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Zanzabum
3 years ago
6

Choose the number of significant figures indicated. 0.078

Chemistry
2 answers:
Feliz [49]3 years ago
5 0

Answer : The number of significant figures are, two

Explanation :

Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Rules for significant figures:

  1. Digits from 1 to 9 are always significant.
  2. All non-zero numbers are always significant. For example: 654, 6.54 and 65.4 all have three significant figures.
  3. All zero’s between integers are always significant. For example: 5005, 5.005 and 50.05 all have four significant figures.
  4. All zero’s preceding the first integers are never significant. For example: 0.0078 has two significant figures.
  5. All zero’s after the decimal point are always significant. For example: 4.500, 45.00 and 450.0 all have four significant figures.

The give number is, 0.078

By using rule 4, we get the number of significant figures which are '7' and '8' that means the number of significant figures are, 2

Hence, the number of significant figures are, two

MAXImum [283]3 years ago
4 0
2 significant figures
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givi [52]

Answer:

Volume of NaOH required = 3.61 L

Explanation:

H2SO3 is a diprotic acid i.e. it will have two dissociation constants given as follows:

H2SO3\leftrightarrow H^{+}+HSO3^{-} --------(1)

where,  Ka1 = 1.5 x 10–2  or pKa1 = 1.824

HSO3^{-}\leftrightarrow H^{+}+SO3^{2-} --------(2)

where,  Ka2 = 1.0 x 10–7 or pKa2 = 7.000

The required pH = 6.247 which is beyond the first equivalence point but within the second equivalence point.

Step 1:

Based on equation(1), at the first eq point:

moles of H2SO3 = moles of NaOH

i.e. \ 5.13\  moles/L*0.385L = moles\  NaOH\\therefore, \ moles\  NaOH = 1.98\ moles\\V(NaOH)\ required = \frac{1.98\ moles}{0.615\ moles/L} =3.22L

Step 2:

For the second equivalence point setup an ICE table:

                  HSO3^{-}+OH^{-}\leftrightarrow H2O+SO3^{2-}

Initial           1.98                    ?                                       0

Change      -x                       -x                                       x

Equil           1.98-x                 ?-x                                    x

Here, ?-x =0 i.e. amount of OH- = x

Based on the Henderson buffer equation:

pH = pKa2 + log\frac{[SO3]^{2-} }{[HSO3]^{-} } \\6.247 = 7.00 + log\frac{x}{(1.98-x)} \\x=0.634 moles

Volume of NaOH required is:

\frac{0.634\ moles}{0.615 moles/L}=0.389L

Step 3:

Total volume of NaOH required = 3.22+0.389 =3.61 L

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(62.94 u) isotope = 69 %

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