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Nikolay
3 months ago
10

A chemical engineer must report the average volume of a certain pollutant produced by the plants under her supervision. Here are

the data she has been given by each plant: plant volume of pollutant Macon 0.519 LAllegheny 44.67 LDoheny 0.826 LWhat average volume should the chemical engineer report? Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
lions [2.9K]3 months ago
5 0

Answer:

The chemical engineer should report an average volume of 15.34 L.

Explanation:

Volume from Macon = 0.519 L

Volume from Allegheny = 44.67 L

Volume from Doheny = 0.826 L

The average concentration of pollutants is calculated as:

\frac{0.519 L+44.67 L+ 0.826L}{3}=\frac{46.015 L}{3}=15.33833 L

When adding or subtracting, the least precise figure after the decimal point dictates the significant figures in the result.

In this case, 44.67 features two significant figures post-decimal, which means the final answer should also retain two decimal places.

15.33833 L\approx 15.34 L

The chemical engineer should report an average volume of 15.34 L.

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it takes 151 kJ/mol to break an iodine-iodine single bond. calculate the maximum wavelength of light for which an iodine-iodine
alisha [2963]

Answer:

To break a single I-I bond, the wavelength of light required is 7.92 × 10⁻⁷ m

Explanation:

The energy needed to break one mole of iodine-iodine single bonds is 151 KJ

The energy necessary to rupture one iodine-iodine bond is calculated as (151 KJ/mol) / 6.02 × 10²³/mol = 2.51 × 10⁻²² KJ

or

2.51 × 10⁻¹⁹ J

Formula:

E = hc / λ    

Where h is Planck's constant    = 6.626 × 10⁻³⁴ js

c is the speed of light = 3 × 10⁸ m/s

λ = wavelength

Solution:

E = hc / λ  

λ   = hc / E

λ   =  (6.626 × 10⁻³⁴ js × 3 × 10⁸ m/s ) / 2.51 × 10⁻¹⁹ J

λ   = 19.878 × 10⁻²⁶ j.m / 2.51 × 10⁻¹⁹ J

λ   = 7.92 × 10⁻⁷ m

6 0
3 months ago
Hydrogen has three isotopes with mass numbers of 1, 2, and 3 and has an average atomic mass of 1.00794 amu. This information ind
VMariaS [2998]

The answer is actually 3, believe me.

Explanation:

7 0
4 months ago
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A 0.821 gram sample of pure NH F was treated with 25.0 mL of 1.00 M NaOH
KiRa [2933]
A total of 0.0222 moles of NaOH are necessary to react with NH4F. \nBased on the reaction NH4F + NaOH --> NaF + NH3 + H2O, we start with: \nMass of NH4F = 0.821 g, NaOH concentration = 1 M, volume of NaOH = 25 mL. \nTo find moles: moles of NaOH = (CV)/1000. Thus, moles of NaOH = (1 * 25)/1000 = 0.025 moles of NaOH used. \nThe molar mass of NH4F is 37 g/mol, making moles of NH4F = 0.821 / 37 = 0.0222 moles. \nThis shows that NaOH is in excess, with 0.025 - 0.0222 = 0.0028 moles of NaOH remaining. Hence, 0.0222 moles of NaOH are needed to react with NH4F.
8 0
3 months ago
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