The answer is B- Low reactivity.
Answer:
b) beta decay
Explanation:
Beta radiations:
Beta radiations are result from the beta decay in which electron is ejected. The neutron inside of the nucleus converted into the proton an thus emit the electron which is called β particle.
The mass of beta particle is smaller than the alpha particles.
They can travel in air in few meter distance.
These radiations can penetrate into the human skin.
The sheet of aluminium is used to block the beta radiation
⁴₆C → ¹⁴₇N + ⁰₋₁e
The beta radiations are emitted in this reaction. The one electron is ejected and neutron is converted into proton.
Answer:
the sodium nitrate in this investigation is
a) the solute
b) undergoing chemical change
<h3>
Explanation:</h3>
generally, solute is the substance which is added in solvent.
The substance which is taken in relatively more volume is the solvent.
so, here sodium nitrate (NaNO₃) is the solute which is dissolved in water (H₂O) , which is solvent.
The reaction between them is :
NaNO₃ + H₂O ⇒ Na⁺ (aq) + NO₃⁻ (aq) ; (aq) denotes aqueous;
so, here sodium nitrate undergoes chemical change into aqueous sodium ions(Na⁺) and aqueous nitrate ions(NO₃⁻).
Answer:
<u>Part -A:</u>
Sulfur released is
.
<u>Part-B:</u>
released of per EPA norms is
.
Explanation:
From the given,
The output = 1100 Mw
Efficiency = η = 0.4
Substitute the values in the following
Input = Output / η
![= \frac{1100}{0.4}= 2750 Mw](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B1100%7D%7B0.4%7D%3D%202750%20Mw)
Let's converts between joules to BTU.
1 BTU = 1056 J
<u>Part-A;</u>
![Energy\,\, required = Power \times time](https://tex.z-dn.net/?f=Energy%5C%2C%5C%2C%20required%20%3D%20Power%20%5Ctimes%20time)
![2750 \times 10^{6} \,\, J/s \times 3600s = 9.9 \times 10^{12}J](https://tex.z-dn.net/?f=2750%20%5Ctimes%2010%5E%7B6%7D%20%5C%2C%5C%2C%20J%2Fs%20%5Ctimes%203600s%20%3D%209.9%20%5Ctimes%2010%5E%7B12%7DJ)
![Energy \,\, required = \frac{9.9 \times 10^{22}}{1056}= 9.375 \times 10^{9} BTU](https://tex.z-dn.net/?f=Energy%20%5C%2C%5C%2C%20required%20%3D%20%5Cfrac%7B9.9%20%5Ctimes%2010%5E%7B22%7D%7D%7B1056%7D%3D%209.375%20%5Ctimes%2010%5E%7B9%7D%20BTU)
![Mass\,of\, coal\,required = \frac{9.375 \times 10^{9}}{14,000}=6.69 \times 10^{5}pounds](https://tex.z-dn.net/?f=Mass%5C%2Cof%5C%2C%20coal%5C%2Crequired%20%3D%20%5Cfrac%7B9.375%20%5Ctimes%2010%5E%7B9%7D%7D%7B14%2C000%7D%3D6.69%20%5Ctimes%2010%5E%7B5%7Dpounds)
But it has 2% sulfur
The mass of sulfur released
![6.69 \times 10^{5}pounds \,coal \times 0.02 = 1.338 \times 10^{4}pound](https://tex.z-dn.net/?f=6.69%20%5Ctimes%2010%5E%7B5%7Dpounds%20%5C%2Ccoal%20%5Ctimes%200.02%20%3D%201.338%20%5Ctimes%2010%5E%7B4%7Dpound)
Therefore, released sulfur is
.
<u>Part -B;</u>
One pound of sulfur produce two pounds of sulfur dioxide
Initial amount of produced sulfur =
![2 \times 1.338 \times 10^{4}pound = 2.676\times 10^{4}pound/hour](https://tex.z-dn.net/?f=2%20%5Ctimes%201.338%20%5Ctimes%2010%5E%7B4%7Dpound%20%3D%202.676%5Ctimes%2010%5E%7B4%7Dpound%2Fhour)
Assuming we added a 80% efficiency then,
Released sulfur dioxide = ![(1-0.8) \times SO_{2} \,produced =0.2 \times 2.676 \times 10^{4}Pounds/hr= 5.352 \times 10^{3}Pounds/hr](https://tex.z-dn.net/?f=%281-0.8%29%20%5Ctimes%20SO_%7B2%7D%20%5C%2Cproduced%20%3D0.2%20%5Ctimes%202.676%20%5Ctimes%2010%5E%7B4%7DPounds%2Fhr%3D%205.352%20%5Ctimes%2010%5E%7B3%7DPounds%2Fhr)
Energy produced in an hour = ![400 mw \times 1 hr = 1.1 \times 10^{6}kwh](https://tex.z-dn.net/?f=400%20mw%20%5Ctimes%201%20hr%20%3D%201.1%20%5Ctimes%2010%5E%7B6%7Dkwh)
![SO_{2}\,released \, as \, per EPA = \frac{5.352 \times 10^{3} \,pounds}{1.1 \times 10^{6}kwh}= 4.865 \times 10^{-3}p\kwh](https://tex.z-dn.net/?f=SO_%7B2%7D%5C%2Creleased%20%5C%2C%20as%20%5C%2C%20per%20EPA%20%3D%20%5Cfrac%7B5.352%20%5Ctimes%2010%5E%7B3%7D%20%5C%2Cpounds%7D%7B1.1%20%5Ctimes%2010%5E%7B6%7Dkwh%7D%3D%204.865%20%5Ctimes%2010%5E%7B-3%7Dp%5Ckwh)
Therefore,
released of per EPA norms is
.
B. the same number of protons.