Just use the Heisenberg Uncertainty principle:
<span>ΔpΔx = h/2*pi </span>
<span>Δp = the uncertainty in momentum </span>
<span>Δx = the uncertainty in position </span>
<span>h = 6.626e-34 J s (plank's constant) </span>
<span>Hint: </span>
<span>to calculate Δp use the fact that the uncertainty in the momentum is 1% (0.01) so that </span>
<span>Δp = mv*(0.01) </span>
<span>m = mass of electron </span>
<span>v = velocity of electron </span>
<span>Solve for Δx </span>
<span>Δx = h/(2*pi*Δp) </span>
<span>And that is the uncertainty in position. </span>
Three resonance structures can be drawn for the allyl cation while two resonance structures can be drawn for the amidate ion.
Sometimes, we cannot fully describe the bonding in a chemical specie using a single chemical structure. In such cases, we have to use a number of structures which cooperatively represent the actual bonding in the molecule. These structures are called resonance or canonical structures.
The resonance structures of the allyl cation and the amidate ion are shown in the images attached to this answer. These structures show the different bonding extremes in these organic ions.
Learn more: brainly.com/question/4933048
Answer:
51207 torr is the new pressure of the gas
Explanation:
We can solve this question using combined gas law that states:
P1V1T2 = P2V2T1
<em>Where P is pressure, V volume and T absolute temperature of 1, initial state and 2, final state of the gas</em>
<em> </em>
Computing the values of the problem:
P1 = 710torr
V1 = 5.0x10²mL
T1 = 273.15 + 30°C = 303.15K
P2 = ?
V2 = 25mL
T2 = 273.15 + 820°C = 1093.15K
Replacing:
710torr*5.0x10²mL*1093.15K = P2*25mL*303.15K
3.881x10⁸torr*mL*K = P2 * 7.579x10³mL*K
P2 = 51207 torr is the new pressure of the gas
Answer:
E = 1.602v
Explanation:
Use the Nernst Equation => E(non-std) = E⁰(std) – (0.0592/n)logQc …
Zn⁰(s) => Zn⁺²(aq) + 2 eˉ
2Ag⁺(aq) + 2eˉ=> 2Ag⁰(s)
_____________________________
Zn⁰(s) + 2Ag⁺(aq) => Zn⁺²(aq) + 2Ag(s)
Given E⁰ = 1.562v
Qc = [Zn⁺²(aq)]/[Ag⁺]² = (1 x 10ˉ³)/(0.150)² = 0.044
E = E⁰ -(0.0592/n)logQc = 1.562v – (0.0592/2)log(0.044) = 1.602v
Answer:
Spiral Galaxy
Explanation:
There are four main categories of galaxies: elliptical, spiral, barred spiral, and irregular. These types of galaxies are further divided into subcategories while at the same time other types of galaxies exist based on their size and other unique features. The most common type of galaxy found throughout the universe is the spiral galaxy.
