How many protons and electrons are in a

1 answer:

Pie3 years ago

3 0

Answer:

B 29 protons, 27 electrons

Explanation:

The given specie is copper ion:

Cu²⁺

We are to find the number of protons and electrons it contains.

The number of protons is the positively charged particles in the atom.

Electrons are the negatively charged particles in the atom.

The number of protons is the same as the atomic number

Atomic number of copper = 29

Number of protons = 29

Now;

Charge = Number of protons - Number of electrons

2 = 29 - number of electrons

-27 = - number of electrons

Number of electrons = 27

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ch4aika [34]

Answer:

Standard enthalpy of formation of Carbon disulfide CS2 = 87.3 KJ/mol

Explanation:

forming CS2 means that it should in the product side

C(graphite) + O2 → CO2 ΔH = -393.5

2S(rhombic) + 2O2 → 2SO2 ΔH = -296.4 x 2

CO2 + 2SO2 → CS2 + 3O2 ΔH = -1073.6 x -1

the second reaction is multiplied by 2 so that the SO2 and O2 can cancel out.

the third reaction is reversed (multiplied by -1) so that CS2 will be on the product side.

after adding the reaction and cancelling out similarities, the final reaction is: C(graphite) + 2S(rhombic) → CS2

Add ΔH to find the enthalpy of formation of CS2

ΔHf = (-393.5) + (-296.4 x 2) + (-1073.6 x -1) = 87.3 KJ/mol

be aware of signs

3 0

3 years ago

Ahat [919]

Answer :

Oxidation number or oxidation state : It represent the number of electrons lost or gained by the atoms of an element in a compound.

Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.

Rules for Oxidation Numbers are :

The oxidation number of a free element is always zero.
The oxidation number of a monatomic ion equals the charge of the ion.
The oxidation number of Hydrogen (H) is +1, but it is -1 in when combined with less electronegative elements.
The oxidation number of oxygen (O) in compounds is usually -2.
The oxidation number of a Group 17 element in a binary compound is -1.
The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.
The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

Now we have to determine the oxidation state of the elements in the compound.

(a) $H_2SO_4$