Answer:
3.2 g O₂
Explanation:
To find the mass of O₂, you need to (1) convert grams H₂O to moles H₂O (via molar mass), then (2) convert moles H₂O to moles O₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles O₂ to grams O₂ (via molar mass). It is important to arrange the ratios/conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 2 sig figs to reflect the sig figs of the given value (3.6 g).
Molar Mass (H₂O): 2(1.008 g/mol) + 15.998 g/mol
Molar Mass (H₂O): 18.014 g/mol
2 H₂O -----> 2 H₂ + 1 O₂
Molar Mass (O₂): 2(15.998 g/mol)
Molar Mass (O₂): 31.996 g/mol
3.6 g H₂O 1 mole 1 mole O₂ 31.996 g
---------------- x --------------- x --------------------- x --------------- = 3.2 g O₂
18.014 g 2 moles H₂O 1 mole
Answer:
33 moles
Explanation:
The given chemical reaction is 2C₄H₁₀(g) + 13 O₂(g) → 10H₂O(g) + 8CO₂(g)
The number of moles of each reactant are as follows;
Butane, C₄H₁₀ = 2 moles of (g) + 13 (g) → 10H₂O(g) + 8CO₂(g)
Oxygen gas, O₂ = 13 moles
Water, H₂O = 10 moles
Carbon dioxide, CO₂ = 8 moles
The total number of moles, n = (2 + 13 + 10 + 8) = 33
∴ The total number of moles involved in the reaction, n = 33 moles.
I’m not 100% sure but I think so
To solve this, we can use two equations.
t1/2 = ln 2 / λ = 0.693 / λ
where, t1/2 is half-life and λ is the decay constant.
t1/2 = 10 min = 0.693 / λ
Hence, λ = 0.693 / 10 min - (1)
Nt = Nο e∧(-λt)
Nt = amount of atoms at t =t time
Nο= initial amount of atoms
t = time taken
by rearranging the equation,
Nt/Nο = e∧(-λt) - (2)
From (1) and (2),
Nt/Nο = e∧(-(0.693 / 10 min) x 20 min)
Nt/Nο = 0.2500
Percentage of remaining nuclei = (nuclei at t time / initial nuclei) x 100%
= (Nt/Nο ) x 100%
= 0.2500 x 100%
= 25.00%
Hence, Percentage of remaining nuclei is 25.00%
Answer:
True
Explanation:
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