40.1g of nitrogen gas is produced.
The equation given is
2 NH₃ + 3 CuO →3 Cu + N₂ + 3 H₂O
This equation is already balanced.
When 3 moles of CuO are consumed, 1 mole of nitrogen gas is produced.
We get 1 mole of nitrogen from 3 moles of copper oxide.
We need to find the number of moles of nitrogen gas produced when 4.3 moles of copper oxide are consumed.
4.3/3 x 1 = 1.433 mols
- 1.433 mols of nitrogen gas are produced
- The molar mass of nitrogen gas is 14+14 = 28g
- The amount of nitrogen gas produced in grams is 28x1.433 = 40.1g
40.1g of nitrogen gas can be made when 4.3 moles of CuO are consumed.
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Elements:
• atom
• cannot
• Periodic Table
Compounds:
• atoms
• chemically
• cannot
Mixtures:
• elements or compounds
• homogeneous
• heterogeneous
Answer; +3
Example Ca3(PO4)2
Answer:In a sample of BrF the strongest interparticle force is the dipole-dipole force.
While Kr is the dispersion force.
And H2SO4 is hydrogen bonding
Explanation:
Brf is polar with a known dipole moment of 1.4D,dipole - dipole force and dispersion force are both present,the dipole-dipole force is the strongest interparticle force action on Brf considering its high boiling points
The bond formed by Kr will be non polar due to greater atomic size and low electronegativity.The force of attraction between the non polar is the dispersion force which is the strongest.
H2SO4 has a great polarity with a very strong hydrogen bonding.
Answer:- 0.138 M
Solution:- The buffer pH is calculated using Handerson equation:
acts as a weak acid and 
as a base which is pretty conjugate base of the weak acid we have.
The acid hase two protons(hydrogen) where as the base has only one proton. So, we could write the equation as:
Phosphoric acid gives protons in three steps. So, the above equation is the second step as the acid has only two protons and the base has one proton.
So, we will use the second pKa value. The acid concentration is given as 0.10 M and we are asked to calculate the concentration of the base to make a buffer of exactly pH 7.00.
Let's plug in the values in the equation:
Taking antilog:
On cross multiply:
[base] = 1.38(0.10)
[base] = 0.138
So, the concentration of the base that is required to make the buffer is 0.138M.
