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Vadim26 [7]
2 years ago
12

In chemistry we did a lab examining iron nails corroding in agar. One of the nails was wrapped in zinc, and the nail did not rus

t. A whitish substance appeared around the nail, what is the whitish substance?
Chemistry
1 answer:
forsale [732]2 years ago
4 0

Answer:

Zn(OH)₂

<em>Explanation: </em>

Zn is above Fe in the activity series, so it is more readily oxidized:

Zn(s) ⟶ Zn²⁺(aq) + 2e⁻

The zinc ions go into solution.

The electrons travel from the Zn to the surface of the iron nail, where they reduce the water to hydrogen and hydroxide ions.

2H₂O(ℓ) + 2e⁻ ⟶ H₂(g) + 2OH⁻(aq)

Neither the zinc ions nor the hydroxide ions can move far through the gel.

Their concentration builds up around the hail. The ions find each other and form a precipitate.

Zn²⁺(aq) + 2OH⁻(aq) ⟶  Zn(OH)₂(s)

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What is the molarity of a solution that contains 0.500 mole of kno3 dissolved in 0.500-liter of solution?
belka [17]

Answer : The molarity of solution is, 1.00 M

Explanation : Given,

Moles of KNO_3 = 0.500 mol

Volume of solution = 0.500 L

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

\text{Molarity}=\frac{\text{Moles of }KNO_3}{\text{Volume of solution (in L)}}

Now put all the given values in this formula, we get:

\text{Molarity}=\frac{0.500mol}{0.500L}=1.00mole/L=1.00M

Therefore, the molarity of solution is, 1.00 M

7 0
2 years ago
If PbI2(s) is dissolved in 1.0MNaI(aq) , is the maximum possible concentration of Pb2+(aq) in the solution greater than, less th
fredd [130]

Answer:

\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

Less than the concentration of Pb2+(aq) in the solution in part ( a )

Explanation:

From the question:

A)

We assume that s to be  the solubility of PbI₂.

The equation of the reaction is given as :

PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq); Ksp = 7 × 10⁻⁹

 [Pb²⁺] =   s

Then [I⁻] = 2s

K_{sp} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (2s)^{2} =  4s^{3}\\s^{3} = \dfrac{K_{sp}}{4}\\\\s =\mathbf{ \sqrt [3]{\dfrac{K_{sp}}{4}}}\\\\\text{The mathematical expressionthat can be used to determine the value of  }\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

B)

The Concentration of Pb²⁺  in water is calculated as :

\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

\mathbf{s =\sqrt [3]{\dfrac{7*10^{-9}}{4}}}

\mathbf{s} =\sqrt[3]{1.75*10^{-9}}

\mathbf{s} =\mathbf{1.21*10^{-3}  \ mol/L }

The Concentration of Pb²⁺  in 1.0 mol·L⁻¹ NaI

\mathbf{PbCl{_2}}  \leftrightharpoons    \ \ \ \ \ \ \  \mathbf{Pb^{2+}}   \ \ \ \  \ +   \ \  \ \ \ \ \ \mathbf{2 I^-}

                             \ \ \ \ \ \ \  \ \   \ \  \ \ \ \ \ \ \  \mathbf0}   \ \ \ \  \ \ \ \ \ \   \ \ \ \ \ \mathbf{1.0}

                            \ \ \ \ \ \ \ \ \ \ \ \ \ \    \ \ \ \ \  \mathbf{+x}   \ \ \ \  \    \ \  \ \ \ \ \ \mathbf{+2x}

                            \ \ \ \ \ \ \ \ \ \ \ \ \ \    \ \ \ \ \  \mathbf{+x}   \ \ \ \  \    \ \  \ \ \ \ \ \mathbf{1.0+2x}

The equilibrium constant:

K_{sp} =[Pb^{2+}}][I^-]^2 \\ \\ K_{sp} = s*(1.0*2s)^2 =7*1.0^{-9} \\ \\ s = 7*10^{-9} \ \  m/L

It is now clear that maximum possible concentration of Pb²⁺ in the solution is less than that in the solution in part (A). This happens due to the  common ion effect. The added iodide ion forces the position of equilibrium to shift to the left, reducing the concentration of Pb²⁺.

3 0
2 years ago
Consider a saturated solution formed when 17.5 g of a solute dissolve in 28.3 g of a solvent, giving a total solution volume of
STALIN [3.7K]

Answer:

a) 38.2 % mass

b) 61.8 g solute/100 g solvent

c) 1.65 g/mL

Explanation:

Given the data:

mass of solute = 17.5 g

mass of solvent= 28.3 g

total solution volume= 27.8 mL

a)- mass percent= mass of solute/mass of solution x 100

mass of solution = mass solute + mass solvent = 17.5 g + 28.3 g = 45.8 g

mass % = 17.5 g/45.8 g x 100 = 38.2 % mass

b)- solubility = grams of solute/ 100 g solvent

                    = 17.5 g x (100 g /28.3 g solvent) = 61.8 g solute/100 g solvent  

c)- density = massof solution/total volumesolution  = 45.8 g/27.8 mL = 1.65 g/mL

7 0
2 years ago
Functional group and bond hybridization of vanillin
lana66690 [7]

Vanillin is the common name for 4-hydroxy-3-methoxy-benzaldehyde.

See attached figure for the structure.

Vanillin have 3 functional groups:

1) aldehyde group:  R-HC=O, in which the carbon is double bonded to oxygen

2) phenolic hydroxide group: R-OH, were the hydroxyl group is bounded to a carbon from the benzene ring

3) ether group: R-O-R, were hydrogen is bounded through sigma bonds to carbons

Now for the hybridization we have:

The carbon atoms involved in the benzene ring and the red carbon atom (from the aldehyde group) have a <u>sp²</u> hybridization because they are involved in double bonds.

The carbon atom from the methoxy group (R-O-CH₃) and the blue oxygen's have a  <u>sp³</u>  hybridization because they are involved only in single bonds.

7 0
2 years ago
Q 11.20: What is the product of the reaction between t-BuCl and MeOH? A : t-BuOH B : MeOCl C : t-BuOMe D : (CH3)2CCH2
Sholpan [36]

Answer:

C : t-BuOMe

Explanation:

The tert -butanol is a tertiary alcohol and when chloride ion attacks the carbocation, it forms t-BuCl.

The reaction of tert-butyl chloride or t-BuCl ((CH3)3C−Cl) with methanol and MeOH (CH3−OH) gives the product tert-Butyl methyl ether or t-BuOMe (CH3)3C−OCH3:

                   (CH3)3C−Cl + CH3−OH => (CH3)3C−OCH3 + HCl

Hence, the correct asnwer is C : t-BuOMe

5 0
2 years ago
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