Using water with an $^{18} \mathrm{O}$ content of $0.85\%$ to dissolve normal potassium carbonate and potassium bicarbonate, a mixed aqueous solution of $\mathrm{K_2CO_3}$ and $\mathrm{KHCO_3}$ was prepared. Under light conditions, this solution was supplied to Chlorella. After a period of time, the $^{18} \mathrm{O}$ content in the carbonate, water, and released oxygen was measured to be $0.61\%$, $0.85\%$, and $0.86\%$, respectively. It is known that oxygen exists in nature in the forms of $^{18} \mathrm{O}$, $^{17} \mathrm{O}$, and $^{16} \mathrm{O}$. The atomic weight of oxygen is 15.9994, while the atomic weights of $^{16} \mathrm{O}$, $^{17} \mathrm{O}$, and $^{18} \mathrm{O}$ are 15.9949, 16.9991, and 17.9992, respectively. The isotopic abundance of $^{16} \mathrm{O}$ is $99.762\%$.

Which of the following statements is correct:
用 $^{18} \mathrm{O}$ 含量为 $0.85 \%$ 的水溶解正常的碳酸钾和碳酸氢钾, 配成 $\mathrm{K_2CO_3}$ 和 $\mathrm{KHCO_3}$ 的混合水溶液, 在光照条件下用此溶液供给小球藻。经过一段时间后, 测得碳酸盐中、水中、放出的氧气中$^{18} \mathrm{O}$的含量分别为 $0.61 \%, 0.85 \%, 0.86 \%$ 。已知氧在
自然界中以$^{18} \mathrm{O}$，$^{17} \mathrm{O}$ ，$^{16} \mathrm{O}$的形式存在, 氧的原子量为 15.9994 , 而$^{16} \mathrm{O}$，$^{17} \mathrm{O}$ ，$^{18} \mathrm{O}$的原子量
分别为 15.9949 , 16.9991 和 17.9992 , $^{16} \mathrm{O}$的同位素丰度为 $99.762 \%$ 。

下列说法正确的是：
At the start of the experiment, the $\mathrm{^{18}O}$ content in the carbonate was 0.31\%.
The gas produced by <i>Chlorella</i> photosynthesis originates from bicarbonate.
In the absence of <i>Chlorella</i>, the $\mathrm{^{18}O}$ content of carbonate does not change.
Adding a small amount of sodium hydroxide solution with a $\mathrm{^{18}O}$ content of $\mathrm{0.85\%}$ to the carbonate solution accelerates the rate of change in the $\mathrm{^{18}O}$ content of the carbonate.
None of the above options are correct.
实验开始时，碳酸盐中$\mathrm{^{18}O}$的含量为0.31\%。
小球藻光合作用产生的气体来源于碳酸氢盐。
无小球藻存在条件下，碳酸盐$\mathrm{^{18}O}$含量不会发生变化。
向碳酸盐溶液中加入少量$\mathrm{^{18}O}$含量为$\mathrm{0.85 \%}$的氢氧化钠溶液，碳酸盐的$\mathrm{^{18}O}$含量变化速率变快。
以上选项均不正确。
Based on the isotopic abundance, let the content of $^{18} \mathrm{O}$ in the carbonate be x, then  

$15.9949\times99.762\%+16.9991\times(1-x-99.762)+17.9992x=15.9994$  

Solving gives $x=0.0021=0.21\%$, so option A is incorrect.  

<Checkpoint>[$^{18} \mathrm{O}$ content in carbonate is $0.21\%$](1)</Checkpoint>  

Since the $^{18} \mathrm{O}$ content in the released oxygen matches that in water but not in bicarbonate, the oxygen produced by Chlorella photosynthesis originates from water, making option B incorrect.  

<Checkpoint>[$^{18} \mathrm{O}$ content in released oxygen matches that in water](1)</Checkpoint>  

<Checkpoint>[Oxygen produced by Chlorella photosynthesis originates from water](1)</Checkpoint>  

Carbonate can hydrolyze to produce carbon dioxide, which then reacts with water or hydroxide to enable $^{18} \mathrm{O}$ exchange between water and carbonate. Thus, even without Chlorella, the $^{18} \mathrm{O}$ content in carbonate would change, making option C incorrect.  

<Checkpoint>[Carbonate can hydrolyze to produce carbon dioxide, which then reacts with water or hydroxide to enable $^{18} \mathrm{O}$ exchange between water and carbonate](1)</Checkpoint>  

After adding sodium hydroxide, the solution's pH increases, reducing the concentrations of bicarbonate and carbonic acid and slowing the reaction, making option D incorrect.  

<Checkpoint>[After adding sodium hydroxide, the solution's pH increases, reducing the concentrations of bicarbonate and carbonic acid and slowing the reaction](1)</Checkpoint>  

In summary, options A-D are all incorrect, and option E is correct.
根据同位素丰度，设$^{18} \mathrm{O}$在碳酸盐的含量为x，可得

$15.9949\times99.762\%+16.9991\times(1-x-99.762)+17.9992x=15.9994$

解得$x=0.0021=0.21\%$，选项A错误。

<Checkpoint>[$^{18} \mathrm{O}$在碳酸盐的含量为$0.21\%$](1)</Checkpoint>

因为放出的氧气中$^{18} \mathrm{O}$的含量与水中$^{18} \mathrm{O}$的含量相符而与碳酸氢盐中$^{18} \mathrm{O}$的含量不符，所以小球藻光合作用产生的氧气来源于水，选项B错误。

<Checkpoint>[放出的氧气中$^{18} \mathrm{O}$的含量与水中$^{18} \mathrm{O}$的含量相符](1)</Checkpoint>

<Checkpoint>[小球藻光合作用产生的氧气来源于水](1)</Checkpoint>

碳酸盐可以通过水解产生二氧化碳，二氧化碳再与水或者氢氧根化合实现水和碳酸盐之间$^{18} \mathrm{O}$的交换，从而碳酸盐即使没有小球藻$^{18} \mathrm{O}$的含量也会发生变化，选项C错误。

<Checkpoint>[碳酸盐水解产生二氧化碳，二氧化碳与水或者氢氧根化合，实现$^{18} \mathrm{O}$交换](1)</Checkpoint>

加入氢氧化钠后溶液pH升高，碳酸氢根和碳酸的浓度降低，反应变慢，选项D错误。

<Checkpoint>[加入氢氧化钠后溶液pH升高，碳酸氢根和碳酸的浓度降低，反应变慢](1)</Checkpoint>

综上，选项A-D均错误，选项E正确。