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3 years ago

The discovery of the electron as a subatomic particle was the result of

Chemistry

1 answer:

Serjik [45]3 years ago

4 0

Answer:

Investigation of cathode ray.

Explanation:

The electron is subatomic particle that revolve around outside the nucleus and has negligible mass. It has a negative charge.

Symbol= e⁻

Mass= 9.10938356×10⁻³¹ Kg

It was discovered by j. j. Thomson in 1897 during the study of cathode ray properties.

He constructed the glass tube and create vacuum in it. He applied electric current between electrodes. He noticed that a ray of particles coming from cathode to wards positively charged anode. This ray was cathode ray.

Properties of cathode ray:

The ray is travel in straight line.

The cathode ray is independent of composition of cathode.

When electric field is applied cathode ray is deflected towards the positively charged plate.

Hence it was consist of negatively charged particles.

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There is 18.066* 10^23 molecules of carbon in 3.50 moles of carbon dioxide.

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nika2105 [10]

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When heat is applied to 80 grams of CaCO3, it yields 39 grams of CO2. Determine

galben [10]

<u>Answer:</u> The percentage yield of $CO_2$ is 90.26%.

<u>Explanation:</u>

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Given mass of $CaCO_3$ = 80 g

Molar mass of $CaCO_3$ = 100 g/mol

Plugging values in equation 1:

$\text{Moles of }CaCO_3=\frac{80g}{100g/mol}=0.8 mol$

For the given chemical equation:

$CaCO_3\rightarow CaO+CO_2$

By the stoichiometry of the reaction:

If 1 mole of $CaCO_3$ produces 1 mole of $CO_2$

So, 0.8 moles of $CaCO_3$ will produce = $\frac{1}{1}\times 0.8=0.8mol$ of $CO_2$

Molar mass of $CO_2$ = 44 g/mol

Plugging values in equation 1:

$\text{Mass of }CO_2=(0.8mol\times 44g/mol)=35.2g$

The percent yield of a reaction is calculated by using an equation:

$\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100$ ......(2)

Given values:

Actual value of $CO_2$ = 35.2 g

Theoretical value of $H_2CO_3$ = 39 g

Plugging values in equation 2:

$\% \text{yield of }CO_2=\frac{35.2g}{39g}\times 100\\\\\% \text{yield of }CO_2=90.26\%$

Hence, the percentage yield of $CO_2$ is 90.26%.

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3 years ago

Xander claims that since the model of the atom has been changed so many times, that means it is a weak model. Willow claims that

uranmaximum [27]

Willow is right because the model of atom is not weak. The model of the atom has not changed. It’s just due to new discoveries due to which atom’s model is changing. The particles were always present in the atom but we were not able to discover it.

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4 years ago

An equilibrium mixture contains 0.750 mol of each of the products (carbon dioxide and hydrogen gas) and 0.200 mol of each of the

sasho [114]

1.302 moles of carbon dioxide would have to be added

<h3>Further explanation</h3>

The equilibrium constant is the value of the product in the equilibrium state of the substance in the right (product) divided by the substance in the left (reactant) with the exponents of each reaction coefficient

The equilibrium constant is based on the concentration (Kc) in a reaction

pA + qB -----> mC + nD

$\large {\boxed {\bold {Kc ~ = ~ \frac {[C] ^ m [D] ^ n} {[A] ^ p [B] ^ q}}}}$