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Fudgin [204]
3 years ago
7

In a sample of pure copper, all atoms have atomic numbers which are

Chemistry
2 answers:
Bumek [7]3 years ago
8 0

Answer:

All atoms have same atomic number because number of electrons in all atoms are same.

Explanation

Every atom consist of nucleus or a positive center. The protons and neutrons are present with in the nucleus while electrons are present out side the nucleus. All these three subatomic particles construct an atom. The number of protons or number of electrons are the atomic number of an atom while the number of protons and number of neutrons are the mass number of an atom. A neutral atom have equal number of protons and electrons. In other words we can say that negative and positive charges are equal in magnitude and cancel the each other.

For example:

Atomic number of copper is 29 it have 29 electrons and 29 protons.

While its atomic mass is 63.546 and have 35 neutrons.

alisha [4.7K]3 years ago
4 0

Explanation:

Atomic number of an atom is actually the sum of total number of protons present in it.  And, atomic mass is the sum of total number of protons and neutrons present in an atom.

And, when an atom is neutral in nature then the number number of electrons present in it is equal to the number of protons in it.

For example, atomic number of copper is 29 and atomic mass of copper is 63.

This means that a sample of pure copper will have all atoms has an atomic number 29, that is, number of protons is 29.

And, number of electrons is also 29 and number of neutrons is 35.

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formic acid, hcooh, is a weak acid with a ka equal to 1.8×10–4. what is the ph of a 0.0115 m aqueous formic acid solution?
kvasek [131]

The pH of a 0.0115 m aqueous formic acid solution is  mathematically given as

pH=2.8424

This is further explained below.

<h3>What is the ph of a 0.0115 m aqueous formic acid solution?</h3>

Generally,   the equation for the chemical equation is  mathematically given as

HCOOH   H^+ + HCOO

\quad C-C \alpha \quad C \alpha

&k a=\frac{c^2 \alpha^2}{e(1-\alpha)}\\\\&K_a=\frac{c \alpha^2}{1-\alpha_4} \quad|\gg\rangle \alpha\\\\&K_a=c \alpha^2\\\\&\alpha=\sqrt{\frac{k a}{c}}\\\\&\alpha=\sqrt{\frac{1.8 \times 10^{-4}}{0.0115}}\\\\&\alpha=\sqrt{156: 527710^{-4}}\\\\&\alpha=\sqrt{1.565 \times 1 \sigma^2}\\\\&\alpha=1.25 \times 10^{-1}\\\\&\alpha=0.125\\

&\left[\mathrm{H}^{+}\right]=\mathrm{C \alpha}\\\\&=0.125 \times 0.0115\\\\&=1.4375 \times 10^{-9}\\\\&P=-\log \left[H^{+}\right]\\\\&=-\log \left[1.4375 \times 10^{-3}\right]\\\\&P H=2.8424

Read more about  chemical equation

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