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3 years ago

Consider the balanced equation for the following reaction:

Chemistry

1 answer:

wel3 years ago

4 0

Answer:

The amount of ammonia needed is 33.3 g

Therefore, we can also say 1.96 moles of NH₃

Explanation:

The reaction is: 3CuO(s) + 2NH₃(g) → 3H₂O(l) + 3Cu(s) + N₂(g)

If we see stoichiometry, 3 moles of water can be produced by 2 moles of NH₃. We propose this rule of three:

3 moles of water can be produced by 2 mole of ammonia

Then, 2.94 moles of water, must be produced by (2.94 . 2) /3 =1.96 moles of NH₃

If we convert the moles to mass. 1.96 mol . 17 g /1mol = 33.3 g

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Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f

N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

<u>Answer:</u> The initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+2B\rightleftharpoons C$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b$

where,

a = order with respect to A

b = order with respect to B

Expression for rate law for first trial:

$5.4\times 10^{-3}=k(0.30)^a(0.050)^b$ ....(1)

Expression for rate law for second trial:

$1.1\times 10^{-2}=k(0.30)^a(0.100)^b$ ....(2)

Expression for rate law for third trial:

$2.2\times 10^{-2}=k(0.50)^a(0.050)^b$ ....(3)

Dividing 2 by 1, we get:

$\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1$

Dividing 3 by 1, we get:

$\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^1$ ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}$

Calculating the initial rate of formation of C by using equation 4, we get:

$k=1.2M^{-2}s^{-1}$

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

$\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}$

Hence, the initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

8 0

3 years ago

The __________ system is made up of the nose, pharynx, trachea, lungs, bronchi, and alveoli. *

Reil [10]

Respiratory system is made of that

4 0

3 years ago

Read 2 more answers

Which label belongs in the region marked X?

lubasha [3.4K]

Answer:

A) involves changes in temperature

Explanation:

The figure is missing, but I assume that the region marked X represents the region in common between Gay-Lussac's law and Charle's Law.

Gay-Lussac's law states that:

"For an ideal gas kept at constant volume, the pressure of the gas is directly proportional to its absolute temperature"

Mathematically, it can be written as

$p\propto T$

where p is the pressure of the gas and T its absolute temperature.

Charle's Law states that:

"For an ideal gas kept at constant pressure, the volume of the gas is directly proportional to its absolute temperature"

Mathematically, it can be written as

$V\propto T$

where V is the volume of the gas and T its absolute temperature.

By looking at the two descriptions of the law, we see immediately that the property that they have in common is

A) involves changes in temperature

Since the temperature is NOT kept constant in the two laws.

7 0

3 years ago

Ano-ano ang mensahe ng grupo ng el gamma penumbra sa kanilang palabas

Alexxandr [17]

Answer:

Pag-alaga ng kalikasan, pagpapakita ng mga tanawin o kagandahan ng Pilipinas

6 0

3 years ago

Locations Cost of land Cost of equipment Cost of mining and reclamation Time taken to mine the area 1 $100,000 $10,000 $5,000 pe

mario62 [17]

Answer:

Location 2 will cost the least to the company to mine.

Explanation:

a) Data and Calculations:

Locations Cost of land Cost of Cost of mining Time taken to

equipment and reclamation mine the area

1 $100,000 $10,000 $5,000 per day 30 days

2 $35,000 $6,000 $4,500 per day 45 days

3 $30,000 $7,500 $3,500 per day 120 days

4 $40,500 $8,000 $7,000 per day 65 days

Locations Cost of land Cost of Cost of mining Total

equipment and reclamation Costs

1 $100,000 $10,000 $150,000 ($5,000 * 30) $260,000

2 $35,000 $6,000 $202,500 ($4,500 * 45) $243,500

3 $30,000 $7,500 $420,000 ($3,500 * 120) $457,500

4 $40,500 $8,000 $445,000 ($7,000 * 65) $503,500

7 0

2 years ago