Cooling and heating system involves the collection and dissipation of heat appropriately. Heat gradient is important for heat transfer between systems because heat travels from the warmer system to the cooler system. The right medium for heat transfer is also important for this exchange – A good conductor of heat is required.
<span>You need to have NAD+ as a source of oxidation for the pyruvate, as well as a supply of coenzyme A. CO2 is released by the pyruvate as a carboxyl group is removed</span>
Answer:
21.02moles of KBr
Explanation:
Parameters given:
Number of moles BaBr₂ = 10.51moles
Complete reaction equation:
BaBr₂ + K₂SO₄ → KBr + BaSO₄
Upon inspecting the given equation, we find out that the atoms are not balanced on both sides of the equation:
The balanced equation is:
BaBr₂ + K₂SO₄ → 2KBr + BaSO₄
From the equation:
1 mole of BaBr₂ produces 2 moles of KBr
∴ 10.51 moles of BaBr₂ will yield (2 x 10.51) moles = 21.02moles of KBr
102 grams.
Equation:
Quantify of heat = mass x specific heat x difference in temperature
We have: quantity of heat : 2300J
specific heat: .449 J/g
difference in t: 80 - 30 = 50
Solve for mass: 2300 = mass x 0.449 x 50
mass = 102.449
2 sig-figs --> 102 grams
<u>Answer:</u> The freezing point of solution is -0.454°C
<u>Explanation:</u>
Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.
The equation used to calculate depression in freezing point follows:

To calculate the depression in freezing point, we use the equation:

Or,

where,
Freezing point of pure solution = 0°C
i = Vant hoff factor = 2
= molal freezing point elevation constant = 1.86°C/m
= Given mass of solute (KCl) = 5.0 g
= Molar mass of solute (KCl) = 74.55 g/mol
= Mass of solvent (water) = 550.0 g
Putting values in above equation, we get:

Hence, the freezing point of solution is -0.454°C