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Sedbober [7]
2 years ago
9

The goal of this lesson is to answer the lesson question, "what does half-life ‘look like' for a radioactive substance?" fill in

the blanks to complete the hypothesis. hypothesis: if an element is radioactive, then the fraction of radioactive remaining after n half-life cycles should be approximately n because…
Chemistry
2 answers:
Serggg [28]2 years ago
6 0

The half life for radioactive can be calculated as:

N /N0 = (1 /2) ^ n

n = T /T half

According to question there are n number of half life are present which would result in remaining amount of element as n.

disa [49]2 years ago
5 0

Atoms

0.5

There are n number of half life are present which would result in remaining amount of element as n.


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It's simple:
rate of change = change in height / time period

= (7600 - 7598) / 40    =   2 / 40 =  0.05 feet / yr</span>
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A voltaic cell consists of a nickel electrode in a solution containing Ni2+ ions, and a copper electrode in a solution containin
Marysya12 [62]
First, we write the half equations for the reduction of the chemical species present:

Cu⁺² + 2e → Cu; E° = 0.34 V
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The second option is correct. (The difference in values is due to different values in literature, and it is negligible)
4 0
2 years ago
Consider the reaction cacn2 3 h2o → caco3 2 nh3 . how much nh3 is produced if 187 g of caco3 are produced?
shusha [124]
B ase from the reaction <span>cacn2 3 h2o → caco3 2 nh3, for every 1 mole of caco3 produced there 2 moles of nh3 being produced. to solved this, we must first convert the caco3 to moles.

mass nh3 = 187 g caco3 (1 mol caco3 / 100 g caco3 ) ( 2 mol nh3 / 1 mol caco3) ( 17 g nh3 / 1 mol nh3)

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7 0
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Show that the Newton has the units of mass times acceleration
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hope this helps :D

7 0
2 years ago
The decomposition of AB given here in this balanced equation 2AB (g)⟶ A2 (g) + B2 (g), has rate constants of 8.58 x 10-9 L/mol s
denis-greek [22]

Answer:

3.24 × 10^5 J/mol

Explanation:

The activation energy of this reaction can be calculated using the equation:

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

Where; Ea = the activation energy (J/mol)

R = the ideal gas constant = 8.3145 J/Kmol

T1 and T2 = absolute temperatures (K)

k1 and k2 = the reaction rate constants at respective temperature

First, we need to convert the temperatures in °C to K

T(K) = T(°C) + 273.15

T1 = 325°C + 273.15

T1 = 598.15K

T2 = 407°C + 273.15

T2 = 680.15K

Since, k1= 8.58 x 10-9 L/mol, k2= 2.16 x 10-5 L/mol, R= 8.3145 J/Kmol, we can now find Ea

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

ln(2.16 x 10-5/8.58 x 10-9) = Ea/8.3145 × (1/598.15 - 1/680.15)

ln(2517.4) = Ea/8.3145 × 2.01 × 10^-4

7.831 = Ea(2.417 × 10^-5)

Ea = 3.24 × 10^5 J/mol

8 0
2 years ago
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