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Sphinxa [80]
3 years ago
13

Water condenses onto grass in the early morning when the air is clear because...

Chemistry
2 answers:
Reika [66]3 years ago
8 0

The answer is b i know what courses you are taking and i can remember it clearly saying in the morning when there is small droplets on the grass and leaves on the trees it is a result of the air close to the ground cooling to the dew point. (I also just took the test and got this question right).

kompoz [17]3 years ago
4 0
The answer is b because as the air gets cooler and the moist in the air will drop as we call it dew
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How much volume would 0.450 mol CO2 gas occupy?
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Answer:

10.08 L.

Explanation:

  • If we assume that CO₂ gas behaves ideally at STP (standard T(0.0 °C) and P(1.0 atm)):

<em>It is known that 1.0 mole of ideal gas occupies 22.4 L at STP conditions.</em>

<em></em>

<u><em>Using cross multiplication:</em></u>

1.0 mole of CO₂ gas occupies → 22.4 L.

0.45 mole of CO₂ gas occupies → ??? L.

<em>∴ The volume occupied by 0.45 mole of CO₂ gas </em>= (0.45 mol)(22.4 L)/(1.0 mol) = <em>10.08 L.</em>

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3 years ago
As methane burns, it reacts with oxygen and forms carbon dioxide and water. The balanced chemical equation below models this che
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Explanation:

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A 0.08541 g sample of gas occupies 10.0-ml at 288.5 k and 1.10 atm. upon further analysis, the compound is found to be 13.068% c
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<span>C2Br2 First, we need to determine how many moles of the gas we have. For that, we'll use the Ideal Gas Law which is PV = nRT where P = pressure (1.10 atm = 111458 Pa) V = volume (10.0 ml = 0.0000100 m^3) n = number of moles R = Ideal gas constant (8.3144598 (m^3 Pa)/(K mol) ) T = Absolute temperature Solving for n, we get PV/(RT) = n Now substituting our known values into the formula. (111458 Pa * 0.0000100 m^3) / (288.5 K * 8.3144598 (m^3 Pa)/(K mol)) = (1.11458/2398.721652) mol = 0.000464656 mol Now let's calculate the empirical formula for this compound. Atomic weight carbon = 12.0107 Atomic weight bromine = 79.904 Relative moles carbon = 13.068 / 12.0107 = 1.08802984 Relative moles bromine = 86.932 / 79.904 = 1.087955547 So the relative number of atoms of the two elements is 1.08802984 : 1.087955547 After dividing all numbers by the smallest, the ratio becomes 1.000068287 : 1 Which is close enough to 1:1 for me to consider the empirical formula to be CBr Now calculate the molar mass of CBr 12.0107 + 79.904 = 91.9147 Finally, let's determine if the compound is actually CBr, or something like C2Br2, or some other multiple. Using the molar mass of CBr, multiply by the number of moles and see if the result matches the mass of the gas. So 91.9147 g/mol * 0.000464656 mol = 0.042708701 g 0.0427087 g is a lot smaller than 0.08541 g. So the compound isn't exactly CBr. Let's divide them to see what the factor is. 0.08541 / 0.0427087 = 1.99982673 1.99982673 is close enough to 2 to within the number of significant digits we have for me to claim that the formula for the unknown gas isn't CBr, but instead is C2Br2.</span>
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