Question

calculate ℅ of sulphur in K2SO4

Answer 1

$\boxed{ Hello Mate...}$

$\boxed{ Your...Answer..}$

$\boxed{ Let ..x ..be...the... sulphur...}$

$\boxed{ Oxidation... no... of ..Oxygen(O) is -2 }$

$\boxed{ Oxidation... no... of ...Potassium (K) is +1}$

$\boxed{ Now ..Acc. To.. Your ...Question.}$

$\boxed{ 2(1) + x + (-2)4 = 0 }$

$\boxed{ 2 + x - 8 = 0 }$

$\boxed{ x = 8 - 2 }$

$\blue{ x= +6 }$

$\boxed{ Be Brainlyy....}$

Answer 2

Answer:

PERCENTAGE OF SULPHUR = 18.39

Explanation:

K=39 S=32 0=16

2 * 37 + 32 + 4 * 16

74 + 32 + 64

= 174

32/174 * 100/1

= 18.39