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beks73 [17]
1 year ago
10

for number less than 0.1, such as 0.06, the zeros to the right of the decimal point but before the first nonzero digit

Chemistry
1 answer:
qaws [65]1 year ago
6 0

For the number less than 0.1 such as 0.006, the zeroes to the right of the decimal point but before the first non zero digit show the decimal place of the first significant digit.

  • The number that is given as digits is established using significant figures.
  • Any two non-zero digits that are separated by a zero are significant figure.
  • Every zero that is both to the right and left of a non-zero digit and the decimal point is not significant figure.
  • The quantity of significant figures frequently reveals the degree of measurement accuracy. From the first non-zero digits in the figure, we may determine the number of significant figures.

There is only one significant figure in the provided number 0.06. The decimal place of the first digit is indicated by the zeros that appear to the right of the decimal point but before the first non-zero digit.

Learn more about significant figure here:

brainly.com/question/11151926

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Answer:

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Explanation:

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In this problem, the weak electrolyte is HNO₂(aq) and the salt is KNO₂(aq). In equation, the buffer solution is 0.55M HNO₂ ⇄ H⁺ + 0.75M KNO₂⁻ . The potassium ion is a spectator ion and does not enter into determination of the pH of the solution. The object is to determine the hydronium ion concentration (H⁺) and apply to the expression pH = -log[H⁺].

Solution using the I.C.E. table:

              HNO₂ ⇄    H⁺   +   KNO₂⁻

C(i)        0.55M       0M      0.75M

ΔC            -x            +x          +x

C(eq)  0.55M - x       x     0.75M + x    b/c [HNO₂] / Ka > 100, the x can be                                    

                                                             dropped giving ...

           ≅0.55M        x       ≅0.75M        

Ka = [H⁺][NO₂⁻]/[HNO₂] => [H⁺] = Ka · [HNO₂]/[NO₂⁻]

=> [H⁺] = 6.80x010⁻⁴(0.55) / (0.75) = 4.99 x 10⁻⁴M

pH = -log[H⁺] = -log(4.99 x 10⁻⁴) -(-3.3) = 3.3

Solution using the Henderson-Hasselbalch Equation:

pH = pKa + log[Base]/[Acid] = -log(Ka) + log[Base]/[Acid]

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= -(-3.17) + 0.14 = 3.17 + 0.14 = 3.31 ≅ 3.3

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