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VMariaS [17]
3 years ago
8

Help asap plzzzz 30 points

Chemistry
1 answer:
Rus_ich [418]3 years ago
4 0

Answer:

Do you need your 0.00 2 oz

Explanation:

Because that is the right answer

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What is the scientific term for rocks formed from magma
gogolik [260]
The answer is Igneous rock<span>. ^-^

</span>
7 0
3 years ago
The two naturally occurring isotopes of bromine are
ch4aika [34]

Answer:

The decreasing order of masses of for different BrCl molecules :

= M_2

= 113.887 amu ,115.884 amu ,115.885 amu ,117.882 amu

Explanation:

The two naturally occurring isotopes of bromine are

81-Br (80.916 amu, 49.31%)   79-Br (78.918 amu, 50.69%).

The two naturally occurring isotopes of chlorine are

37-Cl (36.966 amu, 24.23%) and   35- Cl (34.969 amu, 75.77%).

Molar mass of BrCl molecule composed of  81-Br and  37-Cl.

Mass of 81-Br = 80.916 amu

Mass of  37-Cl = 36.966 amu

Molar mass of this BrCl molecule,M_1 = 80.916 amu + 36.966 amu

= 117.882 amu

Molar mass of BrCl molecule composed of  79-Br and  35-Cl.

Mass of 79-Br = 78.918 amu

Mass of  35-Cl = 34.969 amu

Molar mass of this BrCl  moleculeM_2 = 78.918 amu + 34.969 amu

= 113.887 amu

Molar mass of BrCl molecule composed of  81-Br and  35-Cl.

Mass of 81-Br = 80.916 amu

Mass of  35-Cl = 34.969 amu

Molar mass of this BrCl molecule,M_3 = 80.916 amu + 34.969 amu

= 115.885 amu

Molar mass of BrCl molecule composed of  79-Br and  37-Cl.

Mass of 79-Br = 78.918 amu

Mass of  37-Cl = 36.966 amu

Molar mass of this BrCl molecule,M_4 = 78.918 amu + 36.966 amu

= 115.884 amu

The decreasing order of masses of for different BrCl molecules :

= M_2

= 113.887 amu < 115.884 amu

5 0
3 years ago
What is the coefficient of silver in the final, balanced equation for this reaction?
katovenus [111]

Answer: 3

Explanation:

An oxide-reduction reaction or, simply, redox reaction, is a <u>chemical reaction in which one or more electrons are transferred between the reactants</u>, causing a change in their oxidation states, which is the hypothetical electric charge that the atom would have if all its links with different elements were 100% ionic.

For there to be a reduction-oxidation reaction, in the system there must be an element that yields electrons and another that accepts them:

  • The oxidizing agent picks up electrons and remains with a state of oxidation inferior to that which it had, that is, it is reduced.
  • The reducing agent supplies electrons from its chemical structure to the medium, increasing its oxidation state, ie, being oxidized.

To balance a redox equation you must <u>identify the elements that are oxidized and reduced and the amount of electrons that they release or capture, respectively. </u>

In the reaction that arises in the question the silver (Ag) is reduced <u>because it decreases its oxidation state from +1 to 0</u> and the aluminum (Al) is oxidized because <u>its oxidation state increases from 0 to +3</u>, releasing 3 electrons (e⁻). Then we can raise two half-reactions:

Ag⁺ + e⁻ → Ag⁰

Al⁰ → Al⁺³ + 3e⁻

In order to obtain the balanced equation, we must multiply the first half-reaction by 3 so that, when both half-reactions are added, the electrons are canceled. In this way:

(Ag⁺ + e⁻ → Ag⁰ ) x3

Al⁰ → Al⁺³ + 3e⁻               +

-------------------------------------

3Ag⁺ + Al⁰ → 3Ag⁰ + Al⁺³

So, the coefficient of silver in the final balanced equation is 3.

5 0
4 years ago
A solution is a ____________a liquid____________ mixture.
Tamiku [17]

it is a solute and a solvent


8 0
4 years ago
Read 2 more answers
Phosphorous trichloride and phosphorous pentachloride equilibrate in the presence of molecular chlorine according to the reactio
umka21 [38]

Answer : The value of K_p at this temperature is 66.7

Explanation : Given,

Pressure of PCl_3 at equilibrium = 0.348 atm

Pressure of Cl_2 at equilibrium = 0.441 atm

Pressure of PCl_5 at equilibrium = 10.24 atm

The balanced equilibrium reaction is,

PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{PCl_5})}{(p_{PCl_3})(p_{Cl_2})}

Now put all the values in this expression, we get :

K_p=\frac{(10.24)}{(0.348)(0.441)}

K_p=66.7

Therefore, the value of K_p at this temperature is 66.7

4 0
3 years ago
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