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kirill [66]
3 years ago
14

A mixture contains O2 at 700. torr pressure, F2 at 500. torr pressure, and Cl2 at 500. torr pressure. What is the total pressure

of the gases in the system?
Chemistry
1 answer:
Softa [21]3 years ago
3 0

You have to use Dalton's law of partial pressure for this question. Dalton's law of partial pressure basically states that the total pressure of the system is all of the partial pressures of the components added together. Therefore to answer the question you just need to add all the patial pressures together meaning that the total pressure would be 700+500+500=1700.

The answer would be 1700 torr.

I hope this helps. Let me know if anything is unclear or if you have any further questions.

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Answer:

C5H5N is the base and C5H5NH+ is the conjugate acid

H2O is the acid and OH− is the conjugate base

Explanation:

<u>Hydrogen + is also called a proton</u>

C5H5N is the base because it receives the proton (H+) and C5H5NH+ is its conjugate acid

H2O is the acid  because it gives up the proton and OH− is the conjugate base because it is capable of receiving the proton

Answer:

HNO3 is the acid and NO3- is the conjugate base

H2O is the base and H3O+ is the conjugate acid

Explanation

HNO3 is the acid and NO3− is its conjugate base, capable of receiving a proton

H2O is the base because it receives the proton and H3O+ is a conjugate acid capable of giving up the proton.

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What is the limiting reactant if 4.0 g of Nh3 react with 8.0 g of oxygen?
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Which of the following statements explains why atoms are always neutral in charge
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Answer:

D. They have the same number of protons as electrons.

Explanation:

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Salt is poured from a container at 10 cm³ s-¹ and it formed a conical pile whose height at any time is 1/5 the radius of the abo
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Answer:

\displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}

Explanation:

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We have \displaystyle \frac{dV}{dt} = \frac{10 \ cm^3}{sec} and we want to find \displaystyle \frac{dh}{dt} \Biggr | _{h\ =\ 6}= \ ? when the height is 2 cm.

We can see in our equation for the volume of a cone that we have three variables: V, r, and h.

Since we only have dV/dt and dh/dt, we can rewrite the equation in terms of h only.

We are given that the height of the cone is 1/5 the radius at any given time, 1/5r, so we can write this as r = 5h.

Plug this value for r into the volume formula:

  • \displaystyle V =\frac{1}{3} \pi (5h)^2 h  
  • \displaystyle V =\frac{1}{3} \pi \ 25h^3

Differentiate this equation with respect to time t.

  • \displaystyle \frac{dV}{dt}  =\frac{25}{3} \pi \ 3h^2 \ \frac{dh}{dt}
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Plug known values into the equation and solve for dh/dt.

  • \displaystyle 10 = 25 \pi (2)^2  \ \frac{dh}{dt}
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Divide both sides by 100π to solve for dh/dt.

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The height of the cone is increasing at a rate of 1/10π cm per second.

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