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Lady bird [3.3K]
3 years ago
12

A solution containing potassium bromide is mixed with one containing lead acetate to form a solution that is 0.013 M in KBr and

0.0035 M in Pb(C2H3O2)2.Will a precipitate form in the mixed solution? If so, identify the precipitate.Express answer as a chemical formula.Please explained how you determined the answer.

Chemistry
1 answer:
galben [10]3 years ago
7 0

Answer:

No precipitate will form on mixing potassium bromide and lead acetate

Explanation:

Precipitation occurs when the ionic product of the ions exceeds the solubility product of the corresponding salt in the solution. If, this implies that the solution is over saturated and precipitation of salt will occur.Potassium bromide and lead acetate are strong electrolyte and thus completely dissociate in water.

The dissociation reaction is given as,Potassium bromide and lead acetate react to give lead bromide and potassium acetate.The concentration of lead ions is equal to the concentration of.The concentration of bromide ion is equal to the concentration of.Potassium acetate is soluble in aqueous solution. Lead bromide will precipitate out if the ionic product of lead ion and bromide ion exceeds the solubility product of .

The value of solubility product constant,, for is.The reaction quotient for is calculated as,Substitute the values.The value of the reaction quotient for is less than its solubility product, , therefore, precipitation of will not occur.

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29.5 g of mercury is heated from 32°C to 161°C, and absorbs 499.2 joules of heat in the process. Calculate the specific heat cap
Finger [1]

Answer:

c = 0.13 j/ g.°C

Explanation:

Given data:

Mass of mercury = 29.5 g

Initial temperature = 32°C

Final temperature = 161°C

Heat absorbed = 499.2 j

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Q = m.c. ΔT

ΔT  = T2 - T1

ΔT  = 161°C - 32°C

ΔT  = 129 °C

Q = m.c. ΔT

c = Q / m. ΔT

c = 499.2 j / 29.5 g. 129 °C

c =  499.2 j / 3805.5 g. °C

c = 0.13 j/ g.°C

5 0
3 years ago
Suppose you have a spherical balloon filled with air at room temperature and 1.0 atm pressure; its radius is 17 cm. You take the
Sladkaya [172]
<span>Answer: 17.8 cm
</span>

<span>Explanation:
</span>

<span>1) Since temperature is constant, you use Boyle's law:
</span>

<span>PV = constant => P₁V₁ = P₂V₂


</span><span>=> V₁/V₂ = P₂/P₁</span>
<span>
2) Since the ballon is spherical:


</span><span>V = (4/3)π(r)³</span>
<span>
Therefore, V₁/V₂ = (r₁)³ / (r₂)³
</span>

<span>3) Replacing in the equation V₁/V₂ = P₂/P₁:


</span><span><span>(r₁)³ / (r₂)³ </span>= P₂/P₁</span>
<span>
And you can solve for r₂: (r₂)³ = (P₁/P₂) x (r₁)³


</span>(r₂)³ = (1.0 atm / 0.87 atm) x (17 cm)³ = 5,647.13 cm³
<span>
r₂ = 17.8 cm</span>

4 0
3 years ago
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Leni [432]

Answer:

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Explanation:

Take R =8.314J/mol/K

V=200dm3

PV=nRT

P=0.025×8.314×623/200

=0.65mmHg

7 0
1 year ago
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umka21 [38]
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Answer:

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Explanation:

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