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Naddik [55]
2 years ago
11

The Millikan oil-drop experiment, which was the first experiment to measure the charge of the electron, compared the electrical

force on a small charged oil drop with the force exerted on this drop by gravity.
a. True
b. False
Chemistry
1 answer:
Liono4ka [1.6K]2 years ago
3 0

Answer:

True

Explanation:

The oil drop experiment was carried out by Robert Millikan and Harvey Fletcher in 1909 to determine the charge of an electron. By balancing downward gravitation force with upward drag and electric force, they suspended small charged droplets of oil between two metal electrodes.

The charge over an oil droplet was often an integral value of e, was determined by changing the intensity of the electric field.

So, the given statement is true

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THIS IS A THREE PART QUESTION IF YOU CAN HELP IT WOULD BE REALLY APPRECIATED SO I DONT FAIL.
Nostrana [21]

Answer:

1. B = 1.13M

2. A. 8.46%

3. D = 0.0199

Explanation:

1. Molarity of a a solution = number of moles of solute/ volume of solution in L

Number of moles of solute = mass of solute/molar mass of solute

Molar mass of NaC₂H₃O₂ = 82 g/mol, mass of NaC₂H₃O₂ = 245 g

Number of moles of NaC₂H₃O₂ = 245 g / 82 g/mol = 2.988 moles

Molarity of solution = 2.988 mols/ 2.65 L = 1.13 M

2. Percentage by mass of a substance = mass of substance /mass of solution × 100%

Mass of 2.65 L of water = density × volume

Density of water = 1 Kg/L = 1000 g/L; volume of waterb= 2.65 L

Mass of water = 1000 g/L × 2.65 L = 2650 g

Mass of solution = mass of water + mass of solute = 245 + 2650 =2895 g

Percentage by mass of NaC₂H₃O₂ = 245/2895 × 100% = 8.46%

3. Mole fraction of NaC₂H₃O₂ = moles of NaC₂H₃O₂/moles of solution

Moles of water = mass /molar mass

Mass of water = 2650 g; molar mass of water = 18 g/mol

Moles of water = 2650 g / 18 g/mol = 147.222 moles

Moles of solution = moles of solute + moles of water = 147.222 + 2.988 = 150.21

Moles of NaC₂H₃O₂ =2.988 moles

Moles fraction of NaC₂H₃O₂ =2.988/150.21 = 0.0199

6 0
3 years ago
PLS HELP Students in a chemistry class added 5 g of Zinc (Zn) to 50 g of hydrochloric acid (HCl). A chemical reaction occurred t
Vitek1552 [10]

Answer: In simplest case mass of reactants is same as mass of products.

Without thinking this question deeper, mass of ZnCl2 would be 49, but..

Explanation: Reaction should be  Zn + 2 HCl ⇒ ZnCl2 + H2

Amount of zinc is  5 g / 65,38 g/mol = 0,076476 mol and amount

of Hydrogen Chloride is 50 g / 36.458 g/mol = 1,371 mol.

Althought HCl is needed 0.152 moles, zinc is an limiting reactant.

So it is possible to produce only 0.076476 mol Hydrogen and its mass

is 0.154 g.  Mass of ZnCl2 would be 0.076476 mol · (65.38 + 2·35.45) =

 10.42 g

4 0
2 years ago
An interpenetrating primitive cubic structure like that of CsCl with anions in the corners has an edge length of 664 pm. If the
san4es73 [151]

Answer:

the ionic radius of the anion r^- = 312.52 \ pm

Explanation:

From the diagram shown below :

The anion Cl^- is located at the corners

The cation Cs^+ is located at the body center

The Body diagonal length =  \sqrt{3 \ a }

∴ 2 \ r^+ \ + 2r^- \ = \sqrt{3 \ a}  \\ \\ r^+ +r^- = \frac{\sqrt{3}}{2} a

Given that :

\frac{r^+}{r^-} =0.84   (i.e the  ratio of the ionic radius of the cation to the ionic radius of

                 the anion )

0.84r^- \ + r^- \ = \frac{\sqrt{3}}{2}a  \\ \\  1.84 r^- = \frac{3}{2}a \\ \\ r^- = \frac{\sqrt{3}}{2*1.84}a

Also ; a =  664 pm

Then :

r^- = \frac{\sqrt{3} }{2*1.84}*664 \ pm\\ \\ r^- = 312.52 \ pm

Therefore,  the ionic radius of the anion r^- = 312.52 \ pm

4 0
3 years ago
1. Identify the reactants in this chemical equation:<br> 2H2 + O2 + 2H2O
USPshnik [31]
Oxygen and hydrogen
6 0
3 years ago
If the dosage for a medication is 225mg/lb, how many grams should a 25-lb child be given?
Oksanka [162]

Answer:

5.625 grams

Explanation:

Start your equation with what you have been given.  Place the units you need in your answer on the right side of the equal sign.

225mg

-----------   X  -----------   X ------------- =   ?   g

   lb

Now start to fill in your equation and use a conversions to get rid of the units you don't want.  Convert mg into grams first.  The child's weight (25 lb) is placed over 1 just to get the equation lined up properly so you can see how the units cancel out.

225 mg                 1 g                  25 lb            5.625 g

---------------   X   ---------------  X   -------------  =   ---------------

   lb                    1000 mg                 1                  1

The lb on the top and bottom cancel each other out and you are left with just grams.  Even though it is over one, that is the same at just 5.625 grams.

3 0
2 years ago
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