Question 17:
False; cutting hair would change what it looks like but, braiding it keeps it the same substance it is.
Question 18 :
My best guess would either be A or D. I would lean more towards D because if there are different mixtures then it depends on what you are mixing.
Hope it helps in some way. <3
Answer:
The other electron must have anticlockwise spin.
Explanation:
According to the pauli exclusion principle, the two elecrton present in same orbital must have opposite spin.
If the one electron is clockwise the other must be in anti clockwise direction. The clockwise direction is represented by the sign +1/2 while anti clockwise direction is represented by -1/2.
According the pauli principle, the two electrons must have different fourth electronic quantum number. The electron in same orbital have same first three quantum number i.e, n=1 l=0 and ml =0 in case of first subshell.
Answer:
The molar mass of the liquid 62.89 g/mol
Explanation:
Step 1: Data given
Mass of the sample = 0.1 grams
Temperature = 70°C
Volume = 750 mL
Pressure = 0.05951 atm
Step 2: Calculate the number of moles
p*V = n*R*T
n = (p*V)/(R*T)
⇒ with n = the number of moles gas = TO BE DETERMINED
⇒ with p = The pressure = 0.05951 atm
⇒ with V = The volume of the flask = 750 mL = 0.750 L
⇒ with R = The gasconstant = 0.08206 L*atm/K*mol
⇒with T = the temperature = 70 °C = 343 Kelvin
n = (0.05951 *0.750)/(0.08206*343)
n = 0.00159 moles
Step 3: Calculate molar mass
Molar mass = mass / moles
Molar mass =0.1 gram / 0.00159 moles
Molar mass = 62.89 g/mol
The molar mass of the liquid 62.89 g/mol
Answer:
V2 = 2.88L
Explanation:
P1= 78atm, V1= 2L, T1= 900K, P2= 45atm, V2=? T2= 750K
Applying the general gas equation
P1V1/T1 = P2V2/T2
Substitute the above
(78*2)/900= (45*V2)/750
V2= (78*2×750)/(900*45)
V2= 2.88L
<u>Answer:</u> The equilibrium concentration of
is 1.285 M.
<u>Explanation:</u>
The chemical equation for the decomposition of phosphorus pentachloride follows:

The expression for equilibrium constant is given as:
![K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BPCl_3%5D%5BCl_2%5D%7D%7B%5BPCl_5%5D%7D)
We are given:

![[PCl_3]=0.18M](https://tex.z-dn.net/?f=%5BPCl_3%5D%3D0.18M)
![[Cl_2]=0.30M](https://tex.z-dn.net/?f=%5BCl_2%5D%3D0.30M)
The concentration of solid substances are taken to be 1. Thus, they do not appear in the equilibrium constant expression.
Putting values in above equation, we get:
![0.042=\frac{0.18\times 0.30}{[PCl_5]}](https://tex.z-dn.net/?f=0.042%3D%5Cfrac%7B0.18%5Ctimes%200.30%7D%7B%5BPCl_5%5D%7D)
![[PCl_5]=1.285](https://tex.z-dn.net/?f=%5BPCl_5%5D%3D1.285)
Hence, the equilibrium concentration of
is 1.285 M.