Not always ammonium salts of weak acids form neutral solutions.
When formic acid reacts with ammonia, ammonium formate is produced:
HCO2H + NH3 ----> NH4HCO2
You already know that the weak conjugate bases of NH3 and HCO2H are NH4+ and HCO2, respectively.
How can the pH of the solution be calculated if the salt's anion causes the pH to rise and the salt's cation causes it to fall? The relative intensities of the basic anion and the acidic cation hold the key to the solution.
As was already established, formate is a weak base and will create hydroxide ions in water, whereas ammonium is a weak acid and will make hydronium ions in water.
NH4⁺ + H2O -----> NH3 + H3O⁺
HCO2⁻ + H2O -----> HCO2H + OH⁻
Since the acid ionization of NH4+ is more favored than the base ionization of HCO2-, the solution will be acidic.
To learn more about ammonium salts:
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Answer is: <span>the missing daughter nucleus is rhodium (Rh).</span>
Nuclear
reaction: ¹⁰⁶Ru → ¹⁰⁶Rh + e⁻(electron) +
ve(electron antineutrino).
Beta decay is radioactive decay in which a beta
ray and a neutrino are emitted from an atomic nucleus.There are two types of
beta decay: beta minus and beta plus.
<span>In beta minus decay,
neutron is converted to a proton and an electron and an electron antineutrino.
In beta plus decay, a proton is converted to a neutron and positron and an
electron neutrino, so mass number does not change.</span>
Answer:
The molar mass of the metal is 54.9 g/mol.
Explanation:
When we work with gases collected over water, the total pressure (atmospheric pressure) is equal to the sum of the vapor pressure of water and the pressure of the gas.
Patm = Pwater + PH₂
PH₂ = Patm - Pwater = 1.0079 bar - 0.03167 bar = 0.9762 bar
The pressure of H₂ is:
The absolute temperature is:
K = °C + 273 = 25°C + 273 = 298 K
We can calculate the moles of H₂ using the ideal gas equation.
Let's consider the following balanced equation.
M(s) + H₂SO₄(aq) ⟶ MSO₄(aq) + H₂(g)
The molar ratio of M:H₂ is 1:1. So, 9.81 × 10⁻³ moles of M reacted. The molar mass of the metal is:
Answer:
The answer to your question is:
Vol of NO2 = 11.19 L
Vol of O2 = 2.8 L
Explanation:
Data
N2O5 = 56 g
STP T = 0°C = 273°K
P = 1 atm
MW N2O5 = 216 g
Gases law = PV = nRT
Process
216 g of N2O5 ---------------- 1 mol
54 g ----------------- x
x = (54 x 1) / 216
x = 0.25 mol of N2O5
2 mol of N2O5 ----------------- 4 mol of NO2
0.25 mol ------------------ x
x = (0.25 x 4) / 2 = 0.5 mol of NO2
V = nRT/P
V = (0.5)(0.082)(273) / 1 = 11.19 L
2 mol of N2O5 ----------------- 1 O2
0.25 N2O5 ---------------------- x
x = (0.25 x 1) / 2 = 0.125 mol
Vol = (0.125)((0.082)(273) / 1 = 2.8 L
