The bonds of a glucose molecule store chemical energy
<span>Answer: 0.00649M
The question is incomplete,
</span>
<span>You are told that the first ionization of the sulfuric acid is complete and the second ionization of the sulfuric acid has a constant Ka₂ = 0.012
</span>
<span>
With that you can solve the question following these steps"
</span>
<span>1) First ionization:
</span>
<span>
H₂SO₄(aq) --> H⁺ (aq) + HSO₄⁻ (aq)
Under the fully ionization assumption the concentration of HSO4- is the same of the acid = 0.01 M
2) Second ionization
</span>
<span>HSO₄⁻ (aq) ⇄ H⁺ + SO₄²⁻ with a Ka₂ = 0.012
</span>
<span>Do the mass balance:
</span>
<span><span> HSO₄⁻ (aq) H⁺ SO₄²⁻</span>
</span>
<span /><span /><span> 0.01 M - x x x
</span><span>Ka₂ = [H⁺] [SO₄²⁻] / [HSO₄⁻]</span>
<span /><span>
=> Ka₂ = (x²) / (0.01 - x) = 0.012
</span><span />
<span>3) Solve the equation:
</span><span>x² = 0.012(0.01 - x) = 0.00012 - 0.012x</span>
<span /><span>
x² + 0.012x - 0.0012 = 0
</span><span />
<span>Using the quadratic formula: x = 0.00649
</span><span />
<span>So, the requested concentratioN is [SO₄²⁻] = 0.00649M</span>
Answer: False
Explanation: The nucleus of an atom only contains the protons and neutrons.
The electrons are not found in the nucleus,
they are orbiting the nucleus in different shells.
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>
Hey there!
The answer as well as the explanation is in the image attached. Let me know if there's anything you're unable to see.
Hope this helps!
