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3 years ago

Which of the following is possible for aluminum to have a full outer shell?

Chemistry

2 answers:

lesya [120]3 years ago

8 0

B. Three are lost creating Al+3. Because atoms have and equal amount of Protons and Electons when neutral losing 3 electrons will create a positive Al ion Al+3

liubo4ka [24]3 years ago

4 0

The answer is; B.

This is called an aluminum cation. Aluminum forms bonds with anions in which it can donate its three valence electrons so it can achieve stable electron configuration. Stable electron configuration is having 8 electron in the outer shell. By donating its 3 outer valence electrons, the aluminium will have the more stable 2.8 electron configuration. The +3 means it has lost 3 electrons hence has 3 extra protons (that are not matched by an electron) in its nucleus (also indicating its valency).

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A 46.9 gram sample of a substance has a volume of about 3.5 centimeters3. It is solid at a room temperature of 23ºC. Out of the

horrorfan [7]

Answer : (C) Hafnium is the most likely identity of the given substance.

Solution : Given,

Mass of given substance (m) = 46.9 g

Volume of given substance (V) = 3.5 $Cm^{3}$

First, find the Density of given substance.

Formula used :

$Density=\frac{\text{Mass of given substance}}{\text{Voume of given substance}}$

Now,put all the values in this formula, we get

$Density=\frac{46.9 g}{3.5 Cm^{3} }$ = 13.4 g/ $Cm^{3}$

So, we conclude that the density of given substance (13.4 g/ $Cm^{3}$ ) is approximately equal to the density of Mercury and Hafnium (13.53 and 13.31 g/ $Cm^{3}$ respectively).

According to the question the substance is solid at room temperature but Mercury is liquid at room temperature. So, Mercury is not identical to the given substance.

Another element i.e, Hafnium is the element whose density is approximately equal to the given substance and also solid at room temperature. And we know that the melting point of solid is high.

So, Hafnium is the most likely element which is the identity of the given substance.

3 0

2 years ago

Read 2 more answers

An unknown solution has a pH of 7.2. Which of these chemicals is likely to cause the greatest decrease in the pH of the solution

prohojiy [21]

Answer:

HNO₃.

Explanation:

It is known that acids decrease the pH of the solution, while bases increase the pH of the solution.

So, HF and HNO₃ decrease the pH of the solution as they produce H⁺ in the solution.

While, KOH and NH₃ increase the pH of the solution as they produce OH⁻ in the solution.

HNO₃ will decrease the pH of the solution greater than HF.

Because HNO₃ is strong acid that decomposes completely to produce H⁺ more than the same concentration of HF that is a weak acid which does not decomposed completely to produce H⁺.

7 0

3 years ago

The increasing concentration of a chemical as it moves through the links in a food chain is called ________.

algol13

Answer: Biological Magnification

Explanation:

Organisms acquire toxic substance from the environment along with nutrients and water. Some of the toxins are metabolized and excreted, but others accumulate in specific tissues, especially fat. One of the reasons accumulated toxins are particularly harmful is that the become more concentrated in successive trophic level of the food web, this is the process of biological magnification.

Magnification occurs because the biomass at any given level is produced from a must larger biomass ingested from the level below. Thus the top-level carnivores tend to be the organism most severely affected by toxic compounds in the environment.

Examples of toxins that demonstrate biology magnification are chlorinated hydrocarbons, and many pesticides.

4 0

3 years ago

Find percent yield:

saveliy_v [14]

Answer: The percent yield of the reaction is 91.8 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For $B_5H_9$ :

Given mass of $B_5H_9$ = 4.0 g

Molar mass of $B_5H_9$ = 63.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol$

For oxygen gas:

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol$

The chemical equation for the reaction of $B_5H_9$ and oxygen gas follows:

$2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O$

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of $B_2H_5$

So, 0.3125 moles of oxygen gas will react with = $\frac{2}{12}\times 0.3125=0.052mol$ of $B_2H_5$

As, given amount of $B_2H_5$ is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of $B_2O_3$

So, 0.3125 moles of oxygen gas will produce = $\frac{5}{12}\times 0.3125=0.130moles$ of water

Now, calculating the mass of $B_2O_3$ from equation 1, we get:

Molar mass of $B_2O_3$ = 69.93 g/mol

Moles of $B_2O_3$ = 0.130 moles

Putting values in equation 1, we get:

$0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g$

To calculate the percentage yield of $B_2O_3$ , we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of $B_2O_3$ = 8.32 g

Theoretical yield of $B_2O_3$ = 9.052 g

Putting values in above equation, we get:

$\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%$

Hence, the percent yield of the reaction is 91.8 %

6 0

3 years ago

What type of rays result in shorter days? (science)

OLEGan [10]

Answer:

verical rays

Explanation:

4 0

2 years ago