Answer:
a) Neutralisation
b) Combustion
c) Synthesis
d) Decomposition
e) Neutralisation
f) Double Displacement Reaction
h) Single Displacement Reaction
i) Double Displacement Reaction
j) Combustion
Explanation:
Synthesis is a reaction where various compounds/ elements react to form a totally new compound.
Decomposition is a reaction where a single compound breaks down into several components due to excessive heating or energy applied.
Single Displacement Reaction is a type of chemical reaction where an element reacts with a compound and takes the place of another element in that compound.
Double Displacement Reaction is a type of chemical reaction where two compounds react, and the positive ions (cation) and the negative ions (anion) of the two reactants switch places, forming two new compounds or products.
Combustion is a reaction where a compound/ element oxidises in the presence of Oxygen.
Neutralisation reaction is a reaction where an acid reacts with a base to form a salt.
Answer:
3.37 × 10²³ molecules
Explanation:
Given data:
Mass of C₆H₁₂O₆ = 100 g
Number of molecules = ?
Solution:
Number of moles of C₆H₁₂O₆:
Number of moles = mass/molar mass
Number of moles = 100 g/ 180.16 g/mol
Number of moles = 0.56 mol
Number of molecules:
1 mole contain 6.022 × 10²³ molecules
0.56 mol × 6.022 × 10²³ molecules /1 mol
3.37 × 10²³ molecules
Answer:
B) irreversible process
Explanation:
The process given here is irreversible.
Answer:
Mass = 42.8g
Explanation:
4 NH 3 ( g ) + 5 O 2 ( g ) ⟶ 4 NO ( g ) + 6 H 2 O ( g )
Observe that every 4 mole of ammonia requires 5 moles of oxygen to obtain 4 moles of Nitrogen oxide and 6 moles of water.
Step 1: Determine the balanced chemical equation for the chemical reaction.
The balanced chemical equation is already given.
Step 2: Convert all given information into moles (through the use of molar mass as a conversion factor).
Ammonia = 63.4g × 1mol / 17.031 g = 3.7226mol
Oxygen = 63.4g × 1mol / 32g = 1.9813mol
Step 3: Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.
If all of the 1.9831 moles of oxygen were to be used up, there would need to be 1.9831 × 4 / 5 or 1.5865 moles of Ammonia. We have 3.72226 moles of ammonia - Far excess. Because there is an excess of Ammonia, the Oxygen amount is used to calculate the amount of the products in the reaction.
Step 4: Use the amount of limiting reactant to calculate the amount of H2O produced.
5 moles of O2 = 6 moles of H2O
1.9831 moles = x
x = (1.9831 * 6 ) / 5
x = 2.37972 moles
Mass of H2O = Molar mass * Molar mass
Mass = 2.7972 * 18
Mass = 42.8g
