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3 years ago

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

2 answers:

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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What Makes You Unique??<br> Just bored so yea :))) Free 15 points :)

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3 years ago

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What are five objects that are luminous

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3 years ago

Alex cut a pizza into 8 equal slices. He removed 2 of the slices of pizza. What is the measure of the angle made by the missing

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2 years ago

What pressure, in atm, would be exerted by 0.023 grams of oxygen (O2) if it occupies 31.6 mL at 91

Vikentia [17]

Answer: A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen $(O_2)$ if it occupies 31.6 mL at $91^{o}C$ .

Explanation:

Given : Mass of oxygen = 0.023 g

Volume = 31.6 mL

Convert mL into L as follows.

$1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L$

Temperature = $91^{o}C = (91 + 273) K = 364 K$

As molar mass of $O_2$ is 32 g/mol. Hence, the number of moles of $O_2$ are calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{0.023 g}{32 g/mol}\\= 0.00072 mol$

Using the ideal gas equation calculate the pressure exerted by given gas as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the value into above formula as follows.

$PV = nRT\\P \times 0.0316 L = 0.00072 mol \times 0.0821 L atm/mol K \times 364 K\\P = \frac{0.00072 mol \times 0.0821 L atm/mol K \times 364 K}{0.0316 L}\\= 0.681 atm$

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen $(O_2)$ if it occupies 31.6 mL at $91^{o}C$ .

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3 years ago

A sample of pure calcium fluoride with a mass of 15.0 g contains 7.70 g of calcium. how much calcium is contained in 45.0 g of c

kolbaska11 [484]

In a sample of pure calcium fluoride of mass 15.0 g, 7.70 g of calcium is present. First convert the mass into number of moles as follows:

$n=\frac{m}{M}$

Here, m is mass and M is molar mass.

Molar mass of Ca is 40 g/mol, putting the values,

$n=\frac{7.70 g}{40 g/mol}=0.1925 mol$

Similarly, molar mass of $CaF_{2}$ is 78.07 g/mol thus, number of moles will be:

$n=\frac{15.0 g}{78.07 g/mol}=0.1921 mol$ .

Thus, 0.1921 mol of $CaF_{2}$ have 0.1925 mol of Ca, or 1 mole of $CaF_{2}$ will have approximately 1 mole of Ca.

Now, mass of Ca needs to be calculated in 45.0 g of $CaF_{2}$ . Converting mass into number of moles first,

$n=\frac{45.0 g}{78.07 g/mol}=0.5764 mol$

Thus, number of moles of Ca will also be 0.5764 mol, converting number of moles into mass,

$m=n\times M=0.5764 mol\times 40 g/mol=23.06 g$

Therefore, mass of Ca will be 23.06 g.

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3 years ago