  

pH = 1/2 [ 17.08] = 8.54

6 0
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