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GrogVix [38]
3 years ago
14

What is the electrical power generated from a voltage of 12.3 V if the current is 9.42 amps

Chemistry
1 answer:
miv72 [106K]3 years ago
3 0

Answer:

the electrical power would be 115.86 watts. have a nice day

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Given pH = 8.55 Find: [H3O+] and [OH-] Is this acidic, basic or neutral?
Firlakuza [10]

Answer:

Explanation:

The given pH = 8.55

Unknown:

[H₃O⁺] = ?

[OH⁻] = ?

In order to find these unknowns we must first establish some relationship.

                      pH = -log[H₃O⁺]

                   8.55 = -log[H₃O⁺]

                    [H₃O⁺] = inverse log₁₀(-8.55) = 2.82 x 10⁻⁹moldm⁻³

To find the [OH⁻],

                     pH + pOH = 14

                               pOH = 14 - pH = 14 - 8.55

                                pOH = 5.45

              pOH = -log[OH⁻]

               [OH⁻] = inverse log₁₀ (-5.45) = 3.55 x 10⁻⁶moldm⁻³

The solution is basic because it has more concentration of OH⁻ ions compared to H⁺ ions.

7 0
3 years ago
Write the formulas of the compounds formed by NH4+ with the following anions: NO3−, HCO3−, SO32−, PO43−. Use the given order of
tekilochka [14]
NH₄NO₃
NH₄HCO₃
(NH₄)₂SO₃
(NH₄)₃PO₄
5 0
3 years ago
Which metals have similar chemical properties? Check all that apply. barium
rodikova [14]
Alkina metals all have the similar proteries
6 0
3 years ago
Things an animal does to survive are called...
Elina [12.6K]

Things an animal does to survive are called behavioral adaptations. The correct option is c.

<h3>What is behavioral adaptation?</h3>

The adaptation of behavioral and physical characters that help to survive them.

For example, hibernating during winter, migration, etc. There are two types of behavioral adaptation, learned and instinctive.

Thus, the correct options are c: behavioral adaptations.

Learn more about behavioral adaptation

brainly.com/question/12101480

#SPJ1

3 0
2 years ago
A cell was prepared by dipping a Cu wire and a saturated calomel electrode into 0.10 M CuSO4 solution. The Cu wire was attached
lara [203]

Answer:

a)  cu2+ + 1Hg (l) 1Cl- equilibrium cu (s) + Hg2Cl2 (s)

b)  0.068 V.

Explanation:

A) Cu2+ + 2e- euilibrium cu (s)

 Hg2Cl2 + 2e- equilibrium 2Hg (l) + 1cl-

Cell Reaction: cu2+ + 1Hg (l) 1Cl- equilibrium cu (s) + Hg2Cl2 (s)

B) To calculate the cell voltage

E = E_o Cu2+/Cu - (0.05916 V / 2) log 1/Cu2+

putting values we get

 = 0.339V + (90.05916V/2)log(0.100) = 0.309V

 E_cell = E Cu2+/Cu - E SCE = 0.309 V - 0.241 V = 0.068V.

6 0
2 years ago
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