Answer:
The wavelength of wave is 0.33 ×10²⁰ m.
Explanation:
Given data:
Frequency of wave = 9.12×10⁻¹² Hz
Wavelength of wave = ?
Solution:
Formula:
Speed of light = wavelength × frequency
c = λ × f
λ = c/f
This formula shows that both are inversely related to each other.
The speed of light is 3×10⁸ m/s
Frequency is taken in Hz.
It is the number of oscillations, wave of light make in one second.
Wavelength is designated as "λ" and it is the measured in meter. It is the distance between the two crust of two trough.
Now we will put the values in formula.
λ = 3×10⁸ m/s / 9.12×10⁻¹² Hz
Hz = s⁻¹
λ = 0.33 ×10²⁰ m
The wavelength of radiation is 0.33 ×10²⁰ m .
The piece of unknown metal is in thermal equilibrium with water such that Q of metal is equal to Q of the water. We write this equality as follows:
-Qm = Qw
Mass of metal (Cm)(ΔT) = Mass of water (Cw) (ΔT)
where C is the specific heat capacities of the materials.
We calculate as follows:
-(Mass of metal (Cm)(ΔT)) = Mass of water (Cw) (ΔT)
-68.6 (Cm)(52.1 - 100) = 42 (4.184) (52.1 - 20)
Cm = 1.717 -----> OPTION C
Nitrogen atomic# 7 7protons, 7electrons, 7 neutrons
Oxygen atomic# 8 8protons, 8electrons, 8neutrons
Phosphorous atomic# 15 15protons, 15 electrons, 15 neutrons
Sulfur atomic#16 16protons, 16electrons, 16 neutrons
<u>Answer:</u> The final temperature of the coffee is 43.9°C
<u>Explanation:</u>
To calculate the final temperature, we use the equation:

where,
q = heat released = 
m = mass of water = 10.0 grams
C = specific heat capacity of water = 4.184 J/g°C
= final temperature = ?
= initial temperature = 20°C
Putting values in above equation, we get:

Hence, the final temperature of the coffee is 43.9°C
Answer:-
The reaction of 2-bromopropane reacts with sodium iodide in acetone is an example of Sn2 reaction.
The I - attacks from backside to give the transition state for both.
If we compare the transition state for cyclobromopropane 2-bromopropane then we see in case of cyclobromopropane transition state, one of the H is very close to the incoming I -.
This results in steric strain and less stability of the transition state. Hence 2-bromopropane reacts with sodium iodide in acetone over 104 times faster than bromocyclopropane.