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Montano1993 [528]
3 years ago
5

Write 0.000000093425 in Scientific Notation with 4 significant figures.

Chemistry
2 answers:
worty [1.4K]3 years ago
7 0

Answer:

9.343\times 10^{-8}

Explanation:

Scientific notation is the way of writing numbers which are either large or small. The number is written in the scientific notation when the number is between 1 and 10 and then multiplied by the power of 10.

The given number:

0.000000093425 can be written as 9.3425\times 10^{-8}

Answer upto 4 significant digits = 9.343\times 10^{-8}

Romashka-Z-Leto [24]3 years ago
6 0

<u>Answer:</u> The given number in scientific notation is 9.342\times 10^{-8}

<u>Explanation:</u>

Scientific notation is the notation where a number is expressed in the decimal form. This means that the number is always written in the power of 10 form. The numerical digit lies between 0.1.... to 9.9.....

If the decimal is shifting to right side, the power of 10 is negative and if the decimal is shifting to left side, the power of 10 is positive.

We are given:

A number having value = 0.000000093425

Converting this into scientific notation, we get:

\Rightarrow 0.000000093425=9.342\times 10^{-8}

Hence, the given number in scientific notation is 9.342\times 10^{-8}

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Consider a galvanic cell composed of the SHE and a half-cell using the following reaction: Ag+(aq) + e− → Ag(s)
MissTica

Answer:

a. 0.80V b. 2Ag⁺(aq) + H2(g) ⇄ 2Ag(s) +2H⁺(aq) c. i) 0.88 ii) 1.03 d. Cell is a ph meter with the potential being a function of hydrogen ion concentration

Explanation:

a. The two half cell reactions are

1. 2H⁺(aq) +2e⁻ → H₂(g) Eanode = 0.00V

2. Ag⁺(aq) + e⁻ → Ag(s) Ecathode = 0.80V

The balanced cell reaction is

2Ag(aq)⁺ + H₂(g) ⇄  2Ag(s) + 2H⁺(aq)

therefore Ecell = Ecathode - Eanode = 0.80 - 0.00 = +0.80V

b. Since the Ecell is positive, the spontaneous cell reaction under standar state conditions is

2Ag(aq)⁺ + H₂(g) ⇄  2Ag(s) + 2H⁺(aq)

c. Use Nernst Equation

E = Ecell - (0.0592/n)log([H⁺]/[Ag⁺]²[P H₂]), where n is the number of moles of Ag and P H₂= 1.0 atm

i) E = 0.80 - (0.0592/2)log(4.2x10^-2)/(1.0)²(1.0) = 0.88V

ii) E =  0.80 - (0.0592/2)log(9.6x10^-5)/(1.0)²(1.0) = 1.03V

d . From the above calculation we can conclude that the cell acts as a pH meter as a change in hydrogen ion concentration results in a change in the potential of the cell. A change of ph of 2.64 changes the E of cell by 0.15 V.

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3 years ago
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