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bulgar [2K]
2 years ago
7

How many ions of aluminum oxide (Al²O³) are there in 200 g of Al²O³??​

Chemistry
1 answer:
valentinak56 [21]2 years ago
6 0

= 6.022 × 1020

Explanation<em>;</em>

Mole of aluminium oxide (Al2O3) is

⇒ 2 x 27 + 3 x 16

Mole of aluminium oxide = 102 g

i.e., 102 g of Al2O3= 6.022 x 1023 molecules of Al2O3

Then, 0.051 g of Al2O3 contains = 6.022 x 1023 / (102 x 0.051 molecules)

= 3.011 x 1020 molecules of Al2O3

The number of aluminium ions (Al3+) present in one molecule of aluminium oxide is 2.

Therefore, the number of aluminium ions (Al3+) present in 3.11 × 1020 molecules (0.051g) of aluminium oxide (Al2O3)

= 2 × 3.011 × 1020

=<em> 6.022 × 1020</em>

<em>hope </em><em>it </em><em>helps</em><em>_</em>

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Which choice is an example of an endothermic process?
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Answer: conversion of ice to steam

Explanation: Endothermic process is one in which energy is absorbed by the system.

Conversion of ice to steam is change of solid phase to gaseous phase, thus energy is required to break the strong inter molecular forces of attraction in solids to convert it into gaseous phase.

Conversion of steam to ice, conversion of steam to water  and conversion of water to ice releases energy and are examples of exothermic processes.

7 0
3 years ago
3) What are the long-term effects of the Chernobyl disaster? How many people were displaced?
I am Lyosha [343]

Answer: The issue of long-term effects of the Chernobyl disaster on civilians is very controversial. The number of people whose lives were affected by the disaster is sizable. Over 300,000 people were resettled because of the disaster; millions lived and continue to live in the contaminated area.

Explanation:

5 0
3 years ago
If 21.00 mL of a 0.68 M solution of C6H5NH2 required 6.60 mL of the strong acid to completely neutralize the solution, what was
Andru [333]

Answer:

pH = 2.46

Explanation:

Hello there!

In this case, since this neutralization reaction may be assumed to occur in a 1:1 mole ratio between the base and the strong acid, it is possible to write the following moles and volume-concentrations relationship for the equivalence point:

n_{acid}=n_{base}=n_{salt}

Whereas the moles of the salt are computed as shown below:

n_{salt}=0.021L*0.68mol/L=0.01428mol

So we can divide those moles by the total volume (0.021L+0.0066L=0.0276L) to obtain the concentration of the final salt:

[salt]=0.01428mol/0.0276L=0.517M

Now, we need to keep in mind that this is an acidic salt since the base is weak and the acid strong, so the determinant ionization is:

C_6H_5NH_3^++H_2O\rightleftharpoons  C_6H_5NH_2+H_3O^+

Whose equilibrium expression is:

Ka=\frac{[C_6H_5NH_2][H_3O^+]}{C_6H_5NH_3^+}

Now, since the Kb of C6H5NH2 is 4.3 x 10^-10, its Ka is 2.326x10^-5 (Kw/Kb), we can also write:

2.326x10^{-5}=\frac{x^2}{0.517M}

Whereas x is:

x=\sqrt{0.517*2.326x10^{-5}}\\\\x=3.47x10^-3

Which also equals the concentration of hydrogen ions; therefore, the pH at the equivalence point is:

pH=-log(3.47x10^{-3})\\\\pH=2.46

Regards!

6 0
3 years ago
A city's water supply is contaminated with a toxin at a concentration of 0.63 mg/L. For the water to be safe for drinking, the c
Shalnov [3]

Answer:

Approximately 22.37 days, will it take for the water to be safe to drink.

Explanation:

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

k is rate constant

Given that:- k = 0.27 (day)⁻¹

[A_0] = 0.63 mg/L

[A_t]=1.5\times 10^{-3} mg/L

Applying in the above equation as:-

1.5\times 10^{-3}=0.63e^{-0.27\times t}

63e^{-0.27t}=150\times \:10^{-3}

e^{-0.27t}=\frac{1}{420}

t=\frac{100\ln \left(420\right)}{27}=22.37

<u>Approximately 22.37 days, will it take for the water to be safe to drink.</u>

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