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Harman [31]
3 years ago
11

0.00000000082 - scientific notation

Chemistry
1 answer:
Nadusha1986 [10]3 years ago
6 0

Answer:

8.2 x 106^-11

Explanation:

To begin this problem you must remember the basic rule of scientific notation, which is, must be between 1-10. .000000000082 is much smaller than 1. However by moving the decimal 11 spots to the right, we can make it 8.2

Continue to move the decimal to the right until the value is in the 1-10 range. Make sure to count the moves to the right.

Once the decimal is in the right spot count the spots moved.

Since the number is wayyy smaller than the answer given the number will be negative 10^-11, in order to make it what is was before.

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What is the mass in grams of 9.76 x 10 12 atoms of naturally occurring sodium?
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The number of moles in a substance indicates the amount of the substance that contains the same number of particles as 12 g of the Carbon-12 isotope [or equivalent to 6.02 × 10²³] (which is used as a standard in the world of moles).

Now,

          if     6.02 × 10²³ atoms are found in 1 mole ofsodium
   then let  9.76 × 10¹² atoms are found in     x

⇒     x  =    (9.76 × 10¹² )  ÷  (6.02 × 10²³)

            =  1.619 × 10⁻¹¹ mol


Now, mass = moles × molar mass

∴ mass of Na = 1.619 × 10⁻¹¹ mol ×  23 g/mol
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If a buffer solution is 0.220 M in a weak acid ( Ka=7.4×10−5) and 0.540 M in its conjugate base, what is the pH?
valkas [14]

Answer: the pH of the solution is 4.52

Explanation:

Consider the weak acid as Ha, it is dissociated as expressed below

HA     H⁺  +  A⁻

the Henderson -Haselbach equation can be expressed as;

pH = pKa + log( [A⁻] / [HA])

the weak acid is dissociated into H⁺ and A⁻ ions in the solution.

now the conjugate base of the weak acid HA is

HA(aq) {weak acid}     H⁺(aq)  +  A⁻(aq) {conjugate base}

so now we calculate the value of Kₐ as well as pH value by substituting the values of the concentrations into the equation;

pKₐ = -logKₐ

pKₐ = -log ( 7.4×10⁻⁵ )

pKₐ = 4.13

now thw pH is

pH = pKₐ  + log( [A⁻] / [HA])

pH = 4.13 + log( [0.540] / [0.220])

pH = 4.13 + 0.3899

pH = 4.5199 = 4.52

Therefore the pH of the solution is 4.52

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