Answer:
4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)
This is an oxidation-reduction (redox) reaction:
4 N-III - 20 e- → 4 NII
(oxidation)
10 O0 + 20 e- → 10 O-II
(reduction)
NH3 is a reducing agent, O2 is an oxidizing agent.
Explanation:
good luck
Answer:
I'm converting this if I could remember how
2.882568
2 110321/ 125000
T-T sorry if I'm wrong I have bad memory
so I recommend not using my answer at all,
if that is even how y'all write it.
Answer:
sample B contains the larger density
Explanation:
Given;
volume of sample A, V = 300 mL = 0.3 L
Molarity of sample A, C = 1 M
volume of sample B, V = 145 mL = 0.145 L
Molarity of sample B, C = 1.5 M
molecular mass of sodium chloride, Nacl = 23 + 35.5 = 58.5 g/mol
Molarity is given as;
The reacting mass for sample A = 0.3mol x 58.5 g/mol = 17.55 g
The reacting mass for sample B = 0.2175 mol x 58.5 g/mol = 12.72 g
The density of sample A 
The density of sample B 
Therefore, sample B contains the larger density
<span>To calculate the number of moles of aluminum, sulfur, and oxygen atoms in 4.00 moles of aluminum sulfate, al2(so4)3. We will simply inspect the "number" of aluminum, sulfur, and oxygen atoms available per one mole of the compound. Here we have Al2(SO4)3, which means that for every mole of aluminum sulfate, there are 2 moles of aluminum, 3 (1 times 3) moles of sulfur, and 12 (4x3) moles of oxygen. Since we have four moles of Al2(SO4)3 given, we simply multiply 4 times the moles present per 1 mole of the compound. So we have 4x2 = 8 moles of Al, 4x3 = 12 moles of sulfur, and 4x12 = 48 moles of oxygen.
So the answer is:
8,12,48
</span>
The kinetic energy of an object increases when heat is added.
