Answer: 0.082 atm L k^-1 mole^-1
Explanation:
Given that:
Volume of gas (V) = 62.0 L
Temperature of gas (T) = 100°C
Convert 100°C to Kelvin by adding 273
(100°C + 273 = 373K)
Pressure of gas (P) = 250 kPa
[Convert pressure in kilopascal to atmospheres
101.325 kPa = 1 atm
250 kPa = 250/101.325 = 2.467 atm]
Number of moles (n) = 5.00 moles
Gas constant (R) = ?
To get the gas constant, apply the formula for ideal gas equation
pV = nRT
2.467 atm x 62.0L = 5.00 moles x R x 373K
152.954 atm•L = 1865 K•mole x R
To get the value of R, divide both sides by 1865 K•mole
152.954 atm•L / 1865 K•mole = 1865 K•mole•R / 1865 K•mole
0.082 atm•L•K^-1•mole^-1 = R
Thus, the value of gas constant is 0.082 atm L k^-1 mole^-1
The amount, in mg, of CO present in the room will be 191,520 mg.
<h3>Stoichiometric problem</h3>
The concentration of the gas in the room is 5.7 x 
mg/cm3.
The dimension of the room is 3.5 m x 3.0 m x 3.2 m. This is equivalent to 350 cm x 300 cm x 320 cm.
We can obtain the volume of the room as:
350 x 300 x 320 = 33,600,000 cm3
The concentration is in mg/cm3, meaning that it is mass/volume.
Thus:
mass = concentration x volume = 5.7 x mg/cm3 x 33,600,000 cm3
= 191,520 mg
The mass of CO in the room is 191,520 mg
More on stoichiometric problems can be found here: brainly.com/question/14465605
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Answer:
10.335
Explanation:
An object was carefully weighed on three different balances
Each of these balances were zeroed before weighing
The masses that were weighed are as follows
10.35 g , 10.355 g, 10.30 g
Therefore the average value of these measurements can be calculated as follows
The total number of mass is 3
= 10.30 + 10.355 + 10.30/3
= 31,005/3
= 10.335
Hence the average value of these measurements is 10.335
Answer:
17. tie hair up
wear goggles
18. i don't know
19. you light it on the side of the box
20. it should be orange because that is the safest.
21. you turn down the electricity thing
