Answer:
Mass of nitrogen dioxide produced = 4.6 g
Explanation:
Given data:
Volume of ammonia = 2.30 L
Mass of nitrogen dioxide produced = ?
Solution:
Chemical equation:
4NH₃ + 7O₂ → 4NO₂ + 6H₂O
Number of moles of ammonia at STP:
PV = nRT
n = PV/RT
n = 1 atm × 2.30 L / 0.0821 atm.L/K.mol × 273 K
n = 2.30 atm .L / 22.414 atm.L/mol
n = 0.1 mol
Now we will compare the moles of ammonia with nitrogen dioxide from balance chemical equation.
NH₃ : NO₂
4 : 4
0.1 : 0.1
Mass of NO₂:
Mass = number of moles × molar mass
Mass = 0.1 mol × 46 g/mol
Mass = 4.6 g
Answer:
.0027 M
Explanation:
We must calculate the threshold concentration of PO3−4 using Ksp and the given concentration of Ca2+:
Ca3(PO4)2(s)⇌3Ca2+(aq)+2PO3−4(aq)
Ksp=8.6×10−19=[Ca2+]3[PO3−4]2=(4.9×10−5M)3[PO3−4]2
[PO3−4]=0.0027 M
Answer:
8 electrons
Explanation:
As per the <u>octet rule</u>, the atoms possess a tendency to bond in a manner that every atom must have at least eight(8) electrons in its outermost/valence shell. It can be done either by sharing, gaining, or losing electrons from one atom to another. According to this rule, flourine would have 8 electrons in its outer shell after two shared electrons are given to it.
Answer:
1 mole
Explanation:
A mole of water molecules contains 2 moles of hydrogen atoms and 1 mole of oxygen atoms.
[H+][OH-] = 1x10-14 = Kw (Memorize this relationship)
(2.70x10-2)[OH-] = 1x10-14
[OH-] = 1x10-14/2.70x10-2
[OH-] = 3.70x10-13 M (assuming you ignore the autoionization of H2O)
