What is the % Yield if 3.8 g of Cu was produced?

Two solutions namely, 500 ml of 0.50 m hcl and 500 ml of 0.50 m naoh at the same temperature of 21.6 are mixed in a constant-pre

weeeeeb [17]

24.6 ℃

<h3>Explanation</h3>

Hydrochloric acid and sodium hydroxide reacts by the following equation:

$\text{HCl} \; (aq) + \text{NaOH} \; (aq) \to \text{NaCl} \; (aq) + \text{H}_2\text{O} \; (aq)$

which is equivalent to

$\text{H}^{+} \; (aq) + \text{OH}^{-} \; (aq) \to \text{H}_2\text{O}\; (l)$

The question states that the second equation has an enthalpy, or "heat", of neutralization of $-56.2 \; \text{kJ}$ . Thus the combination of every mole of hydrogen ions and hydroxide ions in solution would produce $56.2 \; \text{kJ}$ or $56.2 \times 10^{3}\; \text{J}$ of energy.

500 milliliter of a 0.50 mol per liter "M" solution contains 0.25 moles of the solute. There are thus 0.25 moles of hydrogen ions and hydroxide ions in the two 0.500 milliliter solutions, respectively. They would combine to release $0.25 \times 56.2 \times 10^{3} = 1.405 \times 10^{4} \; \text{J}$ of energy.

Both the solution and the calorimeter absorb energy released in this neutralization reaction. Their temperature change is dependent on the heat capacity <em>C</em> of the two objects, combined.

The question has given the heat capacity of the calorimeter directly.

The heat capacity (the one without mass in the unit) of water is to be calculated from its mass and <em>specific</em> heat.

The calorimeter contains 1.00 liters or $1.00 \times 10^{3} \; \text{ml}$ of the 1.0 gram per milliliter solution. Accordingly, it would have a mass of $1.00 \times 10^{3} \; \text{g}$ .

The solution has a specific heat of $4.184 \; \text{J} \cdot \text{g}^{-1} \cdot \text{K}^{-1}$ . The solution thus have a heat capacity of $4.184 \times 1.00 \times 10^{3} = 4.184 \times 10^{3} \; \text{J} \cdot\text{K}^{-1}$ . Note that one degree Kelvins K is equivalent to one degree celsius ℃ in temperature change measurements.

The calorimeter-solution system thus has a heat capacity of $4.634 \times 10^{3} \; \text{J} \cdot \text{K}^{-1}$ , meaning that its temperature would rise by 1 degree celsius on the absorption of 4.634 × 10³ joules of energy. $1.405 \times 10^{4} \; \text{J}$ are available from the reaction. Thus, the temperature of the system shall have risen by 3.03 degrees celsius to 24.6 degrees celsius by the end of the reaction.

4 0

2 years ago

Help............???????

balandron [24]

B. .175

Hope it helps!

4 0

2 years ago

rate of a certain reaction is given by the following rate law: rate Use this information to answer the questions below. What is

Sunny_sXe [5.5K]

Complete Question

The rate of a certain reaction is given by the following rate law:

$rate = k [H_2][I_2]$

rate Use this information to answer the questions below.

What is the reaction order in $H_2$ ?

What is the reaction order in $I_2$ ?

What is overall reaction order?

At a certain concentration of H2 and I2, the initial rate of reaction is 2.0 x 104 M / s. What would the initial rate of the reaction be if the concentration of H2 were doubled? Round your answer to significant digits. The rate of the reaction is measured to be 52.0 M / s when [H2] = 1.8 M and [I2] = 0.82 M. Calculate the value of the rate constant. Round your answer to significant digits.

Answer:

The reaction order in $H_2$ is n = 1

The reaction order in $I_2$ is m = 1

The overall reaction order z = 2

When the hydrogen is double the the initial rate is $rate_n = 4.0*10^{-4} M/s$

The rate constant is $k = 35.23 \ M^{-1} s^{-1}$

Explanation:

From the question we are told that

The rate law is $rate = k [H_2][I_2]$

The rate of reaction is $rate = 2.0 *10^{4} M /s$

Let the reaction order for $H_2$ be n and for $I_2$ be m

From the given rate law the concentration of $H_2$ is raised to the power of 1 and this is same with $I_2$ so their reaction order is n=m=1

The overall reaction order is

$z = n +m$

$z =1 +1$

$z =2$

At $rate = 2.0 *10^{4} M /s$

$2.0*10^{4} = k [H_2] [I_2] ---(1)$

= > $k = \frac{2.0*10^{4}}{[H_2] [I_2] }$

given that the concentration of hydrogen is doubled we have that

$rate = k [2H_2] [I_2] ----(2)$

=> $k = \frac{rate_n }{ [2H_2] [I_2]}$

So equating the two k

$\frac{2.0*10^{4}}{[H_2] [I_2] } = \frac{rate_n }{ [2H_2] [I_2]}$

=> $rate_n = 4.0*10^{-4} M/s$

So when

$rate_x = 52.0 M/s$

$[H_2] = 1.8 M$

$[I_2] = 0.82 \ M$

We have

$52 .0 = k(1.8)* (0.82)$

$k = \frac{52 .0}{(1.8)* (0.82)}$

$k = 35.23 M^{-2} s^{-1}$

3 0

2 years ago

What would you observe in a solution of 30 g of KC1O3 in 100 g of water at 10 c?

Mrac [35]

Answer:

Option A is correct. About 5 g of the KClO3 is dissolved

Explanation:

KClO3 is not very good soluble in water.

So, Option C is impossible, because KClO3 is poorly soluble in water.

The low solubility of KClO3 in water causes KClO3 to isolate itself from the reaction mixture by precipitating out of solution.

So, option D will either happen.There will be a part of KClO3 dissolve.

At 10 °C, KClO3 has a solubility of 4.46 g/100 gram (10 °C).

Option A is correct. About 5 g of the KClO3 is dissolved

7 0

2 years ago

Describe the chemical calcium-oxide.

timama [110]

Answer:

Explanation:

Mallabone’s tests suggest there’s no risk the jacket liner will explode or heat up so much that it risks causing burns. The reaction also doesn’t produce acidic byproducts. And because calcium oxide doesn’t react with humidity (water vapor in the air), the liners can be stored for long periods and still work when needed.

Calcium oxide: A substance that gives off heat as it chemically reacts with water. Its chemical formula is CaO (which means each molecule is made up of one calcium atom and one oxygen atom).

Exothermic reaction: A chemical reaction that generates heat as it proceeds. (In Greek, “exo” means outside and “therm” means heat.)

Thought this might help ;)

5 0

2 years ago