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KIM [24]
3 years ago
15

Do you know how to do this? if you have 4 grams of carbon-14, how much carbon 14 remains after 11,460 years? the half life is 5,

730 years.
Chemistry
1 answer:
Delvig [45]3 years ago
6 0
Carbon dating has<span> given archeologists a more accurate method by which they </span>can<span> determine the age of ancient artifacts. The </span>halflife<span> of </span>carbon 14<span> is </span>5730<span> ± 30 </span>years<span>, and the method of dating lies in trying to determine how </span>much carbon 14<span> (</span><span>the radioactive isotope of carbon) is present in the artifact and comparing it to levels</span>
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Where on the table would a group's<br> ionization energy be the greatest?
Ymorist [56]

Answer:

upper left

Explanation:

it is a trend on the periodic table. Ionization energy increases from left to right(->) I t also increases down to up

5 0
3 years ago
2 H2O2(aq) ----&gt; 2 H2O(l) + O2(g) in the presence of I-(aq) is proposed to be: Step 1 (slow): H2O2 + I- -----&gt; H2O + OI- S
andrezito [222]

Answer:

Molecularity of the rate determining step = 2

Explanation:

Step 1 (slow): H₂O₂ + I⁻ -----> H₂O + OI⁻

Step 2 (fast): H₂O₂ + OI⁻ -----> H₂O + O₂ + I⁻

The rate determining step in a reaction mechanism is also considered as slowest step.

Slowest step is also considered its highest activation energy in energy profile diagram.

In this case intermediate  (IO⁻) is formed.

Step 1 considered as a slowest step.

So,  Rate = K [H₂O₂][I⁻]

  Molecularity = 2

6 0
3 years ago
Brainleist if correct<br> ___________is part of the Milky Way Galaxy.
lana [24]

Answer:

Our Sun (a star) and all the planets around it are part of a galaxy known as the Milky Way Galaxy.

4 0
3 years ago
A measurement of 5.685 × 10-6 is equivalent to 0.000 000 568 5.<br><br> True<br> False
Alecsey [184]
The answer is false.
7 0
3 years ago
A particular first-order reaction has a rate constant of 1.35 × 102 s-1 at 25.0°C. What is the magnitude of k at 95.0°C if Ea =
never [62]

Answer:

k ≈ 9,56x10³ s⁻¹

Explanation:

It is possible to solve this question using Arrhenius formula:

ln\frac{k2}{k1} = \frac{-Ea}{R} (\frac{1}{T2} -\frac{1}{T1} )

Where:

k1: 1,35x10² s⁻¹

T1: 25,0°C + 273,15 = 298,15K

Ea = 55,5 kJ/mol

R = 8,314472x10⁻³ kJ/molK

k2 : ???

T2: 95,0°C+ 273,15K = 368,15K

Solving:

ln\frac{k2}{k1} = 4,257

\frac{k2}{k1} = 70,593

{k2} = 9,53x10^3 s^{-1}

<em>k ≈ 9,56x10³ s⁻¹</em>

I hope it helps!

5 0
3 years ago
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