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3 years ago

9

HELP WITH CHEMISTRY!!

1 answer:

poizon [28]3 years ago

5 0

The second one should be correct

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In a combustion reaction, pentane (molar mass (MM) = 72.17 g/mol) burns, in the presence of oxygen (MM = 32.00 g/mol), to produc

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Phosphorus has three electrons in its 3p sublevel and sulfur has four. Phosphorus should have the lower ionization energy but it

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Read 2 more answers

Perform the calculations and determine the absolute and percent relative uncertainty. Express each answer with the correct numbe

Lana71 [14]

Answer:

Explanation:

Given the equation:

$\implies \dfrac{[9.8(\pm0.3)-2.31(\pm 0.01)]}{8.5(\pm0.6)}$

The absolute uncertainty in a measurement is the term used to describe the degree of inaccuracy.

The first step is to determine the algebraic value on the numerator.

Algebraic value = 9.8 - 231

= 7.49

The absolute uncertainty = $\sqrt{(abs. uncertainty_{v_1})^2+(abs. uncertainty_{v_2})^2}$

absolute uncertainty = $\sqrt{(0.3)^3 + (0.01)^2}$

= $\sqrt{0.09 + 0.0001}$

= 0.300167

∴

[9.8(±0.3) - 2.31(±0.01)] = 7.49(±0.300167)

The division process now is:

$\implies \dfrac{[9.8(\pm0.3)-2.31(\pm 0.01)]}{8.5(\pm0.6)}= \dfrac{7.49 (\pm 0.300167)}{8.5 (\pm0.6)}$

Relative uncertainty = $\dfrac{(\pm 0.300167)}{7.49}\times 100 \ , \ \dfrac{(\pm 0.6) }{8.5} \times 100$

Relative uncertainty = ±4.007565% , ±7.058824%

$\text{Relative uncertainty} = \sqrt{(4.007565)^2+(7.058824)^2}$

$\text{Relative uncertainty} = \sqrt{16.06057723+49.82699626}$

$\text{Relative uncertainty} = \sqrt{65.88757349}$

$\text{Relative uncertainty} = 8.117116$

≅ 8%

The algebraic value = $\dfrac{7.49}{8.5}$

= 0.881176

≅ 0.88

The percentage of the relative uncertainty = $\dfrac{\text{Absolute uncertainty }}{\text{calculated value} }\times 100$

By cross multiplying:

$\text{Absolute uncertainty} (\%) = \dfrac{\text{relative uncertainty} \times \text{calculated value}}{100}$

$\text{Absolute uncertainty} (\%) = \dfrac{8.117116\times 0.881176}{100}$

$\text{Absolute uncertainty} (\%) = 0.0715260$

$\mathbf{\text{Absolute uncertainty} (\%) \simeq 0.07}$

Finally:

$\mathbf{\implies \dfrac{[9.8(\pm0.3)-2.31(\pm 0.01)]}{8.5(\pm0.6)}= 0.88 \pm (0.07) \pm 8\%}$

8 0

3 years ago