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2 years ago

What did J. J. Thomson observe when he applied electric voltage to a cathode ray tube in his famous experiment?

Chemistry

2 answers:

svet-max [94.6K]2 years ago

8 0

Explanation:

J.J. Thomson conducted an experiment in which he took a gas at low pressure in a discharge tube and applied high voltage current. When he did so, he noticed that there are some particles emitting from cathode going towards the anode. Then he concluded that they are negatively charged particles and coined them electrons.

irinina [24]2 years ago

7 0

QUICK ANSWER

J.J. Thomson's cathode ray experiment was a set of three experiments that assisted in discovering electrons. He did this using a cathode ray tube or CRT. It is a vacuum sealed tube with a cathode and anode on one side.

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Which of the following is an example of a heterogeneous mixture A.salt-water B.granite C.water(Na)

sladkih [1.3K]

Hi!

B. Granite is an example of a heterogenous mixture.

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2 years ago

Who discovered phosphate?

iris [78.8K]

Answer:

Hennig Brand

Explanation:

4 0

2 years ago

High concentrations of ammonia (NH3), nitrite ion, and nitrate ion in water can kill fish. Lethal concentrations of these specie

kykrilka [37]

Explanation:

It is known that molality is the number of moles present in kg of solution.

Mathematically, Molality = $\frac{\text{no. of moles of solute}}{\text{mass of solvent in Kg}}$

The given data is as follows.

Molar mass of ammonia = 17 g/mol

Concentration = 1.002 mg/L = $\frac{0.001002 g/L}{17 g/mol}$

= $5.89 \times 10^{-4}$ mol/L

Also, density = $\frac{1 g}{mL}$ = 1 kg/L

Therefore, molality will be calculated as follows.

Molality = $\frac{5.89 \times 10^{-4} mol/L}{1 kg/L}$

= $5.89 \times 10^{-4} mol/kg$

And,

Molar mass of nitrite = 46 g/mol

Concentration = 0.387 mg/L = $\frac{0.000412 g/L}{46 g/mol}$

= $8.956 \times 10^{-6}$ mol/L

And, density = $\frac{1 g}{mL}$ = 1 kg/L

Hence, molality = $\frac{8.956 \times 10^{-6} mol/L}{1 kg/L}$

= $8.956 \times 10^{-6}$ mol/kg

Now, Molar mass of nitarte = 62 g/mol

Concentration = 1352.2 mg/L

= $\frac{1.3522 g/L}{62 g/mol}$

= 0.02181 mol/L

Also, density = $\frac{1 g}{mL}$ = 1 kg/L

Hence, molality will be calculated as follows.

Molality = $\frac{0.02181 mol/L}{1 kg/L}$

= 0.02181 mol/kg

Therefore, molality of given species is $5.89 \times 10^{-4} mol/kg$ for ammonia, $8.956 \times 10^{-6}$ mol/kg for nitrite, and 0.02181 mol/kg for nitrate ion.

7 0

3 years ago

Given that a for HBrO is 2. 8×10^−9 at 25°C. What is the value of b for BrO− at 25°C?

lara [203]

If Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).

<h3>What is base dissociation constant? </h3>

The base dissociation constant (Kb) is defined as the measurement of the ions which base can dissociate or dissolve in the aqueous solution. The greater the value of base dissociation constant greater will be its basicity an strength.

The dissociation reaction of hydrogen cyanide can be given as

HCN --- (H+) + (CN-)

Given,

The value of Ka for HCN is 2.8× 10^(-9)

The correlation between base dissociation constant and acid dissociation constant is

Kw = Ka × Kb

Kw = 10^(-14)

Substituting values of Ka and Kw,

Kb = 10^(-14) /{2.8×10^(-9) }

= 3.5× 10^(-6)

Thus, we find that if Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).

DISCLAIMER: The above question have mistake. The correct question is given as

Question:

Given that Ka for HBrO is 2. 8×10^−9 at 25°C. What is the value of Kb for BrO− at 25°C?

learn more about base dissociation constant:

brainly.com/question/9234362

#SPJ4

7 0

1 year ago

If Jonathan went skateboarding from 4:00 PM to 4:30 PM and traveled 450 meters, what was his average speed?

mrs_skeptik [129]

20 m Increase by five

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2 years ago