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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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Balance Chemical Equation for this reaction is,

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According to this eq, 22.4 L (1 moles) of Oxygen requires 44.8 L (2 mole) CH₄ for complete reaction.
So, the volume of CH₄ required to consume 0.66 L of O₂ is calculated as,

 22.4 L O₂ required to consume  =  44.8 L CH₄
0.660 L O₂ will require                =  X L of CH₄

Solving for X,
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                                                X  =  1.320 L of CH₄

Result:
            1.320 L of CH
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Water is flowing in a pipe with a mass flow rate of 100.0 lb/h (M water H,O=18.016). What is the (i) Molar flow rate of H20 (gmo
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Answer:

i) 0,7 molH20/s

ii)11,2 g O/s

iii)1,4 g H/s

Explanation:

i) To find the molar flow rate of water, we just convert the mass of water to moles of water using its molecular weight(g/mol) and changing to the proper units (lb to grames and hours to seconds):

100 \frac{lb}{h}*\frac{453,5g}{1 lb}*\frac{1molH20}{18,016g}*\frac{1h}{3600s}=0,7\frac{molH20}{s}

ii) Now we just consider the oxygen in the water stream (for 1 mole of water there is 1 mole of oxygen):

100\frac{lb}{h} *\frac{453,5g}{1lb}*\frac{1 molH20}{18,016g}*\frac{1molO}{1molH20}*\frac{16gr}{1molO}*\frac{1h}{3600s}=11,2\frac{gO}{s}

iii)Just considering the hydrogen in the stream (for 1 mole of water there is 2 moles of hydrogen):

100\frac{lb}{h} *\frac{453,5g}{1lb}*\frac{1 molH20}{18,016g}*\frac{2molH}{1molH20}*\frac{1gr}{1molH}*\frac{1h}{3600s}=1,4\frac{gH}{s}

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Answer:

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Reaction 6: Single replacement reaction

Reaction 7: Combination reaction.

Reaction 8: Combustion reaction.

Explanation:

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