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!!!
76
114
190
266
Answer:
it B 114 on edge 2020
Explanation:
UwU
Answer:
it B 114
Explanation:
Answer:
2 litres of pure alcohol will be added to make the overall concentration of 9 litres of mixture as 30%.
Explanation:
Suppose x is the number of litres added to the 10% mixture than the quantity of new mixture is given as below
Also the quantity of alcohol is given as
Now the equation is as
So 2 litres of pure alcohol will be added to make the overall concentration of 9 litres of mixture as 30%.
C in CH4 :
H in CH4 :
O in O2 :
C in CO2 :
O in CO2 :
H in H2O :
O in H2O :
Which atom is reduced?
Which atom is oxidized?
The oxidation numbers of the atoms of the specified elements in each of the given atoms are;
1) -4
1) -42) +1
1) -42) +13) 0
1) -42) +13) 04) +4
1) -42) +13) 04) +45) -2
1) -42) +13) 04) +45) -26) +1
1) -42) +13) 04) +45) -26) +17) -2
1) -42) +13) 04) +45) -26) +17) -2Atom oxidized = C
1) -42) +13) 04) +45) -26) +17) -2Atom oxidized = CAtom reduced = O
1) C in CH4
To get the oxidation number of C;
Oxidation state of hydrogen atom is +1 and so if the oxidation state of C is x, then we have;
x + 4(+1) = 0
x + 4 = 0
x = -4
2) H in CH4
Oxidation state on Carbon atom in this case is -4. Thus;
-4 + 4x = 0
4x = 4
x = +1
3) O in O2
This is oxygen gas that exists in it's free state and as such oxidation number is 0.
4) C in CO2
Oxidation state of O here is -2. Thus;
x + 2(-2) = 0
x - 4 = 0
x = +4
5) O in CO2
Oxidation state of C is +4 here. Thus;
4 + 2x = 0
2x = -4
x = -4/2
x = -2
6) H in H2O
Oxidation state of oxygen here is -2. Thus;
2x - 2 = 0
2x = 2
x = 2/2
x = +1
7) O in H2O
Oxidation state of hydrogen here is +1. Thus;
2(1) + x = 0
x = -2
Finally, oxidation number of carbon increased, then it is the atom that was oxidized while the atom reduced is the Oxygen atom.
Read more at; brainly.com/question/22816291
Answer:
1. -4
2. +1
3. 0
4. +4
5. -2
6. +1
7. -2
reduced = H
oxidized = O
Explanation:
Know oxidation rules.
- Hope this helped! Please let me know if you would like to learn this. I could show you the rules and help you work through them.
Answer:
.
Explanation:
Answer:
You could collect the mixture and pour it in water, stir it , ad filter out the sand. This uses the physical property of solubility.
Explanation:
The salt dissolved, the sand didn't.
Alkaline Earth Metals
Alkali metals
Noble Gases
Lanthanides
Answer:
alkali metals- Group 1
Explanation:
they have less valence electrons and therefore are more reactive