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3 years ago

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32.33 mL of 1.031 M potassium hydroxide were required to reach the endpoint of a titration of 50.00 mL of nitric acid. What was

the original concentration of nitric acid in that 50.00 mL sample?

1 answer:

tester [92]3 years ago

4 0

The molar concentration of the nitric acid solution was 0.6666 mol/L.

<em>Balanced equation</em>: KOH + HNO_3 → KNO_3 + H_2O

<em>Moles of KOH</em>: 32.33 mL KOH × (1.031 mmol KOH /1 mL KOH)

= 33.33 mmol KOH

<em>Moles of HNO_3</em>: 33.33 mmol KOH× (1 mmol HNO_3/1 mmol KOH)

= 33.33 mmol HNO_3

<em>Concentration of KOH</em>: <em>c </em>= "moles"/"litres" = 33.33 mmol/50.00 mL

= 0.6666 mol/L

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Some patterns and trend that are present in the periodic table would be

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3 years ago

0.65 moles of O2 originally at 85°C is cooled

soldier1979 [14.2K]

Answer:

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Explanation:

We can solve this problem by using the equation of state for an ideal gas, which relates the pressure, the volume and the temperature of an ideal gas:

$pV=nRT$

where

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V is its volume

n is the number of moles

R is the gas constant

T is the absolute temperature

In this problem we have:

n = 0.65 mol is the number of moles of the gas

V = 8.0 L is the final volume of the gas

$T=40C+273=313 K$ is the temperature of the gas

$R=0.082 atm L mol^{-1} K^{-1}$ is the gas constant

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4 years ago

Text 8.7 Using the virial equation of state for hydrogen at 298 K given in problem 7 (text 8.6), calculate a. The fugacity of hy

gregori [183]

Answer:

a). P = 688 atm

b). P = 1083.04 atm

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Explanation:

a). Fugacity 'f' can be calculated from the following equations :

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$PV=R(1+6.4\times 10^{-4} P)$ ........(1)

Also, PV=ZRT for real gases .......(2)

∴ $ZRT=RT(1+6.4\times 10^{-4} P)$

$Z=1+6.4\times 10^{-4} P$

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$\ln \frac{f}{P}=\int^P_0\left(\frac{1+6.4\times 10^{-4}P-1}{P}\right) dP$

$\ln \frac{f}{P}=\int^P_0\left(\frac{6.4\times 10^{-4}P}{P}\right) dP$

$\ln \frac{f}{P} = 6.4 \times 10^{-4} P$

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$\Delta G=R[\ln\frac{P_2}{P_1}+6.4 \times 10^{-4}(P_2-P_1)]$

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