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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Fusion releases less energy than fission
fusion most commonly combines heavy isotopes of hydrogen into helium.
Explanation:
Nuclear fusion is a form of reaction that involves the combination of two light nuclei to form one that is heavier in mass.
The other form is nuclear fission which is the splitting of heavier nuclei either spontaneously or when bombarded with other nuclei.
- Nuclear fusion reactions in the core of stars powers the universe.
- The reactions produces a huge amount of energy of a greater and massive order than fission reaction.
- Small nuclei are involved in nuclear fusion and not the large ones.
- Nuclear fusion degenerates into series of chain reactions that are extremely difficult to control.
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Answer:
a) rate law1 = k[NO2]²
b) rate law2 = k[NO][O3]
Explanation:
NO2(g) + CO(g) → NO(g) + CO2(g)
NO(g) + O3(g) → NO2(g) + O2(g)
When [NO2] in reaction 1 is doubled, the reaction quadruples
Rxn is second order.
rate law1= [NO2]^a [CO]^b
rate law1= [NO2]² [CO]^0
rate law1 = k[NO2]²
When [NO] in reaction 2 is doubled, the rate doubles.
Rxn is first order
The ratio is 1:1
this makes the rate law2 = k[NO][O3]
