Answer : The ions present in the solution of are and in aqueous state.
Explanation :
When is in aqueous solution then they dissociates into their ions.
The reaction in aqueous medium is,
The charge on potassium ion is +1 and on carbonate ion is -2. To neutralize the charge on carbonate ion, two potassium ion must be used.
Therefore, the ions present in solution of are and [tex]CO^{2-}_3[tex] in aqueous state.
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This reaction is known as respiration and in order to understand the stoichometry better, we first write the reaction equation:
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O
From the equation, we can see that the molar ratio between glucose and oxygen is 1 : 6. Therefore, we can see that if 6 moles of oxygen are consumed, one mole of glucose will be consumed.
Answer:
c- Hot water remains almost same but cold water temperature increases drastically.
Explanation:
In a process of heat exchange where a hot and a cold fluid are present, the amount of heat that is transfered from one to another, the heat lost by the hot fluid, is gained by the cold fluid. That can be calculated by the equation : Q = m * Cp * dT.
Where:
m= mass flowrate
Cp=specific-heat-capacity
dT= difference between the inlet and outlet temperatures
We can state the equality of this equation for both fluids:
(mass-flowrate * specific-heat-capacity * (temperature-in – temperature-out) )<em> for hot medium</em> = (mass-flowrate * specific-heat-capacity * (temperature-out – temperature-in) )<em> for cold medium</em>
Now if we increase the value of the mass flow rate of the hot fluid to its maximum and decrease the mass flowrate of the cold fluid to its minimum. That will mean that the difference between the inlet and outlet temperatures for the cold fluid must increase so as to compensate the increase on the other side of the equation and guarantee the equality. And taking in account that the inlet temperature of the cold fluid is known and what can change is the outlet one, the correct answer is c that says that the outlet temperature of the cold fluid increases drasically.