1. False
2. Magnesium
3. True
4. LiF
Answer:
b energy is being released from the atoms
Explanation:
Seeing light from fireworks implies that energy is being released from the atoms. This is an excitation as the electrons within the atom moves from a higher energy level to a lower energy level.
- When electrons undergo this kind of transition, energy is released.
- The radiation produced is dispersed out and it is seen as colorful lights.
- Most times such processes are exothermic
Answer:
The molecular formula of the compound is C₂H₄F₂ or CH₃CHF₂
Explanation:
From the question given above, the following data were obtained:
Empirical formula of the compound = CH₂F
Fórmula mass of compound = 66 g/mol
Molecular formula of the compound =?
The molecular formula of the compound can be obtained as follow:
Molecular formula = [Empirical formula]ₙ
Molecular formula = [CH₂F]ₙ
[CH₂F]ₙ = 66
[12 + (2×1) + 19] n = 66
[12 + 2+ 19]n = 66
33n = 66
Divide both side by 33
n = 66/33
n = 2
Molecular formula = [CH₂F]ₙ
Molecular formula = [CH₂F]₂
Molecular formula = C₂H₄F₂
Thus, the molecular formula of the compound is C₂H₄F₂ or CH₃CHF₂
Im sorry really dont know the 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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