Answer:
<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>
Explanation:
given amount of salt at time t is A(t)
initial amount of salt =300 gm =0.3kg
=>A(0)=0.3
rate of salt inflow =5*0.4= 2 kg/min
rate of salt out flow =5*A/(200)=A/40
rate of change of salt at time t , dA/dt= rate of salt inflow- ratew of salt outflow

integrating factor

integrating factor 
multiply on both sides by 

integrate on both sides
b)
after long period of time means t - > ∞

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>
Answer:
The answer to your question is 92.7%
Explanation:
Balanced Chemical reaction
3 Zn + Fe₂(SO₄)₃ ⇒ 2Fe + 3ZnSO₄
Molecular weight
Zinc = 65.4 x 3 = 196.2g
Iron (III) = 56 x 2 = 112 g
Proportions
196.2 g of Zinc ------------------ 112 g of Iron
20.4 g of Zinc ----------------- x
x = (20.4 x 112) / 196.2
x = 2284.8/196.2
x = 11.65 g of Iron
% yield = 
% yield = 0.927 x 100
% yield = 92.7
The rows are the one line of elements that goes from left to right horizontally. A column is a line of elements that goes vertically.
Answer:
heat flow
Explanation:
heat flow moves to a higher temperature to a lower temperature
Answer:
0.4 M
Explanation:
Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.
Because there's no O₂ in the beginning, the NO will decompose:
N₂(g) + O₂(g) ⇄ 2NO(g)
0.30 0 0.70 Initial
+x +x -2x Reacts (the stoichiometry is 1:1:2)
0.30+x x 0.70-2x Equilibrium
The equilibrium concentrations are the number of moles divided by the volume (0.250 L):
[N₂] = (0.30 + x)/0.250
[O₂] = x/0.25
[NO] = (0.70 - 2x)/0.250
K = [NO]²/([N₂]*[O₂])
K = 
7.70 = (0.70-2x)²/[(0.30+x)*x]
7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)
4x² - 2.80x + 0.49 = 2.31x + 7.70x²
3.7x² + 5.11x - 0.49 = 0
Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70
x = 0.09 mol
Thus,
[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M