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LekaFEV [45]
3 years ago
8

What is the mass of 2M H2SO4 in a 50ml of solution ​

Chemistry
1 answer:
Cloud [144]3 years ago
8 0

Answer:

It is given in the question that molarity of the sulphuric acid is 2M, the volume of the sulphuric acid is 20 mL, the volume of the solution is 1`L.

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Fill in the following information for the element
e-lub [12.9K]

Answer:

Period: 6 block

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Family: Lanthanides

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The identities of substances are the same before and after which type of change
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Arrange the following compounds from left to right in order of increasing percentage by mass of hydrogen: a) H2O, b) C12H26, c)
Nuetrik [128]

Answer:

(A) N4H6  (B) H2O  (C) LiH  (D) C12H26

Explanation:

The given compounds have been arranged  from left to right in order of increasing percentage by mass of hydrogen.

The percent by mass of hydrogen can be calculated by mass of hydrogen in that compound divided by total mass of that compound and finally multiplying the result with 100 to obtain the required percentage.

4 0
3 years ago
If 50 ml of 0.235 M NaCl solution is diluted to 200.0 ml what is the concentration of the diluted solution
Helen [10]

This is a straightforward dilution calculation that can be done using the equation

M_1V_1=M_2V_2

where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.

Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:

M_2=\frac{M_1V_1}{V_2}.

Substituting in our values, we get

\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].

So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.

5 0
3 years ago
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