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3 years ago

Which is true when a reaction has reached equalibrium

1 answer:

7 0

The answer should be: <span>D. The reaction rate is equal in both directions

In the equilibrium state, the rate of reaction to the right is same as the reaction to the left. Because of this, the concentration of the reactant and product will be kept same.
It might seem like the reaction is stopped because there is no change in the concentration, but it wasn't. Adding a reactant or product will break the equilibrium state.</span>

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Sladkaya [172]

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3 0

3 years ago

Which substance produces hydroxide ions in solution?

Olegator [25]

Answer:

An Arrhenius Base

Explanation:

The definition of this is a base that is a hydroxide ion donor.

8 0

2 years ago

How many parts per million of lead is found in 250 ml of water if there is 1.30g of lead in the water? 1ml=1g

SSSSS [86.1K]

Answer:

5200 ppm

Explanation:

As per the definition, parts per million of a contaminant is a measure of the amount of mass of contaminant present per million amount of the solution. It is denoted by ppm.

Given in the question,

Water = 250 ml = 250 g

Lead = 1.30 g

So,

ppm of Lead = $\frac{Lead}{Water} \times 10^6$ = $\frac{1.30}{250} \times 10^6$ = 5200 ppm

So, as calculated above, there is 5200 ppm of lead present in 250 ml of water.

6 0

3 years ago

When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H

Debora [2.8K]

Answer:

$Y=58.15\%$

Explanation:

Hello,

For the given chemical reaction:

$Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)$

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

$n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3$

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

$n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol Al_2S_3$

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

$m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3$

Finally, we compute the percent yield with the obtained 2.10 g:

$Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%$

Best regards.

7 0

2 years ago

Consider the reaction.

Stella [2.4K]

Answer:

when the rates of the forward and reverse reactions are equal

Explanation:

In a chemical system, the reaction reaches a dynamic equilibrium when the rate of formation of product equals the rate of formation of reactants. This implies that both the forward and revered(backwards) reaction are occurring at the same rate.

5 0

3 years ago