Determine the empirical formula of an oxide of nitrogen containing 70% of oxygen. If the relative molecular mass of the oxide is
1 answer:
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The question is incomplete. There's missing the image, which is shown below.
Answer:
Volume of O₂ = 6 L, volume of mixture: 18 L, volume of H₂O = 12 L, molecule volume of H₂O = 0.667 molecule/L
Explanation:
The reaction between hydrogen gas and oxygen gas to form water is:
2H₂(g) + O₂(g) → 2H₂O(g)
So, for 1 mol of O₂ is necessary 2 moles of H₂ form 2 moles of H₂O. As the images below there's 8 molecules of H₂, 4 molecules of O₂, 12 molecules in the mixture, and 8 molecules of H₂O. Thus, there are stoichiometric values.
All the images are at the same temperature and pressure, so, by the ideal gas law:
PV= nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
The number of moles and molecules are related, so let's substitute it in the equation. For the H₂:
P*12 = 8*RT
RT/P = 12/8 = 1.5
Thus, for O₂:
PV= nRT
V = n*(RT/P)
V = 4*1.5 = 6 L
For the mixture:
V = 12*1.5 = 18 L
For H₂O:
V = 8*1.5 = 12 L
The molecule volume is the number of molecules divided by the volume they occupy, thus for water: 8/12 = 0.667 molecules/L
Answer:
The correct option is;
X, W, Y, Z
Explanation:
The parameters given are;
Spring (S), Spring Constant (N/m)
W, 24
X, 35
Y, 22
Z, 15
The equation for elastic potential energy, , is
The above equation can also be written as
Where:
k = The spring constant in (N/m)
x = The spring extension
Therefore, since the elastic potential energy, , of the spring is directly proportional to the spring constant, k, we have the springs with higher spring constant will have higher elastic potential energy,
, therefore the correct order is as follows;
X > W > Y > Z