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Nata [24]
2 years ago
10

one studrnt rubs a balloon against his hair several times. Another student passes a comb through her hair several times. The stu

dents place the comb near the balloon and observe that the comb repels the balloon. Which is the most likely reason the comb is able to repel the balloon?​
Chemistry
1 answer:
Nookie1986 [14]2 years ago
3 0
Static electricity. they have the same charge and same charges repel each other.
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What is the equilibrium constant of a reaction?
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Answer:

A. It is the ratio of the concentrations of products to the concentrations of reactants.

Explanation:

The equilibrium constant of a chemical reaction is the ratio of the concentration of products to the concentration of reactants.

This equilibrium constant can be expressed in many different formats.

  • For any system, the molar concentration of all the species on the right side are related to the molar concentrations of those on the left side by the equilibrium constant.
  • The equilibrium constant is a constant at a given temperature and it is temperature dependent.
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What is the volume of 0.80 grams of o2 gas at stp? (5 points) group of answer choices 0.59 liters 0.56 liters 0.50 liters 0.47 l
Vladimir [108]

Answer:

0.56L

Explanation:

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Since all of the answer choices are given in units of Liters, it will be convenient to use a value for R that contains "Liters" in its units:R=0.0821\frac{L\cdot atm}{mol\cdot K}

Since the conditions are stated to be STP, we must remember that STP is Standard Temperature Pressure, which means T=273.15K and P=1atm

Lastly, we must calculate the number of moles of O_2(g) there are.  Given 0.80g of O_2(g), we will need to convert with the molar mass of O_2(g).  Noting that there are 2 oxygen atoms, we find the atomic mass of O from the periodic table (16g/mol) and multiply by 2:  32g\text{ }O_2=1mol\text{ }O_2

Thus, \frac{0.80g \text{ }O_2}{1} \frac{1mol\text{ }O_2}{32g\text{ }O_2}=0.25mol\text{ }O_2=n

Isolating V in the Ideal Gas Law:

PV=nRT

V=\frac{nRT}{P}

...substituting the known values, and simplifying...

V=\frac{(0.025 mol \text{ }O_2)(0.0821\frac{L\cdot atm}{mol \cdot K} )(273.15K)}{(1atm)}

V=0.56L \text{ } O_2

So, 0.80g of O_2(g) would occupy 0.56L at STP.

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