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Gwar [14]
3 years ago
10

Consider this reaction: 2Al(s) + 3 CuCl2(aq) → 2AlCl3(aq) + 3 Cu(s) If the concentration of CuCl2 drops from 1.000 M to 0.655 M

in the first 30.0 s of the reaction, what is the average rate of reaction over this time interval?
Chemistry
2 answers:
Yuliya22 [10]3 years ago
6 0

Answer:

r=-0.0115M/s

Explanation:

Hello,

In this case, the rate is computed as:

r=\frac{C_F-C_0}{T_F-T_0}

As the reaction is analyzed during the first 30 seconds, the average rate will be:

r=\frac{0.655M-1.000M}{30s-0s}=-0.0115M/s

Which negative since the cupric chloride is a reagent.

Best regards.

blondinia [14]3 years ago
4 0

Answer:

rate = 0.00383 M/s

Explanation:

The rate of reaction refers to the speed at which the products are formed from the reactants in a chemical reaction

2Al(s) + 3 CuCl2(aq) → 2AlCl3(aq) + 3 Cu(s)

The equation for reaction rate in this chemical reaction (for CuCl₂)  is the following one (The value 1/3 comes from the coefficient in the chemical equation for the reaction):

\text{rate of reaction}  = - \frac{1}{\text{coefficient}} \frac{Reactant}{Time}

Here, the negative sign is used to indicate the decreasing concentration of the reactant.

rate= -\frac{1}{3} \times \frac{C2-C1}{t2-t1}

rate = - 1/3 [0.655 - 1.0 / 30]

rate = -1/3[-0.0115]

rate = 3.83 x 10-3 M/s

rate = 0.00383 M/s

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6 0
3 years ago
Read 2 more answers
A 35.0 ml sample of 0.225 m hbr was titrated with 42.3 ml of koh. What is the concentration of the koh?
Lemur [1.5K]

Answer:

The concentration of KOH is 0.186 M

Explanation:

First things first, we need too write out the balanced equation between HBr and KOH.

This is given as;

KOH (aq) + HBr (aq) → KBr (aq) + H2O (l)

From the reaction above, we can tell that it takes 1 mole of KOH to react with 1 mole of HBr.

We use the acid base formular in calculating unknown concentrations. This is given as;

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Va = Volume of acid

Cb = Concentration of base

Vb = Volume of base

na = Number of moles of acid

nb = Number of moles of base

KOH is the base and HBr is acid.

Hence;

Ca = 0.225

Va = 35

Cb = ?

Vb = 42.3

na = 1

nb = 1

Making Cb subject of formular we have;

Cb = \frac{CaVaNb}{VbNa}

Cb = (0.225 * 35 * 1) / (42.3 * 1)

Cb = 0.186 M

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