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Setler79 [48]
2 years ago
13

10.Using the below valency chart, write chemical formula for aluminum oxide and

Chemistry
1 answer:
sweet [91]2 years ago
5 0

The chemical formula of aluminum chloride is AlCl₃ and the  chemical formula of aluminum oxide is Al₂O₃.

<h3>Aluminum chloride</h3>

The chemical formula of aluminum chloride is written as;

Al³⁺  + Cl⁻   → AlCl₃

<h3>Aluminum oxide</h3>

The chemical formula of aluminum oxide is written as;

Al³⁺  + O²⁻   → Al₂O₃

Thus, the chemical formula of aluminum chloride is AlCl₃ and the  chemical formula of aluminum oxide is Al₂O₃.

Learn more about chemical formula here: brainly.com/question/26388921

#SPJ1

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The element iridium exists in nature as two isotopes: 191Ir has a mass of 190.9606 u, and 193Ir has a mass of 192.9629 u. The av
nlexa [21]

<u>Answer:</u> The percentage abundance of _{77}^{191}\textrm{Ir} and _{77}^{193}\textrm{Ir} isotopes are 37.10% and 62.90% respectively.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i   .....(1)

Let the fractional abundance of _{77}^{191}\textrm{Ir} isotope be 'x'. So, fractional abundance of _{77}^{193}\textrm{Ir} isotope will be '1 - x'

  • <u>For _{77}^{191}\textrm{Ir} isotope:</u>

Mass of _{77}^{191}\textrm{Ir} isotope = 190.9606 amu

Fractional abundance of _{77}^{191}\textrm{Ir} isotope = x

  • <u>For _{77}^{193}\textrm{Ir} isotope:</u>

Mass of _{77}^{193}\textrm{Ir} isotope = 192.9629 amu

Fractional abundance of _{77}^{193}\textrm{Ir} isotope = 1 - x

Average atomic mass of iridium = 192.22 amu

Putting values in equation 1, we get:

192.22=[(190.9606\times x)+(192.9629\times (1-x))]\\\\x=0.3710

Percentage abundance of _{77}^{191}\textrm{Ir} isotope = 0.3710\times 100=37.10\%

Percentage abundance of _{77}^{193}\textrm{Ir} isotope = (1-0.3710)=0.6290\times 100=62.90\%

Hence, the percentage abundance of _{77}^{191}\textrm{Ir} and _{77}^{193}\textrm{Ir} isotopes are 37.10% and 62.90% respectively.

8 0
3 years ago
Would you expect a carbon atom to have the same mass as an oxygen atom
kramer
Its impossible coz carbon has isotopes C-12 and C-13 whereas oygen has isotopes 16<span>O, </span>17<span>O, and </span>18<span>O, so impossible. Hope it helps</span>
3 0
3 years ago
Read 2 more answers
1. If you have a sample of gas at a pressure of 16 atm, what will the pressure be if the volume is halved?
vagabundo [1.1K]

Answer:

1. The pressure will be 32 atm, twice the initial pressure.

2. The pressure will be 1.83 atm, one third of the initial pressure.

Explanation:

Boyle's law is one of the gas laws that relates the volume and pressure of a certain quantity of gas kept at a constant temperature.

This law says that "The volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure." This means that if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

or P * V = k

Ahora es posible suponer que tienes un cierto volumen de gas V1 que se encuentra a una presión P1 al comienzo del experimento. Si varias el volumen de gas hasta un nuevo valor V2, entonces la presión cambiará a P2, y se cumplirá:

P1*V1=P2*V2

1. In this case:

  • P1= 16 atm
  • V1
  • P2= ?
  • V2= V1÷2= \frac{V1}{2} because the volume is halved.

So:

16 atm*V1= P2* \frac{V1}{2}

Solving:

\frac{16 atm*V1*2}{V1}=P2

16 atm*2= P2

32 atm= P2

<u><em>The pressure will be 32 atm, twice the initial pressure.</em></u>

2. Now

  • P1= 5.5 atm
  • V1
  • P2= ?
  • V2= V1*3 because the volume is tripled.

So:

5.5 atm*V1= P2* V1*3

Solving:

\frac{5.5 atm*V1}{3*V1}=P2

\frac{5.5 atm}{3}= P2

1.83 atm= P2

<u><em>The pressure will be 1.83 atm, one third of the initial pressure.</em></u>

7 0
3 years ago
How many liters are present in 31.998 g of O2?
kap26 [50]

Answer:

Explanation:

<em>First calculate how many moles of O2 you have. O is 16g/mole, so O2 is 32g/mole. 50/32 = 1.5625 moles. 1 mole of any gas at stp is 22.4 liters.</em>

<em>1.5625 × 22.4 = 35 liters.</em>

<em>Its been 15 years since I graduated HS, and I still remember how to do this. Do your own homework next time and in 15 years you may be able to do the same.</em>

<em>PLEASE</em><em> </em><em>THANK</em><em>,</em><em> </em><em>RATE</em><em> </em><em>AND</em><em> </em><em>FOLLOW</em><em> </em><em>ME</em><em> </em>

<em>AND</em><em> </em><em>PLEASE</em><em> </em><em>MARK</em><em> </em><em>ME</em><em> </em><em>AS</em><em> </em><em>"</em><em>BRAINLIEST</em><em>"</em><em> </em><em>ANSWER</em><em> </em>

<em>HOPE</em><em> </em><em>IT</em><em> </em><em>HELPS</em><em> </em><em>YOU</em>

8 0
3 years ago
Earth's oceans absorb carbon dioxide emitted by humans and other animals, and eventually turn it into carbonate rocks. Which "sp
Likurg_2 [28]
<span>A. biosphere
</span><span>C. geosphere
</span>E. atmosphere
<span>F. hydrosphere
</span>
Earth's oceans absorb carbon dioxide emitted by humans and other animals, and eventually turn it into carbonate rocks. Which "spheres" are part of this<span>process? Select all that apply.
</span>
NOT:
<span>B. exosphere
</span><span>D. cryosphere</span>
4 0
3 years ago
Read 2 more answers
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