A 25.0 mL sample of a solution of an unknown compound is titrated with a 0.115 M NaOH solution. The titration curve above was ob
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Answer:
28500 years
Explanation:
Applying,
A = A'()............... Equation 1
Where A = Original mass of Carbon-14, A' = Final mass of carbon-14 after decaying, x = total time, y = half-life.
From the question,
Given: A = 1 g, A' = 31.3 mg = 0.0313 g, y = 5700 years.
Substitute these values into equation 1
1 = 0.0313()
= 1/0.0313
= 31.95
≈ 32
≈ 2⁵
Equating the base and solve for x
x/5700 ≈ 5
x ≈ 5×5700
x ≈ 28500 years
Answer:
Explanation:
mole of O₂ =
= .25 moles
mole of CO₂
=
= .1818 moles
moles of SO₂
= .125 moles
Total moles of gas
= .5568 moles.
total volume of gas mixture
= 22.4 x .5568 liter ( volume of one mole of any gas = 22.4 liter)
= 12.47 liter.
gas will exert partial pressure according to their mole fraction
gas having greatest no of moles in the total mole will have greatest mole fraction so
O₂ will have greatest partial pressure.