Answer:
A. 444.5 pm
Explanation:
We know that:
i.e.
in a face-centered cubic crystal, the number of atoms per unit cell is (n) = 4
The molar mass of manganese (II) oxide ![[Mn(11)O] = 70.93 \ g/mol](https://tex.z-dn.net/?f=%5BMn%2811%29O%5D%20%3D%2070.93%20%5C%20g%2Fmol)
Density 
is given as 5.365 g/cm³
Avogadro constant 
= 6.023 × 10²³ atoms/mol
∴
Making th edge length "a" the subject, we get:
![a= \sqrt[3]{8.78 \times 10^{-23} \ cm^3}](https://tex.z-dn.net/?f=a%3D%20%5Csqrt%5B3%5D%7B8.78%20%5Ctimes%2010%5E%7B-23%7D%20%5C%20cm%5E3%7D)
a = 4.445 × 10⁻⁸ cm
a = 444.5 pm
Given :
Rate = 75 m 1 s×60 s 1 min×1 h 60 m.
To Find :
Correct answer after conversion.
Solution :
We know, 1 min = 60 sec.
1 hour = 60 min = 60×60 sec = 3600 sec.
Putting value of min and hour in seconds , we get :
Hence, this is the required solution.
The ability to be dissolved, especially in water.
2 boxes of A
Because C = A + B
2 of A = 20 grams
at the other hand we have 2 of B = 10
So 20 + 10 = 30 grams
Answer:
Q = 3,534.4 lbm/s = 212,062 lbm/min
Explanation:
Mass flowrate of discharge or leakage mass flowrate (Q) is given as
Q = AC₀√(2ρgP)
A = Cross sectional Area of leakage = (πD²/4) = (π×0.7²)/4
A = 0.385 ft²
C₀ = discharge coefficient = 0.98 (For maximum discharge flow rate, the flow is turbulent with discharge coefficient within 1% of 0.98)
ρ = density of butane at 76°F = 35.771 lbm/ft³
g = acceleration due to gravity = 32.2 lbm.ft/lbf.s²
P = Gauge Pressure in the tank = (absolute pressure) - (external pressure) = 19 - 1 = 18 atm = 38091.9 lbf/ft²
Q = AC₀√(2ρgP)
Q = (0.385)(0.98)√(2×35.771×32.2×38091.9)
Q = 3,534.4 lbm/s = 212,062 lbm/min
Hope this Helps!!!
