The answer would be C.
The salt and the water have both undergone physical changes.
Hope this helps!
Explanation:
The given data is as follows.
Heat transfer coefficient (h) = 12
Plate temperature (
) = 
= 303 K
Steady state temperature (
) = ?
Hence, formula applied for steady state is as follows.
= 
Putting the given values into the above formula as follows.
=
= ![5.67 \times 10^{-8} \times [(30 + 273)^{4} - T^{4}_{2}]](https://tex.z-dn.net/?f=5.67%20%5Ctimes%2010%5E%7B-8%7D%20%5Ctimes%20%5B%2830%20%2B%20273%29%5E%7B4%7D%20-%20T%5E%7B4%7D_%7B2%7D%5D)
= 282.66 K
= (282.66 -273)
= 9.66
Thus, we can conclude that the steady state temperature will be 9.66 .
Answer: Francesco Redis Experiment part 1 : The maggots were not observed in setup B and setup C because the jars were covered.
part 2 : He discovered that maggots appeared on the meat in the control jar, the jar left open. No maggots appeared in the jar covered with gauze.
Pasteur Experiment
part 1:In the pastures Experiment the microorganisms come from the air.
part 2:Louis Pasteur's pasteurization experiment illustrates the fact that the spoilage of liquid was caused by particles in the air rather than the air itself.
Explanation:
Explanation:
Heat is a form of thermal energy.
Heat is the sum of all the energy of the molecular motion in an object.
Temperature measures the average heat possessed by each molecule in a given substance.
Molecules at a higher temperature possess more kinetic energy and they will move faster. This kinetic energy form is the heat variant of thermal energy.
Temperature is the measure of this heat energy of molecules.
The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of 
) =-1275.0
enthalpy of combustion of oxygen(Δ of 
) = zero
enthalpy of combustion of carbon dioxide(Δ of 
) = -393.5
enthalpy of combustion of water(Δ of 
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ (products) - ∈Δ (reactants)
(s) +6 (g) → 6 (g)+ 6 (l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
