Answer:
Nucleus
Explanation: it is the center of the atom, and contains protons and nuetrons
The ions are as follows;
- As2O3 - As^3+, O^2-
- MnS - Mn^2+, S^2-
- VCl3 - V^3+, Cl^-
- NiF2 - Ni^2+, F^-
<h3>What is a cation and an anion?</h3>
A cation is a positive ion while an anion is a negative ion. Ionic compounds are composed o cations and anions.
The anions and cations in the ionic compounds in the table are shown below;
- As2O3 - As^3+, O^2-
- MnS - Mn^2+, S^2-
- VCl3 - V^3+, Cl^-
- NiF2 - Ni^2+, F^-
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The pH of a solution which is 0.023 m in weak base and 0 037 m in the conjugate weak acid whose Ka= 7.1 × 10⁻⁶ is 4.93.
pH determines the amount of hydrogen and hydroxide ions in a solution. It is the negative logarithm of hydrogen ion concentration.
Given,
Ka= 7.1 × 10⁻⁶
Weak base = 0.023M
Acid = 0.037M
Using base dissociation constant, Ka we can calculate pKa by:
pKa = -log [Ka]
pKa = -log [7.1 × 10⁻⁶]
pKa = 5.15
A weak base and its conjugate acid are present in equal proportions in buffer solution. Using the Henderson-Hasselbalch equation, we can find out the pH of a buffer solution that constitutes a weak base and its conjugate acid.
Using Henderson-Hasselbalch equation,
pH = pKa + log [Weak base / acid]
pH = 5.15 + log [0.023 / 0.037]
pH = 5.15 + log[0.6]
pH = 5.15 - 0.22
pH = 4.93
Therefore, the pH of the solution is 4.93.
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Answer is: XeF₄.
p₁ (F₂) = 8,0 atm.
p₁ (Xe) = 1,7 atm.
p₂ (F₂) = 4,6 atm.
p₂ (Xe) = 0 atm, all reacted.
p - partial pressure.
pressure of F₂ reacted = Δp (F₂) = 8.0 - 4.6 = 3.4 atm.
pressure of Xe reacted = Δp(Xe) = 1,7 - 0 = 1,7 atm.
Δp (F₂) / Δp(Xe) = 3,4 atm / 1,7 atm = 2:1.
Reaction: 2F₂ + Xe → XeF₄ (xenon-tetrafluoride)
Answer:
Molecular formula: C₂H₄O₂
Empirical formula: CH₂O
Explanation:
40 % C, 6.72 % H and 53.29 % O states the centesimal composition of the compound. These data means that in 100 g of compound we have x grams of a determined element.
We divide the mass by the molar mass of each:
40 g / 12 g/mol = 3.33 moles of C
6.72 g / 1 g/mol = 6.72 moles of H
53.29 g / 16 g/mol = 3.33 moles of O
We can determine rules of three to get, the molecular formula.
In 100 g of compound we have 3.33 moles of C, 6.72 moles of H and 3.33 moles of O; therefore in 60 g (1 mol) we must have
- (60 . 3.33) / 100 = 2 moles of C
- (60 . 6.72) / 100 = 4 moles of H
- (60 . 3.33) / 100 = 2 moles of O
Molecular formula is C₂H₄O₂
Empirical formula has the lowest suscripts; we divide by two, so the empirical formula is CH₂O