Chemistry-mole calculations

What effects does the concentration of reactants have on the rate of a reaction ?

Harlamova29_29 [7]

Question:

What effects does the concentration of reactants have on the rate of a reaction?

Answer:

Reactant concentration. Increasing the concentration of one or more reactants will often increase the rate of reaction. This occurs because a higher concentration of a reactant will lead to more collisions of that reactant in a specific time period.

Increasing the concentration of reactants generally increases the rate of reaction because more of the reacting molecules or ions are present to form the reaction products. ... When concentrations are already high, a limit is often reached where increasing the concentration has little effect on the rate of reaction.

Hope this helps, have a good day. c;

5 0

3 years ago

Boron sulfide, B2S3(s), reacts violently with water to form dissolved boric acid (H3BO3) and hydrogen sulfide gas.Express your a

ioda

Answer:

Explanation:

From the statement of the problem,

B₂S₃ $_{s}$ + H₂O $_{l}$ → H₃BO₃ $_{aq}$ + H₂S $_{g}$

B₂S₃ + H₂O → H₃BO₃ + H₂S

We that the above expression does not conform with the law of conservation of mass:

To obey the law, we need to derive a balanced reaction equation:

Let us use the mathematical method to obtain a balanced equation.

let the balanced equation be:

aB₂S₃ + bH₂O → cH₃BO₃ + dH₂S

where a, b, c and d will make the equation balanced.

Conservating B: 2a = c

S: 3a = d

H: 2b = 3c + 2d

O: b = 3c

if a = 1,

c = 2,

b = 6,

2d = 2(6) - 3(2) = 6, d = 3

Now we can input this into our equation:

B₂S₃ + 6H₂O → 2H₃BO₃ + 3H₂S

B₂S₃ $_{s}$ + 6H₂O $_{l}$ → 2H₃BO₃ $_{aq}$ + 3H₂S $_{g}$

4 0

3 years ago

In a two-step synthesis, C6H11Br is converted into C6H12O. From the structure of the product, molecular formula of the starting

tamaranim1 [39]

Answer:

See explanation below

Explanation:

The question is incomplete. However in picture 1, you have the starting materials and the structure of the product, which you miss in this part.

Now, in picture 2, you have the starting reactant and the product, and the mechanism that is taking place here.

First, all what we have here is an acid base reaction. In the first step, we are using the acid medium to convert the reactant into an alcohol. The bromine there, is not leaving the molecule yet, because it's neccesary for the next step. The starting reactant is an alkene, in that way, we can convert the reactant in the first step into a secondary alcohol. In other words, the first reaction is a alkene hydration.

In the second step, we use a strong base. You may say this is a strong nucleophile and will do a Sn2 reaction to form another alcohol there, but it's not the case, because, before any kind of reaction happens, the priority here is always the acid base, so the base will react with the acidic hydrogen. In this case, it will substract an hydrogen from the OH. When this happens, the lone pair will do an auto condensation here, and attacks the bromine in the molecule. In this way, the molecule will become a cyclomolecule, and that way it form the final product.

See picture 2, for mechanism