Answer:
A
Explanation:
First, let's find the molar mass of CO₂. This is 12 + 2(16) = 44 g/mole.
Now we can write 100g * (1 mole / 44g) = 2.27 mol, or A. Hope this helps!
Answer:
Option (1) Br– is the catalyst, and the reaction follows a faster pathway with Br– than without
Explanation:
Let us consider the equation below:
Step 1:
H2O2(aq) + Br–(aq) → H2O(l) + BrO–(aq)
Step 2:
BrO–(aq) + H2O2(aq) → H2O(l) + O2(g) + Br–(aq)
From the above equation, we can see that Br– is unchanged.
This implies that Br– is the catalyst as catalyst does not take part in a chemical reaction but they create an alternate pathway to lower the activation energy in order for the reaction to proceed at a much faster rate to arrive at the products.
3+
So, compounds of boron contain boron in a positive oxidation state, generally +3. The sum of oxidation numbers of all constituent atoms of a given molecule or ion is equal to zero or the charge of the ion, respectively. ... In most of the stable compounds of boron, its oxidation number is +3
Evaporation happens<span> when atoms or </span>molecules<span> escape from the liquid and turn into a vapor. Not all of the </span>molecules in a liquid have the same energy. <span>Sometimes a </span>liquid<span> can be sitting in one place (maybe a puddle) and its molecules will become a </span>gas<span>. That's the process called </span>evaporation<span>. It can happen when liquids are cold or when they are warm. It happens more often with warmer liquids. You probably remember that when matter has a higher temperature, the molecules have a higher </span>energy<span>. When the energy in specific molecules reaches a certain level, they can have a </span>phase change<span>. Evaporation is all about the energy in individual molecules, not about the average energy of a system. The average energy can be low and the evaporation still continues. </span>
It can be either they can have a negative or positive charge more specifically negatively charged molecule would called an anion and a positive one would be cation.