Calculate the mass of lead(ll)tetraoxosuplhatevi formed when 0.5mol-¹ of tetraoxosulphate vi when react with lead(ll)chloride

The mass of lead(ll)tetraoxosuplhate formed when 0.5mol-¹ of tetraoxosulphate react with lead(ll)chloride is :

151.63 grams

Explanation:

PbCl2 =  lead(ll)chloride

PbSO4 = lead(ll)tetraoxosuplhate

H2SO4 = hydrogen tetraoxosulphate

HCl = Hydrochloric acid

The balanced equation for the reaction is :

The stoichiometric coefficients indicates that :

1 mole PbCl2 = 1 mole H2SO4 =  1 mole PbSO4 = 2 mole HCl

1 mole H2SO4 produces  = 1 mole PbSO4

0.5 mole produce = 0.5 mole of PbSO4

We have 0.5 mole of PbSO4

So , we are asked to calculate the mass of 0.5 mole pf PbSO4

Molar mass of PbSO4 = mass of Pb + Mass of S + 4(mass of O)

= 207.2 + 32.06 +4(16)

=303.26 grams

151.63 grams

where is the following

explanation:

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.