What is the ph of 5.0 l of an aqueous solution that contains 1.0 grams of hbr and 1.0 grams of nitric acid

Hello!:

Molar mass:

HBr = 80.91 g/mol

HNO3 = 63.01 g/mol

Number of moles:

moles HBr =  1.0 / 80.91 => 0.01235 moles of HBr

moles HNO3 = 1.0 / 63.01 => 0.01587 moles of HNO3

So , total moles of H⁺ :

0.01235 + 0.01587 => 0.02822 moles of H⁺

Molarity of solution:

M = n / V

M = 0.02822 / 5.0

M = 0.005644

Therefore:

pH = - log [ H⁺ ]

pH = - log [ 0.005644 ]

pH = 2.25

Hope that helps!

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2

points

explanation: