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MArishka [77]
2 years ago
10

Jupiter's moon lo orbits the planet at a distance of 421,700 km. What is the correct way to write this distance in scientific no

tation?
A. 42.17 x 10^4 km
B. 4.217 x 10^5 km
C. 42.17 x 10-4 km
D. 4.217 x 10-5 km
Chemistry
1 answer:
HACTEHA [7]2 years ago
4 0

Taking into account the definition of Scientific notation, the correct way to write the distance in scientific notation is option B. 4.217×10⁵ km.

First of all, scientific notation is a quick way to represent a number using powers of base 10. This notation is used to be able to express very large or very small numbers very easily.

The numbers are written as a product:

a×10ⁿ

where:

  • a is a real number greater than or equal to 1 and less than 10, to which a decimal point is added after the first digit if it is a non-integer number.
  • n = a whole number, which is called an exponent or an order of magnitude. Represents the number of times the point decimal is shifted. It is always an integer, positive if it is shifted to the left, negative if it is shifted to the right.

In this case, to write the number 421,700 in scientific notation, the following steps are performed:

  • The decimal point is moved to the left as many spaces until it reaches the right of the first digit. This number will be the value of a in the previous expression. Then a = 4.21700= 4.217
  • The base 10 is written with the exponent equal to the number of spaces that the decimal point moves. This is a positive number because the decimal point is shifted to the left, and it will have a value of n = 5.

So, the correct way to write the distance in scientific notation is option B. 4.217×10⁵ km.

Learn more about scientific notation:

brainly.com/question/18073768?referrer=searchResults

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2 years ago
determine the ph of a buffer that is 0.55 M HNO2 and 0.75 M KNO2. tha value of Ka for HNO2 is 6.8*10^-4
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Answer:

pH = 3.3

Explanation:

Buffer solutions minimize changes in pH when quantities of acid or base are added into the mix. The typical buffer composition is a weak electrolyte (wk acid or weak base) plus the salt of the weak electrolyte. On addition of acid or base to the buffer solution, the solution chemistry functions to remove the acid or base by reacting with the components of the buffer to shift the equilibrium of the weak electrolyte left or right to remove the excess hydronium ions or hydroxide ions is a way that results in very little change in pH of the system. One should note that buffer solutions do not prevent changes in pH but minimize changes in pH. If enough acid or base is added the buffer chemistry can be destroyed.

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]

= -log(6.8 x 10⁻⁴) + log[(0.75M)/(0.55M)]

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

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