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Dafna1 [17]
2 years ago
6

131i has a half-life of 8.04 days. assuming you start with a 1.53 mg sample of 131i, how many mg will remain after 13.0 days ___

_______?
a.0.835
b.0.268
c.0.422
d.0.440
e.0.499
Chemistry
1 answer:
inn [45]2 years ago
8 0
For this problem we can use half-life formula and radioactive decay formula.

Half-life formula,
t1/2 = ln 2 / λ

where, t1/2 is half-life and λ is radioactive decay constant.
t1/2 = 8.04 days

Hence,         
8.04 days    = ln 2 / λ                         
λ   = ln 2 / 8.04 days

Radioactive decay law,
Nt = No e∧(-λt)

where, Nt is amount of compound at t time, No is amount of compound at  t = 0 time, t is time taken to decay and λ is radioactive decay constant.

Nt = ?
No = 1.53 mg
λ   = ln 2 / 8.04 days = 0.693 / 8.04 days
t    = 13.0 days 

By substituting,
Nt = 1.53 mg e∧((-0.693/8.04 days) x 13.0 days))
Nt = 0.4989 mg = 0.0.499 mg

Hence, mass of remaining sample after 13.0 days = 0.499 mg

The answer is "e"

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5 0
2 years ago
Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce ammonia. N 2 ( g ) + 3 H 2 ( g )
Kisachek [45]

Answer:

After complete reaction, 0.280 moles of ammonia are produced

Explanation:

Step 1: Data given

Number of moles N2 = 0.140 moles

Number of moles H2 = 0.434 moles

Step 2: The balanced equation

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Step 3: Calculate the limiting reactant

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

N2 is the limiting reactant. It will completely be consumed (0.140 moles).

H2 is in excess. There will react 3*0.140 = 0.420 moles

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Step 4: Calculate moles NH3

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After complete reaction, 0.280 moles of ammonia are produced

5 0
3 years ago
What is the coefficient for hydrogen in the balanced equation for the reaction of solid molybdenum(iv) oxide with gaseous hydrog
anastassius [24]

The  coefficient  for hydrogen  in  the  balanced  equation   of solid  molybdenum(iV)  oxide  with gaseous  hydrogen  is  2


 Explanation

Coefficient   is defined  to  as  a number  in front   of a chemical formula in a  balanced chemical equation.

   The  reaction   of   molybdenum (iv) oxide  with  gaseous  hydrogen  is  as below,


MoO2  + 2 H2→  Mo  +2 H2O

From   balanced equation above  the coefficient   for  H2  is  2  since  the number in  front of  H2  is 2



3 0
3 years ago
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Answer:

Molarity = 0.4M

Explanation:

Molar mass of NaOH (M)= 40

m= 8g, V= 500ml=0.5L

n= m/M=[8/40]= 0.2mol

Applying

n= CV

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8 0
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