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Alika [10]
3 years ago
15

How can criteria be used to help define the problem?

Chemistry
1 answer:
Andre45 [30]3 years ago
4 0

Answer:

It helps get you to the proof of the matter

Explanation:

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Name two things that reduce friction
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Oil and a smooth surface
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In the following equation, ______ is being oxidized and ______ is being reduced.
Mashcka [7]

oxidation \: number \: of \: oxygen =  \\ before \: rxn =  - 2 \\ after \: rxn =  - 2

oxidation \: number \: of \: hydrogen = \\ before \: rxn =  + 1 \\ after \: rxn =  \\ 2x - 2 = 0 \\ x =  + 1

oxidation \: number \: of \: carbon =  \\ before \: rxn =  \\ x  - 6 =  - 2 \\ x = 4 \\ after \: rxn =  \\ x - 4 = 0 \\ x = 4

<h2>Option A</h2>

oxidation \: numbers \: remain \: constant \\ so \: none \:a re \: undergoing \: oxidation \: \\ nor \: reduction \:

6 0
1 year ago
Use enthalpies of formation to determine the ΔHreaction for the reaction
Daniel [21]

Answer:

The enthalpy is increased by the increased heat of the reaction.

Explanation:

In this reaction, as the transition from solid to liquid state, enthalpy increases, that is, the heat applied to change its state is greater and this increases, reaching a mayor disorder.

If the reaction increases its heat, and a liquid state passes, the enthalpy increases, therefore the disorder also and the entropy will also be increased.

5 0
3 years ago
Calculate [H3O+] and [OH−] for each of the following solutions at 25 ∘C given the pH. pH= 8.74, pH= 11.38, pH= 2.81
Gnom [1K]

Answer:

Explanation:

Given parameters;

pH  = 8.74

pH = 11.38

pH = 2.81

Unknown:

concentration of hydrogen ion and hydroxyl ion for each solution = ?

Solution

The pH of any solution is a convenient scale for measuring the hydrogen ion concentration of any solution.

It is graduated from 1 to 14

      pH = -log[H₃O⁺]

      pOH = -log[OH⁻]

 pH + pOH = 14

Now let us solve;

   pH = 8.74

             since  pH = -log[H₃O⁺]

                           8.74 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{8.74}

                             [H₃O⁺]  = 1.82 x 10⁻⁹mol dm³

       pH + pOH = 14

                 pOH = 14 - 8.74

                  pOH = 5.26

                  pOH = -log[OH⁻]

                     5.26  = -log[OH⁻]

                     [OH⁻] = 10^{-5.26}

                      [OH⁻] = 5.5 x 10⁻⁶mol dm³

2.  pH = 11.38

             since  pH = -log[H₃O⁺]

                           11.38 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{11.38}

                             [H₃O⁺]  = 4.17 x 10⁻¹² mol dm³

           pH + pOH = 14

                 pOH = 14 - 11.38

                  pOH = 2.62

                  pOH = -log[OH⁻]

                     2.62  = -log[OH⁻]

                     [OH⁻] = 10^{-2.62}

                      [OH⁻] =2.4 x 10⁻³mol dm³

3. pH = 2.81

             since  pH = -log[H₃O⁺]

                           2.81 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{2.81}

                             [H₃O⁺]  = 1.55 x 10⁻³ mol dm³

           pH + pOH = 14

                 pOH = 14 - 2.81

                  pOH = 11.19

                  pOH = -log[OH⁻]

                     11.19  = -log[OH⁻]

                     [OH⁻] = 10^{-11.19}

                      [OH⁻] =6.46 x 10⁻¹²mol dm³

5 0
3 years ago
5. Compare the threats faced by the Everglades and the Louisiana wetan
SOVA2 [1]
The Everglades and the Louisiana wetan are the same
5 0
3 years ago
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