Answer:
a) 90 kg
b) 68.4 kg
c) 0 kg/L
Explanation:
Mass balance:
w is the mass flow
m is the mass of salt
v is the volume flow
C is the concentration
![-[ln(2000L+3*L/min*t)-ln(2000L)]=ln(m)-ln(90kg)](https://tex.z-dn.net/?f=-%5Bln%282000L%2B3%2AL%2Fmin%2At%29-ln%282000L%29%5D%3Dln%28m%29-ln%2890kg%29)
![-ln[(2000L+3*L/min*t)/2000L]=ln(m/90kg)](https://tex.z-dn.net/?f=-ln%5B%282000L%2B3%2AL%2Fmin%2At%29%2F2000L%5D%3Dln%28m%2F90kg%29)
![m=90kg*[2000L/(2000L+3*L/min*t)]](https://tex.z-dn.net/?f=m%3D90kg%2A%5B2000L%2F%282000L%2B3%2AL%2Fmin%2At%29%5D)
a) Initially: t=0
![m=90kg*[2000L/(2000L+3*L/min*0)]=90kg](https://tex.z-dn.net/?f=m%3D90kg%2A%5B2000L%2F%282000L%2B3%2AL%2Fmin%2A0%29%5D%3D90kg)
b) t=210 min (3.5 hr)
![m=90kg*[2000L/(2000L+3*L/min*210min)]=68.4kg](https://tex.z-dn.net/?f=m%3D90kg%2A%5B2000L%2F%282000L%2B3%2AL%2Fmin%2A210min%29%5D%3D68.4kg)
c) If time trends to infinity the division trends to 0 and, therefore, m trends to 0. So, the concentration at infinit time is 0 kg/L.
Mass of KCl= 1.08 g
<h3>Further explanation</h3>
Given
1 g of K₂CO₃
Required
Mass of KCl
Solution
Reaction
K₂CO₃ +2HCl ⇒ 2KCl +H₂O + CO₂
mol of K₂CO₃(MW=138 g/mol) :
= 1 g : 138 g/mol
= 0.00725
From the equation, mol ratio K₂CO₃ : KCl = 1 : 2, so mol KCl :
= 2/1 x mol K₂CO₃
= 2/1 x 0.00725
= 0.0145
Mass of KCl(MW=74.5 g/mol) :
= mol x MW
= 0.0145 x 74.5
= 1.08 g
since the concentration of Carbon Dioxide will increase, it would make Q > K, cause equilibrium to shift in the direction with less moles of gas to alleviate the extra pressure. In this case, the reaction will shift left because there are fewer moles of gas present.
Answer:
option (B) is correct
Explanation:
In case of nuclear reactors first the nuclear energy is emitted due to the nuclear fission of heavy elements.
This nuclear energy is emitted in the form of heat energy.
This heat energy is used to rotate the turbines, that means it is converted in the form of mechanical energy and then finally this mechanical energy is converted into electrical energy.
Answer:
you must add 50 mL
Explanation:
Hi
KOH is a strong base and by adding 100mL 0.05M you will have an amount of 5 millimol.
NaCN is a base and by adding 50 mL 0,150 M you will have an amount of 7,5 mmol.
HCl is a acid and by adding 200 mL 0,075 M you will have an amount of 15 mmol.
The acid reacts with the bases leaving 2.5 mmol unreacted.
Na3PO4 is a base and by adding 50 mL 0,1 M you will have an amount of 5 mmol.
The 2.5 mmol of acid react with the base PO4 ^ -3 forming a regulatory solution of PO4 ^ -3 and HPO4 ^ -2 of pKa 2.12
5 mmol of acid (HNO3) must be added to obtain a regulatory solution formed by the same amount of HPO4 ^ -2 (2.5 mmol) and H2PO4 ^ -1 (2.5 mmol) with pKa 7.21
Considering a quantity of 5 mmol of HNO3 of concentration 0.1 M, 50 mL must be added.
To calculate the pH of the regulatory solution you should consider pH = pKa × Ca / Cb pH = 7.21 × 2.5 / 2.5 = 7.21 Being in the same solution the volume is the same and can be simplified to achieve a faster calculation.
successes with your homework
