pOH is calculated by taking negative logarithm of hydroxide ion concentration and pH is calculated by taking negative logarithm of hydrogen ion concentration

Putting in the values:

$pOH=-\log[7.609\times 10^{-13}]$

$pOH=12.1$

Now , $pH+pOH=14$

$pH=14-12.1=1.9$

$pH=-\log [H^+]$

$[H^+]=0.013M$

The hydrogen ion concentration in molarity is 0.013

3 0

2 years ago

How do you balance Sb+I2=SbI3 show work please

egoroff_w [7]

Answer:

2Sb + 3I₂ → 2SbI₃

Explanation:

The reaction expression we need to balance is:

Sb + I₂ → SbI₃

To balance this expression, we need to assign the variables a,b and c to each specie as the coefficient that will balance the expression.

aSb + bI₂ → cSbI₃

Conserving Sb : a = c

I : 2b = 3c

let a = 1, c = 1, b = $\frac{3}{2}$

Multiply through by 2;

a = 2, b = 3 and c = 2

2Sb + 3I₂ → 2SbI₃

7 0

2 years ago