20 points!! Please help ASAP!

How many electrons does an element have if its atomic number is 20

olga55 [171]

Answer:

20 electrons

A neutral atom with atomic number 20 will have 20 electrons.

The atomic number is, by definiton, the number of protons in an atom's nucleus but for a neutral atom it's also equal to the number of electrons. Each element has a different, unique number of protons that determines its identity.

Explanation:

6 0

2 years ago

Calculate the mass defect for the formation of phosphorus-31. The mass of a phosphorus-31 nucleus is 30.973765 amu. The masses o

Nata [24]

<u>Answer:</u> The mass defect for the formation of phosphorus-31 is 0.27399

<u>Explanation:</u>

Mass defect is defined as the difference in the mass of an isotope and its mass number.

The equation used to calculate mass defect follows:

$\Delta m=[(n_p\times m_p)+(n_n\times m_n)]-M$

where,

$n_p$ = number of protons

$m_p$ = mass of one proton

$n_n$ = number of neutrons

$m_n$ = mass of one neutron

M = mass number of element

We are given:

An isotope of phosphorus which is $_{15}^{31}\textrm{P}$

Number of protons = atomic number = 15

Number of neutrons = Mass number - atomic number = 31 - 15 = 16

Mass of proton = 1.00728 amu

Mass of neutron = 1.00866 amu

Mass number of phosphorus = 30.973765 amu

Putting values in above equation, we get:

$\Delta m=[(15\times 1.00728)+(16\times 1.00866)]-30.973765\\\\\Delta m=0.27399$

Hence, the mass defect for the formation of phosphorus-31 is 0.27399

8 0

2 years ago

The standard cell potential Ec for the reduction of silver ions with elemental copper is 0.46V at 25 degrees celsius. calculate

Cloud [144]

Answer : The $\Delta G$ for this reaction is, -88780 J/mole.

Solution :

The balanced cell reaction will be,

$Cu(s)+2Ag^+(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)$

Here, magnesium (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half oxidation-reduction reaction will be :

Oxidation : $Cu\rightarrow Cu^{2+}+2e^-$