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zhuklara [117]
3 years ago
9

Identify the substances that will appear in the equilibrium constant expression for the equation: 2Ag+(aq)+Zn(s)<->Zn2+(aq

)+2Ag(s)
Chemistry
2 answers:
Simora [160]3 years ago
8 0

Hey there!


We Know that:



 2 Ag⁺(aq) + Zn(s) <-> Zn²⁺(aq)+2 Ag(s)


The equilibrium expression for the reaction is:



Kc =  [ Zn⁺² ]  /  [Ag⁺ ]²


Hope that helps!

astraxan [27]3 years ago
6 0

Answer : The expression for equilibrium constant for the given reaction is,

K_c=\frac{[Zn^{2+}]}{[Ag^+]^2}

Explanation :

The given balanced chemical reaction is,

2Ag^+(aq)+Zn(s)\rightleftharpoons Zn^{2+}(aq)+2Ag(s)

In the expression of equilibrium constant, only aqueous and gaseous substances are included. pure liquids and solids are not included.

Thus, the expression for equilibrium constant for the given reaction is,

K_c=\frac{[Zn^{2+}]}{[Ag^+]^2}

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How many mL of a stock 50% (w/v) KNO3 solution are needed to prepare 250 mL of a 20% (w/v) KNO3 solution?
Andre45 [30]

Answer:

100ml of a stock 50% KNO3 solutions are needed to prepare 250ml of a 20% KNO3 solution.

Explanation:

In the given question it is mentioned that

     S1=50%

      V2=250ml

      S2= 20%

We all know that

                     V1S1=V2S2

                     ∴V1=  V2×S2÷S1

                     ∴V1=  V2S2×1/S1

                      ∴V1= 250×20÷50

                       ∴V1= 100ml

 

6 0
3 years ago
A. Describe the molecule chlorine dioxide, CIO in terms of three possible resonance structures.
cupoosta [38]

Answer:

See explanation

Explanation:

The compound ClO2 has 19 valence electrons.  ClO2 is a bent molecule with tetrahedral electron pair geometry but has two  lone pairs of electrons. This is indicated by the presence of four electron pairs on the outermost shell of the central atom.

The molecule has an odd number of valence electrons, hence, it is generally regarded as a paramagnetic radical. None of the proposed Lewis structures for the molecule is satisfactory because none of them obeys the octet rule.

From the images attached, one can easily see that the electron dots around the oxygen and chlorine atoms does not satisfy the octet rule in all the resonance structures shown.

5 0
3 years ago
A 25.0-mL sample of 0.150 M hydrocyanic acid is titrated with a 0.150 M NaOH solution. The Ka of hydrocyanic acid is 4.9 × 10-10
lara [203]

Answer:

The pOH = 1.83

Explanation:

Step 1: Data given

volume of the sample = 25.0 mL

Molarity of hydrocyanic acid = 0.150 M

Molarity of NaOH = 0.150 M

Ka of hydrocyanic acid = 4.9 * 10^-10

Step 2: The balanced equation

HCN + NaOH → NaCN + H2O

Step 3: Calculate the number of moles hydrocyanic acid (HCN)

Moles HCN = molarity * volume

Moles HCN = 0.150 M * 0.0250 L

Moles HCN = 0.00375 moles

Step 3: Calculate moles NaOH

Moles NaOH = 0.150 M * 0.0305 L

Moles NaOH = 0.004575 moles

Step 4: Calculate the limiting reactant

0.00375 moles HCN will react with 0.004575 moles NaOH

HCN is the limiting reactant. It will completely be reacted. There will react 0.00375 moles NaOH. There will remain 0.004575 - 0.00375 = 0.000825 moles NaOH

Step 5: Calculate molarity of NaOH

Molarity NaOH = moles NaOH / volume

Molarity NaOH = 0.000825 moles / 0.0555 L

Molarity NaOH = 0.0149 M

Step 6: Calculate pOH

pOH = -log [OH-]

pOH = -log (0.0149)

pOH = 1.83

The pOH = 1.83

6 0
3 years ago
The maximum number of electrons in a single d-subshell is:
Fantom [35]
<span>The maximum number of electrons in a single d-subshell is:
10</span>
8 0
3 years ago
Read 2 more answers
Help pls , question is in picture
Nata [24]

Explanation:

IM PRETTY SURE IT IS D !! IF ITS WRONG IM SORRY THAT WHAT

I GOT

8 0
3 years ago
Read 2 more answers
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