Answer:
3.711 L
Explanation:
The formula you need to use is the following:
3.4L / 298 K = V2 / 273 K
V2 = 3.711 L
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:
49.35 mL
Explanation:
Given: 56.2 mL of gas
To find: volume that 56.2 mL of gas at 820 mm of Hg would occupy at 720 mm of Hg
Solution:
At 820 mm of Hg, volume of gas is 56.2 mL
At 1 mm of Hg, volume of gas is 
At 720 mm of Hg, volume of gas is 
Answer:
Zn 3
s 1
o 4
Zn. zine
s. sulphate
o oxygen
3znso4. zine sulphateoxide
