1) Excess reagent
1 mol N2 / 3 mol H2
6.0 mol N2 *3 mol H2 / 1 mol N2 = 18 mol H2
18mol H2 > 12 mol H2 => H2 is limiting (you need 18 mol H2 to use all the 6 mol N2), then N2 is in excees.
12.0 mol H2 * 1mol N2/ 3 mol H2 = 4 mol N2 is the quantity that will react, then the excess is 6 mol N2 - 4 mol N2 = 2 mol N2
2) NH3 produced
12 mol H2 * [2 mol NH3 / 3 mol H2] = 8 mol NH3
Aslso, 4 mol N2 *[2molNH3 / 1 molN2] = 8 mol NH3, the same result.
3) Yield
80% * 8 mol NH3 = 6.4 mol NH3
Answer:
Eªcell > 0; n = 2
Explanation:
The reaction:
I2 (s) + Pb (s) → 2 I- (aq) + Pb2+ (aq)
Is product favored.
A reaction that is product favored has ΔG < 0 (Spontaneous)
K > 1 (Because concentration of products is >>>> concentration reactants).
Eªcell > 0 Because reaction is spontaneous.
And n = 2 electrons because Pb(s) is oxidizing to Pb2+ and I₂ is reducing to I⁻ (2 electrons). Statements that are true are:
<h3>Eªcell > 0; n = 2</h3>
Answer:
226.8 mg of mupirocin powder are required
Explanation:
Given that;
weight of standard pack = 22 g
mupirocin by weight = 2%
so weight of mupirocin = 2% × 22 = 2/100 × 22 = 0.44 g
so by adding the needed quantity of mupirocin powder to prepare a 3% w/w mupirocin ointment
mg of mupirocin powder are required = ?, lets rep this with x
Total weight of ointment = 22 + x g
Amount of mupirocin = 0.44 + x g
percentage of mupirocin in ointment is 3?
so
3/100 = 0.44 + x g / 22 + x g
3( 22 + x g ) = 100( 0.44 + x g )
66 + 3x g = 44 + 100x g
66 - 44 = 100x g - 3x g
97 x g = 22
x g = 22 / 97
x g = 0.2268 g
we know that; 1 gram = 1000 Milligram
so 0.2268 g = x mg
x mg = 0.2268 × 1000
x mg = 226.8 mg
Therefore, 226.8 mg of mupirocin powder are required
You have a. 3, b. 2, e. 6, f. 3, i. 1
Answer:
Continuing on across the periodic table we see that fluorine is the next element after oxygen. ... Rather than forming seven bonds fluorine only forms a single bond for basically the same reasons that oxygen only forms two bonds. Hydrogen fluoride, HF, has one bond, but four centers of electron density around the fluorine.
