Answer:
2.22 g/L
Explanation:
There's a relationship using the ideal gas law between molar mass and density: 
, where MM is the molar mass, d is the density, R is the gas constant, T is the temperature, and P is the pressure.
We know from the problem that MM = 32.49 g/mol, T = 458 Kelvin, and P = 2.569 atm. The gas constant, R, in terms of the units atm and Kelvin is 0.08206. Let's substitute these values into the formula:
Solve for d:
d * 0.08206 * 458 K = 32.49 * 2.569
d = (32.49 * 2.569) / (0.08206 * 458 K) ≈ 2.22 g/L
The answer is thus 2.22 g/L.
<em>~ an aesthetics lover</em>
Answer:
A) a new type of refrigerant that is less damaging to the environment is developed. Applied development
B) a new element is synthesized in a particle accelerator. Basic development.
C) a computer is redesigned to increase the speed of the computer. Technological development.
Explanation:
BaCl2 reacts with KOH forming KCl which is a salt and Ba(OH)2 which is the precipitate.
The initial UNBALANCED equation expressing this reaction would be:
BaCl2 + KOH .............> KCl + Ba(OH)2
Now, we need to balance this equation:
we have two moles of Cl as reactants and one mole only as product. Therefore, we will multiply the KCl in the product by 2 and the KOH in the reactants by 2.
This will balance the equation as follows:
BaCl2 + 2KOH ......> 2KCl + Ba(OH)2
Noticing this equation, we will find that:
The precipitate was formed due to the combination of the Ba2+ ion with 2 OH- ions as follows:
Ba2+ + OH- ............> Ba(OH)2
Answer: a) 0.070 moles of oxygen were produced.
b) New pressure due to the oxygen gas is 2.4 atm
Explanation:
According to ideal gas equation:
P = pressure of gas = 2.7 atm
V = Volume of gas = 700 ml = 0.7 L
n = number of moles = ?
R = gas constant =
T =temperature = 329 K
Thus 0.070 moles of oxygen were produced.
When the 700 mL flask cools to a temperature of 293K.
The new pressure due to the oxygen gas is 2.4 atm
Answer:
The correct answer is "430 kJ/kg". A further explanation is given below.
Explanation:
The given values are:
T₁ = 400 k
T₂ = 800 k
The average temperature will be:
= 
= 
= 
From table,
At 600 k the 
will be = 1.075
Now,
⇒ The specific enthalpy = 
⇒ 
⇒ 
