Answer:
6 moles of SO₃ formed.
Explanation:
Given data:
Number of moles of SO₃ formed = ?
Number of moles of oxygen react = 3 mol
Solution:
Chemical equation;
2SO₂+ O₂ → 2SO₃
now we will compare the moles of oxygen and sulfur trioxide.
O₂ : SO₃
1 : 2
3 : 2/1×3 = 6 moles
Thus, six moles of SO₃ will formed.
CS2 + 3O2 = CO2 + 2SO2
1 mole of CS2 gives 1 mole of CO2
12 + 2(32) = 76g of CS2 yields 44 g of CO2
Theoretically 1 g of CS2 yields 44/76 g CO2
Therefore 50 g CS2 should yield 50*44 / 76 = 28.95 g
So % yield = 103.6 % ( which is not possible because you can't create matter from nothing).
The 30g cannot be right . This is experimental err.
<h3>
Answer:</h3>
5.71 × 10² nm
<h3>
Explanation:</h3>
The product of wavelength and frequency of a wave gives the speed of the wave.
Therefore;
Velocity of wave = Wavelength × Frequency
c = f ×λ
In our case;
Frequency = 5.25 × 10^14 Hz
Speed of light = 2.998 × 10^8m/s
But;
λ = c ÷ f
= 2.998 × 10^8m/s ÷ 5.25 × 10^14 Hz
= 5.71 × 10^-7 m
But; 1 M = 10^9 nm
Therefore;
wavelength = 5.71 × 10^-7 × 10^9
= 5.71 × 10² nm
The wavelength of light wave 5.71 × 10² nm
Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
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