Answer:
To find the number of neutrons in an isotope, subtract the number of protons from the atomic mass of the isotope. The atomic number of the element equals the number of protons.
Answer: 7.8 moles of NaCl result from the complete reaction of 3.9 mol of 
Explanation:
To calculate the moles :
As 
is the excess reagent, is the limiting reagent as it limits the formation of product.
According to stoichiometry :
1 mole of gives = 2 moles of 
Thus 3.9 moles of will give=
of
7.8 moles of NaCl result from the complete reaction of 3.9 mol of
Answer:
Rubidium and cesium
Explanation:
It is noteworthy to say here that larger cations have more stable superoxides. This goes a long way to show that large cations are stabilized by large cations.
Let us consider the main point of the question. We are told in the question that the reason why potassium reacts with oxygen to form a superoxide is because of its low value of first ionization energy.
The implication of this is that, the other two metals that can be examined to prove this point must have lower first ionization energy than potassium. Potassium has a first ionization energy of 419 KJmol-1, rubidium has a first ionization energy of 403 KJ mol-1 and ceasium has a first ionization energy of 376 KJmol-1.
Hence, if we want to validate the hypothesis that potassium's capacity to form a superoxide compound is related to a low value for the first ionization energy, we must also consider the elements rubidium and cesium whose first ionization energies are lower than that of potassium.
The balanced reaction: C2H6 + 7/2 O2 -> 2 CO2 + 3 H2O
We first convert volume of C2H6 to no. of moles. We use the conditions at STP where 1 mol = 22.4 L thus,
Moles C2H6 = 16.4 L/ 22.4 L =0.7321 mol
In order to determine the limiting reagent, we look at the given amounts of the reactants.
0.7321 mol C2H6 (7/2 mol O2 / 1 mol C2H6) = 2.562 mol O2
From the given amounts of the reactants, we can say that O2 is the limiting reactant since we need 2.562 mol O2 to completely react the given amount of C2H6. The excess reagent is C2H6
To calculate for the amount of products and excess reactants:
0.980 mol O2 (2 mol CO2 / (7/2 mol O2)) = 0.56 mol CO2 (22.4 L / 1 mol ) =12.544 L CO2
<span>0.980 mol O2 (1 mol C2H6 / (7/2 mol O2)) = 0.28 mol C2H6
Excess C2H6 = 0.7321 mol - 0.28 mol C2H6 = 0.4521 mol C2H6
We then use the molecular weight of C2H6 to convert the excess amount to grams.
0.4521 mol C2H6 (30.08 g C2H6 / 1 mol C2H6) = <span>13.60 g C2H6
</span></span>
<span>Since the limiting reagent is O2 there will be no oxygen atoms that will be left after the reaction.</span>
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