Answer:
The molar mass of the gas is 36.25 g/mol.
Explanation:
- To solve this problem, we can use the mathematical relation:
ν = 
Where, ν is the speed of light in a gas <em>(ν = 449 m/s)</em>,
R is the universal gas constant <em>(R = 8.314 J/mol.K)</em>,
T is the temperature of the gas in Kelvin <em>(T = 20 °C + 273 = 293 K)</em>,
M is the molar mass of the gas in <em>(Kg/mol)</em>.
ν =
(449 m/s) = √ (3(8.314 J/mol.K) (293 K) / M,
<em>by squaring the two sides:</em>
(449 m/s)² = (3 (8.314 J/mol.K) (293 K)) / M,
∴ M = (3 (8.314 J/mol.K) (293 K) / (449 m/s)² = 7308.006 / 201601 = 0.03625 Kg/mol.
<em>∴ The molar mass of the gas is 36.25 g/mol.</em>
Here we have to choose the property of the liquid, among the given, which the engineer should primary look to use as the fluid of the engine.
The engineer should primarily looked the liquid which have B. low specific heat capacity.
The Specific heat capacity of a substance is equivalent to the heat needed to increase 1⁰ temperature of the unit mass of the substance. Now if the specific heat capacity of the liquid is low then it will need very little amount of energy to transfer the heat.
A. The low thermal conductivity will cause more time to exchange the energy. Which is not required for the engine.
C. There is no relation between the reactivity of the liquid with the use of the same in the engine.
D. The internal energy will also not affect the rate of the cooling or heating of the liquid.
E. The high density also cannot affect the low rate of exchange energy.
Assuming the balloon initially has volume of 0 when deflated, the total P(deltaV) = (1.00 atm)(550,000 ft^3) = 550,000 atm-ft^3. To convert into work units, we can first convert ft^3 to L:
(550,000 atm-ft^3)(1 m/3.28 ft)^3
= (15,586.2 atm-m^3)
Then we convert to L:
(15,586.2 atm-m^3)(1000 L/m^3)
= 15,586,200 atm-L
Then we convert to J:
(15,586,200 atm-L)(101.325 J / 1 atm-L)
= 1.579 x 10^9 J
Answer:
127.0665 amu
Explanation:
Firstly, to answer the question correctly, we need to access the percentage compositions of the iodine and the contaminant iodine. We can do this by placing their individual masses over the total and multiplying by 100%.
We do this as follows. Since the mass of the contaminant iodine is 1.00070g, the mass of the 129I in that particular sample will be 12.3849 - 1.00070 = 11.3842g
The percentage abundances is as follows:
Synthetic radioisotope % = 1.0007/12.3849 * 100% = 8.1%
Since there are only two constituents, the percentage abundance of the 129I would be 100 - 8.1 = 91.9%
Now, we can use these percentages to get the apparent atomic mass. We get this by multiplying the percentage abundance’s by the atomic masses of both and adding together.
That is :
[8.1/100 * 128.9050] + [91.9/100 * 126.9045] = 10.441305 + 116.6252355 = 127.0665 amu
