The 7160 cal energy is required to melt 10. 0 g of ice at 0. 0°C, warm it to 100. 0°C and completely vaporize the sample.
Calculation,
Given data,
Mass of the ice = 10 g
Temperature of ice = 0. 0°C
- The ice at 0. 0°C is to be converted into water at 0. 0°C
Heat required at this stage = mas of the ice ×latent heat of fusion of ice
Heat required at this stage = 10 g×80 = 800 cal
- The temperature of the water is to be increased from 0. 0°C to 100. 0°C
Heat required for this = mass of the ice×rise in temperature×specific heat of water
Heat required for this = 10 g×100× 1 = 1000 cal
- This water at 100. 0°C is to be converted into vapor.
Heat required for this = Mass of water× latent heat
Heat required for this = 10g ×536 =5360 cal
Total energy or heat required = sum of all heat = 800 +1000+ 5360 = 7160 cal
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Answer:The functional groups in an organic compound can frequently be deduced from its infrared absorption spectrum. A compound, C5H10O2, exhibits strong, broad absorption across the 2500-3200 cm^1 region and an intense absorption at 1715 cm'^-1. Relative absorption intensity: (s)=strong, (m)-medium, (w) weak. What functional class(cs) docs the compound belong to List only classes for which evidence is given here. Attach no significance to evidence not cited explicitly. Do not over-interpret exact absorption band positions. None of your inferences should depend on small differences like 10 to 20 cm^1. The functional class(es) of thla compound is(are) alkane (List only if no other functional class applies.) alkene terminal alkyne internal alkyne arene alcohol ether amine aldehyde or ketone carboxylic acid ester nitr
Step 1 - Discovering the ionic formula of Chromium (III) Carbonate
Chromium (III) Carbonate is formed by the ionic bonding between Chromium (III) (Cr(3+)) and Carbonate (CO3(2-)):
Step 2 - Finding the molar mass of the substance
To find the molar mass, we need to multiply the molar mass of each element by the number of times it appears in the formula of the substance and, finally, sum it all up.
The molar masses are 12 g/mol for C; 16 g/mol for O and 52 g/mol for Cr. We have thus:
The molar mass will be thus:
Step 3 - Finding the percent composition of carbon
As we saw in the previous step, the molar mass of Cr2(CO3)3 is 284 g/mol. From this molar mass, 36 g/mol come from C. We can set the following proportion:
The percent composition of Carbon is thus 12.7 %.
Answer: Sorry I try to download it but it not showing anything, if you take a picture of the work maybe I would do it....
Explanation:
