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Svetlanka [38]
3 years ago
15

I NEED HELP KNOW LAST QUESTION ON MY TEST

Chemistry
2 answers:
RoseWind [281]3 years ago
8 0

Answer:  

Explanation: In my opinion  law is stated as fact because it is passes through and approved by many peoples but theory is an idea someone has and is not approved and is set on a base of beliefs.

solniwko [45]3 years ago
7 0
ANSWER: a law is proven through multiple trials, experiments, and approved by many scientists. a theory is not proven, as it is only an idea or hypothesis of a situation
You might be interested in
Predict the formula of the ionic compound that would result from combining the monatomic ions formed by each pair of elements. M
skad [1K]

Answer:

E - Be and O

A - Mg and N

E - Li and Br

F - Ba and Cl

B - Rb and O

Explanation:

Be and O

Be is a metal that loses 2 e⁻ to form Be²⁺ and O is a nonmetal that gains 2 e⁻ to form O²⁻. For the ionic compound to be neutral, it must have the form BeO (E-MX).

Mg and N

Mg is a metal that loses 2 e⁻ to form Mg²⁺ and N is a nonmetal that gains 3 e⁻ to form O³⁻. For the ionic compound to be neutral, it must have the form Mg₃N₂ (A-M₃X₂).

Li and Br

Li is a metal that loses 1 e⁻ to form Li⁺ and Br is a nonmetal that gains 1 e⁻ to form Br⁻. For the ionic compound to be neutral, it must have the form LiBr (E-MX).

Ba and Cl

Ba is a metal that loses 2 e⁻ to form Ba²⁺ and Cl is a nonmetal that gains 1 e⁻ to form Cl⁻. For the ionic compound to be neutral, it must have the form BaCl₂ (F-MX₂).

Rb and O

Rb is a metal that loses 1 e⁻ to form Rb⁺ and O is a nonmetal that gains 2 e⁻ to form O²⁻. For the ionic compound to be neutral, it must have the form Rb₂O (B-M₂X).

4 0
3 years ago
Can some one help me with this soon as possible please will give brainliest
lana66690 [7]

1. C

2. C

3. In elastic deformation, the deformed body returns to its original shape and size after the stresses are gone. In ductile deformation, there is a permanent change in the shape and size but no fracturing occurs. In brittle deformation, the body fractures after the strength is above the limit. 

4. Normal faults are faults where the hanging wall moves in a downward force based on the footwall; they are formed from tensional stresses and the stretching of the crust. Reverse faults are the opposite and the hanging wall moves in an upward force based on the footwall; they are formed by compressional stresses and the contraction of the crust.  Thrust faults are low-angle reverse faults where the hanging wall moves in an upward force based on the footwall; they are formed in the same way as reverse faults. Last, Strike-slip faults are faults where the movement is parallel to the crust of the fault; they are caused by an immense shear stress. 


I hope this helped :D

7 0
3 years ago
Given K = 3.61 at 45°C for the reaction A(g) + B(g) equilibrium reaction arrow C(g) and K = 7.19 at 45°C for the reaction 2 A(g)
Firlakuza [10]

Answer:

K = 0.55

Kp = 0.55

mol fraction B = 0.27

Explanation:

We need to calculate the equilibrium constant for the reaction:

C(g) + D(g) ⇄ 2B(g)              K₁= ?                       (1)

and we are given the following equilibria with their respective Ks

A(g) + B(g) ⇄ C(g)                 K₂= 3.61                 (2)

2 A(g) + D(g)  ⇄ C(g)             K₃= 7.19                 (3)

all at 45 ºC.

What we need to do to solve this question is to manipulate equations (2) and (3)  algebraically  to get our desired equilibrium (1).

We are allowed to reverse  reactions, in that case we take the reciprocal of K as our new K' ; we can also  add two equilibria together, and the new equilibrium constant will be the product of their respective Ks .

Finally if we multiply by a number then we raise the old constant to that factor to get the new equilibrium constant.

With all this  in mind, lets try to solve our question.

Notice A is not in our goal equilibrium (3)  and we want D as a reactant . That  suggests we should reverse the first equilibria and multiply it by two since we have 2 moles of B  as product in our  equilibrium (1) . Finally we would add (2) and (3) to get  (1) which is our final  goal.

2C(g)             ⇄  2A(g) + 2B(g)  K₂´= ( 1/ 3.61 )²  

                                   ₊

2 A(g) + D(g)  ⇄     C(g)               K₃ = 7.19  

<u>                                                                                    </u>

C(g) + D(g)     ⇄    2B(g)       K₁ = ( 1/ 3.61 )²   x  7.19

                                             K₁ = 0.55

Kp is the same as K = 0.55 since the equilibrium constant expression only involves  gases.

To compute the last part lets setup the following mnemonic  ICE table to determine the quantities at equilibrium:

pressure (atm)        C             D           B

initial                     1.64          1.64         0

change                    -x             -x        +2x

equilibrium          1.64-x         1.64-       2x

Thus since

Kp =0.55 = pB²/ (pC x pD) = (2x)²/ (1.64 -x)²  where p= partial pressure

Taking square root to both sides of the equation we have

√0.55 = 2x/(1.64 - x)

solving for x  we obtain a value of 0.44 atm.

Thus at equilibrium we have:

(1.64 - 0.44) atm = 1.20 atm = pC = p D

2(0.44) = 0.88 = pB

mole fraction of B = partial pressure of B divided into the total gas pressure:

X(B) = 0.88 / ( 1.20 + 1.20 + 0.88 ) = 0.27

8 0
3 years ago
Which of the following arrangements of the visible colors of light is in the correct order of increasing energy?
IrinaK [193]
Violet, indigo, blue, green, yellow, orange, red
4 0
2 years ago
Someone answer this please...
boyakko [2]

Answer:

120 mol Mg

General Formulas and Concepts:

<u>Chemistry - Stoichiometry</u>

  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

120 moles H₂

<u>Step 2: Identify Conversions</u>

RxN:   3 mol H₂ = 3 mol Mg

<u>Step 3: Stoichiometry</u>

<u />120 \ mol \ H_2(\frac{3 \ mol \ Mg}{3 \ mol \ H_2} ) = 120 mol Mg

7 0
3 years ago
Read 2 more answers
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