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1 year ago

what volume of a 0.138 m potassium hydroxide solution is required to neutralize 26.0 ml of a 0.205 m nitric acid solution?

1 answer:

4 0

A neutralization titration is a chemical response this is used to decide the composition of an answer and what kind of acid or base is in it. This is a way of volumetric analysis and the formula is ( $M1V1 = M2V2$ ).

Utilize the titration method of $M1V1 = M2V2$ in view that we're given the concentrations of every compound and the quantity of $KOH$ . Let: M1 = 0.138M, V1 = x, M2 = 0.205M, V2 = 26.0 ML.

M1 = initial mass
V1= initial volume
M2 = final mass
V2= final volume

$(M1V1 = M2V2)$
(0.138)(V1) = (0.205)x(26.0)
V2=(0.205)x(26.0)\ 0.138
V2 = 47.10 M/L
The final value of Volume needed for neutralization of nitric acid solution is V2 = 47.10 M/L

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Trimix 10/50 is a gas mixture that contians 10% oxygen and 50% helium, and the rest is nitrogen. If a tank of trimix 10/50 has a

Nataly_w [17]

Answer: 1.61 x 10⁴ kPa

Dalton's law states that the sum of the partial pressures of each gas equals the total pressure of the gas mixture. According to this law,

Pi = xi P

where Pi is the partial pressure of the gas i, xi is the mole fraction of the gas i in the gas mixture and P is the total pressure.

The mole fraction is defined as the quotient between the moles of solute (ni) and the total moles of the mixture (nt), which is calculated by adding the moles of all its components:

xi = $\frac{n_{i} }{n_{t} }$

In the Trimix 10/50 mix you have 10% oxygen, 50% helium and 40% nitrogen.

To calculate the total number of moles of the mixture and thus determine the molar fraction of helium, we consider 100 g and calculate the number of moles that represent 10 g of O₂ (n₁), 50 g of He (n₂) and 40 g of N₂ (n₃):

n₁ = 10 g x $\frac{1 mol}{31.998 g}$ = 0.313 mol

n₂ = 50 g x $\frac{1 mol}{8.005 g}$ = 6.246 mol

n₃ = 40 g x $\frac{1 mol}{28.013 g}$ = 1.428 mol

Then the total number of moles (nt) will be:

nt = n₁ + n₂ + n₃ = 0.313 mol + 6.246 mol +1.428 mol

nt = 7,987 mol

Then, the mole fraction of helium (x₂) in the mixture will be,

x₂ = $\frac{6.246 mol}{7.987 mol}$ = 0.78

and the partial pressure of helium in the mixture, according to Dalton's law, will be:

P₂ = x₂ P = 0.78 x 2.07 x 10⁴ kPa

P₂= 1.61 x 10⁴ kPa

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3 years ago

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