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e-lub [12.9K]
3 years ago
5

Round 9.9100881 × 101 to 6 significant digits in decimal form.

Chemistry
1 answer:
vichka [17]3 years ago
8 0

Answer:

Round to six significant = 1000.92

Explanation:

Given data:

Given value = 9.9100881×101

Round given value to six significant figures

Solution:

Significant figures:

All non-zero digits are consider significant figures like 1, 2, 3, 4, 5, 6, 7, 8, 9.

Leading zeros are not consider as a significant figures. e.g. 0.03 in this number only one significant figure present which is 3.

Zero between the non zero digits are consider significant like 104 consist of three significant figures.

The zeros at the right side e.g 2400 are also significant. There are four significant figures are present.

How to round the digits:

First we will look at the six digit of given values.

we will draw a line after six digit then we will look the digit after line.

It the digit after line which is 7th digit is 5 or more than 5 then one will be added to the six digit.

If it it is less than five then six digit remain same.

After this we will add zeros to the right till decimal point.

In given measurement:

9.9100881×101  = 1000.9188981

1000.9188981

Six digit is 1 and 7th is 8 thus 1 is converted into 2.

Round to six significant = 1000.92

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At 66.0 ∘c , what is the maximum value of the reaction quotient, q, needed to produce a non-negative e value for the reaction so
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Answer:

\boxed{5.5 \times 10^{-28}}

Explanation:

We must use the Nernst equation

E = E^{\circ} - \frac{RT}{zF}\ln Q

Step 1. Calculate E°

SO₄²⁻(aq) + 4H⁺(aq) + 2e⁻ ⇌ SO₂(g) + 2H₂O(ℓ)

<u>2Br⁻⇌ Br₂(aq) + 2e⁻                                                                </u>

SO₄²⁻(aq) + 4H⁺(aq) + 2Br⁻(aq) ⇌ Br₂(aq) + SO₂(g) + 2H₂O(ℓ)

E° = 0.17 - 1.0873 = -0.92 V

Step 2. Calculate Q

E   =  0      V

E° = -0.92 V

R = 8.314 J·K⁻¹mol⁻¹

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F = 96 485 C/mol

Calculations:  

T = 66.0 + 273.15 = 339.15 K

0 = -0.92 - \dfrac{8.314 \times 339.15}{2 \times 96 485}\ln Q\\\\0.92 = -0.0146\times \ln Q \\\\\ln Q = -\dfrac{0.92}{0.0146} = -62.8\\\\Q = e^{-62.8} =\boxed{5.5 \times 10^{-28}}\\

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Explanation:

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The relation between Moles, Number of Atoms and Avogadro's Number is given as,

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Putting values,

              Number of Moles  =  1.0 × 10⁶ Atoms ÷ 6.022 × 10²³ Atoms/mol

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