Answer:
the initial concentration of SCN- in the mixture is 0.00588 M
Explanation:
The computation of the initial concentration of the SCN^- in the mixture is as follows:
As we know that
As it is mentioned in the question that KSCN is present 10 mL of 0.05 M
So, the total milimoles of SCN^- is
= 10 × 0.05
= 0.5 m moles
The total volume in mixture is
= 45 + 10 + 30
= 85 mL
Now the initial concentration of the SCN^- is
= 0.5 ÷ 85
= 0.00588 M
hence, the initial concentration of SCN- in the mixture is 0.00588 M
I thinlk it's by radiation?......
Explanation:
Given parameters:
Number of moles of the sulfur trioxide = 1.55kmol = 1.55 x 10³mole
Unknown:
Mass of the sulfur trioxide = ?
Solution:
To solve for the mass of the sample of sulfur trioxide:
- Find the molar mass of the compound i.e SO₃
atomic mass of S = 32g
O = 16g
molar mass = 32 + 3(16) = 80g/mol
mass of SO₃ = number of moles x molar mass
mass of SO₃ = 1.55 x 10³ x 80 = 124000g or 124kg
Learn more:
mole calculation brainly.com/question/13064292
#learnwithBrainly
The answer
first of all, we should know that NaOH is a strong base. For such a product, the conentration of the OH - is equivalent to the concentration of the NaOH itself.
that means:
[ OH -] = [ NaOH] =<span>0.001 62
and for a strong basis, pH can be calculated as pH = 14 + log </span>[ OH -]
first we compute log [ OH -] :
log [ OH -] = log (0.001 62)= -2.79
finally pH = 14 -2.79 = 11.20
Predators? I just looked it up lol