From the information presented in the question, the number of molecules present of water present is obtained 2.41 × 10^21 molecules.
From the information we have;
Volume of the damp air = 1 L
Pressure of the damp air = 741.0 torr or 0.975 atm
Temperature of the gas = 20 oC + 273 = 293 K
R = 0.082 atm LK-1mol-1
Number of moles = ?
n =PV/RT
n = 0.975 × 1/0.082 × 293
n = 0.041 moles
Volume of water vapor = 1 L
Temperature of water = -10 oC + 273 = 263 K
Pressure of the gas = 607.1 torr or 0.799 atm
R = 0.082 atm LK-1mol-1
n= PV/RT
n = 0.799 × 1/ 0.082 × 263
n = 0.037 moles
Number of moles of water = 0.041 moles - 0.037 moles = 0.004 moles
If 1 mole = 6.02 × 10^23 molecules
0.004 moles = 0.004 moles × 6.02 × 10^23 molecules/1 mole
= 2.41 × 10^21 molecules
Learn more: brainly.com/question/2510654
Explanation:
Defining law of definite proportions, it states that when two elements form more than one compound, the ratios of the masses of the second element which combine with a fixed mass of the first element will always be ratios of small whole numbers.
A. One of the oxides (Oxide 1) contains 63.2% of Mn.
Mass of the oxide = 100g
Mass of Mn = 63.2 g
Mass of O = 100 - 63.2
= 36.8 g
Ratio of Mn to O = 63.2/36.8
= 1.72
Another oxide (Oxide 2) contains 77.5% Mn.
Mass of oxide = 100 g
Mass of Mn = 77.5 g
Mass of O = 100 - 77.5
= 22.5 g
Ratio of Mn to O = 77.5/22.5
= 3.44
Therefore, the ratio of the masses of Mn and O in Oxide 1 and Oxide 2 is in the ratio 1.72 : 3.44, which is also 1 : 2. So the law of multiple proportions is obeyed.
B.
Oxide 1
Mass of Mn per 1 g of O = mass of Mn/mass of O
= 77.5/22.5
= 3.44 g/g of Oxygen.
Oxide 2
Mass of Mn per 1 g of O = mass of Mn/mass of O
= 77.5/22.5
= 3.44 g/g of Oxygen.
Based on the information provided, it appears that you will need to calculate the amount of heat absorbed by the water from the peanut that was burned. We are given the following information:
specific heat capacity, c = 1.0 cal/g°C
mass of water = 76 g
Ti = 22°C
Tf = 46°C
change in temperature, ΔT = 24°C
We can use the formula q = mcΔT to measure the amount of energy absorbed by the water to increase in tempature:
q = (76 g)(1.0 cal/g°C)(24°C)
q = 1824 cal
Therefore, the water absorbed 1824 calories from the peanut that was burned.
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