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2 years ago

What characteristic of the light is affected by the amplitude of the light

Chemistry

2 answers:

valkas [14]2 years ago

4 0

Answer: Amplitude of the light wave impacts brightness; the wavelength is what impacts/is associated with color.

Vadim26 [7]2 years ago

3 0

Answer:

I think B but I'm not sure.

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An aqueous solution containing 10 g of an optically pure compound was diluted to 500 mL with water and was found to have a speci

lapo4ka [179]

Answer:

<u>Optical purity = 76.9231 %</u>

<u>Specific rotation of mixture = - 97.6923 °</u>

Explanation:

The mass of the racemic mixture = 3 g

It means it contains R enantiomer = 1.5 g

S enantiomer = 1.5 g

Amount of Pure R = 10 g

Total R = 11.5 g

Total volume = 500 mL + 500 mL = 1000 mL = 1 L

[R] = 11.5 g/L

[S] = 1.5 g/L

Enantiomeric excess = $\frac {Excess}{Total\ Concentration}\times 100$ = $\frac {11.5-1.5}{11.5+1.5}\times 100$ = 76.9231 %

<u>Optical purity = 76.9231 %</u>

Also,

Optical purity = $\frac {optical\ rotation\ of\ mixture}{optical\ rotation\ of\ pure\ enantiomer}\times 100$

Optical rotation of pure enantiomer = −127 °

$76.9231=\frac {optical\ rotation\ of\ mixture}{-127^0}\times 100$

<u>Specific rotation of mixture = - 97.6923 °</u>

6 0

2 years ago

would you expect potassium to have a high electronegativity or a low electronegativity? explain your answer

GalinKa [24]

A low electronegativity

Explanation:

Potassium is a metal that is expected to have a very low electronegativity value.

Electronegativity is the relative tendency by which an atom attracts valence electrons in a chemical bond.

Potassium is an element in the first group on the periodic table.

The common trend is that electronegativity increases from left to right and decreases down a group.

Potassium as metal will prefer to lose electrons rather than attracting because that will make it achieve the octet configuration that will ensure its stability.
This is why it will have low electronegativity.

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4 0

2 years ago

Why did workers form labor unions?

USPshnik [31]

Answer:

Labor unions were created in order to help the workers with work-related difficulties such as low pay, unsafe or unsanitary working conditions, long hours, and other situations. ... Sometimes the unions organized strikes in order to try to change the conditions of the workers.

Explanation:

5 0

2 years ago

Read 2 more answers

Calculate the maximum numbers of moles and grams of H₂S that can form when 158 g of aluminum sulfide reacts with 131 g of water:

Phantasy [73]

What is Chemical Reaction?

A chemical reaction is the chemical transformation of one set of chemical components into another.

Main Content

Mass of aluminium sulfide is 158g

Mass of water is 131g

The chemical reaction: $Al_{2}S_{3} +H_{2}O _\to Al(OH)_{3} + H_{2}S$

First, balance the chemical equation

$Al_{2}S_{3} + 6H_{2}O \to 2Al(OH)_{3} + 3H_{2}S$

Aluminium sulfide has a molar mass of 150.16 g/mol and water has a molar mass of 18.02 g/mol. As a result, the moles of aluminum sulfide are computed as follows:

$n_{Al_{2}S_{3} } = \frac{Mass}{Molar mass}\\n_{Al_{2} S_{3} } = \frac{158g}{150.16g/mol} \\n_{Al_{2}S_{3} }=1.05 mol$

From the chemical reaction , the ratio of molar is 3mol $H_{2}S/1 mol Al_{2}S_{3}.$ So, the moles of hydrogen sulfide are:

$n_{H_{2} O} =\frac{131g}{18.02g/mol}$

= 7.26mol

From the chemical reaction, the molar ratio is 3 mol $H_{2}S/6$ mol $H_{2}O.$ So, the moles of hydrogen sulfide are:

Moles of $H_{2}S$ formed = 7.26 mol $H_{2}O \times \frac{3 mol H_{2}S }{6 mol H_{2} O} }$

Th liming reactant is $Al_{2}S_{3}$ beacuse the mass of $Al_{2}S_{3}$ forms less product than water. Therefore, the maximum number of moles of $H_{2}S$ is 3.15 mol. We know that molar mass of $H_{2}S$ is 34.10g/mol. So, the maximum mass of $H_{2}S$ formed is,

$m_{H_{2}S } = n_{H_{2}S } \times$ Molar mass of $H_{2}S$

= 3.15 mol $\times$ 34.10g/mol

= 107.4g

Now, multiplying the number of moles of $Al_{2}S_{3}$ by the molar ratio between $Al_2S_3$ and $H_2O$ which is 6mol $H_2O/1mol$ $Al_2S_3$ we get the number of moles of $H_2O$ reacted.

Moles of $H_2O$ reacted = 1.05 mol $Al_{2}S_3 \times \frac{6 mol H_2O}{1 mol Al_2S_3}$

= 6.31 mol $H_2O$

The mass of $H_2O$ is,

$m_{H_{2} O}$ = 6.31 mol $\times$ 18.02g/ mol

= 114g

On subtracting, the mass of $H_2O$ reacted from the given mass of $H_2O$ is,

$m_{H_2O}$ = (131-114)g

= 17g

Hence, the excess remaining reactant is 17g

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8 0

2 years ago

Write a balanced half-reaction for the reduction of permanganate ion mno−4 to solid manganese dioxide mno2 in acidic aqueous sol

Svetlanka [38]

The half-reaction includes either the reduction or the oxidation reaction of the redox reactions. In acidic solution permanganate ion will react with hydrogen ion to yield manganese ion and water.

<h3>What are Redox reactions?</h3>

Redox or oxidation-reduction reactions are the chemical reactions in which the oxidation and the reduction of the chemical species occur simultaneously.

Permanganate (VII) ion is a strong oxidizing agent and gets easily reduced to manganese ion in presence of the hydrogen ion in an acidic solution.

The balanced half-reaction for reduction is shown as,

$\rm MnO_{4}^{-} \; (aq)+ 8H^{+} \; (aq)+ 5e^{-} \rightarrow Mn^{2+} \; (aq)+ 4H_{2}O \; (l)$

Learn more about reduction reactions here:

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8 0

1 year ago