Answer:
Esters have lower boiling point than alcohols.
Explanation:
Esters are the fruity smelling compounds which are formed from carboxylic acid and alcohol with the removal of water.
The general formula for the ester is RCOOR' which is prepared from RCOOH acid and R'OH alcohol.
Ester can not form strong hydrogen bond as there is no hydrogen attached to the electronegative atom in the ester and thus cannot form hydrogen bonds with each other.<u> Due to this factor, the interactions within the molecules of the ester is lower than that of alcohols which exist in strong hydrogen bonding. As a result, ester can be easily boiled when compared to the alcohols and thus they have lower value of boiling points.</u>
Answer:
Conversion factor;
Molar mass;
Avogadro's constant and molar mass
Explanation:
- Firstly, an intermediate step is to define the conversion factor that will be then used in a conversion technique called dimensional analysis in order to convert from one unit to another. An example of a conversion factor would be, for example, 1 L = 1000 mL, which can be manipulated as a fraction, either
or 
; - Secondly, in order to convert mass to moles, we need to know the molar mass of a compound which has a units of g/mol (that is, it shows how many grams we have per 1 mole of substance.
- Thirdly, Avogadro's constant,
tells us that there is 
number of molecules or atoms in 1 mole of substance. We need two conversion factors to convert the number of molecules to a mass: firstly, we need to convert the number of molecules into the number of moles using Avogadro's constant and then we need to use the molar mass to convert the moles obtained into mass.
Formula = Ca3P2
RAM of; Ca = 40 x 3 = 120g, P = 30 x 2 = 60g. 120 + 60 = 180g
Mole = 493.4g / 180g = 2.74mols
Answer:
a. 50KCal
b. 400KCal
c. Same as (a) above
Explanation:
Given
To raise the temperature of 1kg of liquid water at 1°C requires 1KCal
To raise the temperature of 1kg of ice or water vapour by 1°C requires 0.5KCal
To melt 1kg of ice at 0°C requires 80KCal
To evaporate 1kg of liquid water sitting at 100°C requires 540KCal
a. How much heat is required to raise the temperature of 5 kg of liquid water by 20 C?
To raise the temperature of 5 kg of ice by 20°C requires:
5 kg * (0.5 kcal / kgC) * 20C
= 50 KCal
b. How much heat is required to melt 5 kg of ice at 0 C?
To melt an ice of 5 kg of ice at 0 C requires:
5 kg * (80 kcal / kg)
= 400 KCal
c. Same as (a) above
Answer:
I believe 1+
Explanation:
when Na loses 1 electron in the outer shell it has 8 valence protons on it's new most outer shell. so now it has 11 protons and 10 electrons. that extra proton (positively charged) adds one extra charge. so +1
