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3 years ago

Calculate the number of moles in 3.01 x 1024 atoms of Krypton.

Chemistry

1 answer:

Kaylis [27]3 years ago

6 0

Answer:

So 3.01×10236.022×1023mol−1 is clearly 1/2 a mole of silicon atoms. You should be quickly able to tell me the mass of this quantity of silicon atoms.

I think or 1/3 a mole Id k I’m just starting to learn this ok

Explanation:

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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

Additional solute is added to a saturated solution of a solid and a liquid after being heated above room temperature. Once no mo

PSYCHO15rus [73]

Answer:

After the solution is heated, but before additional solute is added

Explanation:

An unsaturated solution is a solution that contains less solute than it can normally hold at a given temperature. Hence an unsaturated solution can still dissolve more solute.

When the solution is heated, the saturated cold solution becomes an unsaturated hot solution which is capable of dissolving more solute at this point.

Once more solute is dissolved, the solution becomes saturated again just before it begins to cool since no more solute dissolves in the solution at some point before cooling and addition of seed crystals.

4 0

3 years ago

Which of the following latitudes receive the most direct solar energy?

Misha Larkins [42]

Answer:

The answer to your question is D.

Explanation:

The latitudes near the equator receives the most direct solar energy.

Hope this helps :)

8 0

3 years ago

a 125 g chunk of aluminum at 182 degrees Celsius was added to a bucket filled with 365 g of water at 22.0 degrees Celsius. Ignor

Diano4ka-milaya [45]

<h3>Answer:</h3>

32.98°C

<h3>Explanation:</h3>

We are given the following;

Mass of Aluminium as 125 g

Initial temperature of Aluminium as 182°C

Mass of water as 265 g

Initial temperature of water as 22°C

We are required to calculate the final temperature of the two compounds;

First, we need to know the specific heat capacity of each;

Specific heat capacity of Aluminium is 0.9 J/g°C

Specific heat capacity of water is 4.184 J/g°C

<h3>Step 1: Calculate the Quantity of heat gained by water.</h3>

Assuming the final temperature is X°C

we know, Q = mcΔT

Change in temperature, ΔT = (X-22)°C

therefore;

Q = 365 g × 4.184 J/g°C × (X-22)°C

= (1527.16X-33,597.52) Joules

<h3>Step 2: Calculate the quantity of heat released by Aluminium </h3>

Using the final temperature, X°C

Change in temperature, ΔT = -(X°- 182°)C (negative because heat was lost)

Therefore;

Q = 125 g × 0.90 J/g°C × (182°-X°)C

= (20,475- 112.5X) Joules

<h3>Step 3: Calculating the final temperature</h3>

We need to know that the heat released by aluminium is equal to heat absorbed by water.

Therefore;

(20,475- 112.5X) Joules = (1527.16X-33,597.52) Joules

Combining the like terms;

1639.66X = 54072.52

X = 32.978°C

= 32.98°C

Therefore, the final temperature of the two compounds will be 32.98°C

7 0

3 years ago

How many grams of carbon dioxide are produced from 0.98 mol of Fe3O4?

nexus9112 [7]

Hello

the answer is 43.129310000000004

Have a nice day

4 0

3 years ago