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levacccp [35]
3 years ago
11

To which third period element do these ionization values belong?

Chemistry
2 answers:
Arlecino [84]3 years ago
7 0

Answer:

Aluminum, Al

Explanation:

aleksklad [387]3 years ago
4 0

IE1 = 578 kJ/mol 


IE2 = 1820 kJ/mol 


IE3 = 2750 kJ/mol 


IE4 = 11600 kJ/mol


If the following set of successive ionization energies are your ionization values this would likely belong to Aluminum. Since there is a huge point between the third and fourth ionization energies, which designates that the atom reached noble gas configuration after the third electron was removed. The element which has 3 valence electrons in the third period is aluminum.

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Explanation:

Plant cells have a cell wall, as well as a cell membrane. In plants, the cell wall surrounds the cell membrane. This gives the plant cell its unique rectangular shape. Animal cells simply have a cell membrane, but no cell wall.

3 0
2 years ago
Identify the acid and conjugate base pair in the following equation:
Serjik [45]

Answer:

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2 years ago
How many grams of sodium metal are needed to make 29.3 grams of sodium chloride? given the reaction: 2na + cl2 → 2nacl?
Bess [88]
2Na + Cl2 → 2NaCl
46 g. ➡️ 117 g
x. ➡️ 29.3 g
x =  \frac{46 \times 29.3}{117}  \\ x = 11.5 \: g
5 0
3 years ago
Read 2 more answers
Which mechanisms is most significant in returning the ion concentrations to the resting state?
IceJOKER [234]

Answer:

Active transport by the Na+-K+ pump

Explanation:

Active transport by the Na+-K+ pump

Maintenance (and restoration) of the resting ion concentrations depends on the Na+-K+ pump. Once gated ion channels are closed, the combined action of the pump and ion leakage (particularly that of K+) establishes a resting membrane potential in a typical neuron of around âˆ'70 mV.

5 0
2 years ago
Gasoline and kerosene (specific gravity 0.820) are blended to obtain a mixture with a specific gravity of 0.770. Calculate the v
Dmitrij [34]

Answer:

The volumetric ratio is 0,71

Explanation:

Let's begin with the equation:

Db = Mb/Vb (1)

Where:

Db: Blend Density, Mb: Blend Mass and Vb: Blend Volume

And we know: Vb = Vg + Vk (2)

Where:

Vg: Gasoline Volume and Vk: Kerosene Volume

Therefore replacing (2) into (1):

Db = (Mg + Mk) / (Vg + Vk)

Db = (Dg * Vg + Dk * Vk)/(Vg + Vk) (3)

Where:

Dg: Gasoline Density and Dk: Kerosene Density

The specific gravity is defined as:

SG = Substance Density / Reference Density

Therefore:

Db = SGb * Dref\\Dg = SGg * Dref\\Dk = SGk * Dref

Where:

Dref: Reference Density

SGb: Blend Specific Gravity

SGg: Gasoline Specific Gravity (which is 0.7 approximately)

SGk: Kerosene Specific Gravity

Replacing these equations into (3) we get:

SGb * Dref = (SGg * Dref * Vg + SGk * Dref * Vk)/(Vg + Vk)

SGb * Dref = Dref * (SGg * Vg + SGk * Vk)/(Vg + Vk)

SGb = (SGg * Vg + SGk * Vk)/(Vg + Vk)

SGb * (Vg + Vk) = SGg * Vg + SGk * Vk

SGb * Vg + SGb* Vk = SGg * Vg + SGk * Vk

Replacing with the Specific Gravity data, we obtain:

0.77 * Vg + 0.77 * Vk = 0.7 * Vg + 0.82 * Vk

0.77 * Vg - 0.7 * Vg = 0.82 * Vk - 0.77 * Vk

0.07 * Vg = 0.05 * Vk

Vg/Vk = 0.05/0.07

Vg/Vk = 0.71

8 0
3 years ago
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