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lozanna [386]
3 years ago
6

The weak acid HA is 5 % ionized (dissociated) in a 0.35 M solution.

Chemistry
1 answer:
ddd [48]3 years ago
5 0

Answer:idk

Explanation:of a 0.26 M triethylamine, (C2H5)3N, solution. ... 5. Predict if a decrease in temperature is favorable for the following reactions: (5 pts) a. H2O(s)

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What do an electron and neutron have in common?
CaHeK987 [17]

Answer:

each particle exists inside an atom

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7 0
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8.41 x 10 to the -7 grams to microgram
mezya [45]
0.841 micrograms is the answer.

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Two posssible lewis structures for the molecule ch2s are given. determine the formal charge on each atom in both structures
jok3333 [9.3K]

The formal charge on each atom can be known by counting up the number of valence electrons that is present on each atom.

<h3>What is Lewis structure?</h3>

The term Lewis structure is the representation of a compound using its symbols and showing the valence electrons as dots. This question is incomplete as the Lewis structures were not shown.

If it is shown, we can know the formal charge on each atom by counting up the number of valence electrons that is present on each atom.

Learn more about Lewis structure: brainly.com/question/4144781

8 0
2 years ago
Students conduct an experiment:
sweet-ann [11.9K]

Answer:

The balloon becomes inflated

Explanation:

The equation of the reaction between baking soda (sodium bicarbonate) and vinegar(ethanoic acid) is shown below;

NaHCO3 + HC2CH3O2 ------> NaC2H3O2 + H2O + CO2

The gas (CO2) evolved in the process leads to the inflation of the balloon dropped on the bottle in which the reaction is taking pace.

This observation provides evidence that a gas was really evolved in the reaction.

7 0
3 years ago
A 10.0 mL sample of 0.25 M NaOH(aq) is titrated with 0.10 M HCl(aq) (adding HCl to NaOH). Determine which region on the titratio
Anna11 [10]

Answer:

1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.

2) The pH of the solution after adding HCl is 12.6

Explanation:

10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.

nNaOH=\frac{0.25mol}{L} .10.0 \times 10^{-3} L=2.5 \times 10^{-3}mol

nHCl=\frac{0.10mol}{L} \times 15.0 \times 10^{-3} L=1.5 \times 10^{-3}mol

There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.

                    NaOH       +       HCl       ⇒       NaCl      +         H₂O

Initial          2.5 × 10⁻³         1.5 × 10⁻³               0                      0

Reaction    -1.5 × 10⁻³        -1.5 × 10⁻³          1.5 × 10⁻³          1.5 × 10⁻³

Final            1.0 × 10⁻³               0                 1.5 × 10⁻³          1.5 × 10⁻³

The concentration of NaOH is:

[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M

NaOH is a strong base so [OH⁻] = [NaOH].

Finally, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log 0.040 = 1.4

pH = 14 - pOH = 14 - 1.4 = 12.6

5 0
3 years ago
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