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ivanzaharov [21]
3 years ago
8

You need to make 25 microliters of a 3M NaOh solution for a test reagent. Your laboratory routinely stocks 500 milliliters of a

10M NaOh solution. How would you prepare your solution? 

Chemistry
1 answer:
MrRissso [65]3 years ago
8 0

Answer:

0.0075 milliliters (7.5 microliters) will be taken from the stock solution and then diluted with 0.0249925 milliliters (24.9925 microliters) of water to make 0.025 milliliters (25 microliters) of NaOH solution.

Explanation:

This is a problem of dilution using the equation:

<em>initial concentration x initial volume = final concentration x final volume.</em>

The final volume to be prepared is 25 microliters.

The final concentration to be prepared is 3 M.

The initial volume to be taken is not known yet.

The initial concentration is 10 M.

Now, let's substitute these parameters into the the equation above.

10 x initial volume = 3 x 25

Initial volume = 3 x 25/10

     = 7.5 microliters

Note that: 1 microliter = 0.001 milliliters

Hence,

7.5 microliters = 0.0075 milliliters

<u>This means that an initial volume of 0.0075 milliliters (7.5 microliters) will be taken from the stock solution. This amount will then be diluted with 0.0249925 milliliters (24.9925 microliters) of water to make 0.025 milliliters (25 microliters) of NaOH solution.</u>

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The titration of Na2CO3 with HCl has the following qualitative profile: a. Identify the major species in solution at points A-F.
exis [7]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Note: This question is incomplete and lacks very important data to solve this question. But I have found the similar question which shows the profiles about which question discusses. Using the data from that question, I have solved the question.

a) We need to find the major species from A to F.

Major Species at A:

1. Na_{2} CO_{3}

Major Species at B:

1. Na_{2} CO_{3}

2. NaHCO_{3}

Major Species at C:

1. NaHCO_{3}

Major Species at D:

1. NaHCO_{3}

2. H_{2}CO_{3}

Major Species at E:

1. H_{2}CO_{3}

Major Species at F:

1. H_{2}CO_{3}

b) pH calculation:

At Halfway point B:

pH = pKa_{1} + log[CO_{3}.^{-2}]/[HCO_{3}.^{-1}]

pH = pKa_{1} = 6.35

Similarly, at halfway point D.  

At point D,

pH = pKa_{2} + log [HCO_{3}.^{-1}]/[H2CO_{3}]

pH = pKa_{2} = 10.33

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An aluminum ion has a +3 charge. What change did the atom undergo to attain this charge?
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3 years ago
The density of H2O2 is 1.407 g/mL, and the density of O2 is 1.428 g/L. How many liters of O2 can be made from 55 mL H2O2
balandron [24]

Explanation:

mass H2O2 = 55 mL(1.407 g/mL) = 80.85 g

molar mass H2O2 = 2(1.01 g/mol) + 2(16.00 g/mol) = 34.02 g/mol

moles H2O2 = 80.85 g/34.02 g/mol = 2.377 moles H2O2

For each mole of H2O2 you obtain 0.5 mole of O2 (see the equation).

moles O2 = 2.377 moles H2O2 (1 mole O2)/(2 moles H2O2) = 1.188 moles O2

Now, you need the temperature.  If you are at STP (273 K, and 1.00 atm) then 1 mole of an ideal gas at STP has a volume of 22.4 L.  Without temperature you are not really able to continue.  I will assume you are at STP.

Volume O2 = 1.188 moles O2(22.4 L/mole) = 0.0530 L of O2.

which is  53 mL.

8 0
3 years ago
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