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Nuetrik [128]
1 year ago
13

2. what volume of 0.0150-m hcl solution is required to titrate 150. ml of a 0.0100-m ca(oh)2 solution? (10 points)

Chemistry
1 answer:
laila [671]1 year ago
3 0

100ml volume of  0.0150m hcl solution is requires to titrate 150ml of a 0.0100m caoh2 solution.

Dilution is a solution of decreasing the concentration of a solute in the solution by adding more solvent to the solution. We can use the expression for dilute formula,

    C1 V1 =C2 V2

where C1 is the initial concentration,C2 is the final concentration,V1 is the initial volume and V2 is the final volume. Here given, volume of 0.0150M(C1)  HCL solution is required to titrate 150ml(V2) of a 0.0100M(C1)  Caoh2 solution.

While diluting a solution from a high concentration substance to a low concentration substance we always use the formula of dilution.so, putting all value give  in the expression we get the volume of the final concentration.

          V1=  C2 V2/ C1

               = 0.0100m . 150ml /0.0150M

               = 100ml

The volume of the hcl solution is 100ml.

To learn more about dilution formula please visit:

brainly.com/question/7208939

#SPJ4

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Based on the sign of E cell, classify these reactions as spontaneous or non spontaneous as written.? assume standard conditions.
sammy [17]
A electrochemical reaction is said to be spontaneous, if E^{0} cell is positive. 

Answer 1:
Consider reaction: <span>Ni^2+ (aq) + S^2- (aq) ----> + Ni (s) + S (s) 

The cell representation of above reaction is given by;
    </span>S^{2-}/S //  Ni^{2+}/Ni

Hence, E^{0}cell =  E^{0} Ni^{2+/Ni} -  E^{0} S/S^{2-}
we know that, {E^{0} Ni^{2+}/Ni  = -0.25 v
and {E^{0} S/ S^{2-}  = -0.47 v

Therefore, E^{0} cell = - 0.25 - (-0.47) = 0.22 v

Since,  E^{0} cell is positive, hence cell reaction is spontaneous
.....................................................................................................................

Answer 2: 
Consider reaction: <span>Pb^2+ (aq) +H2 (g) ----> Pb (s) +2H^+ (aq)
</span>
The cell representation of above reaction is given by;
    H_{2} /  H^{+} //  Pb^{2+} /Pb

Hence, E^{0}cell = E^{0} Pb/Pb^{2+} - E^{0} H_{2}/H^{+}
we know that, {E^{0} Pb^{2+}/Pb = -0.126 v
and {E^{0} H_{2}/ H^{+} = -0 v

Therefore, E^{0} cell = - 0.126 - 0 = -0.126 v

Since,  E^{0} cell is negative, hence cell reaction is non-spontaneous.

....................................................................................................................

Answer 3: 
Consider reaction: <span>2Ag^+ (aq) + Cr(s) ---> 2 Ag (s) +Cr^2+ (aq)
</span>
The cell representation of above reaction is given by;
    Cr/Cr^{2+} // Ag^{+}/Ag

Hence, E^{0}cell = E^{0} Ag^{+}/Ag - E^{0} Cr/Cr^{2+}
we know that, {E^{0} Ag^{+}/Ag = -0.22 v
and {E^{0} Cr/ Cr^{2+} = -0.913 v

Therefore, E^{0} cell = - 0.22 - (-0.913) = 0.693 v

Since,  E^{0} cell is positive, hence cell reaction is spontaneous
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When ethyl acetoacetate (ch3coch2co2ch2ch3) is treated with one equivalent of ch3mgbr, a gas is evolved from the reaction mixtur
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The protons of methylene group between the two carbonyl groups in ethylacetoacetate are acidic in nature. When compounds containing such acidic protons are treated with bases the loose proton and form enolates.

In this particular example when ethylacetoacetate is reacted with methyl magnesium bromide, the methyl group abstracts the acidic proton and converts into methane gas. The enolate when hydrolyzed is again converted into ethylacetoacetate as shown below,

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