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Genrish500 [490]
3 years ago
15

Identify the type of reaction and the products. Fe(NO3)2 + Na ->

Chemistry
1 answer:
Vinvika [58]3 years ago
5 0

Answer:

Substitution and Na(No3) + Fe

Explanation:

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For the reaction between aqueous solutions of acetic acid (CH3COOH) and barium hydroxide, Ba(OH)2, 1. Write the balanced molecul
Crazy boy [7]

Answer:

2CH_3COOH(aq)+Ba(OH)_2(aq)\rightarrow Ba(CH_3COO)_2(aq)+2H_2O(l)

Explanation:

When acetic acid solution and barium hydroxide solution react together to give an aqueous solution of barium acetate and water

The balanced chemical reaction will be given by

2CH_3COOH(aq)+Ba(OH)_2(aq)\rightarrow Ba(CH_3COO)_2(aq)+2H_2O(l)

8 0
3 years ago
The theory of plate tectonics describes the movement of continental plates across the Earth's surface as if they were riding on
emmainna [20.7K]

Answer:

B and E

Explanation:

These two options support the theory of plate tectonics.

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3 years ago
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Formed when an amine is combined with a carboxyl group
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Hello.

The answer is C.Amine

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7 0
3 years ago
Balancing Chemical Equation<br> C=H2=CH4
svlad2 [7]

Answer:

C + 2H2 ⇒ CH4

Explanation:

In order to balance a chemical equation you need to make sure that the number of atoms on both sides are equal

C + H2 = CH4

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Products:

C = 1

H = 4

H2 = 2 × 2 = 4

C + 2H2 ⇒ CH4

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4 0
2 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
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