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Ilya [14]
3 years ago
5

Consider the first-order reaction shown here. the yellow spheres in the pictures to the right represent the reactant,

Chemistry
1 answer:
Luda [366]3 years ago
8 0
The rate constant of the reaction K we can get it from this formula:

K=㏑2/ t1/2 and when we have this given (missing in question):
that we have one jar is labeled t = 0 S and has 16 yellow spheres inside and the jar beside it labeled t= 10 and has 8 yellow spheres and 8 blue spheres and the yellow spheres represent the reactants A and the blue represent the products B

So when after 10 s  and we were having 16 yellow spheres as reactants and becomes 8 yellow and 8 blue spheres as products so it decays to the half amount so we can consider T1/2 = 10 s
a) by substitution in K formula:
∴ K = ㏑2 / 10 = 0.069 
The amount of A (the reactants) after N half lives = Ao / 2^n
b) so no.of yellow spheres after 20 s (2 half-lives) =  16/2^2 = 4
and the blue spheres = Ao - no.of yellow spheres left = 16 - 4 = 12
c) The no.of yellow spheres after 30 s (3 half-lives) = 16/2^3 = 2
and the blue spheres = 16 - 2 = 14

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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
3 years ago
14.A sample of fluorine gas has a density of _____
yarga [219]

Answer:

d = 0.793 g/L

Explanation:

Given data:

Density of fluorine gas = ?

Pressure of gas = 0.554 atm

Temperature of gas = 50 °C (50+273.15K = 323.15 K)

Solution:

Formula:

PM = dRT

M = molar mass of gas

P = pressure

R = general gas constant

T = temperature

d = PM/RT

d = 0.554 atm × 37.99 g/mol / 0.0821 atm.L /mol.K × 323.15 K

d = 21.05 atm.g/mol/26.53 atm.L /mol

d = 0.793 g/L

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Potential to Kinetic Energy. 
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How are pressure units converted ?
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