The average atomic mass of the imaginary element : 47.255 amu
<h3>Further explanation </h3>
The elements in nature have several types of isotopes
Isotopes are elements that have the same Atomic Number (Proton)
Atomic mass is the average atomic mass of all its isotopes
Mass atom X = mass isotope 1 . % + mass isotope 2.% ..
isotope E-47 47.011 amu, 87.34%
isotope E-48 48.008 amu, 6.895
isotope E-49 50.009 amu, 5.77%
The average atomic mass :

Answer:
a. Oxygen gas is limiting
Explanation:
hydrogen gas and oxygen gas are reacted to form water
2H₂ + O₂ → 2H₂O
the above balanced equation shows that 2 moles of H₂ is required for 1 mole of O₂
Given equal masses of H₂ and O₂
assuming 'x' gm for each, no. of moles of each gas =
no. of moles of H₂ = x/2 = 0.5x moles
no.of moles of O₂ = x/32 = 0.031x moles
This shows that no. of moles of O₂ is very less so O₂ will become the limiting reagent.
<u>Answer:</u>
Nitrogen gas be a mineral only, if it is in organic forms.
<u>Explanation:</u>
Most of the forms of organic nitrogen is not be taken by plants, with the exception in the form of small organic molecules. Also plants can promptly take the nitrogen when it is in other forms like ammonia and nitrate.
The microorganisms in the soil converts the organic forms of nitrogen to mineral form when they decompose organic matters and also fresh plant residues. This type of process is called mineralisation.
4 moles of water are produced
Explanation:
- 4 moles of water are produced when 5 moles of hydrogen is reacted with 2 moles of oxygen gas
- The balanced equation given is when 2 moles of hydrogen reacts with 1 mole of oxygen and it forms 2 moles of water.
- The equation we have to solve is the 5 moles of hydrogen is reacting with 2 moles of oxygen gas, we can write the equation as
- This is the balanced equation when 5 moles of hydrogen reacts with 2 moles of oxygen. The balanced equation means the number of hydrogen atoms and oxygen atoms on both sides would be equal in number.
Answer:
1) The power needed to process 50 ton/hr is 135.4 HP.
2) The void fraction of the bed is 0.37.
Explanation:
1) For this type of milling operations, we can estimate the power needed for an operation according to the work index (Ei), the passing size of the circuit feed (F80) and the passing size of the product (P80).
We assume the units of Ei are kWh/t.
The equation that relates this parameters and the power is (size of particles in μm):

The power needed to process 50 ton/hor is

2) The density of the packed bed can be expressed as

being f the fraction and ρ the density of every fraction. We know that the density of the void is 0 (ρv=0) and that fv=1-fs (the sum of the fractions ois equal to the total space).
Then we can rearrange

The void fraction of the bed is 0.37.