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3 years ago

What pressure is required to achieve a co2 concentration of 5.80×10−2 m at 20∘c?

Chemistry

1 answer:

skelet666 [1.2K]3 years ago

6 0

The concentration (c) of a gas at pressure (p) at a given temperature (T) through the Henry's constant kH

c = kH * p

at 20 C or 293 K, the henry's constant kH for CO2 = 0.037 mol/L-atm

Given concentration c = 0.0580 moles/L

Hence the pressure is given as:

p = c/kH = 0.0580 molesL-1/0.037 molL-1atm-1 = 1.57 atm

Pressure required is 1.57 atm

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Answer:

Name Phosphorus

Symbol P

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Explanation:

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3 years ago

50.00 mL of 0.10 M HNO 2 (nitrous acid, K a = 4.5 × 10 −4) is titrated with a 0.10 M KOH solution. After 25.00 mL of the KOH sol

boyakko [2]

Answer:

b. 3.35

Explanation:

To calculate the pH of a solution containing both acid and its salt (produced as a result of titration) we need to use Henderson’s equation i.e.

pH = pKa + log ([salt]/[acid]) (Eq. 01)

Where

pKa = -log(Ka) (Eq. 02)

[salt] = Molar concentration of salt produced as a result of titration

[acid] = Molar concentration of acid left in the solution after titration

Let’s now calculate the molar concentration of HNO2 and KOH considering following chemical reaction:

HNO2 + KOH ⇆ H2O + KNO2 (Eq. 03)

This shows that 01 mole of HNO2 and 01 mole of KOH are required to produce 01 mole of KNO2 (salt). And if any one of them (HNO2 and KOH) is present in lower amount then that will be considered the limiting reactant and amount of salt produced will be in accordance to that reactant.

Moles of HNO2 in 50 mL of 0.01 M HNO2 solution = 50/1000x0.01 = 0.005 Moles

Moles of KOH in 25 mL of 0.01 M KOH solution = 25/1000x0.01 = 0.0025 Moles

As it can be seen that we have 0.0025 Moles of KOH therefore considering Eq. 03 we can see that 0.0025 Moles of KOH will react with only 0.0025 Moles of HNO2 and will produce 0.0025 Moles of KNO2.

Therefore

Amount of salt produced i.e [salt] = 0.0025 moles (Eq. 04)

Amount of acid left in the solution [acid] = 0.005 - 0.0025 = 0.0025 moles (Eq.05)

Putting the values in (Eq. 01) from (Eq.02), (Eq. 04) and (Eq. 05) we will get the following expression:

pH= -log(4.5x10 -4) + log (0.0025/0.0025)

Solving above we get

pH = 3.35

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3 years ago