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Kazeer [188]
3 years ago
15

Potassium reacts with aluminum bromide to produce aluminum and potassium bromide. You will generate ____ moles of aluminum when

you react 0.372 moles of potassium.
Chemistry
1 answer:
SpyIntel [72]3 years ago
6 0

Answer:

You will generate 0.124 moles of aluminum when you react 0.372 moles of potassium

Explanation:

Determine the reaction:

3K + AlBr₃ → Al + 3KBr

Ratio is 3:1

3 moles of K can produce 1 mol of Al

So, 0.372 moles of K, may produce (0.372 .1) /3 = 0.124 moles of Al

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Mamont248 [21]

Answer:

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So, here is used categorical thinking because is a abstract concept that is hard to understand sometimes. It's abstract when we say that matter is all, not  just what we see, but even what we don't see like ''Dark Matter'', which is the majority in the Universe.

4 0
3 years ago
What is the mass of a 5.00 cm^3 piece of copper having a density of 8.96 g/cm^3
erik [133]

Answer:

44.8 g

Explanation:

Density = mass / Volume

Mass = density x Volume = 8.96x 5 = 44.8 g

5 0
3 years ago
A mixture of XO2 (P = 3.00 atm) and O2 (P = 1.00 atm) is placed in a container. This elementary reaction takes place at 27 °C: 2
sukhopar [10]

Answer:

a) \triangle G^{0} = 7.31 kJ/mol

b) K_{-1} = 0.0594 m^{-1} s^{-1}

Explanation:

Equation of reaction:

                                     2 XO_{2} (g) + O_{2} (g) \rightleftharpoons 2XO_{3} (g)

Initial pressure                  3              1              0

Pressure change             2P           1P             2P

Total pressure = (3-2P) + (1-P) + (2P)

Total Pressure = 3.75 atm

(3-2P) + (1-P) + (2P) = 3.75

4 - P = 3.75

P = 4 - 3.75

P = 0.25 atm

Let us calculate the pressure of each of the components of the reaction:

Pressure of XO2 = 3 - 2P = 3 - 2(0.25)

Pressure of XO2 =2.5 atm

Pressure of O2 = 1 - P = 1 -0.25

Pressure of O2 = 0.75 atm

Pressure of XO3 = 2P = 2 * 0.25

Pressure of XO3 = 0.5 atm

From the reaction, equilibrium constant can be calculated using the formula:

K_{p} = \frac{[PXO_{3}] ^{2} }{[PXO_{2}] ^{2}[PO_{2}] }

K_{p} = \frac{0.5^2}{2.5^2 *0.75} \\K_{p} = 0.0533 = K_{eq}

Standard free energy:

\triangle G^{0} = - RT ln k_{eq} \\\triangle G^{0} = -(0.008314*300* ln0.0533)\\\triangle G^{0} = 7.31 kJ/mol

b) value of k−1 at 27 °C, i.e. 300K

K_{1} = 7.8 * 10^{-2} m^{-2} s^{-1}

K_{c} = K_{p}RT\\K_{c} = 0.0533* 0.0821 * 300\\K_{c} = 1.313 m^{-1}

K_{-1} = \frac{K_{1} }{K_{c} } \\K_{-1} = \frac{7.8 * 10^{-2}  }{1.313 }\\K_{-1} = 0.0594 m^{-1} s^{-1}

6 0
3 years ago
A 13.00 g sample of a compound contains 4.15 g potassium (k), 3.76 g chlorine (cl), and oxygen (o). calculate the empirical form
MrRissso [65]

To solve this problem, let us all convert the mass of each element into number of moles using the formula:

moles = mass / molar mass

Where,

molar mass K = 39.10 g / mol

<span>molar mass Cl = 35.45 g / mol</span>

molar mass O = 16 g / mol

<span>and mass O = 13 g – 4.15 g – 3.76 g  = 5.09 g</span>

 

moles K = 4.15 g / (39.10 g / mol) = 0.106 mol

<span>moles Cl = 3.76 g / (35.45 g / mol) = 0.106 mol</span>

moles O = 5.09 g / (16 g / mol) = 0.318 mol

 

The ratio becomes:

0.106 K: 0.106 Cl: 0.318 O

We divide all numbers with the smallest number, in this case 0.106. This becomes:

K: Cl: 3O

 

Therefore the empirical formula is:

KClO_{3}

7 0
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