The catalyst is what appears exactly the same at the end and appears early in the equation set. In this case Cl(g).The intermediate appears "intermediately" not at the beginning or at the end, but is made and consumed in the middle. Like ClO(g). A substance that is regenerated in the next is a catalyst and is consumed in the first step. In contrast, when a substance is formed in the first step and is consumed in the next step, then it is known as an intermediate.
Answer:
2KClO3 —> 2KCl + 3O2
The coefficients are 2, 2, 3
Explanation:
From the question given above, we obtained the following equation:
KClO3 —> 2KCl + 3O2
The above equation can be balance as follow:
There are 2 atoms of K on the right side and 1 atom on the left side. It can be balance by putting 2 in front of KClO3 as shown below:
2KClO3 —> 2KCl + 3O2
Now, the equation is balanced.
Thus, the coefficients are 2, 2, 3
Half-life is defined as the amount of time it takes a given quantity to decrease to half of its initial value. The equation to describe the decay is
Nt=N0(1/2)

where N0 is the initial quantity, Nt is the remaining quantity after time t, t1/2 is the half-time. So work out the equation, t1/2 = t (-ln2)/ln(Nt/N0) = 11.5*(-ln2)/ln(12.5/100) = 3.83 days
Answer:
74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.
Explanation:
The balanced reaction is:
Na₂CO₃ + Ca(NO₃)₂ ⟶ CaCO₃ + 2 NaNO₃
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 in the reaction:
- Na₂CO₃: 1 mole
- Ca(NO₃)₂: 1 mole
- CaCO₃: 1 mole
- NaNO₃: 2 mole
Being the molar mass of the compounds:
- Na₂CO₃: 106 g/mole
- Ca(NO₃)₂: 164 g/mole
- CaCO₃: 100 g/mole
- NaNO₃: 85 g/mole
then by stoichiometry the following quantities of mass participate in the reaction:
- Na₂CO₃: 1 mole* 106 g/mole= 106 g
- Ca(NO₃)₂: 1 mole* 164 g/mole= 164 g
- CaCO₃: 1 mole* 100 g/mole= 100 g
- NaNO₃: 2 mole* 85 g/mole= 170 g
You can apply the following rule of three: if by stoichiometry 106 grams of Na₂CO₃ produce 100 grams of CaCO₃, 79.3 grams of Na₂CO₃ produce how much mass of CaCO₃?

mass of CaCO₃= 74.81 grams
<u><em>74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.</em></u>
Answer:
I guess B They have a charge