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Maru [420]
3 years ago
4

The student performs a second titration using the 0.10MNaOH(aq) solution again as the titrant, but this time with a 20.mL sample

of 0.20MHCl(aq) instead of 0.10MHCl(aq) . (f) The box below to the left represents ions in a certain volume of 0.10MHCl(aq) . In the box below to the right, draw a representation of ions in the same volume of 0.20MHCl(aq) . (Do not include any water molecules in your drawing.)

Chemistry
1 answer:
marishachu [46]3 years ago
6 0

Answer:

Here's what I get  

Explanation:

In the figure below, assume the black dots are H⁺ ions  and the open circles are Cl⁻ ions.

Also, assume that the square for 0.10 mol·L⁻¹ HCl contains four dots and four circles.

Then, the square for 0.20 mol·L⁻¹ HCl will contain eight dots and eight circles.

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3 years ago
A chemistry teacher adds 50.0 mL of 1.50 M H2SO4 solution to 200 mL of water. What is the concentration of the final solution? U
Aleks [24]
Hello!

The concentration of the final solution when a<span> chemistry teacher adds 50.0 mL of 1.50 M H2SO4 solution to 200 mL of water is 0,3 M

To calculate that, you'll need to use the dilution law, where initial and final concentrations are M1 and M2 respectively, and initial and final volumes are V1 and V2, as shown below. Keep in mind that the final volume is the sum of the 200 mL of water and the 50 mL of H</span>₂SO₄ that were added by the teacher. 

M2= \frac{M1*V1}{V2}= \frac{1,50 (mol H_2SO_4/L)*50mL}{(50 mL + 200 mL)}=0,3(mol H_2SO_4/L)

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6 0
3 years ago
Read 2 more answers
A sample of cesium carbonate, weighing 3.80 g, requires 1.90 g of hydrogen bromide gas to completely decompose to water, cesium
vladimir1956 [14]

Answer:

Choice A. 0.50 g.

Explanation:

According to the question, the reaction here converts

  • caesium (cesium) carbonate and
  • hydrogen bromide

to

  • cesium bromide,
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By the Law of Conservation of Mass, matter can neither be created nor destroyed in a chemical reaction. (Shrestha et. al, Introductory & GOB Chemistry, Chemistry Libretexts, 2019.)

In other words, the mass of the reactants, combined, shall be the same as the mass of the products, combined.

What's the mass of the reactants?

\rm \underbrace{\rm 3.80\;g}_{\mathrm{Cs_2CO_3}} + \underbrace{\rm 1.90\; g}_{\mathrm{HBr}} = 5.70\;g.

What's the mass of the products?

Let m(\mathrm{CO_2}) represent the mass of carbon dioxide produced in this reaction.

The mass of the products will be:

\rm \underbrace{\rm 5.20\;g}_{\mathrm{CeBr}\text{ and }\mathrm{H_2O}} + \underbrace{m(\mathrm{CO_2})}_{\mathrm{CO_2}}.

The two masses shall be equal. That is:

\rm 5.20\; g + \mathnormal{m}(\mathrm{CO_2}) = 5.70\;g.

m(\mathrm{CO_2}) = \rm 0.50\;g.

In other words, by the Law of Conservation of Mass, the mass of carbon dioxide produced in this reaction will be \rm 0.50\;g.

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