Answer:
70.0°C
Explanation:
We are given;
- Amount of heat generated by propane as 104.6 kJ or 104600 Joules
- Mass of water is 500 g
- Initial temperature as 20.0 ° C
We are required to determine the final temperature of water;
Taking the initial temperature is x°C
We know that the specific heat of water is 4.18 J/g°C
Quantity of heat = Mass × specific heat × change in temperature
In this case;
Change in temp =(x-20)° C
Therefore;
104600 J = 500 g × 4.18 J/g°C × (x-20)
104600 J = 2090x -41800
146400 = 2090 x
x = 70.0479
=70.0 °C
Thus, the final temperature of water is 70.0°C
The relative humidity shows us saturation of water in air. Since it is 58% and not 100% we have to scale density that we were given when humidity is 100%.
At 100% humidity, density is 20grams/cubic meter
at 58% it will be:
0.58 * 20 = 11.6 g/m^3
Since the question is to find density, the answer cannot be in %.
Answer is 11.6 g/m^3
Answer:1.
Explanation: This reaction is catalyzes by pyruvate dehydrogenase. Pyruvate being the end product of glycolysis has many fates after glycolysis,one of which is to enter the TCA(Tricarboxylic acid cycle) cycle. It is first converted to actetate by the action of pyruvate dehydrogenase. This enzyme converts pyruvate to acetate releasing CO2 and NADH because this oxidative decarboxylation of pyruvate is coupled with reduction of NAD+ which can feed into the electron transport chain.
Answer:
Rolling down a steep hill
Explanation:
