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.
- C_5H_8+13/2O_2—»5CO_2+4H_2O
Balanced one
- 2C_5H_8+13O_2—»10CO_2+8H_2O
Moles of Pentyne
- Given mass/Molarmass
- 34/68
- 0.5mol
Moles of H_2O
1mol releases 241.8KJ
2mol releases 241.8(2)=483.6KJ
To present his data on the <span>solubility of three different salts in water at 22°C, bar graph should be used since there are different salts and the only variable is the type of salt used. Line graph and scatter plot use two coordinates or variables and is common to comparing data using the same sample while using histogram to find out data distribution is irrelevant.</span>
<span>The solution to the problem is as follows:
125/58.69 = 2.12 mol
</span>
Therefore, there are 2.12 moles of <span>nickel (Ni) atoms are in 125 g Ni.
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