Hat is the % of each element in the compound mg3(po4)2?

1. First calculate the total molar mass of the entire compound.

Mg = 3 x 24.31 = 72.93 g/mol

P = 2 x 30.97 = 61.94 g/mol

O = 8 x 16.00 = 128.00 g/mol

* Add them up;

72.93 + 61.94 + 128.00 = 262.87 g/mol

2. Take the individual masses of each element that you calculated, divide them by the total mass (262.87 g/mol) and then multiply by 100.

Mg = 72.93/262.87 = 0.2774 x 100 = 27.74%

P = 61.94/262.87 = 0.2356 x 100 = 23.56%

O = 128.00/262.87 = .4870 x 100 = 48.70%

3. All of the percentages should add up to equal 100%.

* Hope this helps

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the   per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:                   1 - α               α               α           = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

=  

or,

or,   = (at v = 1l)

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:               (1 - α)/0.5           α/0.5         α/0.5       = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

or,  

=  

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%

it is a good conductor

explanation: