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Alchen [17]
3 years ago
14

How many grams of PbBr2 will precipitate when excess CuBr2 solution is added to 77.0 mL of 0.595 M Pb(NO3)2 solution?Pb(NO3)2(aq

) + CuBr2(aq) PbBr2(s) + Cu(NO3)2(aq)
Chemistry
1 answer:
Alexxx [7]3 years ago
8 0

Answer:

16.89g of PbBr2

Explanation:

First, let us calculate the number of mole of Pb(NO3)2. This is illustrated below:

Molarity of Pb(NO3)2 = 0.595M

Volume = 77mL = 77/1000 = 0.077L

Mole =?

Molarity = mole/Volume

Mole = Molarity x Volume

Mole of Pb(NO3)2 = 0.595x0.077

Mole of Pb(NO3)2 = 0.046mol

Convert 0.046mol of Pb(NO3)2 to grams as shown below:

Molar Mass of Pb(NO3)2 =

207 + 2[ 14 + (16x3)]

= 207 + 2[14 + 48]

= 207 + 2[62] = 207 +124 = 331g/mol

Mass of Pb(NO3)2 = number of mole x molar Mass = 0.046 x 331 = 15.23g

Molar Mass of PbBr2 = 207 + (2x80) = 207 + 160 = 367g/mol

Equation for the reaction is given below:

Pb(NO3)2 + CuBr2 —> PbBr2 + Cu(NO3)2

From the equation above,

331g of Pb(NO3)2 precipitated 367g of PbBr2

Therefore, 15.23g of Pb(NO3)2 will precipitate = (15.23x367)/331 = 16.89g of PbBr2

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  • The concentration of hydronium ion is equal to the concentration of HNO₃:

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3 0
2 years ago
A 215-g sample of copper metal at some temperature is added to 26.6 g of water. The initial water temperature is 22.22 oC, and t
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The initial temperature of the copper metal was 27.38 degrees.

Explanation:

Data given:

mass of the copper metal sample = 215 gram

mass of water = 26.6 grams

Initial temperature of water = 22.22 Degrees

Final temperature of water = 24.44 degrees

Specific heat capacity of water = 0.385 J/g°C

initial temperature of copper material , Ti=?

specific heat capacity of water = 4.186 joule/gram °C

from the principle of:

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Now heat lost by metal = heat gained by water

82.77Ti - 2019.71 = 247.19

Ti = 27.38 degrees

8 0
3 years ago
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