B.)<span>A doctor would use technetium-99m because a short half-life indicates a quick measurement, and a geologist would use rubidium-87 because a longer half-life means that there is a longer time to measure old rock. </span>
The correct reaction equation is:
Answer:
b) 1 mole of water is produced for every mole of carbon dioxide produced.
Explanation: <u>CONVERT EVERYTHING TO MOLES OR VOLUME, THEN COMPARE IT WITH THE COMPOUND'S STOICHIOMETRY IN CHEMICAL EQUATION.</u>
a) <u>22.4 L of 
gas</u> is produced only when <u>
L of 
</u> is reacted with 22.4 L of 
. So it is wrong.
b) Since in the chemical equation the stoichiometric coefficient of 
and 
are same so the number of moles or volume of each of them will be same whatever the amount of reactants taken. <u>Therefore it is correct option.</u>
c) 
molecules is equal 1 mole of 
if produced then 3 moles of is required, which is not given in the option. So it is wrong.
d) 54 g of water or 3 moles of (<em>Molecular Weight of water is 18 g</em>) is produced when 3 moles of is used but in this option only one mole of is given. So it is wrong.
The chemical formula for ammonia is NH3. So first, you need to find the molar mass of ammonia (how many grams in one mole).
N=14g
H3=3g
So one mole of NH3 is 17 grams, you can divide 82.9 grams by 17 grams to find the number of molecules. The answer should be 4.876 moles (molecules) of ammonia. Hope this helps!
Answer
is: <span>the
percent ionizationof formic acid is 1,82%.
Chemical reaction: HCOOH(aq) </span>⇄ H⁺(aq)
+ HCOO⁻(aq).<span>
pKa(</span>HCOOH) = 3,77.
Ka(HCOOH) = 1,7·10⁻⁴.
c(HCOOH) = 0,5 M.
<span>
[H</span>⁺]
= [HCOO⁻] = x; equilibrium
concentration.<span>
[HA] = 0,1 M - x.
Ka = [H</span>⁺] · [HCOO⁻] / [HCOOH].<span>
0,00017 = x² / 0,5 M - x.
Solve quadratic equation: x = 0,0091 M.
α = 0,0091 M ÷ 0,5 M · 100% = 1,82%.</span>
Generally speaking, organic molecules tend to dissolve in solvents that have similar physical properties. A good rule of thumb is that "like dissolves like". Meaning, polar compounds can dissolve polar compounds and nonpolar compounds can dissolve nonpolar compounds.
To apply this to the current problem, we are told that the brushes are being cleaned with vegetable oil or mineral oil. In this case, the oils are used as solvents. In order for these solvents to be effective, the compounds they are trying to dissolve must be similar in structure and properties to other oils. Therefore, vegetable oil or mineral oil will be most effective in removing oil-based paints, as these will have the similar properties needed to dissolve in the oil solvents.
