Answer:
diffusion
Explanation:
Diffusion can be defined as the net movement of particles from a high concentration to a low concentration. This sounds like the most plausible word in this case.
It's likely what's wanted is
<span><span>Li</span>→<span><span>Li</span><span>2+</span></span>+2<span>e−
</span></span>
The reason is because IEs are usually reported from the neutral atom, that is, IE2 is the energy required to remove two electrons from a neutral Li atom, as above, rather than the additional energy required to remove one more electron from an Li+ cation.
Answer:
<u>1. Net ionic equation:</u>
- Cl⁻(aq) + Ag⁺(aq) → AgCl(s)
<u />
<u>2. Volume of 1.0M AgNO₃</u>
Explanation:
1. Net ionic equation for the reaction of NaCl with AgNO₃.
i) Molecular equation:
It is important to show the phases:
- (aq) for ions in aqueous solution
- (s) for solid compounds or elements
- (g) for gaseous compounds or elements
- NaCl(aq) + AgNO₃(aq) → AgCl(s) + NaNO₃(aq)
ii) Dissociation reactions:
Determine the ions formed:
- NaCl(aq) → Na⁺(aq) + Cl⁻(aq)
- AgNO₃(aq) → Ag⁺(aq) + NO₃⁻(aq)
- NaNO₃(aq) → Na⁺(aq) + NO₃⁻(aq)
iii) Total ionic equation:
Substitute the aqueous compounds with the ions determined above:
- Na⁺(aq) + Cl⁻(aq) + Ag⁺(aq) + NO₃⁻(aq) → AgCl(s) + Na⁺(aq) + NO₃⁻(aq)
iv) Net ionic equation
Remove the spectator ions:
- Cl⁻(aq) + Ag⁺(aq) → AgCl(s) ← answer
2. How many mL of 1.0 M AgNO₃ will be required to precipitate 5.84 g of AgCl
i) Determine the number of moles of AgNO₃
The reaction is 1 to 1: 1 mole of AgNO₃ produces 1 mol of AgCl
The number of moles of AgCl is determined using the molar mass:
- number of moles = mass in grams / molar mass
- molar mass of AgCl = 143.32g/mol
- number of moles = 5.84g / (143.32g/mol) = 0.040748 mol
ii) Determine the volume of AgNO₃
- molarity = number of moles of solute / volume of solution in liters
- V = 0.040748mol / (1.0M) = 0.040748 liter
- V = 0.040748liter × 1,000ml / liter = 40.748 ml
Round to two significant figures: 41ml ← answer
The percent yield of the reaction between ammonia gas with oxygen gas is 90.52%.
A chemical reaction between ammonia gas (NH3) with oxygen gas (O2)
NH₃ + O₂ → NO₂ + H₂O
The balanced reaction 4NH₃ + 7O₂ → 4NO₂ + 6H₂O
Calculate the number of moles from the reactant
- Ammonia gas
Molar mass N = 14 gr/mol
Molar mass H = 1 gr/mol
Molar mass NH₃ = 14 + (3 × 1) = 14 + 3 = 17 gr/mol
mass = 28.5 grams
n = m ÷ molar mass = 28.5 ÷ 17 = 1.68 mol - Oxygen gas
Molar mass O = 16 gr/mol
Molar mass O₂ = 16 × 2 = 32 gr/mol
mass = 83.4 grams
n = m ÷ molar mass = 83.4 ÷ 32 = 2.61 mol - n O₂ ÷ coefficient O₂ = 2.61 ÷ 7 = 0.37
n NH₃ ÷ coefficient NH₃ = 1.68 ÷ 4 = 0.42
0.42 > 0.37 it means that the ammonia gas is in excess and the O₂ is limiting.
According to stoichiometry, the number of moles NO₂ with the number of moles O₂ has the ratio with the coefficient in reaction.
- Theoretically the number moles of NO₂
n O₂ : n NO₂ = 7 : 4
2.61 : n NO₂ = 7 : 4
n NO₂ = 4 x 2.61 : 7 = 1.49 mol - The actual number of moles NO₂
Molar mas NO₂ = 14 + (16 × 2) = 14 + 32 = 46 gr/mol
n NO₂ = m ÷ molar mass = 61.9 ÷ 46 = 1.35 mol
The percent yield NO₂ is the ratio of the actual number of moles NO₂ with the theoretical number of moles NO₂ times 100%.
P = (1.35 ÷ 1.49) × 100%
P = 0.9052 × 100%
P = 90.52%
Learn more about stoichiometry here: brainly.com/question/13691565
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