Answer:
fewer collisions occur between particles or lowering the temperature
Explanation:
Answer:
32.4 mol
Explanation:
Given data:
Number of moles of C atom present = ?
Number of moles of glucose = 5.4 mol
Solution:
Glucose formula = C₆H₁₂O₆
There are 6 moles of C atoms are present in one mole of glucose.
In 5.4 moles of glucose:
5.4 mol × 6 = 32.4 mol
Answer:
Explanation:
a) In an exothermic reaction, the energy transferred to the surroundings from forming new bonds is ___more____ than the energy needed to break existing bonds.
b) In an endothermic reaction, the energy transferred to the surroundings from forming new bonds is ___less____ than the energy needed to break existing bonds.
c) The energy change of an exothermic reaction has a _____negative_______ sign.
d) The energy change of an endothermic reaction has a ____positive________ sign.
The energy changes occur during the bonds formation and bonds breaking.
There are two types of reaction endothermic and exothermic reaction.
Endothermic reactions:
The type of reactions in which energy is absorbed are called endothermic reactions.
In this type of reaction energy needed to break the bond are higher than the energy released during bond formation.
For example:
C + H₂O → CO + H₂
ΔH = +131 kj/mol
it can be written as,
C + H₂O + 131 kj/mol → CO + H₂
Exothermic reaction:
The type of reactions in which energy is released are called exothermic reactions.
In this type of reaction energy needed to break the bonds are less than the energy released during the bond formation.
For example:
Chemical equation:
C + O₂ → CO₂
ΔH = -393 Kj/mol
it can be written as,
C + O₂ → CO₂ + 393 Kj/mol
Answer:
The amount left after 49.2 years is 3mg.
Explanation:
Given data:
Half life of tritium = 12.3 years
Total mass pf tritium = 48.0 mg
Mass remain after 49.2 years = ?
Solution:
First of all we will calculate the number of half lives.
Number of half lives = T elapsed/ half life
Number of half lives = 49.2 years /12.3 years
Number of half lives = 4
Now we will calculate the amount left after 49.2 years.
At time zero 48.0 mg
At first half life = 48.0mg/2 = 24 mg
At second half life = 24mg/2 = 12 mg
At 3rd half life = 12 mg/2 = 6 mg
At 4th half life = 6mg/2 = 3mg
The amount left after 49.2 years is 3mg.
Umm I’ll figure it out rn! Will come back in 1 min
