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4 years ago

15

The outermost layer of Earth is called the mantle. Please select the best answer from the choices provided T F

2 answers:

frozen [14]4 years ago

7 0

I think it would be false

Elden [556K]4 years ago

5 0

The answer is false.

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When group 2A element from ion they

Anarel [89]

Well, knowing that group 2A elements are metals, and metals lose electrons to gain a full electron shell, we can eliminate B and D. Knowing that these are 2 electrons, that the group 2A element provides to another atom, to form an ion.

The solution is C. Lose 2 electrons.

4 0

4 years ago

What would be the correct answers

horsena [70]

Answer:

A chemical bond is formed and energy is absorbed I do believe

Explanation:

4 0

3 years ago

At every depth, soil composition remains the same.<br> O<br> True<br> O False

pickupchik [31]

Answer:

false

Explanation:

Hope the answer was usefull for us

7 0

4 years ago

Read 2 more answers

When aqueous solutions of Li3PO4 and MgSO4 are mixed, Mg3(PO4)2 is a precipitate

Lesechka [4]

Answer:

Li2SO4

Explanation:

Balance the reaction of Li3PO4 + MgSO4 = Li2SO4 + Mg3(PO4)2 using this chemical equation balancer!

8 0

3 years ago

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