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Elina [12.6K]
3 years ago
9

You purchase a bottle of concentrated sulfuric acid from a chemical supplier. The bottle reads Sulfuric acid (95% w/w) plastic c

oated safety bottle. The label lists the density of the acid as 1.85 g/mL and the molar mass as 98.08 g/mol, but the label fails to list the molarity of the concentrated acid! Calculate the molarity of the sulfuric acid based upon the information given.
Chemistry
1 answer:
My name is Ann [436]3 years ago
4 0

Answer:

The molarity of the sulfuric acid is 0.018 M

Explanation:

The molarity of a solution is the number of moles of the solute (sulfuric acid in this case) in a 1-liter solution.

Every 100 g of the solution, we have 95 g sulfuric acid because its concentration is 95% w/w.

With the density, we can calculate how many liters are 100 g of solution:

density = mass / volume

1.85 g / ml = 100 g / volume

volume = 100 g / 1.85 g/ml

volume = 54.1 ml or 0.0541 l

Now, we know that we have 95 g sulfuric acid in 0.0541 l solution. In 1 l, we have then:

1 l * 95g / 0.0541 l = 1.756 g sulfuric acid.

But we want to know how many moles sulfuric acid we have per liter. Then, using the molar mass, we can calculate how many moles there are in 1.756 g sulfuric acid:

1.756 g * 1 mol / 98.08 g = 0.018 mol

The molarity is 0.018 M

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What is the molecule shown below?
FrozenT [24]

Answer:

2,3–dimethylpentane

Explanation:

To know which option is correct, we shall determine the name of the compound.

To obtain the name of the compound, do the following:

1. Determine the longest continuous carbon chain. This gives the parent name of the compound.

2. Identify the substituent group attached to the compound.

3. Locate the position of the substituent group by giving it the lowest possible count.

4. Combine the above to obtain the name of the compound.

Now, we shall determine the name of the compound as follow:

1. The longest continuous carbon chain is 5. Thus, the parent name of the compound is pentane.

2. The substituent group attached is methyl (–CH₃)

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8 0
2 years ago
Two equilibrium reactions of nitrogen with oxygen, with their corresponding equilibrium constants (Kc) at a certain temperature,
7nadin3 [17]

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Kc = 1.54e - 31 / 2.61e - 24

Explanation:

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2)   N_{2}(gas) + 1/2O_{2}(gas)\rightarrow N_{2}O(gas)  ; Kc = 2.16e - 24

   upon reversing  ( 2 )  equation

     N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)   Kc = 1/2.16e - 24  

    now adding 1 and reversed equation (2)

       N_{2}(gas) + O_{2}(gas)\rightarrow 2NO(gas)

      N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)

   we get ,

                  N_{2}O(gas) + 1/2O_{2}(gas)\rightarrow 2NO(gas)  Kc = 1.54e-31 × 1/2.61e - 24

       equilibrium constant of equation (3) is -

            Kc = 1.54e - 31 / 2.61e - 24

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