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Leviafan [203]
3 years ago
5

What is the wavelength of radiation with a frequency of 3.68 X 10^12 /s?

Chemistry
1 answer:
konstantin123 [22]3 years ago
6 0

Answer:

The wavelength of radiation = 8.2 x 10⁻⁵

Explanation:

by the formula

v x λ = C

v = frequency  

λ = wavelength

C = speed of light (3 x 10⁸ m/s)

λ = C / v

λ = 3 x 10⁸ / 3.68 x 10^{12}

λ = 8.2 x 10⁻⁵

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Iodine-131 decays with a half-life of 8.02
dlinn [17]
Radioactive material undergoes 1st order decay kinetics.

For 1st order decay, half life = 0.693/k

where k = rate constant

k = 0.693/half life = 0.693/8.02 = 0.0864 day-1

Now, for 1st order reaction,
k = \frac{2.303}{t} X log \frac{initial.conc}{final.conc}

Given: t = 6.01d, initial conc. = 5mg

∴0.0864 = \frac{2.303}{6.01} X log \frac{5}{final.conc}
∴ final conc. = 2.975 mg
3 0
3 years ago
Compound a is an alkene that was treated with ozone (followed by dms) to yield only 4-heptanone. Draw the major product that is
olga nikolaevna [1]

Alkenes on reacting with ozone results in the formation of ozonide which undergo reductive  cleavage in presence of dimethyl sulfide to form carbonyl compounds (aldehyde or ketone). Whereas in presence of hydrogen peroxide it undergoes oxidative cleavage to form carboxylic acids or ketones.

Since, A alkene yields 4-heptanone only on treatment with ozone and DMS thus, it implies that both the chains on the side of the double-bond are similar the product is 4-heptanone that means the double bond is present between the chains at the 4th carbon. Therefore the structure of compound A is 4,5-dipropyloct-4-ene.

The reaction is as shown in the image.

The reaction of A with m-CPBA (meta-perchlorobenzoic acid) followed by aqueous acid H_3O^{+} is shown in the image.

m-CPBA (meta-perchlorobenzoic acid) is a peracid and forms epoxides on reacting with alkenes.

5 0
3 years ago
< please help!!! >
svlad2 [7]

Answer:

n = 0.573mol

Explanation:

PV = nRT => n = PV/RT

P = 1.5atm

V = 8.56L

R = 0.08206Latm/molK

T = 0°C = 273K

n = (1.5atm)(8.56L)/(0.08206Latm/molK)(273K) = 0.573mol

8 0
3 years ago
Which one is a decomposition reaction?​
vodomira [7]

Answer:

b no

Explanation:

because it is decomposing into two elements

8 0
2 years ago
Read 2 more answers
The normal boiling point of bromine is 58.8°C, and its enthalpy of vaporization is 30.91 kJ/mol. What is the approximate vapor p
saul85 [17]

Answer : The vapor pressure of bromine at 10.0^oC is 0.1448 atm.

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of bromine at 10.0^oC = ?

P_2 = vapor pressure of propane at normal boiling point = 1 atm

T_1 = temperature of propane = 10.0^oC=273+10.0=283.0K

T_2 = normal boiling point of bromine = 58.8^oC=273+58.8=331.8K

\Delta H_{vap} = heat of vaporization = 30.91 kJ/mole = 30910 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{1atm}{P_1})=\frac{30910J/mole}{8.314J/K.mole}\times (\frac{1}{283.0K}-\frac{1}{331.8K})

P_1=0.1448atm

Hence, the vapor pressure of bromine at 10.0^oC is 0.1448 atm.

4 0
3 years ago
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