A Brønsted-Lowry base is a base is a proton acceptor.
In the only case where this is done is when HCO3- accepts a proton and becomes H2CO3.
In the other cases, HCO3- is donating a proton which makes it an acid.
Since there is no weight, I would assume that this is a 100g of pure compound.
Okay so I would be changing the percentage to gram to solve for the mole.
So
40.0g C (1 mol C/12.01 g C) = 3.33 mol C
6.73g H (1 mol H/1.01 g H ) = 6.66 mol H
53.3g O (1 mol O/16.00 g O) = 3.33 mol O
With that, two of our moles is 3.33, so we consider that are our 1, as it is also the lowest. Therefore the empirical formula is CH2O
Rutherford's gold foil experiment proved that there was a small, dense, positively charged nucleus at the center, which contained most of the mass of the atom. Which contained electrons orbiting the nucleus.
Answer:
MnO4⁻ (aq) + 8H⁺ (aq) + 5Fe³⁺ (aq) →Mn(aq)²⁺ + 4H2O (l) + 5Fe²⁺(aq)
Explanation:
a)
MnO4⁻ (aq) + 8H⁺ (aq) + 5e⁻ → Mn(aq)²⁺ + 4H2O (l)
b)
5Fe³⁺ (aq) +5e⁻ → 5Fe²⁺(aq)
c)
MnO4⁻ (aq) + 8H⁺ (aq) + 5Fe³⁺ (aq) →Mn(aq)²⁺ + 4H2O (l) + 5Fe²⁺(aq)
Answer:
The ΔHrxn for the above equation = 179 kJ/mol
Explanation:
The reaction bond enthalpies are for the reactant;
3 × N-H = 3 × 390 = 1,170 kJ/mol
2 × O=O = 2 × 502 = 1004 kJ/mol
The reaction bond enthalpies are for the product;
3 × N-O = 3 × 201 = 603 kJ/mol
3 × O-H = 3 × 464 = 1,392 kJ/mol
The ΔHrxn for the above equation is therefore;
ΔHrxn = 1,170 + 1,004 - (603 + 1,392) = 179 kJ/mol