There are 3.98 × 10^23 atoms of oxygen in the sample.
Given that;
1 mole of Mo(NO3)6 contains 6.02 × 10^23 atoms of Nitrogen
x moles of Mo(NO3)6 contains 2.22 x 10^22 atoms of nitrogen
x = 1 mole × 2.22 x 10^22 atoms/6.02 × 10^23 atoms
x = 0.0368 moles
The number of oxygen atoms in the sample is given by; 0.0368 × 6.02 × 10^23 × 18
Therefore, there are 3.98 × 10^23 atoms of oxygen in the sample.
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Answer:
The answer to your question is: 0.25 l
Explanation:
Data
P1 = 1 atm
V1 = 0.5 l
P2 =2 atm
V2 = ?
T = constant
Formula
V1P1 = V2P2
Clear V2 from the formula
V2 = V1P1/P2
Substitution
V2 = (0.5)(1)/2 substitution
= 0.25 l result
It’s c I believe because it says 209 years to complete an orbit so hint yourself ?.....
Answer:
ΔG° = -533.64 kJ
Explanation:
Let's consider the following reaction.
Hg₂Cl₂(s) ⇄ Hg₂²⁺(aq) + 2 Cl⁻(aq)
The standard Gibbs free energy (ΔG°) can be calculated using the following expression:
ΔG° = ∑np × ΔG°f(products) - ∑nr × ΔG°f(reactants)
where,
ni are the moles of reactants and products
ΔG°f(i) are the standard Gibbs free energies of formation of reactants and products
ΔG° = 1 mol × ΔG°f(Hg₂²⁺) + 2 mol × ΔG°f(Cl⁻) - 1 mol × ΔG°f(Hg₂Cl₂)
ΔG° = 1 mol × 148.85 kJ/mol + 2 mol × (-182.43 kJ/mol) - 1 mol × (-317.63 kJ/mol)
ΔG° = -533.64 kJ
