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Naddik [55]
3 years ago
7

From his experiments, J. J. Thomson concluded that

Chemistry
2 answers:
In-s [12.5K]3 years ago
7 0
<span>atoms contain small negatively charged particles that are called electrons.</span>
sergiy2304 [10]3 years ago
6 0
Answer: option D. Atoms containd small negatively charged particles that are called electrons.

JJ Thompson carried out three important experiments with cathode rays. With them he proveed that cathode rays were negative charged particles and that the ratio charge/mass was so large that the particles either had a huge charge or had a very small mass, and he came up with the latter, defining that these negative charged particle were very small.
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castortr0y [4]

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They can eat meal worms! :)

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3 years ago
What are the coefficients that will balance the skeleton equation below?alcl3 + naoh → al(oh)3 + nacl?
Karo-lina-s [1.5K]
AlCl3 + 3NaOH -> Al(OH)3 + 3NaCl

1 3 1 3
3 0
3 years ago
Read 2 more answers
Please help thank you
Fynjy0 [20]

Answer:

D

Explanation:

8 0
3 years ago
HELP!!!! A glass flask containing 100 milliliters of water is sealed with a rubber stopper and placed on a burner. After a while
kondor19780726 [428]

Answer: A closed system, because energy can enter or leave the container, but the water molecules cannot

Explanation:

Open system: In this system energy and matter both have access to their surroundings beyond the boundaries of system. .

Closed system :In this type system only energy has an access to its surroundings beyond the boundaries of system but not matter.

Isolated system:In this type system exchange of both energy and matter are restricted to move outside the boundaries of system.

According to question, the system given is a closed system because energy is transferred from the burner to glass flask and from the glass flask to the water (matter). But water molecules are only getting condensed on the inside surface of the flask that is exchange of matter beyond the boundaries of the system is restricted. Hence, closed system ,A closed system, because energy can enter or leave the container, but the water molecules cannot.

7 0
3 years ago
An analytical chemist is titrating of a solution of propionic acid with a solution of 224.9 ml of a 0.6100M solution of propioni
Svetllana [295]

<u>Answer:</u> The pH of acid solution is 4.58

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}    .....(1)

  • <u>For KOH:</u>

Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

1.1000M=\frac{\text{Moles of KOH}\times 1000}{41.04}\\\\\text{Moles of KOH}=\frac{1.1000\times 41.04}{1000}=0.04514mol

  • <u>For propanoic acid:</u>

Molarity of propanoic acid solution = 0.6100 M

Volume of solution = 224.9 mL

Putting values in equation 1, we get:

0.6100M=\frac{\text{Moles of propanoic acid}\times 1000}{224.9}\\\\\text{Moles of propanoic acid}=\frac{0.6100\times 224.9}{1000}=0.1372mol

The chemical reaction for propanoic acid and KOH follows the equation:

                 C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O

<u>Initial:</u>          0.1372         0.04514  

<u>Final:</u>           0.09206          -                0.04514

Total volume of solution = [224.9 + 41.04] mL = 265.94 mL = 0.26594 L     (Conversion factor:  1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[\text{salt}]}{[acid]})

pH=pK_a+\log(\frac{[C_2H_5COOK]}{[C_2H_5COOH]})

We are given:  

pK_a = negative logarithm of acid dissociation constant of propanoic acid = 4.89

[C_2H_5COOK]=\frac{0.04514}{0.26594}

[C_2H_5COOH]=\frac{0.09206}{0.26594}

pH = ?  

Putting values in above equation, we get:

pH=4.89+\log(\frac{(0.04514/0.26594)}{(0.09206/0.26594)})\\\\pH=4.58

Hence, the pH of acid solution is 4.58

7 0
4 years ago
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