My Apoligies for not being sooner.
Answer: The closer the particles get or the further apart they get, the greater the sound's amplitude. Sound amplitude causes a sound's loudness and intensity. The bigger the amplitude is, the louder and more intense the sound. - (This is copied)
General Answer : Its a change in sound, bassically the bigger the waves are, the louder the sound is.
Answer:
The density of Lithium β is 0.5798 g/cm³
Explanation:
For a face centered cubic (FCC) structure, there are total number of 4 atoms in the unit cell.
we need to calculate the mass of these atoms because density is mass per unit volume.
Atomic mass of Lithium is 6.94 g/mol
Then we calculate the mass of four atoms;
⇒next, we estimate the volume of the unit cell in cubic centimeter
given the edge length or lattice constant a = 0.43nm
a = 0.43nm = 0.43 X 10⁻⁹ m = 0.43 X 10⁻⁹ X 10² cm = 4.3 X 10⁻⁸cm
Volume of the unit cell = a³ = (4.3 X 10⁻⁸cm)³ = 7.9507 X 10⁻²³ cm³
⇒Finally, we calculate the density of Lithium β
Density = mass/volume
Density = (4.6097 X 10⁻²³ g)/(7.9507 X 10⁻²³ cm³)
Density = 0.5798 g/cm³
I know they are fishing and there is an alligator in the lake. The alligator might grab the bait and attack the fishers.
Hope I helped!
Let me know if you need anything else!
~ Zoe
Answer:
[Zn²⁺] = 4.78x10⁻¹⁰M
Explanation:
Based on the reaction:
ZnBr₂(aq) + K₂CO₃(aq) → ZnCO₃(s) + 2KBr(aq)
The zinc added produce the insoluble ZnCO₃ with Ksp = 1.46x10⁻¹⁰:
1.46x10⁻¹⁰ = [Zn²⁺] [CO₃²⁻]
We can find the moles of ZnBr₂ added = Moles of Zn²⁺ and moles of K₂CO₃ = Moles of CO₃²⁻ to find the moles of CO₃²⁻ that remains in solution, thus:
<em>Moles ZnB₂ (Molar mass: 225.2g/mol) = Moles Zn²⁺:</em>
6.63g ZnBr₂ * (1mol / 225.2g) = 0.02944moles Zn²⁺
<em>Moles K₂CO₃ = Moles CO₃²⁻:</em>
0.100L * (0.60mol/L) = 0.060 moles CO₃²⁻
Moles CO₃²⁻ in excess: 0.0600moles CO₃²⁻ - 0.02944moles =
0.03056moles CO₃²⁻ / 0.100L = 0.3056M = [CO₃²⁻]
Replacing in Ksp expression:
1.46x10⁻¹⁰ = [Zn²⁺] [0.3056M]
<h3>[Zn²⁺] = 4.78x10⁻¹⁰M</h3>
