put 8 in front of the oxygen in the reactants side to make it 16 molecules then put a 5 in front of the co2 in the product side to balance the carbon atoms then put a 6 in front of the H20 on the product side this balances both the hydrogen and oxygen atoms here is a representation
C5H12(g)+8O2(g)=5CO2(g)+6H20
Answer:
yes answer os Na because it's electronic configuration is 1s^2,2s^2,2p^6,3s^1
Answer:
NO would form 65.7 g.
H₂O would form 59.13 g.
Explanation:
Given data:
Moles of NH₃ = 2.19
Moles of O₂ = 4.93
Mass of NO produced = ?
Mass of produced H₂O = ?
Solution:
First of all we will write the balance chemical equation,
4NH₃ + 5O₂ → 4NO + 6H₂O
Now we will compare the moles of NO and H₂O with ammonia from balanced chemical equation:
NH₃ : NO NH₃ : H₂O
4 : 4 4 : 6
2.19 : 2.19 2.19 : 6/4 × 2.19 = 3.285 mol
Now we will compare the moles of NO and H₂O with oxygen from balanced chemical equation:
O₂ : NO O₂ : H₂O
5 : 4 5 : 6
4.93 : 4/5×4.93 = 3.944 mol 4.93 : 6/5 × 4.93 = 5.916 mol
we can see that moles of water and nitrogen monoxide produced from the ammonia are less, so ammonia will be limiting reactant and will limit the product yield.
Mass of water = number of moles × molar mass
Mass of water = 3.285 mol × 18 g/mol
Mass of water = 59.13 g
Mass of nitrogen monoxide = number of moles × molar mass
Mass of nitrogen monoxide = 2.19 mol × 30 g/mol
Mass of nitrogen monoxide = 65.7 g
Answer:
λ = 1*10⁻¹⁰m
Explanation:
Frequency (f) = 3.0*10¹²MHz = 3.0*10¹⁸Hz
Speed (v) = 3.0*10⁸m/s
Speed (v) of a wave = frequency (f) * wavelength (λ)
V = fλ
Solve for λ,
λ = v / f
λ = 3.0*10⁸ / 3.0*10¹⁸
λ = 1*10⁻¹⁰m
λ = 0.
