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2 years ago

What was important about the Hydrogen bright line emission spectrum?

Chemistry

1 answer:

Art [367]2 years ago

5 0

Answer:

The emission spectrum lines are the changes in the quantum energy levels of the single electron in the Hydrogen atom.

Explanation:

Electrons move around the hydrogen atom in electron waves patterns. These waves occur in distinct quantum energy levels.

The change from one quantum energy level to another energy level has a definite energy level change. There is a wave length associated with each change in energy levels. These wave lengths have a color associated with them.

Low energy level changes have long wave lengths associated with them giving red and orange colors.

High energy level changes have short wave lengths associated with them giving blue and violet colors.

Note each atom has a unique emission spectrum associated with the energy levels and electron structure of the atom. Helium was first discovered by looking at its emission spectrum as seen in the sun.

Explanation:

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A blood sample with a known glucose concentration of 102.0 mg/dL is used to test a new at home glucose monitor. The device is us

Zinaida [17]

Answer:

A. 96.3 mg/dL

Absolute error: 5.7 mg/dL

Relative error: 5.6%

B. 97.2 mg/dL

Absolute error: 4.8 mg/dL

Relative error: 4.7%

C. 104.8 mg/dL

Absolute error: 2.8 mg/dL

Relative error: 2.7%

D. 111.5 mg/dL

Absolute error: 9.5 mg/dL

Relative error: 9.3%

E. 110.5 mg/dL

Absolute error: 8.5 mg/dL

Relative error: 8.3%

Explanation:

The formula for the absolute error is:

Absolute error = |Actual Value - Measured Value|

The formula for the relative error is:

Relative error = |Absolute error/Actual value|

In your exercise, we have that

Actual Value = 102.0 mg/dL

A. 96.3 mg/dL:

$E_{ABS} = |102.0 - 96.3| = 5.7mg/dL$

$E_{R} = \frac{5.7}{102} = 0.056 = 5.6%$

B. 97.2 mg/dL

$E_{ABS} = |102.0 - 97.2| = 4.8mg/dL$

$E_{R} = \frac{4.8}{102} = 0.047 = 4.7%$

C. 104.8 mg/dL

$E_{ABS} = |102.0 - 104.8| = 2.8mg/dL$

$E_{R} = \frac{2.8}{102} = 0.027 = 2.7%$

D. 111.5 mg/dL

$E_{ABS} = |102.0 - 111.5| = 9.5mg/dL$

$E_{R} = \frac{9.5}{102} = 0.093 = 9.3%$

E. 110.5 mg/dL

$E_{ABS} = |102.0 - 111.5| = 8.5mg/dL$

$E_{R} = \frac{8.5}{102} = 0.083 = 8.3%$

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tigry1 [53]

Answer:

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Find the reagent to sythesize 1- phenyl-1-propanol

vova2212 [387]

Answer:

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Explanation:

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3 years ago