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rusak2 [61]
3 years ago
9

Iodine-131 decays with a half-life of 8.02

Chemistry
1 answer:
dlinn [17]3 years ago
3 0
Radioactive material undergoes 1st order decay kinetics.

For 1st order decay, half life = 0.693/k

where k = rate constant

k = 0.693/half life = 0.693/8.02 = 0.0864 day-1

Now, for 1st order reaction,
k = \frac{2.303}{t} X log \frac{initial.conc}{final.conc}

Given: t = 6.01d, initial conc. = 5mg

∴0.0864 = \frac{2.303}{6.01} X log \frac{5}{final.conc}
∴ final conc. = 2.975 mg
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The atomic particle with a charge of -1.6 x 10-19 C is
kobusy [5.1K]

no sefdrcdftrgfkjj jhhhgfd

7 0
3 years ago
Read 2 more answers
Where is the kinetic energy GREATEST?
Zarrin [17]
Kinetic Energy is greatest at the bottom of the highest point
4 0
3 years ago
Given the following data: Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ΔH = −23 kJ 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) ΔH = −39
nalin [4]

Answer:

ΔH° = -11 kj

Explanation:

Step 1: Data given

1)   Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)    ΔH = −23 kJ

2)   3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g)   ΔH = −39 kJ

3)   Fe3O4(s) + CO(g) → 3 FeO(s) + CO2(g) ΔH = +18 kJ

Step 2: The balanced equation

FeO + CO → Fe + CO2

Step 3:  Calculate ΔH for the reaction FeO(s) + CO(g) → Fe(s) + CO2(g).

To get this equation, we need to combine the 3 equations

We have to multiply the third equation by 2.

2Fe3O4(s) + 2CO(g) → 6 FeO(s) + 2CO2(g) ΔH = +54 kJ

<u>This equation we add to the second equation</u>

3Fe2O3(s) + CO(g) + 2Fe3O4(s) + 2CO(g) → 2Fe3O4(s) + CO2(g) + 6FeO(s) + 2CO2 (g)

3Fe2O3(s) + 3CO(g) →  3CO2(g) + 6FeO(s)

ΔH°  = 2*18 + (-39) = 36 - 39 =  -3 kJ

<u>This new equation we will divide by 3 </u>

3Fe2O3(s) + 3CO(g) →  3CO2(g) + 6FeO(s)

Fe2O3(s) + CO(g) →  CO2(g) + 2FeO(s)

ΔH°  =-3/3 = -1 kJ

<u>Now we will substract this new equation from the first equation</u>

Fe2O3(s) + 3CO(g) - Fe2O3(s) - CO(g) → 2Fe(s) + 3CO2(g) -2FeO(s) - CO2(g)

2CO(g)  + 2FeO(s)  → 2Fe(s) + 2CO2(g)

ΔH° = -23kJ +1kJ

ΔH° = -22 kj

<u>The next equation we will divide by 2</u>

2CO(g)  + 2FeO(s)  → 2Fe(s) + 2CO2(g)

CO(g)  + FeO(s)  → Fe(s) + CO2(g)

ΔH° = -22kJ /2

ΔH° = -11 kj

7 0
3 years ago
Automotive airbags inflate when a sample of sodium azide, NaN3, is very rapidly decomposed.
Kobotan [32]

Answer:

148 g

Explanation:

Step 1: Write the balanced equation for the decomposition of sodium azide

2 NaN₃ ⇒ 2 Na + 3 N₂

Step 2: Calculate the moles corresponding to 95.8 g of N₂

The molar mass of N₂ is 28.01 g/mol.

95.8 g × 1 mol/28.01 g = 3.42 mol

Step 3: Calculate the moles of NaN₃ needed to form 3.42 moles of N₂

The molar ratio of NaN₃ to N₂ is 2:3. The moles of NaN₃ needed are 2/3 × 3.42 mol = 2.28 mol.

Step 4: Calculate the mass corresponding to 2.28 moles of NaN₃

The molar mass of NaN₃ is 65.01 g/mol.

2.28 mol × 65.01 g/mol = 148 g

8 0
2 years ago
Suppose a 1.0 L solution contains 0.020 M in Cu(NO3)2, then 0.40 moles of NH3 are added. Assuming no change in volume, what is t
Rasek [7]

Answer:

4.6*10^{-14} M

Explanation:

Concentration of Cu^{2+} = [Cu(NO_3)_2] = 0.020 M

Constructing an ICE table;we have:

                                 Cu^{2+}+4NH_3_{aq} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)}

Initial (M)             0.020          0.40                        0

Change (M)         -  x                - 4 x                       x

Equilibrium (M)  0.020 -x        0.40 - 4 x              x

Given that: K_f =1.7*10^{13}

K_f } = \frac{[Cu(NH_3)_4]^{2+}}{[Cu^{2+}][NH_3]^4}

1.7*10^{13} = \frac{x}{(0.020-x)(0.40-4x)^4}

Since x is so small; 0.40 -4x = 0.40

Then:

1.7*10^{13} = \frac{x}{(0.020-x)(0.0256)}

1.7*10^{13} = \frac{x}{(5.12*10^{-4}-0.0256x)}

1.7*10^{13}(5.12*10^{-4} - 0.0256x) = x

8.704*10^9-4.352*10^{11}x =x

8.704*10^9 = 4.352*10^{11}x

x = \frac{8.704*10^9}{4.352*10^{11}}

x = 0.0199999999999540

Cu^{2+}= 0.020 - 0.019999999999954

Cu^{2+} = 4.6*10^{-14} M

8 0
3 years ago
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