The oxidation state numbercan aid in identifying the unknown element present in both compounds. They denote the number of electrons that are either donated, received, or shared to yield compounds.
Remember the fundamental principles governing oxidation numbers.
1. In a neutral compound, the total of all oxidation numbers is zero.
2. Chlorine, bromine, and iodine typically exhibit an oxidation number of -1(unless paired with fluorine and oxygen)
Assume the oxidation state for element Mis designated as x.
Referring to rule 2, chlorine possesses an oxidation state of -1.
Now, for the compound MCl₂ (which is neutral), the equation can be formulated as
x + (2 * -1)= 0 ⇒ x₁= +2
For MCl₃, the corresponding equation is
x + (3 * -1)= 0 ⇒ x₂= +3
This indicates that the elementhas two distinct oxidation statesin its compounds, which are +2and +3.
The identified element is iron (Fe), as it shows +2 and +3 oxidation states across these compounds.
Memorizing this is essential. Regrettably, there isn't a simpler method to tackle these oxidation states.
The final answer is iron (Fe).
Answer:
The original halide's formula is SrCl₂.
Explanation:
- The chemistry reaction's balanced equation is:
SrX₂ + H₂SO₄ → SrSO₄ + 2 HX, where X indicates the halide.
- Based on the equation's stoichiometry, 1.0 mole of strontium halide yields 1.0 mole of SrSO₄.
- The moles of SrSO₄ (n = mass/molar mass) = (0.755 g) / (183.68 g/mole) = 4.11 x 10⁻³ mole.
- The moles of SrX can thus be calculated as 4.11 x 10⁻³ moles based on stoichiometry from the balanced equation.
- n = mass / molar mass, thus n = 4.11 x 10⁻³ moles and mass = 0.652 g.
- The molar mass of SrX₂ is calculated using mass / n = (0.652) / (4.11 x 10⁻³ moles) = 158.62 g/mole.
- The molar mass of SrX₂ (158.62 g/mole) = Atomic mass of Sr (87.62 g/mole) + (2 x Atomic mass of halide X).
- Calculating the atomic mass of halide X, we find = (158.62 g/mole) - (87.62 g/mole) / 2 = 71 / 2 g/mole = 35.5 g/mole.
- This identifies the atomic mass of Cl.
- Consequently, the original halide's formula is SrCl₂.
Sagot:
0.1 M NaCl
Paliwanag:
Ang tanong na ito ay nagpapaalala sa atin ng mga patakaran sa solubility. Alalahanin natin na ang lahat ng chlorides ay natutunaw maliban sa mga ng lead, mercury II at silver na hindi natutunaw sa tubig.
Ang sumusunod na reaksyon ay mangyayari na humahantong sa pagbuo ng isang precipitate;
Pb(NO3)2(aq) + 2NaCl(aq) -------> 2NaNO3(aq) + PbCl2(s)
Ang puting precipitate na nabuo ay PbCl2.
The energy needed to vaporize 1.5 kg of aluminum amounts to 16.345 GJ. The heat of vaporization for aluminum is given as ΔHvap = 294000 kJ/mol. The mass of aluminum in this case equals 1.5 kg which converts to 1500 g. We can calculate the number of moles of aluminum using the formula: Mass of aluminum/(Molar Mass of aluminum). The Molar Mass of aluminum stands at 26.98 g/mol. Using this information, Number of moles calculates to 1500/26.98, which equals 55.6 moles. The total energy required can be expressed as the product of the heat of vaporization and the number of moles of aluminum, so the energy required calculates to 294000 × 55.6, resulting in 16345441.0675 kJ or approximately 16.345 GJ.
