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UNO [17]
3 years ago
6

What is the mass percent concentration of a saline solution prepared by dissolving 1.00 mol of Na2SO4 in 1.00 L of water?

Chemistry
1 answer:
Iteru [2.4K]3 years ago
8 0

Answer:

good luck hope you do good

Explanation:

good luck

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Answer the following questions for H2CrO4:
Sholpan [36]

If this molecule is one half of a buffer, then the formula of the second half of the buffer is M2CrO4 where M is a univalent metal.

<h3>What is a strong acid?</h3>

A weak acid is one that is able to ionize completely in solution. The acid called chromic acid H2CrO4 is not  able to ionize completely in solution.

We know that a buffer is composed of a weak acid and its salt or a weak base and its salt hence if the acid H2CrO4 is present in a buffer then the other half must be  salt of the acid.

If this molecule is one half of a buffer, then the formula of the second half of the buffer is M2CrO4 where M is a univalent metal.

Learn more about buffer:brainly.com/question/22821585

#SPJ1

7 0
2 years ago
1. What is the equality between mL and L?
Mice21 [21]

Answer:

1 liter (L) = 1000 milliliters (mL)

Explanation:

5 0
3 years ago
First, get about 20 mL of the quinine stock solution in a clean beaker to prepare 250 mL of about 2 ppm quinine in 0.05 M H2SO4.
iris [78.8K]

Answer:

Volume of acid, Va=250mL; Volume of quinine,Vb=20mL; Molarity of acid, Ma=0.05M.

Molar mass of acid= H2+S+O4= 2+32+(16X4)= 2+32+64=98g

Concentration of acid, Ca= Molar mass of acid/ Ma =98/0.05=1960g/mol

Explanation: To calculate concentration of quinine, Cb is as follow

Va*Ca=Vb*Cb

∴ Cb=Va*Ca/Vb =250*1960/20 =24500g/mol

5 0
3 years ago
A certain metal M forms a soluble sulfate salt MSO, Suppose the left half cell of a galvanic cell apparatus is filled with a 3.0
dimaraw [331]

Explanation:

It is known that for high concentration of M^{2+}, reduction will take place. As, cathode has a positive charge and it will be placed on left hand side.

Now, E^{o}_{cell} = 0 and the general reaction equation is as follows.

         M^{2+} + M \rightarrow M + M^{2+}

                3.00 M        n = 2       30 mM

         E = 0 - \frac{0.0591}{2} log \frac{50 \times 10^{-3}}{1}

            = -\frac{0.0591}{2} log (5 \times 10^{-2})

            = 0.038 V

Therefore, we can conclude that voltage shown by the voltmeter is 0.038 V.

5 0
3 years ago
Which is true about the dissolving process in water?
amid [387]
Water molecules move througout the solute
5 0
3 years ago
Read 2 more answers
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