Answer:
9.2 L is the average volume should the chemical engineer report.
Explanation:
Volume of pollutant from Cross creek plants 10.88 L
Volume of pollutant from Oglala plants = 0.92 L
Volume of pollutant from Platte plants = 15.82 L
Average volume pollutants will be given by :
9.2 L is the average volume should the chemical engineer report.
RIPPLE MARKS are produced by flowing water or wave action, analogous to cross-bedding
Answer:
A. 2.4 x 10¹⁴ J
Explanation:
m (mass) = 2.7 x 10⁻³ kg
c = 3.0 x 10⁸ m/s
Plug these numbers into the equation E=mc²
E=mc²
E= (2.7 x 10⁻³)(3.0 x 10⁸)²
E= 2.43 x 10¹⁴
So, the answer is A. 2.4 x 10¹⁴ J.
Answer:
Moles of NO = 0.18
Moles of Br2 = 0.18
Explanation:
The given reaction with ICE table is:
Initial 0.64 - -
Change -2x +2x +x
Equilb (0.64-2x) 2x 2x
At equilibrium:
[NOBr] = 0.64-2x
NO = Br2 = 2x
The equilibrium moles of NOBr = 0.46
Moles of NO = Moles of Br2 = 2x = 2(0.09) = 0.18 moles
Oh ok thanks so where the picture