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Kaylis [27]
2 years ago
6

The specific heats at constant pressure of some common gases are provided as a thirdorder polynomial: �;<<< = � + �� +

��C + ��E, with units J/(mol K). For methane (CH4) the coefficients are � = 19.89, � = 5.024 × 10NC, � = 1.269 × 10NO, � = −11.01 × 10NQ. At a production facility, the gas in a 50-liter tank is compressed to 3,600 psi (gage) during which time the temperature rises to 200°F. How much heat in J is given off as the gas cools to room temperature? Suppose the compressed gas is helium, how much heat would be given off in this case?

Chemistry
1 answer:
3241004551 [841]2 years ago
3 0

Answer:

1.991 kJ

Explanation:

Calculate the amount of heat ( J )

CH4 ;

coefficients are :  a = 19.89 , b = 5.02 * 10^-2 , c = 1.269 * 10^-5 , d = -11.01 * 10^-9

attached below is the detailed solution

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Which of the following gas samples would contain the same amount of gas as 200 mL of helium, He(g), at 25° C and 1 atm?
monitta

Answer:

200\; \rm mL of neon \rm Ne\, (g) at 25^{\circ} \rm C and 1\; \rm atm (the second choice) would contain an equal number of gas particles as 200\; \rm mL\! of \rm He \, (g) at 25^{\circ} \rm C\! and 1\; \rm atm\! (assuming that all four gas samples behave like ideal gases.)

Explanation:

By Avogadro's Law, if the temperature and pressure of two ideal gases is the same, the number of gas particles in each gas would be proportional to the volume of that gas.

All four gas samples in this question share the same temperature and pressure. Hence, if all these gases are ideal gases, the number of gas particles in each sample would be proportional to the volume of that sample. Two of these samples would contain the same number of gas particles if and only if the volume of the two samples is equal to one another.

The second choice, 200\; \rm mL of neon \rm Ne\, (g) at 25^{\circ} \rm C and 1\; \rm atm, is the only choice where the volume of the sample is also 200\; \rm mL \!. Hence, that choice would be the only one with as many gas particles as 200\; \rm mL\! of \rm He \, (g) at 25^{\circ} \rm C\! and 1\; \rm atm\!.

7 0
2 years ago
What mass of oxygen gas is consumed in a reaction that produces 4.60 mol sulfur dioxide?
babunello [35]

Answer:

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7 0
2 years ago
How to change 5 % W/V of NaCl to ppm , M ? molar mass = 58.5<br>please clear explain​
11Alexandr11 [23.1K]

Answer:

50000ppm and 0.855M.

Explanation:

ppm is an unit of chemistry defined as the ratio between mg of solute (NaCl) and Liters of solution. Molarity, M, is the ratio between moles of NaCl and liters

A 5% (w/v) NaCl contains 5g of NaCl in 100mL of solution.

To solve the ppm of this solution we need to find the mg of NaCl and the L of solution:

<em>mg NaCl:</em>

5g * (1000mg / 1g) = 5000mg

<em>L Solution:</em>

100mL * (1L / 1000mL) = 0.100L

ppm:

5000mg / 0.100L = 50000ppm

To find molarity we need to obtain the moles of NaCl in 5g using its molar mass:

5g * (1mol / 58.5g) = 0.0855moles NaCl

Molarity:

0.0855mol NaCl / 0.100L = 0.855M

7 0
3 years ago
In a titration of 0.35 M HCl and 0.35 M NaOH, how much NaOH should be added to 45.0 ml of HCl to completely neutralize the acid?
Bezzdna [24]
It would be the same amount. So, 45 ml of NaOH is required to be added to the 45 ml of HCI to neutralize the acid fully. Here is a brief calculation:

Firstly, here is your formula: M(HCI) x V(HCI) = M(NaOh) x V(NaOH) 
With the values put in: 0.35 x 45 = 0.35 x V(NaOH) 
= 45 ml. 
There is 45 ml of V(NaOH)

Let me know if you need anything else. :)

           - Dotz
6 0
3 years ago
What is the molar mass of fluorine gas, F2?
Paladinen [302]

Answer:

37.99681 g/mol

Explanation:

Hope it helps!

6 0
3 years ago
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