Answer:
![\boxed {\boxed {\sf 10.2 \ kJ}}](https://tex.z-dn.net/?f=%5Cboxed%20%7B%5Cboxed%20%7B%5Csf%2010.2%20%5C%20kJ%7D%7D)
Explanation:
We are asked to find how many kilojoules of energy would be required to heat a block of aluminum.
We will use the following formula to calculate heat energy.
![q=mc \Delta T](https://tex.z-dn.net/?f=q%3Dmc%20%5CDelta%20T)
The mass (m) of the aluminum block is 225 grams and the specific heat (c) is 0.897 Joules per gram degree Celsius. The change in temperature (ΔT) is the difference between the final temperature and the initial temperature.
- ΔT = final temperature - inital temperature
The aluminum block was heated from 23.0 °C to 73.5 °C.
- ΔT= 73.5 °C - 23.0 °C = 50.5 °C
Now we know all three variables and can substitute them into the formula.
- m= 225 g
- c= 0.897 J/g° C
- ΔT= 50.5 °C
![q= (225 \ g )(0.897 \ J/g \textdegree C)(50.5 \textdegree C)](https://tex.z-dn.net/?f=q%3D%20%28225%20%5C%20g%20%29%280.897%20%5C%20J%2Fg%20%5Ctextdegree%20C%29%2850.5%20%5Ctextdegree%20C%29)
Multiply the first two numbers. The units of grams cancel.
![q= (225 \ g * 0.897 \ J/g \textdegree C)(50.5 \textdegree C)](https://tex.z-dn.net/?f=q%3D%20%28225%20%5C%20g%20%20%2A%200.897%20%5C%20J%2Fg%20%5Ctextdegree%20C%29%2850.5%20%5Ctextdegree%20C%29)
![q= (225 * 0.897 \ J / \textdegree C)(50.5 \textdegree C)](https://tex.z-dn.net/?f=q%3D%20%28225%20%20%20%2A%200.897%20%5C%20J%20%2F%20%5Ctextdegree%20C%29%2850.5%20%5Ctextdegree%20C%29)
![q= (201.825\ J / \textdegree C)(50.5 \textdegree C)](https://tex.z-dn.net/?f=q%3D%20%28201.825%5C%20J%20%2F%20%5Ctextdegree%20C%29%2850.5%20%5Ctextdegree%20C%29)
Multiply again. This time, the units of degrees Celsius cancel.
![q= 201.825 \ J * 50.5](https://tex.z-dn.net/?f=q%3D%20201.825%20%5C%20J%20%2A%2050.5)
![q= 10192.1625 \ J](https://tex.z-dn.net/?f=q%3D%2010192.1625%20%5C%20J)
The answer asks for the energy in kilojoules, so we must convert our answer. Remember that 1 kilojoule contains 1000 joules.
![\frac { 1 \ kJ}{ 1000 \ J}](https://tex.z-dn.net/?f=%5Cfrac%20%7B%201%20%20%5C%20kJ%7D%7B%201000%20%5C%20J%7D)
Multiply by the answer we found in Joules.
![10192.1625 \ J * \frac{ 1 \ kJ}{ 1000 \ J}](https://tex.z-dn.net/?f=10192.1625%20%5C%20J%20%2A%20%5Cfrac%7B%201%20%5C%20kJ%7D%7B%201000%20%5C%20J%7D)
![10192.1625 * \frac{ 1 \ kJ}{ 1000 }](https://tex.z-dn.net/?f=10192.1625%20%20%2A%20%5Cfrac%7B%201%20%5C%20kJ%7D%7B%201000%20%7D)
![\frac {10192. 1625}{1000} \ kJ](https://tex.z-dn.net/?f=%5Cfrac%20%7B10192.%201625%7D%7B1000%7D%20%5C%20kJ)
![10.1921625 \ kJ](https://tex.z-dn.net/?f=10.1921625%20%5C%20kJ)
The original values of mass, temperature, and specific heat all have 3 significant figures, so our answer must have the same. For the number we found, that is the tneths place. The 9 in the hundredth place tells us to round the 1 up to a 2.
![10.2 \ kJ](https://tex.z-dn.net/?f=10.2%20%5C%20kJ)
Approximately <u>10.2 kilojoules</u> of energy would be required.
D) if stuff is changing then the reaction is hardly in equilibrium is it? Everything is just chilling at equilibrium so there would be constant concentration
Many atoms if they are radioactive isotopes will loss protons and neutrons as radiation in order to gain and become more stable
hope that helps
Answer: The density of chloroform is 1.47 g/mL
Explanation : Given,
Volume = 40.5 mL
Mass of cylinder = 85.16 g
Mass of cylinder and liquid = 145.10 g
First we have to calculate the mass of liquid (chloroform).
Mass of liquid = Mass of cylinder and liquid - Mass of cylinder
Mass of liquid = 145.10 g - 85.6 g
Mass of liquid = 59.5 g
Now we have to calculate the density of liquid (chloroform).
Formula used:
![Density=\frac{Mass}{Volume}](https://tex.z-dn.net/?f=Density%3D%5Cfrac%7BMass%7D%7BVolume%7D)
Now putting g all the given values in this formula, we get:
![Density=\frac{59.5g}{40.5mL}](https://tex.z-dn.net/?f=Density%3D%5Cfrac%7B59.5g%7D%7B40.5mL%7D)
![Density=1.47g/mL](https://tex.z-dn.net/?f=Density%3D1.47g%2FmL)
Therefore, the density of chloroform is 1.47 g/mL