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Nadusha1986 [10]
3 years ago
7

Which of the following will increase the pressure of a gas in a closed container? I. part of the gas is removed II. the containe

r size is decreased III. temperature is increased
Chemistry
2 answers:
maks197457 [2]3 years ago
7 0
I would go with III. If i’m correct, Brainliest?
Anastasy [175]3 years ago
3 0
There are 2 correct answers I suppose, II and III.

Gas pressure is increased if the molecules inside the container hit the inner wall of the container faster (more frequently), or harder. These are 2 ways of how the gas pressure of some gas is increased. We can figure out if the options will increase the gas pressure. 

For I, removing part of the gas, this will remove part of the gas molecules. In result, fewer molecules will hit the wall because some are removed, like the number density is reduced. This does not increase the gas pressure, but instead decreasing it. So, option I is not correct. 

For II, if the container size is decreased, the gas molecules have a shorter distance to travel before they hit the inner wall of the container, so, they hit the wall faster and more frequently. This can help increase the gas pressure, so option II is correct.

For III, if the temperature of the molecules is increased, the molecules will have a higher kinetic energy. They tend to move faster and of course eventually, they hit the inner wall of the container more frequently. In addition, if they have a higher kinetic energy, they'll even hit the inner wall harder, which can increase the gas pressure too. Therefore, option III is correct too. 

The answer is II and III. 

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Match the following.
Andru [333]

Answer: Please see below for answers

Explanation: Matching appropriate labels , we have

1)3/4 of the way to second equivalence point of a diprotic acid/strong base titration-- pH=pka₂

equivalence point of a weak base/strong acid titration=pH<7

equivalence point of a strong acid/strong base titration= pH=7

equivalence point of a weak acid/strong base titration=pH>7

half-way to equivalence point of a weak acid/strong base titration pH =pka

where

pH gives the measure of the amount of concentration of hydrogen ions in an aqueous solution.

pKa  is known as acid dissociation constant which explains the equilibrum at which a chemical species can give out or receive proton

pka₂ is the acid dissociation constant for the second ionization energy.

5 0
2 years ago
What is the period of a wave with a frequancy of 3 Hz and an amplitude off 0.01
lara31 [8.8K]

0.33 seconds is the period of a wave with a frequency of 3 Hz and an amplitude off 0.01 .

<u>Explanation:</u>

We have , period of a wave with a frequency of 3 Hz and an amplitude off 0.01 . We know that period of a wave is amount of time needed to complete one oscillation . In order to calculate period of wave we use frequency and the formula use is period = \frac{1}{frequency} . We are given that frequency = 3 Hz:

period = \frac{1}{frequency}

⇒ period = \frac{1}{frequency}

⇒ period = \frac{1}{3 Hz}

⇒ period = 0.33 seconds

Therefore, 0.33 seconds is the period of a wave with a frequency of 3 Hz and an amplitude off 0.01 .

4 0
3 years ago
A balloon has a pressure of 3.1 atm at a volume of 155 ml. if the temperature is held constant, what is the volume (ml) if the p
GarryVolchara [31]

Answer:

45.8 mL

Explanation:

If all variables are held constant, the new volume can be found using the Boyle's Law equation. The equation looks like this:

P₁V₁ = P₂V₂

In this equation, "P₁" and "V₁" represent the initial pressure and volume. "P₂" and "V₂" represent the final pressure and volume. You can find the new volume by plugging the given values into the equation and simplifying.

P₁ = 3.1 atm                        P₂ = 10.5 atm

V₁ = 155 mL                       V₂ = ? mL

P₁V₁ = P₂V₂                                                 <----- Boyle's Law equation

(3.1 atm)(155 mL) = (10.5 atm)V₂                <----- Insert values

480.5 = (10.5 atm)V₂                                  <----- Multiply 3.1 and 155

45.8 = V₂                                                    <----- Divide both sides by 10.5

4 0
2 years ago
Explain why there might be a change in the density of a forged product as compared to that of the cast blank.
kkurt [141]

Answer:

Forged parts are often tougher than cast parts. This can be determined by performing tensile tests on various areas on the parts. Additionally, the microstructures of forged and cast parts can be used to determine if a part was forged or cast. The microstructure of a cast part will have a more uniform grain structure.

Explanation:

4 0
2 years ago
Help! How is this incorrect?
Ainat [17]
On the first one it is supposed to be 18. when you have a +1 charge you subtract it once. how i got 18 tho was from the protons. there was 19 so i subtracted that with 1 and got 18. hope that helped! :)

btw i’m not the best at explaining, i’m sorry :/
8 0
2 years ago
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