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Digiron [165]
3 years ago
11

A 232-lb fullback runs the 40-yd dash at a speed of 19.8 ± 0.1 mi/h.

Chemistry
1 answer:
Neporo4naja [7]3 years ago
5 0

Answer:

(a)  7.11 x 10⁻³⁷ m

(b)  1.11 x 10⁻³⁵ m

Explanation:

(a)  The de Broglie wavelength is given by the expression:

λ = h/p = h/mv

where h is plancks constant, p is momentum which is equal to mass times velocity.

We have all the data required to calculate the wavelength, but first we will have to convert the velocity to m/s, and the mass to kilograms to work in metric system.

v = 19.8 mi/h x ( 1609.34 m/s ) x ( 1 h / 3600 s ) = 8.85 m/s

m = 232 lb x ( 0.454 kg/ lb ) = 105.33 kg

λ = h/ mv = 6.626 x 10⁻³⁴ J·s / ( 105.33 kg x 8.85 m/s ) = 7.11 x 10⁻³⁷ m

(b) For this part we have to use the uncertainty principle associated with wave-matter:

ΔpΔx > = h/4π

mΔvΔx > = h/4π

Δx = h/ (4π m Δv )

Again to utilize this equation we will have to convert the uncertainty in velocity to m/s for unit consistency.

Δv = 0.1 mi/h x ( 1609.34 m/mi ) x ( 1 h/ 3600 s )  

     = 0.045 m/s

Δx = h/ (4π m Δv ) = 6.626 x 10⁻³⁴ J·s / (4π x 105.33 kg x 0.045 m/s )

     = 1.11 x 10⁻³⁵ m

This calculation shows us why we should not be talking of wavelengths associatiated with everyday macroscopic objects for we are obtaining an uncertainty of 1.11 x 10⁻³⁵ m for the position of the fullback.

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Answer:

410mg/day *\frac{1000 ug}{1 mg}

Recommended daily Amount (RDA) of magnesium is  410,000 μg/day.

Explanation:

There are 1000μg in 1 mg and 1000 mg in 1g

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The Recommended daily Amount (RDA) is 410 mg/day of magnesium. Converting 410 mg/day into μg/day

410mg/day *\frac{1000 ug}{1 mg}

It will become 410,000 μg/day

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The powder could be acetaminophen, analgesic  having chemical formula C_{17}H_{21}NO_{4}

<h3>What is an empirical formula?</h3>

A chemical formula showing the simplest ratio of elements in a compound rather than the total number of atoms in the molecule.

We are given:

Percentage of C = 67.31 %

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Percentage of O = 21.10 %

Let the mass of the compound be 100 g. So, the percentages given are taken as mass.

Mass of C = 67.31 g

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To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of carbon = \frac{mass}{molar \;mass}

Moles of carbon =\frac{67.31g}{12g/mole}

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=6.978 moles

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For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.329 moles.

We get the ratio of C : H : N : O = 17 : 21 : 1 : 4

The empirical formula for the given compound is C_{17}H_{21}NO_{4}.

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<h3>What is percentage by mass?</h3>

The term percentage by mass refers to the amount of a particular moiety by mass in a molecule.

In this case, we know that the molar mass of Na2CO3.10H2O is 286 g/mol the molar mass of the 10 moles of water is 180 g/mol hence the percentage by mass of water in the compound is; 180 g/mol /286 g/mol  * 100/1 = 62.9%

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