Answer:
removing the methanol as it is formed
Explanation:
One of the ways to drive the equilibrium position towards the right is to remove one of the products formed.
According to Me Chatelier's principle, the imposition of a constraint on a system in a equilibrium causes the equilibrium position to shift towards a new position that annuls the constraint. Hence, removing the methanol causes the equilibrium position to shift to the far right in order to reestablish equilibrium according to Le Chatelier's principle.
Assuming that the ammonium sulfide formula is (NH4)2S then you can see that there are 2 nitrogen, 8 hydrogen and 2 sulfur atoms for every ammonium sulfide. If the amount of ammonium sulfide is 8.9 moles, then the number of hydrogen atoms should be: 8/1 * 8.9 mol= 71.2 moles
Answer:
BaBr2 (aq) + H2SO4 (aq) → BaSO4 (s) + 2 HBr (aq)
Explanation:
This is a precipitation reaction: BaSO4 is the formed precipitate.
Explanation:
P1= 44 kpa
P2= 50 kpa
V1= 4.50 L
V2= ?
P1 V1= P2 V2
44 × 4.50 = 50 × V2
198= 50 × V2
V2 = 198/ 50
V2= 3.96 L "the new volume"
Answer:
(a) 7.11 x 10⁻³⁷ m
(b) 1.11 x 10⁻³⁵ m
Explanation:
(a) The de Broglie wavelength is given by the expression:
λ = h/p = h/mv
where h is plancks constant, p is momentum which is equal to mass times velocity.
We have all the data required to calculate the wavelength, but first we will have to convert the velocity to m/s, and the mass to kilograms to work in metric system.
v = 19.8 mi/h x ( 1609.34 m/s ) x ( 1 h / 3600 s ) = 8.85 m/s
m = 232 lb x ( 0.454 kg/ lb ) = 105.33 kg
λ = h/ mv = 6.626 x 10⁻³⁴ J·s / ( 105.33 kg x 8.85 m/s ) = 7.11 x 10⁻³⁷ m
(b) For this part we have to use the uncertainty principle associated with wave-matter:
ΔpΔx > = h/4π
mΔvΔx > = h/4π
Δx = h/ (4π m Δv )
Again to utilize this equation we will have to convert the uncertainty in velocity to m/s for unit consistency.
Δv = 0.1 mi/h x ( 1609.34 m/mi ) x ( 1 h/ 3600 s )
= 0.045 m/s
Δx = h/ (4π m Δv ) = 6.626 x 10⁻³⁴ J·s / (4π x 105.33 kg x 0.045 m/s )
= 1.11 x 10⁻³⁵ m
This calculation shows us why we should not be talking of wavelengths associatiated with everyday macroscopic objects for we are obtaining an uncertainty of 1.11 x 10⁻³⁵ m for the position of the fullback.
