Answer : The limiting reagent is 
Solution : Given,
Moles of methane = 2.8 moles
Moles of = 5 moles
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 2 mole of react with 1 mole of 
So, 5 moles of react with 
moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.
Hence, the limiting reagent is
Answer:
sulfide ion
Explanation:
however, when non-metalic elements gain electrons to form anions the end of their name is change to "-ide".foe example, a flourin atom gains one electronto become a flouride ion(F`) and sulfur gains two electrons to become a sulfide ion(s^2)
Answer : The correct option is, 
Explanation :
- Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. That means, the loss of electrons takes place.
Or we can say that, oxidation reaction occurs when a reactant losses electrons in the reaction.
- Reduction reaction : It is defined as the reaction in which a substance gains electrons. That means, the gain of electrons takes place.
Or we can say that, reduction reaction occurs when a reactant gains electrons in the reaction.
According to the electrochemical series, most likely to be reduced because
Hence, the ion most likely to be reduced is .
Answer:
Pure compounds include elements and compounds where a combination of two or more pure substances is a mixture. Only one type of atom comes in the form of a pure element. Multiple atoms consist of a molecule and different atoms consist of a compound. These are all pure substances and individually.
Answer:
The answer to your question is AgCl
Explanation:
Data
Silver = Ag = 75%
Chlorine = Cl = 25%
1.- Convert percent numbers to grams
Silver = 75 g
Chlorine = 25 g
2.- Calculate the moles of each element
Mass number Ag = 108 g
Mass number Cl = 35.5 g
108 g of Ag ------------------ 1 mol
75 g of Ag ------------------ x
x = (75 x 1) / 108
x = 75 / 108
x = 0.69 moles
35.5 g of Cl ------------------ 1 mol
25 g of Cl ----------------- x
x = (25 x 1) / 35.5
x = 0.70 moles
3.- Divide by the lowest number of moles
Ag = 0.69/0.69 = 1
Cl = 0.70 / 0.69 = 1.02 ≈ 1
4.- Write the empirical formula
Ag₁ Cl₁ = AgCl
