To find the number of molecules in 9.0 g of steam (water vapor), first determine the number of moles. The molar mass of water (H₂O) is approximately 18.02 g/mol. Therefore, 9.0 g of steam is equivalent to ( \frac{9.0 \text{ g}}{18.02 \text{ g/mol}} \approx 0.5 ) moles. Since one mole contains Avogadro's number of molecules ((6.022 \times 10^{23}) molecules/mol), the total number of molecules is (0.5 \text{ moles} \times 6.022 \times 10^{23} \text{ molecules/mole} \approx 3.01 \times 10^{23} ) molecules.
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1.Convert to common units then can divide 90g by 0.09 kg:1 kg = 1000 g→ 0.09 kg = 0.09 x 1000 g= 90 g→ 90 g ÷ 0.09 kg = 90 g ÷ 90 g= 1.
90 g is less than 9 kg
3 g under the 4 kg then go over to the next side write 90 g under the 05 g that equals 915 g because you are not going to get 9 in it so don't think i did it wrong.
4.05 kg - 3.09 kg = 0.96 kg (960 g)
The latent heat of condensation of steam is 2260 Joules per gram (539.3 cals/g). So the amount of heat released by 12.4 g = 12.4*2260 Joules = 28,024 Joules or 6687 cals.