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lesya [120]
3 years ago
14

What is 4,678,679 dL in scientific notation

Chemistry
1 answer:
vesna_86 [32]3 years ago
8 0
4.7e^6 is how I would write it. You simply count in from the right side where the decimal is until you reach the point just before the 4.
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An unknown compound, X is thought to have a carboxyl group with a pKa of 2.0 and another ionizable group with a pKa between 5 an
Westkost [7]

Answer:

7.3

Explanation:

By Henderson Hasselbalch equation we can calculate the pH or the pOH of a solution by its pKa. Remember that pH = -log[H^{+}], and pKa = -logKa. Ka is the equilibrium constant of the acid.

Henderson Hasselbalch equation :

pH = pKa - log \frac{[HA]}{[A^{-}]}

Where [HA] is the concentration of the acid, and [A^{-}] is the concentration of the anion which forms the acid.

So, acid X, has two ionic forms, the carboxyl group and the other one. First, we have 0.1 mol/L of the acid, in 100 mL, so the number of moles of X

n1 = (0.1 mol/L)x(0.1 L) = 0.01 mol

When it dissociates, it forms 0.005 mol of the carboxyl group and 0.005 mol of the other group. Assuming same  stoichiometry.

Adding NaOH, with 0.1 mol/L and 75 mL, the number of moles of OH^- will be

n2 = (0.1 mol/L)x(0.075 L) = 0.0075 mol

So, the 0.0075 mol of OH^- reacts with 0.005 mol of carboxyl, remaining 0.0025 mol of OH^-, which will react with the 0.005 mol of the other group. So, it will remain 0.0025 mol of the other group.

The final volume of the solution will be 175 mL, but both concentrations (the acid form and ionic form) have the same volume, so we can use the number of mol in the equation.

Note that, the number of moles of the acid form is still 0.01 mol because it doesn't react!

So,

6.72 = pKa - log \frac{0.01}{0.0025}

6.72 = pKa - log 4

pKa - log4 = 6.72

pKa = 6.72 + log4

pKa = 6.72 + 0.6

pKa = 7.3

8 0
3 years ago
How can a match light without a spark
Serhud [2]

Answer:

Use friction

Explanation:

Friction is the force that stops something in motion. It also heats up something, creating a gravitational heat! A match striked against the lighter will create a flame, because of friction :)

8 0
3 years ago
Read 2 more answers
How many joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point?
Korolek [52]

Answer: 1123000 Joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point

Explanation:

Latent heat of vaporization is the amount of heat required to convert 1 mole of liquid to gas at atmospheric pressure.

Amount of heat required to vaporize 1 mole of lead =  177.7 kJ

Molar mass of lead = 207.2 g

Mass of lead given = 1.31 kg = 1310 g       (1kg=1000g)

Heat required to vaporize 207.2 of lead = 177.7 kJ

Thus Heat required to vaporize 1310 g of lead =\frac{177.7}{207.2}\times 1310=1123kJ=1123000J

Thus 1123000 Joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point

7 0
3 years ago
2.00 gallons is equal to how many liters?
Naily [24]

Answer:

4 to 5 . is the answer

Explanation:

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2 years ago
Use ocean trenches in a sentence
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Just do this for the sentence lol "Ocean Trenches are trenches in the Ocean".
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