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3 years ago

How did Thomson discover the electron?

1 answer:

4 0

J. J. Thomson discovered electron by performing an experiment using cathode ray tubes. High voltage across is applied across two electrodes at that causes a beam of particles to flow from the the negatively-charged electrode that is cathode to the positively-charged electrode that is anode. Properties of the particles, are tested using two oppositely-charged electric plates around the cathode ray. The cathode ray was deflected away from the negatively-charged electric plate and towards the positively-charged plate. This indicated that the cathode ray was composed of negatively-charged particles. And these negatively charged particles are called electrons.

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What would be the pH of a solution of a solution with a hydrogen ion concentration of 0.001 M?

son4ous [18]

0001 M HCl is the same as saying that 1 *10-4 moles of H+ ions have been added to solution. The -log[. 0001] =4, so the pH of the solution =4.

7 0

2 years ago

As the speed of a particle increases, does its associated wavelenght increase or decrease

Zina [86]

The de Broglie hypothesis proposed that all particles have wave-like properties, with the wavelength being inversely proportional to the velocity of the particle.

Therefore as the velocity (speed in this question) increases, the wavelength *decreases*.

7 0

3 years ago

A student Increases the temperature of a 417 cubic centimeter balloon from 278 K to 231 K. Assuming constant pressure what shoul

WITCHER [35]

Initial Conditions:
Volume= v1= 417 cm³
Temperature= T1 = 278 K

Final Conditions:
Temperature= T2 = 231K
Volume = v2 =?

Use the general gas equation;

P1*v1/T1 = P2*v2/T2
As, the temperature is constant;
So,
v1/T1 = v2/T2
417/278 = v2/231
v2= 346.5 cm³

5 0

3 years ago

A solution of rubbing alcohol is 68.6 % (v/v) isopropanol in water. How many milliliters of isopropanol are in a 88.2 mL sample

KonstantinChe [14]

From the information given, the total volume of rubbing alcohol is 88.2 ml

68.6 % of this volume is isopropanol.

We will assume 88.2 ml represents 100% volume, so the volume of water will be 31.4 %

The volume of isopropanol is

68.6/100 x 88.2 → 0.686 × 88.2 = 60.505 ml

The volume of isopropanol is 60.5 ml.

Volume of water will be 88.20 - 60.5 = 27.7 ml

(27.7 / 88.2 × 100 = 31.4% )

Adding 60.5 ml of isopropanol to 27.7 ml of water to make up 88.2 ml will give 68.6 % v/v isopropanol to water solution.

8 0

3 years ago

A 0.105 L sample of an unknown HNO 3 solution required 35.7 mL of 0.250 M Ba ( OH ) 2 for complete neutralization. What is the c

oksano4ka [1.4K]

Answer: 0.17M

Explanation:

The equation for the reaction is :

2HNO3 + Ba(OH)2 —> Ba(NO3)2 + 2H2O

From the balanced equation, we obtain :

nA = mole of acid = 2

nB = mole of the base = 1

From the question, we obtain:

Va = Vol. Of acid = 0.105L

Ma = conc. Of acid =?

Vb = Vol of base = 35.7 mL = 0.0357L

Mb = conc. of base = 0.25M.

We solve for the conc. of the acid using:

MaVa / Mb Vb = nA / nB

(Ma x 0.105) / (0.25x0.0357) = 2

Cross multiply to express in linear form. We have:

Ma x 0.105 = 0.25 x 0.0357 x 2

Divide both side by 0.105. We have

Ma = (0.25 x 0.0357 x 2) / 0.105

Ma = 0.17M

Therefore the concentration of the acid(HNO3) is 0.17M

3 0

2 years ago