Upon adding solid potassium hydroxide pellets to water the following reaction takes place: KOH(s) → KOH(aq) + 43 kJ/mol Answer the following questions regarding the addition of 14.0 g of KOH to water: Does the beaker get warmer or colder? Is the reaction endothermic or exothermic? What is the enthalpy change for the dissolution of the 14.0 grams of KOH?

Answer:

Balanced equation:

Explanation:

The chemical reaction between Lead(II) Nitrate and potassium carbonate is as follows.

Ionic equation:

Cancel the same ions on the both sides of the reaction.

The net ionic equation is as follows.

Answer:

FCC.

Explanation:

Hello,

In this case, since the density is defined as:

Whereas n accounts for the number of atoms per units cell (2 for BCC and 4 for FCC), M the atomic mass of the element, Vc the volume of the cell and NA the Avogadro's number. Thus, for both BCC and FCC, the volume of the cell is:

Hence, we compute the density for each crystal structure:

Therefore, since the density computed as a FCC crystal structure matches with the actual density, we conclude rhodium has a FCC crystal structure.

Regards.

Reactants take 504.87 yr to reach 12.5% of their original value in first-order decomposition reaction.

Equation for the first-order decomposition reaction:-

....(1)

Here,   is the final concentration, t is the time,   is the initial concentration, and k is the rate constant.

Given:-

k=

Substitute the above value in equation (1) as follows:-

So, 504.87 yr does it take for the reactant to reach 12.5% of its original value.

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Final answer:

The time required for a reactant to reach 12.5% of its original value in a first-order reaction is approximately 1482 years, obtained by applying the formula for the half-life of a first-order reaction and multiplying by 3.

Explanation:

In a first-order reaction, the half-life of the reaction, which is the time it takes for half of the reactant to be consumed, is independent of the concentration of the reactant. Also, for a first-order reaction, it would take approximately 3 half-lives for the reactant to be reduced to 12.5% of its original value. The Integrated Rate Law for a First-Order Reaction can be applied to determine the time it will take.

Given the rate constant (k) is 0.00140 yr¯¹, we will use the formula for the half-life of a first-order reaction: t₁/₂ = 0.693 / k. After calculating the half-life (t₁/₂), multiply it by 3 to determine the time for the reactant concentration to reach 12.5% of its original value. Hence, it would take approximately 1482 years to reach 12.5% of the original value when rounded to the correct number of significant figures.

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0.085 moles of Al are required to form 23.6 g of AlBr₃.

Let's consider the following balanced equation for the synthesis reaction of AlBr₃.

2 Al(s) + 3 Br₂(l) → 2 AlBr₃(s)

First, we will convert 23.6 g to moles using the molar mass of AlBr₃ (266.69 g/mol).

The molar ratio of Al to AlBr₃ is 2:2. The moles of Al required to form 0.0885  moles of AlBr₃ are:

0.085 moles of Al are required to form 23.6 g of AlBr₃.

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Answer:

0.088 mole of Al.

Explanation:

First, we shall determine the number of mole in 23.6 g of AlBr₃.

This is illustrated below:

Mass of AlBr₃ = 23.6 g

Molar Mass of AlBr₃ = 27 + 3(80) = 267 g/mol

Mole of AlBr₃ =.?

Mole = mass/Molar mass

Mole of AlBr₃ = 23.6 / 267

Mole of AlBr₃ = 0.088 mol

Next, we shall writing the balanced equation for the reaction.

This is given below:

2Al(s) + 3Br₂(l) → 2AlBr₃(s)

From the balanced equation above,

2 moles of Al reacted with 3 mole of Br₂ to 2 moles AlBr₃.

Finally, we shall determine the number of mole of Al needed for the reaction as follow:

From the balanced equation above,

2 moles of Al reacted to 2 moles AlBr₃.

Therefore, 0.088 mole of Al will also react to produce 0.088 mole of AlBr₃.

Answer:

0.1 M

Explanation:

The overall balanced reaction equation for the process is;

IO3^- (aq)+ 6H^+(aq) + 6S2O3^2-(aq) → I-(aq) + 3S4O6^2-(aq) + 3H2O(l)

Generally, we must note that;

1 mol of IO3^- require 6 moles of S2O3^2-

Thus;

n (iodate) = n(thiosulfate)/6

C(iodate) x V(iodate) = C(thiosulfate) x V(thiosulfate)/6

Concentration of iodate C(iodate)= 0.0100 M

Volume of iodate= V(iodate)= 26.34 ml

Concentration of thiosulphate= C(thiosulfate)= the unknown

Volume of thiosulphate=V(thiosulfate)= 15.51 ml

Hence;

C(iodate) x V(iodate) × 6/V(thiosulfate) = C(thiosulfate)

0.0100 M × 26.34 ml × 6/15.51 ml = 0.1 M

Final answer:

To determine the moles of KIO_3 titrated, use the balanced equation 2 KIO_3 + 5 Na_2S_2O_3 + 6 HCl → 3 I_2 + 6 NaCl + 6 NaClO + 3 H_2O. Therefore, 0.001551 mol of KIO_3 were titrated.

Explanation:

To determine the moles of KIO3 titrated, we need to use the balanced equation for the reaction:

2 KIO3 + 5 Na2S2O3 + 6 HCl → 3 I2 + 6 NaCl + 6 NaClO + 3 H2O

From the equation, we can see that 2 moles of KIO3 react with 5 moles of Na2S2O3. Therefore, the moles of KIO3 titrated can be calculated using the following proportion:

(0.0100 M KIO3 / 1 L) * (15.51 mL / 1000 mL) * (2 mol KIO3 / 5 mol Na2S2O3) = 0.001551 mol KIO3

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