This process is called meiosis! good luck!
Answer:1.123 x 10^-31cm
Explanation:
mass of humming bird= 11.0g
speed= 1.20x10^2mph
but I mile = 1.6m
1km=1000
I mile = 1.6x10^3m
1.20x10^2mph= 1.6x10^3m /1mile x at 1.20 x 10^2
=1.932 x10^5m
recall that
1 hr= 60 min
1 min=60 secs, 1hr=3600s
Speed = distance/ time
=1.932 x10^5 / 3600= 5.366 x 10 ^1 m/s
m= a 11.0g= 11.0 x 10^-3kg
h=6.626*10^-34 (kg*m^2)/s
Wavelength = h/mu
= 6.626*10^-34/(11 x 10^-3 x 5.366x 10^1)
6.63x10^-34/ 590.26x 10 ^-3= 1.123 x10^-33m
but 1m = 100cm
1.123 x 10 ^-33 x 100 = 1.123 x 10^-31cm
de broglie wavelength of humming bird = 1.123 x 10 ^-31cm
0.040 mol / dm³. (2 sig. fig.)
<h3>Explanation</h3>
in this question acts as a weak base. As seen in the equation in the question, produces 
rather than 
when it dissolves in water. The concentration of will likely be more useful than that of for the calculations here.
Finding the value of ![[\text{OH}^{-}]](https://tex.z-dn.net/?f=%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D)
from pH:
Assume that 
,
.
![[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}](https://tex.z-dn.net/?f=%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D10%5E%7B-%5Ctext%7BpOH%7D%7D%20%3D10%5E%7B-2.80%7D%20%3D%201.59%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7Bdm%7D%5E%7B-3%7D)
.
Solve for ![[(\text{CH}_3)_3\text{N}]_\text{initial}](https://tex.z-dn.net/?f=%5B%28%5Ctext%7BCH%7D_3%29_3%5Ctext%7BN%7D%5D_%5Ctext%7Binitial%7D)
:
![\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D_%5Ctext%7Bequilibrium%7D%5Ccdot%5B%28%5Ctext%7BCH%7D_3%29_3%5Ctext%7BNH%7D%5E%7B%2B%7D%5D_%5Ctext%7Bequilibrium%7D%7D%7B%5B%28%5Ctext%7BCH%7D_3%29_3%5Ctext%7BN%7D%5D_%5Ctext%7Bequilibrium%7D%7D%20%3D%20%5Ctext%7BK%7D_b%20%3D%201.58%5Ctimes%2010%5E%7B-3%7D)
Note that water isn't part of this expression.
The value of Kb is quite small. The change in is nearly negligible once it dissolves. In other words,
![[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}](https://tex.z-dn.net/?f=%5B%28%5Ctext%7BCH%7D_3%29_3%5Ctext%7BN%7D%5D_%5Ctext%7Binitial%7D%20%3D%20%5B%28%5Ctext%7BCH%7D_3%29_3%5Ctext%7BN%7D%5D_%5Ctext%7Bfinal%7D)
.
Also, for each mole of produced, one mole of 
was also produced. The solution started with a small amount of either species. As a result,
![[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}](https://tex.z-dn.net/?f=%5B%28%5Ctext%7BCH%7D_3%29_3%5Ctext%7BNH%7D%5E%7B%2B%7D%5D%20%3D%20%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%2010%5E%7B-2.80%7D%20%3D%201.58%5Ctimes%2010%5E%7B-3%7D%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7Bdm%7D%5E%7B-3%7D)
.
![\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D_%5Ctext%7Bequilibrium%7D%5Ccdot%5B%28%5Ctext%7BCH%7D_3%29_3%5Ctext%7BNH%7D%5E%7B%2B%7D%5D_%5Ctext%7Bequilibrium%7D%7D%7B%5B%28%5Ctext%7BCH%7D_3%29_3%5Ctext%7BN%7D%5D_%5Ctextbf%7Binitial%7D%7D%20%3D%20%5Ctext%7BK%7D_b%20%3D%201.58%5Ctimes%2010%5E%7B-3%7D)
,
![[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}](https://tex.z-dn.net/?f=%5B%28%5Ctext%7BCH%7D_3%29_3%5Ctext%7BN%7D%5D_%5Ctextbf%7Binitial%7D%20%3D%5Cdfrac%7B%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D_%5Ctext%7Bequilibrium%7D%5Ccdot%5B%28%5Ctext%7BCH%7D_3%29_3%5Ctext%7BNH%7D%5E%7B%2B%7D%5D_%5Ctext%7Bequilibrium%7D%7D%7B%5Ctext%7BK%7D_b%7D)
,
![[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}](https://tex.z-dn.net/?f=%5B%28%5Ctext%7BCH%7D_3%29_3%5Ctext%7BN%7D%5D_%5Ctext%7Binitial%7D%20%3D%5Cdfrac%7B%281.58%5Ctimes10%5E%7B-3%7D%29%5E%7B2%7D%7D%7B6.3%5Ctimes10%5E%7B-5%7D%7D%20%3D%200.040%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7Bdm%7D%5E%7B-3%7D)
.
Answer:
2.78 x 10²³
Explanation:
1 mole contains 6.02 x 10²³ hydrogen atoms => 0.46 mole contains 0.46(6.02 x 10²³) hydrogen atoms or 2.78 x 10²³ atoms.
Caution => When to use H vs H₂ => This problem is specific for 'hydrogen atoms' but some may simply say hydrogen. In such cases use H₂ or 'molecular hydrogen' is the focus. it's a matter of semantics, H vs H₂.
Answer:
the energy of an electron and the most probable distance of the electron from the nucleus.
Explanation:
The principal quantum number, n, describes the energy of an electron and the most probable distance of the electron from the nucleus. In other words, it refers to the size of the orbital and the energy level an electron is placed in. The number of subshells, or l, describes the shape of the orbital.
Glad I could help!!
