<span>The correct answer is b. Radon. Oxygen, Hyrdrogen, and Boron, are not radioactivel ike Radon and it regenerates quickly meaning that even though it has a short half-life period, it stays for a long time once released. It also has no taste so it's difficult to notice without proper gear because you can't feel it.</span>
Explanation:
When we increase the temperature of a substance then there will occur an increase in the kinetic energy of its molecules.
Also, K.E = 
So, kinetic energy is directly proportional to the temperature.
Hence, when temperature and pressure are kept the same for both oxygen and hydrogen gas then values of their kinetic energy will be the same irrespective of their masses.
Thus, we can conclude that kinetic energy of oxygen molecule is the same as compared to hydrogen molecule.
Answer:
2NaOH (aq) + CaCl2 (aq) -> 2NaCl(aq) + Ca(OH)2(s)
Formula of precipitate: Ca(OH)2 <em>(s)</em>
Explanation:
First, we do the double replacement reaction to determine our chemical equation between the reactants and products. Once we have our products, with a solubility chart (I added one below) we can determine which of the products is soluble or insoluble.
In this case NaCl is soluble or aqueous (meaning it can dissolve in water) and Ca(OH)2 is insoluble (meaning that when the reactions takes place, these two will form a solid/precipitate)
Carbon is the element at the heart of all organic compounds, and it is such a versatile element because of its ability to form straight chains, branched chains, and rings. Because these chains and rings can have all sorts of different functional groups in all sorts of different ways (giving the compond all sorts of different physical and chemical properties), carbon's ability to form the backbone of these large structures is critial to the existence of most chemical compounds known to man. Above all, the organic molecules crucial to the biochemical systems that govern living organisms depend on carbon compounds.
Answer:
E = 0.062 V
Explanation:
(a) See the attached file for the answer
(b)
Calculating the voltage (E) using the formula;
E = - (2.303RT/nf)log Cathode/Anode
Where,
R = 8.314 J/K/mol
T = 35°C = 308 K
F- Faraday's constant = 96500 C/mol,
n = number of moles of electron = 2
Substituting, we have
E = -(2.303 * 8.314 *308/2*96500) *log (0.03/3)
= -0.031 * -2
= 0.062V
Therefore, the voltmeter will show a voltage of 0.062 V
