How many grams of nitric acid (HNO3) must be dissolved in 740 mL of water to make a solution with a pOH = 8.3?

The system N2O4 <--->2NO2 maintained in a closed vessel at 60º C & a pressure of 5 atm has an average (i.e. observed) molecular weight of 69, calculate Kp.

At what pressure at the same temperature would the observed molecular weight be (230/3)

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The system N2O4 <--->2NO2 maintained in a closed vessel at 60º C & a pressure of 5 atm has an average (i.e. observed) molecular weight of 69, calculate Kp.
At what pressure at the same temperature would the observed molecular weight be (230/3)

just change the value

i found it on y!A

find the [H+} in pH of 4.5 ,
10^-4.5 =
[H+] = 3.16 e-5 molar

find the moles acid in 350 ml
0.350 litres @ 3.16 e-5 mol/litre =
1.11 e-5 moles of HNO3

find the grams using molar mass:
1.11 e-5 moles of HNO3 @ 63 g/mol =
0.00070 grams of HNO3
======================================…

@ pH of 2.9, find the H+:
10^-2.9 =
0.00126 molar H+

use the dilution formula:
C1V1 = C2V2
(4.2M) V1 = (0.00125M) (500ml)
V1 = 0.15 ml of 4.2 Molar is required

see same question

how many grams nitric acid HNO3 must be dissolved in 350 ml water to make a solution with pH of 4.5"?

or "how many ml of a 4.2M nitric acid solution are required to make 500 ml of solution with pH of 2.9

Well, let me calculate it using my clown calculator... Just kidding! I don't have a calculator, but I can still make you smile.

To find the grams of nitric acid needed, we need to use some chemistry magic.

First, let's convert the pOH to OH- concentration using the formula: pOH = -log[OH-]. So, 8.3 = -log[OH-].

Now, let's convert that to [OH-]. Take the inverse of both sides, and we get [OH-] = 10^(-8.3).

Since nitric acid (HNO3) is a strong acid, it will completely dissociate in water to form H+ and NO3-. That means the H+ concentration will be equal to the OH- concentration, which is 10^(-8.3) M.

Finally, let's figure out the moles of HNO3 needed. Multiply the H+ concentration by the volume of water in liters:

moles of HNO3 = (10^(-8.3)) M * 0.74 L.

But we're not done clowning yet! We need to convert moles to grams using the molar mass of nitric acid (HNO3). One mole of HNO3 weighs 63 grams.

So, moles of HNO3 = 63 g/mol * (10^(-8.3)) M * 0.74 L.

Now, calculate that and you'll find the answer. And remember, laughter is the best solution!

To solve this problem, we need to use the relationship between pOH and concentration of hydroxide ions (OH-). Here are the steps:

Step 1: Calculate the concentration of hydroxide ions (OH-) using the pOH value given. Remember that pOH is the negative logarithm (base 10) of the hydroxide ion concentration.

pOH = -log[OH-]
8.3 = -log[OH-]

To remove the negative sign, we can take the antilog of both sides:
[OH-] = 10^(-pOH)

Calculate the concentration of hydroxide ions (OH-) using the above equation.

Step 2: Convert the concentration of hydroxide ions (OH-) to the concentration of nitric acid (HNO3). The reaction between nitric acid and water is:

HNO3 + H2O ⇌ H3O+ + NO3-

This means that for every nitric acid molecule, one hydrogen ion (H+) and one nitrate ion (NO3-) are produced. Since pOH refers to the concentration of hydroxide ions (OH-), we need to find the concentration of hydrogen ions (H+), which is equal to the concentration of nitric acid (HNO3).

Therefore, the concentration of nitric acid (HNO3) will be equal to the concentration of hydrogen ions (H+), which is the same as the concentration of hydroxide ions (OH-).

Step 3: Calculate the moles of nitric acid (HNO3) using the concentration and volume information provided.

moles of HNO3 = concentration of HNO3 × volume of solution (in liters)

Step 4: Convert moles of nitric acid (HNO3) to grams using the molar mass of HNO3.

mass of HNO3 = moles of HNO3 × molar mass of HNO3

Now let's apply these steps to solve the problem:

Step 1: Calculate the concentration of hydroxide ions (OH-):

[OH-] = 10^(-pOH)
[OH-] = 10^(-8.3)

Step 2: Convert the concentration of hydroxide ions (OH-) to the concentration of nitric acid (HNO3):

[HNO3] = [OH-]

Step 3: Calculate the moles of nitric acid (HNO3):
moles of HNO3 = [HNO3] × volume of solution (in liters)
The volume of solution is given as 740 mL, which should be converted to liters by dividing by 1000.

Step 4: Convert moles of nitric acid (HNO3) to grams using the molar mass of HNO3. The molar mass of nitric acid (HNO3) is 63.01 g/mol.

mass of HNO3 = moles of HNO3 × molar mass of HNO3

Now you can substitute the values into the equations and calculate the final answer.