The event which is most likely occurring in this scenario is effusion because there is a movement of a gas through a small opening into a larger volume and is denoted as option C.
<h3>What is Effusion?</h3>
This is referred to as the process in which a gas or a substance escapes from a container through a hole of diameter which is usually smaller.
The type of event which is most likely occurring is effusion because of the presence of the small holes in which the balls are made to pass through the center which is why option C was chosen.
Read more about Effusion here brainly.com/question/2097955
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The options are:
- diffusion because particles move from regions of high concentration to regions of low concentration.
- diffusion because particles move from regions of low concentration to regions of high concentration.
- effusion because there is a movement of a gas through a small opening into a larger volume.
- effusion because there is a movement of a gas through a large opening into a smaller volume
As stated on google it says “ Farmers and breeders allowed only the plants and animals with desirable characteristics to reproduce, causing the evolution of farm stock. This process is called artificial selection because people (instead of nature) select which organisms get to reproduce. ... This is evolution through artificial selection.”
The combustion reaction is as expressed,
CxHy + O2 --> CO2 + H2O
The mass fraction of carbon in CO2 is 3/11. Hence,
mass of C in CO2 = (3.14 g)(3/11) = 0.86 g C.
Given that we have 1 g of the hydrocarbon, the mass of H is equal to 0.14 g.
moles of C = 0.86 g C / 12 g = 0.0713
moles of H = 0.14 g H / 1 g = 0.14
The empirical formula for the hydrocarbon is therefore, CH₂.
Answer:
B
Explanation:
since isotopeA has bigger mass number
<span><span>N2</span><span>O3</span><span>(g)</span>→NO<span>(g)</span>+<span>NO2</span><span>(g)</span></span>
<span><span>[<span>N2</span><span>O3</span>]</span> Initial Rate</span>
<span>0.1 M r<span>(t)</span>=0.66</span> M/s
<span>0.2 M r<span>(t)</span>=1.32</span> M/s
<span>0.3 M r<span>(t)</span>=1.98</span> M/s
We can have the relationship:
<span>(<span><span>[<span>N2</span><span>O3</span>]/</span><span><span>[<span>N2</span><span>O3</span>]</span>0</span></span>)^m</span>=<span><span>r<span>(t)/</span></span><span><span>r0</span><span>(t)
However,
</span></span></span>([N2O3]/[N2O3]0) = 2
Also, we assume m=1 which is the order of the reaction.
Thus, the relationship is simplified to,
r(t)/r0(t) = 2
r<span>(t)</span>=k<span>[<span>N2</span><span>O3</span>]</span>
0.66 <span>M/s=k×0.1 M</span>
<span>k=6.6</span> <span>s<span>−<span>1</span></span></span>
