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defon
3 years ago
12

Which methods will help you properly identify pH with indicator paper? Check all that apply. Set a piece of pH paper in front of

each solution before testing to keep organized. Dip the pH paper into each test tube and lay it on the table to let it change color. Use a transfer pipet to remove a few drops from the solution to drop onto the paper. Wait a minute or so before reading color. Use the same piece of indicator paper for each solution to conserve paper.
Chemistry
2 answers:
AnnyKZ [126]3 years ago
7 0

Answer:

A) Set a piece of pH paper in front of each solution before testing to keep organized.

C) Use a transfer pipet to remove a few drops from the solution to drop onto the paper.

D) Wait a minute or so before reading color.

monitta3 years ago
4 0

Set a piece of pH paper in front of each solution before testing to keep organized.

Use a transfer pipet to remove a few drops from the solution to drop onto the paper.

Wait a minute or so before reading color.

Is the correct answer...

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If I have a 78c and I fail a test that’s worth 15% of my grade what would it drop to ?
koban [17]
Depends on the grade of the test. But for the most part if it’s below a 78, it’ll drop, If it’s above it’ll raise. It shouldn’t drop or raise it more than 3 point/percent.
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3 years ago
An acid containing the cooh group is called a carbo-oxy acid.
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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

5 0
3 years ago
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The excerpt from an essay about higher education that most clearly shows an example of logos is:

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Learn more about logos from

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Using the equation, <span>ΔS(surr)=-(-283kJ/298K)=0.9497kJ/K or 949.7J/K

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I hope that helps.</span>
8 0
3 years ago
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