Answer:
In order to initiate most fission reactions, an atom is bombarded by a neutron to produce an unstable isotope, which undergoes fission. When neutrons are released during the fission process, they can initiate a chain reaction of continuous fission which sustains itself.
The volume of the gas at a temperature of 405.0 K would be 607.5 mL. Making option D the right answer to the question.
What is the volume of the gas?
To find the volume of the gas, the equation to be used would have to be combine gas law.
Combine gas law as the name suggest uses the combination of Charles law which measures Volume against temperature, and Gay-Lussac's law which measures Pressure/Temperature, and Boyle's law which measures pressure X volume where k is constant.
Using the combine law to find the volume, we have:
P₁V₁/T₁=P₂V₂/T₂
Where P₁ = initial pressure
V₁ = initial volume
T₁ = initial temperature
P₂ = final pressure
V₂ = final volume
T₂ = final temperature
P₁ = 2.25atm
V₁ = 450.0 mL
T₁ = 300 K
T₂ = 405.0 K
V₂ = ?
D) 607.5 mL
= [2.25(450)]÷300=[2.25(V₂]÷405
Making V₂ the subject
3.375=2.25 V₂ ÷ 405
V₂ = 3.375 x 405 ÷ 2.25
V₂ = 607.5 mL
In summary, a gas with an initial pressure of 2.25atm, an initial pressure of 450.0 mL and an initial temperature of 300 K would have a final volume of 607.5 mL if the temperature is increased to 405.0 K.
Learn more about Combine gas law here: brainly.com/question/13538773
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Yes they are stable because they follow octet rule but am not sure if they are exist separetly
Answer:
ok
Explanation:
why I needed to give a answer to a question. I want to ask the question. brainly solve this issue. i cant ask questions without answering to 1 questions.
As per the given chemical formula- Na2CO3.10H2O, one mole of the chemical compound contains 13 moles of oxygen atoms. Hence
Number of moles of oxygen atoms in one mole of Na2CO3.10H2O = 13
number of moles of oxygen atoms in 0.2 moles of Na2CO3.10H2O = 13 X 0.2 = 2.6
Now, one mole of a substance contains 6.022 X 10^23 particles of the substance. Thus
number of atoms of oxygen in one mole of oxygen atom = 6.022 X 10^23
number of moles of oxygen atoms in 2.6 moles of oxygen atoms = 2.6 X 6.022 X 10^23 = 15.657 X 10^23
= 1.566 X 10^24
Thus, there are 1.566 X 10^24 atoms of oxygen in 0.2 moles of Na2CO3.10H2O.
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