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Contact [7]
3 years ago
8

ALUMINUM HAS THREE OUTER ELECTRONS Will the element share, gain, or lose electrons when bonding with another element? explain​

Chemistry
1 answer:
bearhunter [10]3 years ago
7 0

Answer:

aluminum will lose 3 electrons.

Explanation:

to be stable and losing 3 is easier than gaining 5.

anddd its a metal. :)

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Which substance can not be broken down by a
Tomtit [17]

Answer:  (3) silicon

Explanation:

Element is a pure substance which is composed of atoms of similar elements.It can not be decomposed into simpler constituents using chemical reactions.Example: Silicon (Si)

Compound is a pure substance which is made from atoms of different elements combined together in a fixed ratio by mass.It can be decomposed into simpler constituents using chemical reactions. Example: ethane (C_2H_6), propanone (CH_3COCH_3) and water (H_2O)

7 0
3 years ago
Suppose a compound is involved in three different reactions denoted R1, R2, and R3. Tripling the concentration of this reactant
pickupchik [31]

Answer:

The order of reaction is as follows, R1 = 1; R2 = 2; R3 = 0

Explanation:

The rate of a chemical reaction is the number of moles of reactants consumed per unit time or the number of moles of products formed per unit. the rate of a chemical reaction is affected by the concentration of reactants

The relationship between the rate of a chemical reaction and the concentration of its reactants is given by the rate law or equation.

Generally, the rate equation is given as;

Rate = k[A]ᵃ[B]ᵇ..., where k = rate constant which is independent of concentration of the reactants, [A] = concentration of reactant A, a = order of reaction A, [B] = concentration of reaction B, b = order of reaction B.

For the given reactions R1, R2 and R3

For R1; rate = 3, Concentration = 3[A]

3 = k[A]3ˣ

3¹ = k[A]3ˣ

Since rate is proportional to concentration, therefore, the order of reaction, x = 1

For R2; rate = 9, Concentration = 3[A]

9 = k[A]3ˣ

3² = k[A]3ˣ

Since rate is proportional to concentration, therefore, the order of reaction, x = 2

For R1; rate = 1, Concentration = 3[A]

1 = k[A]3ˣ

3⁰ = k[A]3ˣ

Since rate is proportional to concentration, therefore, the order of reaction, x = 0

Therefore, the order of reaction is as follows, R1 = 1; R2 = 2; R3 = 0

6 0
3 years ago
Assuming the reaction is first order in sucrose, determine the mass of sucrose that is hydrolyzed when 2.75 L of a 0.130 M sucro
Advocard [28]

Answer:125.84g

Explanation:Sucrose is dissacharides an organic compound in the class of carbonhydrate with the chemical formula C11H22O11.molar concentration is given by number of moles/Volume,this implies that moles=molar concentration ×Volume=0.130M×2.75L=0.3575moles.

Furthermore,number of moles=Mass of Sucrose/molecular Mass of Sucrose.

From it's formular C11H22O11, molecular Mass is the addition of the mass number which is 12 for C,2 for H and 16 for oxygen,O.so molecular Mass of Sucrose is (12×11)+(2×22)+(16×11)=352.

So mass =moles ×molecular mass=0.3575moles×352g/moles=125.84g

3 0
3 years ago
Why and how do ions form?
Zanzabum

Answer and Explanation:

Ions are electrically charged particles that are formed from the removing and addition of electrons. It can be a positively or negatively charged atom.

8 0
3 years ago
A sample initially contained 150 mg of radon-222. After 11.4 days only 18.75mg of the radon-222 in the sample remained. How many
beks73 [17]

<u>Answer: </u>

A sample initially contained 150 mg of radon-222. After 11.4 days only 18.75mg of the radon-222 in the sample remained where 3 half-lives have passed

<u>Explanation:</u>

Given, the initial value of the sample, A_0 = 150mg

Final value of the sample or the quantity left, A = 18.75mg

Time = 11.4 days

The amount left after first half life will be ½.

The number of half-life is calculated by the formula

\frac{A}{A_{0}}=\left(\frac{1}{2}\right)^{N}

where N is the no. of half life

Substituting the values,

\frac{18.75}{150}=\left(\frac{1}{2}\right)^{N}

\left(\frac{1}{2}\right)^{3}=\left(\frac{1}{2}\right)^{N}

On equating, we get, N = 3

Therefore, 3 half-lives have passed.

7 0
4 years ago
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