Does The nucleus of an atom contain all of the protons in the atom

Chemistry

1 answer:

insens350 [35]3 years ago

5 0

Answer:

The nucleus, that dense central core of the atom, contains both protons and neutrons.

So to answer your question yes it does contain all of the protons in the atom.

* Hopefully this helps:) Mark me the brainliest:)!

You might be interested in

A drop of vinegar ( a weak acid ) is placed on a sample of each of the materials below. Which will show the most active reaction

professor190 [17]

Answer: Option (B) is the correct answer.

Explanation:

Marble is also known as calcium carbonate and its chemical formula is $CaCO_{3}$ .

When it combines with vinegar then it results in the formation of calcium acetate which is a water soluble compound along with the release of carbon dioxide which occurs in the form of bubbles and water.

The reaction is as follows.

$CaCO_{3} + CH_{3}COOH \rightarrow Ca(CH_{3}COOH)_{2} + H_{2}O + CO_{2}$

Whereas vinegar being a weak acid will not react with glass, copper and steel.

8 0

3 years ago

Read 2 more answers

The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 2.60 g of naphthalene (C10H8) in 675 g of be

Mrac [35]

<u>Answer:</u> The freezing point of solution is 5.35°C

<u>Explanation:</u>

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 4.90°C/m

$m_{solute}$ = Given mass of solute (naphthalene) = 2.60 g

$M_{solute}$ = Molar mass of solute (naphthalene) = 128.2 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 675 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 4.90^oC/m\times \frac{2.60\times 1000}{128.2g/mol\times 675}\\\\\text{Freezing point of solution}=5.35^oC$