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3 years ago

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Why does it take less

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KengaRu [80]3 years ago

8 0

Answer:I’m not for sure

Explanation:

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If 12.5 grams of the original sample of cesium-137 remained after 90.6 years, what was the mass of the original sample?

myrzilka [38]

Answer:

Mass of original sample = 100 g

Explanation:

Half life of cesium-137 = 30.17 years

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac{\ln2}{t_{1/2}}$

$k=\frac{\ln2}{30.17}\ year^{-1}$

The rate constant, k = 0.02297 year⁻¹

Time = 90.6 years

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Initial concentration $[A_0]$ = ?

Final concentration $[A_t]$ = 12.5 grams

Applying in the above equation, we get that:-

$12.5\ g=[A_0]e^{-0.02297\times 90.6}$

$[A_0]=\frac{12.5}{e^{-0.02297\times 90.6}}\ g=100\ g$

<u>Mass of original sample = 100 g</u>

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3 years ago

The Toucan has a long, narrow beak that allows it to reach fruit that is hard to reach for other birds. Plants, like the Monster

gladu [14]

<h2>Please mark me as brainliest please yaa!!!^_^</h2>

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2 years ago

Will Mark Brainliest <br>Aswer for<br>box 1 and box 2<br>On top of the oval says <br>A-U-G-G-C-A

DerKrebs [107]

<span>A-U-G - G-C-A
Met - Ala

If AUG is going to be in the beginning of the chain, </span>A-U-G, also will mean a start codon. For eukaryotes, codon for Met means start of the amino acid chain.

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3 years ago

How does Kinetic energy and how does it relates to diffusion and temperature.

olya-2409 [2.1K]

Answer:

ok

Explanation:

khfdnkv Klein pick pick or en la Web outta

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3 years ago

Consider the reaction FeO (S) + CO(g) <-----> Fe(s) + CO2(g) for which KP is found to have the following values:

Svetach [21]

Explanation:

$\Delta G^o=-RT\ln K_1$

where,

R = Gas constant = $8.314J/K mol$

T = temperature = $600^oC=[273.15+600]K=873.15 K$

$K_p$ = equilibrium constant at 600°C = 0.900

Putting values in above equation, we get:

$\Delta G^o=-(8.314J/Kmol)\times 873.15 K\times \ln (0.900 )$

$\Delta G^o=764.85 J/mol$

The ΔG° of the reaction at 764.85 J/mol is 764.85 J/mol.

Equilibrium constant at 600°C = $K_1=0.900$

Equilibrium constant at 1000°C = $K_2=0.396$

$T_1=[273.15+600]K=873.15 K$

$T_2=[273.15+1000]K=1273.15 K$

$\ln \frac{K_2}{K_1}=\frac{\Delta H^o}{R}\times [\frac{1}{T_1}-\frac{1}{T_2}]$

$\ln \frac{0.396}{0.900}=\frac{\Delta H^o}{8.314 J/mol K}\times [\frac{1}{873.15 K}-\frac{1}{1273.15 K}]$

$\Delta H^o=-18,969.30 J/mol$

The ΔH° of the reaction at 600 C is -18,969.30 J/mol.

ΔG° = ΔH° - TΔS°

764.85 J/mol = -18,969.30 J/mol - 873.15 K × ΔS°

ΔS° = -22.60 J/K mol

The ΔS° of the reaction at 600 C is -22.60 J/K mol.

$FeO (s) + CO(g)\rightleftharpoons Fe(s) + CO_2(g)$

Partial pressure of carbon dioxide = $p_1=P\times \chi_1$

Partial pressure of carbon monoxide = $p_2=P\times \chi_2$

Where $\chi_1\& \chi_2$ mole fraction of carbon dioxide and carbon monoxide gas.

The expression of $K_p$ is given by:

$K_p=\frac{p_1}{p_2}=\frac{P\times \chi_1}{P\times \chi_2}$

$0.900=\frac{\chi_1}{\chi_2}$

$\chi_1=0.900\times \chi_2$

$\chi_1+\chi_2=1$

$0.9\chi_2+\chi_2=1$

$1.9\chi_2=1$

$\chi_2=\frac{1}{1.9}=0.526$

$\chi_1=1-\chi_2=1-0.526=0.474$

Mole fraction of carbon dioxide at 600°C is 0.474.

6 0

3 years ago