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NNADVOKAT [17]
2 years ago
14

Base your answers to questions 74 through 76 on the data table below and on your knowledge of Earth

Chemistry
1 answer:
Serjik [45]2 years ago
6 0

Answer:

See explanation

Explanation:

Now we have, the graph attached.the stable disintegration product of C-14 is N-14.

Then;

Since the mass of C-14 originally present is 64g, at a time t= 17100 years, we will have;

N/No = (1/2)^t/t1/2

N = mass of C-14  at time t

No= mass C-14 originally present

t = time taken for  N amount of C-14 to remain

No = mass of C-14 originally present

t1/2 = half life of C-14

N/64 = (1/2)^17,100/5730

N/64 = (1/2)^3

N/64 = 1/8

8N = 64

N = 8 g

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Carbon dioxide and helium which has a higher van der waals force of attraction between molecules ​
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2 years ago
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
Alexxx [7]

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

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Mole =?

Molarity = mole/Volume

Mole = Molarity x Volume

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

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

Molar Mass of Pb(NO3)2 =

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

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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

8 0
3 years ago
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