Answer: melting, freezing, sublimation, deposition, condensation, and vaporization
Answer:
Equilibrium concentration of 
is 12.5 M
Explanation:
Given reaction: 
Here, ![K_{c}=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}][H_{2}O]}](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%7B%5BC_%7B2%7DH_%7B5%7DOH%5D%7D%7B%5BC_%7B2%7DH_%7B4%7D%5D%5BH_%7B2%7DO%5D%7D)
where 
represents equilibrium constant in terms of concentration and species inside third bracket represent equilibrium concentrations
Here, ![[C_{2}H_{4}]=0.015M](https://tex.z-dn.net/?f=%5BC_%7B2%7DH_%7B4%7D%5D%3D0.015M)
, ![[C_{2}H_{5}OH]=1.69M](https://tex.z-dn.net/?f=%5BC_%7B2%7DH_%7B5%7DOH%5D%3D1.69M)
and 
So, ![[H_{2}O]=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}]\times K_{c}}=\frac{1.69}{0.015\times 9.0}=12.5M](https://tex.z-dn.net/?f=%5BH_%7B2%7DO%5D%3D%5Cfrac%7B%5BC_%7B2%7DH_%7B5%7DOH%5D%7D%7B%5BC_%7B2%7DH_%7B4%7D%5D%5Ctimes%20K_%7Bc%7D%7D%3D%5Cfrac%7B1.69%7D%7B0.015%5Ctimes%209.0%7D%3D12.5M)
Hence equilibrium concentration of is 12.5 M
Variance, standard deviation and interquartile range
Answer:
2.16 MeV
Explanation:
To determine the amount of work done that is needed to assemble the atomic mass; we need to apply the equation;
U = 
where:
= proportionality constant = 
e = magnitude of the charge of each electron = 
r = length of each side of the vertex = 
So; replacing our values into above equation; we have:
U = 
U = 3.456 × 10 ⁻¹³ J
If we have to convert our unit from J to Mev; then we are going to have:
U = 3.456 × 10 ⁻¹³ J 
U = 2.16 MeV
Therefore, the amount of work done needed to assemble an atomic nucleus = 2.16 MeV
Answer:
3.86x10^24 molecules
Explanation:
From the question given, the following data were obtained:
Volume of ice = 0.5cup
1 cup = 240mL
0.5cup = 0.5 x 240 = 120mL
Density of ice = 0.9167 g/mL
molar mass of water = 18.015g/mol
First let us calculate the mass of ice. This is illustrated below:
Density = Mass /volume
Mass = Density x volume
Mass = 0.9167 x 120
Mass of ice = 110.004g = Mass of water
From Avogadro's hypothesis, we understood that 1mole of any substance contains 6.02x10^23 molecules. This means that 1mole of water will also contain 6.02x10^23 molecules.
1mole of water = 18.015g
If 1 mole (18.015g) contains 6.02x10^23 molecules,
Therefore, 110.004g of water will contain = (110.004x6.02x10^23)/18.015 = 3.86x10^24 molecules
