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Vesnalui [34]
3 years ago
5

Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of th

e following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected. (a) AgCl(s) in 0.025 M NaCl (b) CaF2(s) in 0.00133 M KF (c) Ag2SO4(s) in 0.500 L of a solution containing 19.50 g of K2SO4 (d) Zn(OH)2(s) in a solution buffered at a pH of 11.45\
Chemistry
1 answer:
kirill [66]3 years ago
8 0

Answer:

Explanation:

<u>a) AgCl(s) in 0.025 M NaCl</u>

Equation:  AgCl(s) ⇄ Ag⁺ (aq) + Cl⁻ (aq)

Initial conc :    S            O               O

equili conc :    O            S                S

                  NaCl(s) ⇒ Na⁺ (aq) + Cl⁻ (aq)

Initial conc :  0.025      0           0

equili conc :     0          0.025    0.025

Therefore the concentration:  Ag⁺ = 6.4 * 10^-9 M,  Cl⁻  = 0.025 M

attached below is the detailed solution of the

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Major organic products are- (a) propan-1-ol and (b) 2-methylpropan-2-ol

Explanation:

methyl magnesium bromide gives nucleophilic addition reaction with carbonyl group. Because methyl magnesium bromide is a strong nucleophile and carbonyl group is a strong electrophilic center.

Propanal contains an aldehyde group and propanone contains a ketone group. hence they both give nucleophilic addition with methyl magnesium bromide.

Dilute acid is added to protonate the alkoxide produced during nucleophilic addition.

Reactions are shown below.

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How many moles of Hcl can be produced from 1.21 moles of zn?
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3 years ago
To aid in the prevention of tooth decay, it is recommended that drinking water contains 0.900 ppm fluoride (F-). A) How many g o
Romashka-Z-Leto [24]

Answer:

a) <u>1.740 g</u> of F- must be added to a cylindrical water reservoir

b) Grams of sodium fluoride, NaF, that contain this much fluoride:

3.84 g

Explanation:

Step 1. calculate the volume of the tank:

Volume of cylinder =

\pi  r^{2}h ,

Here r = radius of the cylinder = d/2

h = depth = 21.80m

r=\frac{d}{2}

=\frac{3.36x10^{2}}{2}

= 168 m

Volume =

=\frac{22\times 168^{2}\times 21.80}{7}

=1.93\times 10^{6} m^{3}

2.Convert ppm to g/m3 and Solve for mass of F-

1ppm = 1g/m^{3}

0.9ppm = 0.9g/m^{3}

Because both ppm and g/m3 are same quantity .

g/m^{3} =\frac{mass\ of\ F-(g)}{Volume\ m^{3}}\times 10^{6}

0.9 =\frac{mass\ of\ F-}{1.93\times 10^{6} m^{3}}\times 10^{6}

mass\ of\ F- =1.740g

mass of F- required = 1.740 g

3. Apply <u>mole concept </u>to calculate grams of sodium fluoride produced

mass of 1 mole of F2 = 38 g

mass of 1 mole of NaF = 42 g

(from periodic table calculate molar mass)

2Na+F_{2}\rightarrow 2NaF

Here 1 mole of F2 produce = 2 mole of NaF

So,

38 g  of F2 produce = 2 x 42 g of NaF

38 g of F2 produce = 84 g of NaF

1 g of F2 produce = 84/38 g of NaF

1.74 g F2 produce =

\frac {84}{38}\times 1.74

1.74 g F2 produce = 3.84 g of NaF

3.84 g of NaF is produced

3 0
3 years ago
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