Answer:
It is less dense than air so it will help the airship float, and it is not flammable so the airship is less likely to catch on fire. None of the other noble gases have similar or better reasons.
Explanation:
12.3 mol of NO reacts with ______ mol of ammonia
5.87 mol of NO yields _______ mol nitrogen
12.3 moles of nitrogen oxide (NO) react with 8.2 moles of ammonia (NH3) in the given reaction. The reaction of 5.87 moles of NO yields 4.89 moles of nitrogen (N2).
The overall reaction equation represents an ideal stoichiometric ratio where 4 moles of ammonia (NH3) react with 6 moles of nitrogen oxide (NO) to produce 5 moles of nitrogen (N2) and 6 moles of water (H2O). Let's work with this ratio to fill in the blanks of your question.
Given that 6 moles of NO react with 4 moles of NH3, we can set up a ratio to find the unknown quantity of NH3 that reacts with 12.3 moles of NO: (4 NH3 / 6 NO) x 12.3 NO = 8.2 moles of NH3
Similarly, 5 moles of nitrogen are produced per 6 moles of NO reacted, so we can find out how much nitrogen is produced from 5.87 moles of NO: (5 N2 / 6 NO) x 5.87 NO = 4.89 moles of nitrogen.
#SPJ12
12.3 moles of NO reacts with 8.2 moles of Ammonia. 5.87 moles of NO yields 4.89 moles of Nitrogen.
In these types of problems, we use the concept of mole ratios from the balanced chemical equation. According to the reaction, 4 moles of NH3 react with 6 moles of NO to yield 5 moles of N2. Therefore, for 12.3 moles of NO, we apply the ratio 4/6 to determine that it reacts with 8.2 moles of NH3.
Similarly, 5.87 moles of NO, based on the ratio 5/6, yields 4.89 moles of N2. So, these are the missing quantities for the reaction.
#SPJ12
(2) the relative abundance of each naturally occurring isotope of chlorine, only
(3) the atomic mass and the relative abundance of each naturally occurring isotope of chlorine
(4) the atomic mass and the relative abundance of each naturally occurring and artificially produced isotope of chlorine
Answer is: (3) the atomic mass and the relative abundance of each naturally occurring isotope of chlorine.
m(³⁵Cl) = 34.97 amu; the average atomic mass of chlorine-35.
m(³⁷Cl) = 36.97 amu; the average atomic massof chlorine-37.
ω(³⁵Cl) = 75.76% ÷ 100% = 0.7576; fractional abudance of chlorine-35.
ω(³⁷Cl) = 24.24% ÷ 100% = 0.2424; fractional abudance of chlorine-37.
m(Cl) = m(³⁵Cl) · ω(³⁵Cl) + m(³⁷Cl) · ω(³⁷Cl).
m(Cl) = 34.97 amu · 0.7576 + 36.97 amu · 0.2424.
m(Cl) = 35.4548 amu; average atomic mass of chlorine.
The atomic mass of chlorine is determined by both the atomic mass and the relative abundance of each naturally occurring isotope of chlorine. It's based on an average of the masses of the isotopes according to their relative abundance.
To determine the atomic mass of the element chlorine, option (3), both the atomic mass and the relative abundance of each naturally occurring isotope of chlorine is required. The atomic mass of chlorine is not just the mass of one particular atom, or isotope; instead, it's an average of the masses for all the isotopes according to their relative abundance. For instance, Chlorine has two isotopes Chlorine-35 and Chlorine-37. If the relative abundance of Chlorine-35 is 75% and that of Chlorine-37 is 25%, the atomic mass is calculated by the formula [(0.75 x 35) + (0.25 x 37)].
#SPJ2
b. 35 g
c. 65 g
d. 6.02 × 1023 g
b. the ozone layer
c. banded iron formations
d. stromatolites
Banded iron formations
Before free oxygen gas became abundant in the oceans and atmosphere, it reacted with a metal and produced banded iron formations.