Silicon is the element having a mass of 28.09 g
<u>Explanation</u>:
- Silicon is the element having an atomic mass of 28.09 g / mol. So 28.09 g of silicon contains 6.023
10^23 atoms. One mole of each element can produce one mole of compound.
- The Atomic weight of an element can be determined by the number of protons and neutrons present in one atom of that element. So atomic weight expressed in grams always contain the same number of atoms( 6.023 10^23).
- Avagadro number is the number of atoms of 1 mole of any gas at standard temperature and pressure. It has been determined that 6.023 10^23 atoms of an element are equal to the average atomic mass of that element.
Answer:
Covalent Bonds are formed when two non-metals share electrons
Hope this helps
Their lungs would try to expand to about 4 timed the normal volume which would force air into the various body tissues. this can cause a lung expansion injury and it could case air embolism. Air embolism is when air bubbles get trapped in blood vessels. This can lead to a blockage which will could be fatal.
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.
