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Delvig [45]
3 years ago
12

A weather balloon is filled with helium that occupies a volume of 5.37 104 L at 0.995 atm and 32.0°C. After it is released, it r

ises to a location where the pressure is 0.720 atm and the temperature is -11.7°C. What is the volume of the balloon at that new location?
Chemistry
1 answer:
svp [43]3 years ago
8 0

Answer:

The new volume is 63583 L

Explanation:

Step 1: Data given

The initial volume of the balloon = 5.37 * 10^4 L

The initial pressure = 0.995 atm

The initial temperature = 32.0 °C = 305.15 K

The pressure decreased to 0.720 atm

The temperature decreased to -11.7 °C = 261.45 K

Step 2: Calculate the new volume

P1*V1 / T1 = P2*V2/T2

⇒with P1 = the initial pressure = 0.995 atm

⇒with V1 = the initial volume = 5.37 *10^4 L

⇒with T1 = the initial temperature = 305.15 K

⇒with P2 = the decreased pressure = 0.720 atm

⇒with V2 = the new volume = TO BE DETERMINED

⇒with T2 = the decreased temperature : 261.45 K

(0.995 * 5.37*10^4)/305.15 = (0.720 * V2) / 261.45

V2 = 63583 L

The new volume is 63583 L

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Sorry bro but idk maybe its :1 L=<span>103</span><span>mL</span>
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Answer:

Activation energy is needed so reactants can move together, overcome forces of repulsion, and to begin breaking bonds.

Explanation:

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Combustion analysis of 0.300 g of an unknown compound containing carbon, hydrogen, and oxygen produced 0.5213 g of co2 and 0.283
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First, we have to get how many grams of C & H & O in the compound:
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3 years ago
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Calculate the standard potential for the following galvanic cell: Ni(s) | Ni2+(aq) || Ag+(aq) | Ag(s) which has the overall bala
mylen [45]

Answer:

1.06  V  

Explanation:

The standard reduction potentials are:

Ag^+/Ag     E° =  0.7996 V  

Ni^2+/Ni     E° = -0.257   V

The half-cell and cell reactions for Ni | Ni^2+ || Ag^+ | Ag are

Ni → Ni^2+ + 2e-                     E° = 0.257   V

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To three significant figures, the standard potential for the cell is 1.06 V .

8 0
3 years ago
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