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ozzi
1 year ago
12

5. A Dumas bulb is filled with chlorine gas at the ambient pressure and is found to contain 7.1 g of chlorine when the temperatu

re is T (in Kelvin). The bulb is then placed in a second thermostat bath whose temperature is 300C hotter than the first one. The stopcock on the bulb is opened so that the chlorine gas pressure returns to its original value. The bulb is now found to contain 6.4 g of chlorine. (a) Calculate the value of the original temperature, T. (b) If the volume of the bulb was 2.24 liters, what was the ambient pressure?
Chemistry
1 answer:
kati45 [8]1 year ago
7 0

Answer:

a. The original temperature of the gas is 2743K.

b. 20atm.

Explanation:

a. As a result of the gas laws, you can know that the temperature is inversely proportional to moles of a gas when pressure and volume remains constant. The equation could be:

T₁n₁ = T₂n₂

<em>Where T is absolute temperature and n amount of gas at 1, initial state and 2, final states.</em>

<em />

<em>Replacing with values of the problem:</em>

T₁n₁ = T₂n₂

X*7.1g = (X+300)*6.4g

7.1X = 6.4X + 1920

0.7X = 1920

X = 2743K

<h3>The original temperature of the gas is 2743K</h3><h3 />

b. Using general gas law:

PV = nRT

<em>Where P is pressure (Our unknown)</em>

<em>V is volume = 2.24L</em>

<em>n are moles of gas (7.1g / 35.45g/mol = 0.20 moles)</em>

R is gas constant = 0.082atmL/molK

And T is absolute temperature (2743K)

P*2.24L = 0.20mol*0.082atmL/molK*2743K

<h3>P = 20atm</h3>

<em />

You might be interested in
Unit Conversion Help Thank you
AlekseyPX

Answer : 1721.72 g/qt are in 18.2 g/cL

Explanation :

As we are given: 18.2 g/cL

Now we have to convert 18.2 g/cL to g/qt.

Conversions used are:

(1) 1 L = 100 cL

(2) 1 L = 1000 mL

(3) 1 qt = 946 qt

The conversion expression will be:

\frac{18.2g}{1cL}\times \frac{100cL}{1L}\times \frac{1L}{1000mL}\times \frac{946mL}{1qt}

=1721.72\text{ g/qt}

Therefore, 1721.72 g/qt are in 18.2 g/cL

5 0
1 year ago
Determine whether the stopcock should be completely open, partially open, or completely closed for each activity involved with t
densk [106]

Answer:

Close to the calculated endpoint of a titration - <u>Partially open</u>

At the beginning of a titration - <u>Completely open</u>

Filling the buret with titrant - <u>Completely closed</u>

Conditioning the buret with the titrant - <u>Completely closed</u>

Explanation:

'Titration' is depicted as the process under which the concentration of some substances in a solution is determined by adding measured amounts of some other substance until a rection is displayed to be complete.

As per the question, the stopcock would remain completely open when the process of titration starts. After the buret is successfully placed, the titrant is carefully put through the buret in the stopcock which is entirely closed. Thereafter, when the titrant and the buret are conditioned, the stopcock must remain closed for correct results. Then, when the process is near the estimated end-point and the solution begins to turn its color, the stopcock would be slightly open before the reading of the endpoint for adding the drops of titrant for final observation.

3 0
1 year ago
A driver with a nearly empty fuel tank may say she is "running on fumes."
tensa zangetsu [6.8K]

Answer:

The farthest the vehicle could travel (if it gets 20.0 miles per gallon on liquid gasoline) is 1.62 miles.

Explanation:

The automobile gas tank has a volume capacity of 15 gallons which can be converted to liters: 15 × 3.7854 = 56.781 liters

We can find the moles of gasoline by using the ideal gas equation: PV = nRT.

Make n (number of moles) the subject of the formula: n = PV/RT, where:

P = 747 mmHg

V = 56.781 liters

R (universal gas constant) = 0.0821 liter·atm/mol·K

T = 25 ∘C = (273 + 25) K = 298 K

1 atm (in the unit of R) = 760 mmHg

Therefore n = 747 × 56.781/(0.0821 × 760 × 298) = 2.281 mol.

Given that the molar mass of the gasoline = 101 g/mol,

the mass of gasoline = n × molar mass of gasoline = 2.281 mol × 101 g/mol = 230.38 g

the density of the liquid gasoline = 0.75 g/mL

In order to calculate the distance the vehicle can travel, we have to calculate volume of gasoline available = mass of the liquid gasoline ÷ density of liquid gasoline

= 230.38 g ÷ 0.75 g/mL = 307.17 mL = 0.3071 liters = 0.3071 ÷ 3.7854 = 0.0811 gallons

since the vehicle gets 20.0 miles per gallon on liquid gasoline, the distance traveled by the car = gallons available × miles per gallon = 0.0811 × 20 = 1.62 miles.

6 0
2 years ago
If a large marshmallow has a volume of 2.75 in3 and density of 0.242 g/cm3, how much would it weigh in grams? 1 in3=16.39 cm3.
Anestetic [448]
<span>
•   </span>Volume of the marshmallow:

V = 2.75 in^3          (but, 1 in^3 = 16.39 cm^3)

V = 2.75 × 16.39 cm^3

V = 2.75 × 16.39 cm^3

V = 45.0725 cm^3


•   Density:

d = 0.242 g/cm^3


<span>•  </span>Mass:

m = d × V

m = (0.242 g/cm^3) × (45.0725 cm^3)

m = (0.242 g/cm^3) × (45.0725 cm^3)

m = 10.907545 g

m ≈ 10.9 g   <——<span>—  this is the answer.


I hope this helps. =)
</span>
4 0
2 years ago
IF you’ve eaten 45% of a box of chocolate cereal that has 13oz when you buy it how many oz of the cereal have you eaten
weqwewe [10]
To determine the number of ounces of the chocolate cereal that you have eaten, the operation that needs to be performed is simply the multiplication of the initial number of ounce in the box of cereal and the decimal equivalent of the given percentage. If we let n be the answer, 
   
    n = (13 oz)(0.45) 
 
Simplifying, this will give us the answer which is equal to 5.85 oz.

<em>Answer: 5.85 oz. </em>
6 0
1 year ago
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