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Studentka2010 [4]
3 years ago
6

A sample of H2 gas (2.0 L) at 3.5 atm was combined with 1.5 L of N2 gas at 2.6 atm pressure at a constant temperature of 25 °C i

nto a 7.0 L flask. The total pressure in the flask is __________ atm. Assume the initial pressure in the flask was 0.00 atm and the temperature upon mixing was 25 °
c.
Chemistry
1 answer:
Tamiku [17]3 years ago
4 0
For this system, we use Dalton's law of partial pressures where the total pressure of a gas mixture is said to be equal to the sum of the partial pressures of the gases. The partial pressure of each gas would be calculated by the product of the mole fraction and the original pressure of the gas. We do as follows:

Total pressure = x1P1 + x2P2
Total pressure = (2.0 / 7.0 )(3.5) + (1.5/7.0)(2.6)
Total pressure = 1.56 atm
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Identify the correct mole ratio for each substance. Sodium chloride (NaCl) Na:Cl = 1: Ammonium nitrate (NHNO) H:O = 4:
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<u>Answer:</u> The mole ratio of H : O in ammonium nitrate is 4 : 3.

<u>Explanation:</u>

We are given a compound named ammonium nitrate having formula NH_4NO_3

There are 3 elements in this compound which are nitrogen, hydrogen and oxygen.

To calculate the mole ratio, we write the ratio of their subscripts. For this compound, it is:

N:H:O::2:4:3

The mole ratio of H and O for this compound is 4 : 3.

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3 years ago
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A compound containing chromium, Cr; chlorine, Cl; and oxygen, O, is analyzed and found to be 33.6% chromium, 45.8% chlorine, and
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Answer

The empirical formula is CrO₂Cl₂

Explanation:

Empirical formula is the simplest whole number ratio of an atom present in a compound.

The compound contain, Chromium=33.6%

                                         Chlorine=45.8%

                                          Oxygen=20.6%

And the molar mass of Chromium(Cr)=51.996 g mol.

                 Chlorine containing molar mass (Cl)= 35.45    g mol.

                 Oxygen containing molar mass (O)=15.999  g mol.

Step-1

 Then,we will get,

Cr=\frac{1}{51.996} \times33.6=0.64 mol

Cl= \frac{1}{35.45} \times45.8=1.29 mol.

O=\frac{1}{15.99} \times=1.28 mol.

Step-2

Divide the mole value with the smallest number of mole, we will get,

Cr= \frac{0.64}{0.64} =1

Cl= \frac{1.29}{0.64} =2

O= \frac{1.28}{0.64} =2

Then, the empirical formula of the compound is CrO₂Cl₂ (Chromyl chloride)

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I think it's Chlorine but, not 100% sure. so C.
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3 years ago
A ball is equipped with a speedometer and launched straight upward. The speedometer reading four seconds after launch is shown a
Andrew [12]

Answer:

Question 1: <u>1 s after the motion starts</u>

Question 2: <u>0 (just when the motion starts)</u>

Explanation:

You will need to work with approximates values because the precision of the speedometers is low and you are requested to find approximate times.

<u>1. From the speedometer shown at the right.</u>

You can obtain how long the ball has been falling from the highest altitute it reached using the speed of 10 m/s shown by the speedometer at the right.

  • Free fall equation: Vf = Vo - gt

  • Vo = 0 ⇒ Vf = gt ⇒ t = Vf / g

For this problem, I recommend to work with a rough estimate of g: g = 10 m/s² ( I will tell you why soon)/

  • t = [10 m/s] / [10 m/s²] = 1 s

That is the time falling. Since four seconds after launch have elapsed, the upward time was 3 seconds. This will let you to calculate the launching speed.

<u>2. Time when the speedometer displays a reading of 20 m/s</u>

First, calculate the launching speed:

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Since the ball was 3 seconds going upward and the speed at the maximum altitude is 0 you get:

  • 0 = Vo - gt

   

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Now, use the initial velocity to calculate when the ball is going upward with the speedometer reading is 20 m/s

  • 20 m/s = 30 m/s - 10 m/s² × t

  • t = [ 30 m/s - 20 m/s] / [10 m/s²] = 1 s

Thus, the first answer is t = 1 s.

<u />

<u>3. Time when the speedometer displays a reading of 30 m/s</u>

This is the same speec estimated for the launching: 30 m/s.

So, this reading corresponds to the moment when the ball was launched.

Thus time is 0, i.e. it is the same instant of the launch.

If you had worked with g = 9.80 m/s², the time had been negative. This is due to the precision of the instruments.

That is why I recommended to work with g = 10 m/s².

6 0
3 years ago
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