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Aleks [24]
3 years ago
10

Which one is more metallic? A) Gold or B) Bismuth

Chemistry
1 answer:
Arlecino [84]3 years ago
6 0

Answer:

Gold

Explanation:

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What is an ion? What are the two ways in which an ion can be made?
denpristay [2]
Answer: Ions are formed by the addition of electrons to, or the removal of electrons from, neutral atoms or molecules or other ions; by combination of ions with other particles; or by rupture of a covalent bond between two atoms in such a way that both of the electrons of the bond are left in association with one of the ...
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What two molecules were condensed in an aldol condensation to produce (ch3)3cch=chcoch3?
inessss [21]
The given compound is being synthesized by condensing Acetone and Pivaldehyde (Trimethylacetaldehyde).

First Acetone is treated with Base, the base abstracts the mildly acidic proton present at alpha position to carbonyl group. The resulting specie called enolate act as a nucleophile and attacks on highly reactive aldehyde which upon dehydration yields the Aldol Product. The Reaction is as follow,

5 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
7.26 of a hydrate, Cu(NO3)2.xH2O, formed 2.4 g copper(II) oxide.
goldfiish [28.3K]

Number of moles= mass/ molar mass

Or n=m/MM

n = number of moles

m = mass

MM = molar mass

1) n CuO = 2.4g / 79.54g/mol = 0.03 mol CuO

2) n Cu(NO3)2.xH2O = 7.26 g / 205.6 = 0.035 moles of Cu(NO3)2.xH2O

3) 205.6 g

Cu = 63.5 g

N = 14g

O =16g

H= 1 g

63.5+ (14+(16*3))*2+1*2+16 =205.6 g

4) yes is 188g

5) I don’t know, I assume was 1

6 0
3 years ago
Read 2 more answers
Webb has calculated the percent composition of a compound. How can he check his result?
goldenfox [79]

by making sure they are in the lowest ratio. by adding them to see if they total 100. by checking that they are whole-number multiples. by dividing them by the molar mass.

7 0
2 years ago
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