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Verdich [7]
3 years ago
10

describe how the discoveries of scientist over time have contributed to our current understanding of atom

Chemistry
1 answer:
Olenka [21]3 years ago
5 0

Answer:

In 1897, the British physicist J. J. Thomson (1856–1940) proved that atoms were not the most basic form of matter. He demonstrated that cathode rays could be deflected, or bent, by magnetic or electric fields, which indicated that cathode rays consist of charged particles (Figure  2.2.2 ). More important, by measuring the extent of the deflection of the cathode rays in magnetic or electric fields of various strengths, Thomson was able to calculate the mass-to-charge ratio of the particles. These particles were emitted by the negatively charged cathode and repelled by the negative terminal of an electric field. Because like charges repel each other and opposite charges attract, Thomson concluded that the particles had a net negative charge; these particles are now called electrons. Most relevant to the field of chemistry, Thomson found that the mass-to-charge ratio of cathode rays is independent of the nature of the metal electrodes or the gas, which suggested that electrons were fundamental components of all atoms.

Explanation:

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A solution is made by mixing 0.82 grams of sodium acetate, 1 mL of 12 M acetic acid (pKa=4.76) and water to give a final volume
kompoz [17]

Answer:

This solution acts as an efficient buffer

Explanation:

the pH of a buffer solution can be described like this: pH=pKa+log\frac{[base]}{[acid]}

[acid]=[acetic acid]=1mL.\frac{12mol}{1000mL} . \frac{1}{1L} = 0.012M

[base]=[sodium acetate]=0.82g.\frac{1mol}{82g} .\frac{1}{1l} = 0.01M

replacing, pH=4.76+log\frac{0.01M}{0.012M} =4.84

If we add an acid, pH will decrease a little bit and if we add a base, pH wil increase a little bit.

lets supose that we change the rate by increasing [base] to 0.1, then

pH=4,76+log\frac{0.1}{0.012} = 5.68

and now lets supose that we increase [acid] to 0.1 pH=4.76+log\frac{0.01}{0.1} = 3.76

Big changes in concentration of base or acid doesn´t produce big changes in pH, in that way the mix of sodium acetate with acetic acid is a good buffer solution.

4 0
2 years ago
Number of grams of hcl that can react with 0.500 grams of Al(OH)3
vlabodo [156]

Answer:

0.7g of HCl

Explanation:

First, let us write a balanced equation for the reaction between HCl and Al(OH)3.

This is illustrated below:

Al(OH)3 + 3HCl —> AlCl3 + 3H2O

Next, let us obtain the masses of Al(OH)3 and HCl that reacted together according to the equation. This can be achieved as shown below:

Molar Mass of Al(OH)3 = 27 + 3(16+1)

= 27 + 3(17) = 27 + 51 = 78g/mol.

Molar Mass of HCl = 1 + 35.5 = 36.5g/mol

Mass of HCl from the balanced equation = 3 x 36.5 = 109.5g

Now we can obtain the mass of HCl that would react with 0.5g of Al(OH)3. This can be achieved as follow:

Al(OH)3 + 3HCl —> AlCl3 + 3H2O

From the equation above,

78g of Al(OH)3 reacted with 109.5g of HCl.

Therefore, 0.5g of Al(OH)3 will react with = (0.5 x 109.5)/78 = 0.7g of HCl

7 0
3 years ago
How are the isotopes of a particular element alike?
zhannawk [14.2K]
The isotopes of a particular elements means that every element has there own isotopes like H has three isotopes protium, dutrium,tritium. like that cl has also thier own isotopes.
6 0
2 years ago
Define not matter and give examples​
VLD [36.1K]

Answer:

Everything that has mass and takes up space is matter. Every day, you encounter phenomena that either don't have mass or don't take up space. They are non-matter. Basically, any type of energy or any abstract concept is an example of something that is not matter.

An apple.

A person.

A table.

Air.

Water.

A computer.

Paper.

Iron.

hope this helped you

3 0
2 years ago
A 7.12 L cylinder contains 1.21 mol of gas A and 4.94 mol of gas B, at a temperature of 28.1 °C. Calculate the partial pressure
GenaCL600 [577]

Answer:

P_A=4.20atm\\\\P_B=17.1atm

Explanation:

Hello!

In this case, since the equation for the ideal gas is:

PV=nRT

For each gas, given the total volume, temperature (28.1+273.15=301.25K) and moles, we can easily compute the partial pressure as shown below:

P_A=\frac{n_ART}{V} =\frac{1.21mol*0.082\frac{atm*L}{mol*K}*301.25K}{7.12L} \\\\P_A=4.20atm\\\\P_B=\frac{n_BRT}{V} =\frac{4.94mol*0.082\frac{atm*L}{mol*K}*301.25K}{7.12L} \\\\P_B=17.1atm

Best regards!

8 0
2 years ago
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