A physical change is when something don't permanently change
2NaBr + Ca(OH)2 ➡️ CaBr2 + 2NaOH
In this compound (Phosgene) the central atom (carbon is Sp² Hybridized).
Sp, Sp² and Sp³ can be calculated very simply by doing three steps,
Step 1:
Assume triple bond and double bond as one bond and assign s or p to it. In this example carbon double bond oxygen is considered once and let suppose it is s. Now we are having our s.
Step 2:
Count lone pair of electron, each lone pair counts for s and p. In this case there is no lone pair of electron on carbon, so not included.
Step 3:
Count single bonds for s and p. As we have already assigned s to the double bond, now one p for one single bond, and other p for the other single bond.
Result:
So, we counted 1 s for double bond, 1 p for one single and other p for second single bond. As a whole we got,
Sp²
Practice:
You can practice for hybridization of Oxygen in this molecule. Oxygen has 2 lone pair of electrons. (Hint: Sp² Hybridization)
This problem is providing the heating curve of ethanol showing relevant data such as the initial and final temperature, melting and boiling points, enthalpies of fusion and vaporization and specific heat of solid, liquid and gaseous ethanol, so that the overall heat is required and found to be 1.758 kJ according to:
<h3>Heating curves:</h3>
In chemistry, we widely use heating curves in order to figure out the required heat to take a substance from a temperature to another. This process may involve sensible heat and latent heat, when increasing or decreasing the temperature and changing the phase, respectively.
Thus, since ethanol starts off solid and end up being a vapor, we will find five types of heat, three of them related to the heating-up of ethanol, firstly solid, next liquid and then vapor, and the other two to its fusion and vaporization as shown below:

Hence, we begin by calculating each heat as follows, considering 1 g of ethanol is equivalent to 0.0217 mol:
![Q_1=0.0217mol*111.5\frac{J}{mol*\°C}[(-114.1\°C)-(-200\°C)] *\frac{1kJ}{1000J} =0.208kJ\\ \\ Q_2=0.0217mol*4.9\frac{kJ}{mol} =0.106kJ\\ \\ Q_3=0.0217mol*112.4\frac{J}{mol*\°C}[(78.4\°C)-(-114.1\°C)] *\frac{1kJ}{1000J} =0.470kJ\\ \\ Q_4=0.0217mol*38.6\frac{kJ}{mol} =0.838kJ\\ \\ Q_5=0.0217mol*87.5\frac{J}{mol*\°C}[(150\°C)-(78.4\°C)] *\frac{1kJ}{1000J} =0.136kJ](https://tex.z-dn.net/?f=Q_1%3D0.0217mol%2A111.5%5Cfrac%7BJ%7D%7Bmol%2A%5C%C2%B0C%7D%5B%28-114.1%5C%C2%B0C%29-%28-200%5C%C2%B0C%29%5D%20%2A%5Cfrac%7B1kJ%7D%7B1000J%7D%20%3D0.208kJ%5C%5C%0A%5C%5C%0AQ_2%3D0.0217mol%2A4.9%5Cfrac%7BkJ%7D%7Bmol%7D%20%3D0.106kJ%5C%5C%0A%5C%5C%0AQ_3%3D0.0217mol%2A112.4%5Cfrac%7BJ%7D%7Bmol%2A%5C%C2%B0C%7D%5B%2878.4%5C%C2%B0C%29-%28-114.1%5C%C2%B0C%29%5D%20%2A%5Cfrac%7B1kJ%7D%7B1000J%7D%20%3D0.470kJ%5C%5C%0A%5C%5C%0AQ_4%3D0.0217mol%2A38.6%5Cfrac%7BkJ%7D%7Bmol%7D%20%3D0.838kJ%5C%5C%0A%5C%5C%0AQ_5%3D0.0217mol%2A87.5%5Cfrac%7BJ%7D%7Bmol%2A%5C%C2%B0C%7D%5B%28150%5C%C2%B0C%29-%2878.4%5C%C2%B0C%29%5D%20%2A%5Cfrac%7B1kJ%7D%7B1000J%7D%20%3D0.136kJ)
Finally, we add them up to get the result:

Learn more about heating curves: brainly.com/question/10481356