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PilotLPTM [1.2K]
3 years ago
10

What is the term for a type of reaction in which an acid and a base react to produce a salt and water?

Chemistry
1 answer:
WINSTONCH [101]3 years ago
5 0
Neutralization 
is the probable answer, which I don't know how to explain, but I know that's the answer because I learned about it a few years ago and I'm pretty smart. =]

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50 POINTS
Trava [24]
My guess is b for the question
3 0
2 years ago
Read 2 more answers
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
I MARK AS BRAINLIEST IF YOU ANSWER QUICK!
Sedaia [141]

Answer:

Earth's average albedo is about 0.3. In other words, about 30 percent of incoming solar radiation is reflected back into space and 70 percent is absorbed. A sensor aboard NASA's Terra satellite is now collecting detailed measurements of how much sunlight the earth's surface reflects back up into the atmosphere.

Explanation:

hope this helps :)

5 0
3 years ago
For a molecule of fluorous acid, the atoms are arranged as HOFO. (Note: In this oxyacid, the placement of fluorine is an excepti
Maksim231197 [3]

Answer:

0,0,0,0

Explanation:

The formal charge formula:

Formal charge= Valence electrons - lone pair electrons - 0.5 shared electrons

So:

Hydrogen: 1 elec. of valence and shares two electrons with the O

Formal charge= 1 - 0 - 0.5*2=0

Oxygen: 6 elec. of valence, 2 lone pairs and shares two electrons with the H and two with the F

Formal charge= 6 - 4 - 0.5*4=0

Fluorine: 7 elec. of valence, 6 lone pairs and shares two electrons with the O

Formal charge= 7 - 6 - 0.5*2=0

Oxygen: 6 elec. of valence, 3 lone pairs

Formal charge= 6 - 6 - 0.5*0=0

Note: the dative bond between F and the second O doesn't count as shared electrons.

8 0
3 years ago
Trick to change word equation to chemical equation​
Elden [556K]

Answer:

actually word equation has the names and we need to write molecular formula in chemical eqn. so I don't think there is sort of trick.You just need to memorise the molecular formula of chemicals.

6 0
3 years ago
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