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3 years ago

Which types of particles are charged and located within the nucleus of an atom?

Chemistry

2 answers:

exis [7]3 years ago

6 0

Answer : The correct option is, protons

Explanation :

As we know that an atom is the smallest unit of a matter that consist of three subatomic particles which are electrons, protons and neutrons.

The protons and the neutrons are located inside the nucleus or center of the nucleus where the mass of the an atom is concentrated and the electrons are located around the nucleus.

The protons are positively charged, the electrons are negatively charged and the neutrons are neutral that means it has no charge.

Hence, the particle protons are positively charged and located within the nucleus of an atom.

Readme [11.4K]3 years ago

3 0

Answer:

Protons, electron, and neutrons

Explanation:

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Two molecules of one reactant combine with 3 molecules of another to produce 5 molecules of a product.

maksim [4K]

Answer:

A = 2A + 3B → 5C

Explanation:

The two molecule of A and three molecules of B will react to form the five molecules of C.

2A + 3B → 5C

Other options are incorrect because,

B = A₂ + B₃ → C₅

in this reaction one molecule of A₂ and one molecule of B₃ combine to form one molecule of C₅.

C = 2A + 5B → 3C

in this reaction two molecules of A and five molecules of B combine to form three molecule of C.

D = A₂ + B₃ → C₃

in this reaction one molecule of A₂ and one molecule of B₃ combine to from one molecule of C₃.

3 0

2 years ago

In the chemical formula 3nh3, how many atoms of hydrogen (h) are there? 3 6 9 11

ANTONII [103]

Your answer is 3! Hope you get a good grade :)♥

7 0

3 years ago

Read 2 more answers

What volume of 0.307 m naoh must be added to 200.0ml of 0.425m acetic acid (ka = 1.75 x 10-5 ) to produce a buffer of ph = 4.250

Blababa [14]

The buffer solution target has a pH value smaller than that of pKw (i.e., pH < 7.) The solution is therefore acidic. It contains significantly more protons $\text{H}^{+}$ than hydroxide ions $\text{OH}^{-}$ . The equilibrium equation shall thus contain protons rather than a combination of water and hydroxide ions as the reacting species.

Assuming that $x \; \text{L}$ of the 0.307 $\text{mol} \cdot \text{dm}^{-3}$ sodium hydroxide solution was added to the acetic acid. Based on previous reasoning, $x$ is sufficiently small that acetic acid was in excess, and no hydroxide ion has yet been produced in the solution. The solution would thus contain $0.2000 \times 0.425 - 0.307 \; x = 0.085 - 0.307 \; x$ moles of acetic acid and $0.307 \; x$ moles of acetate ions.

Let $\text{HAc}$ denotes an acetic acid molecule and $\text{Ac}^{-}$ denotes an acetate ion. The RICE table below resembles the hydrolysis equilibrium going on within the buffer solution.

$\begin{array}{lccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\end{array}$

The buffer shall have a pH of 4.250, meaning that it shall have an equilibrium proton concentration of $10^{4.250}\; \text{mol}\cdot \text{dm}^{-3}$ . There were no proton in the buffer solution before the hydrolysis of acetic acid. Therefore the table shall have an increase of $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ in proton concentration in the third row. Atoms conserve. Thus the concentration increase of protons by $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ would correspond to a decrease in acetic acid concentration and an increase in acetate ion concentration by the same amount. That is:

$\begin{array}{lcccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\text{C} & - 10^{-4.250} & & +10^{-4.250} & & +10^{-4.250} \\\text{E} & 0.085 - 10^{-4.250} - 0.307 \; x& & 10^{-4.250} & & 10^{-4.250} + 0.307 \; x\end{array}$

By definition:

$\text{K}_{a} = [\text{H}^{+}] \cdot [\text{Ac}^{-}] / [\text{HAc}]\\\phantom{\text{K}_{a}} = 10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x)$

The question states that

$\text{K}_{a} = 1.75 \times 10^{-5}$

such that

$10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x) = 1.75 \times 10^{-5}\\6.16 \times 10^{-5} \; x = 1.48 \times 10^{-6}\\x = 0.0241$