Answer: 3.69 × 10^27
Explanation:
Amount of energy required = 7.06 × 10^4 J
Frequency of microwave (f) = 2.88 × 10^10 s−1
Planck's constant (h) = 6.63 × 10^-34 Jᐧs/quantum
Recall ;
Energy of photon = hf
Therefore, energy of photon :
(6.63 × 10^-34)j.s× (2.88 × 10^10)s^-1
= 19.0944 × 10^(-34 + 10) = 19.0944×10^-24 J
Hence, number of quanta required :
(7.06 × 10^4)J / (19.0944 × 10^-24)J
= 0.369 × 10^(4 + 24) = 0.369×10^28
= 3.69 × 10^27
Answer:
38.3958 °C
Explanation:
As,
1 gram of carbohydrates on burning gives 4 kilocalories of energy
1 gram of protein on burning gives 4 kilocalories of energy
1 gram of fat on burning gives 9 kilocalories of energy
Thus,
27 g of fat on burning gives 9*27 = 243 kilocalories of energy
20 g of protein on burning gives 4*20 = 80 kilocalories of energy
48 gram of carbohydrates on burning gives 4*48 = 192 kilocalories of energy
Total energy = 515 kilocalories
Using,

Given: Volume of water = 23 L = 23×10⁻³ m³
Density of water= 1000 kg/m³
So, mass of the water:
Mass of water = 23 kg
Initial temperature = 16°C
Specific heat of water = 0.9998 kcal/kg°C

Solving for final temperature as:
<u>Final temperature = 38.3958 °C </u>
The answer is Index fossils its the only answer it could be
Using ideal gas equation, PV = nRT, and since there is no volume change and amount change, the equation is now P = kT, where k =nR/V. Temperature must be in kelvin
From the given, k = (0.82)/ (21 + 273) = 2.78 x 10^-3
Substituting T = -3.5+273, P = 0.75 atm