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Maksim231197 [3]
3 years ago
14

1. A 50 gram sample of a radioisotope decays to 12.5 grams in 14.4 seconds. What is

Chemistry
2 answers:
stiv31 [10]3 years ago
5 0

Answer:

Half life = 7.2 seconds

Explanation:

from \: decay \: equation \\ 2.303 log( \frac{n.}{nt} )  = kt \\ 2.303 log( \frac{50}{12.5} )  = k \times 14.4 \\ k =  \frac{2.303 log( \frac{50}{12.5} )}{14.4}  \\ k = 0.096288 {s}^{ - 1}  \\ since \: half \: life =  \frac{ ln(2) }{k}  \\ half \: life  =   \frac{ ln(2) }{0.096288}  \\ half \: life = 7.2 \: seconds

sp2606 [1]3 years ago
3 0
14.2 because basically you just divide it by 2 which would be half of it

That will work with all of those problems

So if it’s is a quarter of an isotope you’d divide by 4
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Answer:

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Explanation:

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6 0
3 years ago
How many atoms are in 0.909 moles of silver?
BabaBlast [244]

Answer:

5.47 x 10^23 atoms in Ag (sliver)

Explanation:

0.909 mol of Ag x 6.02 x 10^23 atoms of Ag/ 1 mol of Ag

8 0
3 years ago
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3 years ago
The combustion of a sample of butane, C4H10 (lighter fluid), produced 2.46 grams of water.
avanturin [10]

a. 0.137

b. 0.0274

c. 1.5892 g

d. 0.1781

e. 5.6992 g

<h3>Further explanation</h3>

Given

Reaction

2 C4H10 + 13O2 -------> 8CO2 + 10H2O

2.46 g of water

Required

moles and mass

Solution

a. moles of water :

2.46 g : 18 g/mol = 0.137

b. moles of butane :

= 2/10 x mol water

= 2/10 x 0.137

= 0.0274

c. mass of butane :

= 0.0274 x 58 g/mol

= 1.5892 g

d. moles of oxygen :

= 13/2 x mol butane

= 13/2 x 0.0274

= 0.1781

e. mass of oxygen :

= 0.1781 x 32 g/mol

= 5.6992 g

6 0
3 years ago
When heated a sample consisting of only CaCO3 and MgCO3 yields a mixture of CaO and Mgo. If the weight of the combined oxides is
ExtremeBDS [4]

Answer:

38.83 %  of  CaCO3

61.17 %  of  MgCO3

Explanation:

where Moles of CaCO3 is equals to x and MgCO3 is y we have that...

CaCO3 molar mass = 100.09 g / mol  = 100.09 x

MgCO3 molar mass = 84.31 g / mol  = 84.31 y

decomposition reactions :

CaCO3 ---> CaO + CO2

MgCO3 ---> MgO + CO2  

So we have that , Moles of CaO = Moles of CaCO3 = x

and Moles of MgO = Moles of MgCO3 = y

CaO molar mass = 56.08 g / mol

MgO molar mass = 40.30 g / mol

CaO = 56.08 x

 MgO = 40.30 y

"If the weight of the combined oxides is equal to 51.00% of the initial sample weight,"

total mass of MgO and CaO = 51.00 % of Total Mass of MgCO3 and CaCO3  

thus

56.08 x + 40.30 y = 0.51 ( 100.09 x + 84.31 y )

56.08 x + 40.30 y = 51.04 x + 42.99y

5.04 x = 2.7 y

y = 1.87 x    

CaCO3 % in the sample

= 100.09 x × 100 / ( 100.09 x + 84.31 y )

= 10009 x / ( 100.09 x + 84.31 × 1.87 x )

= 10009 x / ( x ( 100.09 + 157.66 ) )

= 10009 / 257.75

= 38.83 %

MgCO3 % in the sample

= 100 - 38.83  

=   61.17 %  

7 0
3 years ago
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