The picture of Au₃N is attached below.
The first part of the picture shows the formation of Au and N ions.
Formation of Au⁺¹ :
As Gold has one valence electron in 6s¹ therefore, it will loose it to form Au⁺¹. In case of Au₃N three atoms of Au looses three electrons to form three Au⁺¹ ions.
Formation of N⁻³ :
As Nitrogen has 5 valence elctrions therefore, it will gain three electrons that lost by Au to form Nitrite (i.e. N⁻³)
Formation of Au₃N:
Three cations of Au⁺ combines with one anion of N⁻³ to form a neutral ionic compound i.e. Au₃N as shown in second part of the picture.
Answer:
Therefore when 1.26 moles of H₂O is reacted 0.63 mole of O₂ are produced.
Explanation:
The required reaction is
H₂O → O₂+H₂
The reaction is not balance reaction. First we have to balance the reaction.
2H₂O → O₂+2H₂
From the above reaction we conclude that,
2 mole of H₂O produces 1 mole of O₂ and 2 moles of H₂.
∴When 2 mole of H₂O reacts then it produces 1 mole of O₂
1 mole of H₂O reacts then it produces mole of O₂.
1.26 mole of H₂O reacts then it produces mole of O₂.
=0.63 mole
Therefore when 1.26 moles of H₂O is reacted 0.63 mole of O₂ are produced.
Sn2+
Your protons equal the atomic number of an element: Sn (Tin).
The amount of protons and electrons will be the same if the atom is neutral. In this case, tin has lost two electrons and so tin is no longer neutral and becomes Sn2+
To double check, you would subtract the atomic number from the mass number of tin to see if the number of neutrons is equal to what has been provided.
mass number - atomic number = number of neutrons
118 - 50 = 68
Answer:
AgNO₃ will be the limiting reagent.
Explanation:
The balanced reaction is:
AgNO₃ + Li → LiNO₃ + Ag
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate:
- AgNO₃: 1 mole
- Li: 1 mole
- LiNO₃: 1 mole
- Ag: 1 mole
To determine the limiting reagent, you can use a simple rule of three as follows: if by stoichiometry AgNO₃, how much moles of AgNO₃ will be needed if 0.35 moles of Li react?
moles of AgNO₃= 0.35
But 0.35 moles of AgNO₃ are not available, 0.30 moles are available. Since you have less moles than you need to react with 0.35 moles of Li, <u><em>AgNO₃ will be the limiting reagent.
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Answer:
4. 1HCL + 1NAOH- I NACL+1H2O
Explanation:
there is 2 hydrogens on the left side of the equation, balancing out to the other side of the equation. everything else would be balanced out too