1. How many calories are required to warm 20.4 g of water from80°C to 100.0°C an
ID: 688856 • Letter: 1
Question
1. How many calories are required to warm 20.4 g of water from80°C to 100.0°C and convert it to steam at 100.0°C? Theheat of vaporization of water is 540. cal/g.2. Compounds of boron and hydrogen are remarkable, not onlyfor their unusual bonding, but also for their reactivity. With themore reactive halogens, for example, diborane(B2H6) forms trihalides even at lowtemperatures: B2H6(g) +6Cl2(g) 2BCl3(g) + 6HCl(g)
Horxn = -755.4 kJ
How much heat is released when 694.4 kg diborane reacts?
You can answer either one that you know.. help isappreciated.. THANKS!
2. Compounds of boron and hydrogen are remarkable, not onlyfor their unusual bonding, but also for their reactivity. With themore reactive halogens, for example, diborane(B2H6) forms trihalides even at lowtemperatures: B2H6(g) +6Cl2(g) 2BCl3(g) + 6HCl(g)
Horxn = -755.4 kJ
How much heat is released when 694.4 kg diborane reacts? 2. Compounds of boron and hydrogen are remarkable, not onlyfor their unusual bonding, but also for their reactivity. With themore reactive halogens, for example, diborane(B2H6) forms trihalides even at lowtemperatures:
You can answer either one that you know.. help isappreciated.. THANKS!
Explanation / Answer
20.4 g *(1 cal/g/C) *(100 C - 80 C) + 20.4 g *(540 cal/g) = 11424 cal ~ 1.14 x 104 cal 2) 694.4 g diborane *(1 mol diborane/ 27.67 g )*(755.4 kJ/1 moldiborane) = 1.896 x 104 kJ
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