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barxatty [35]
3 years ago
10

If you expose a photosynthesizing plant to water that contains both radioactive h and radioactive o, in which of the products of

photosynthesis will the radioactive h and o show up? if you expose a photosynthesizing plant to water that contains both radioactive h and radioactive o, in which of the products of photosynthesis will the radioactive h and o show up? h in glucose and water; o in o2 h in glucose; o in water h in water; o in glucose h and o both in glucose
Chemistry
1 answer:
ValentinkaMS [17]3 years ago
3 0
I think the answer is H in glucose and water and O in O2. Photosynthesis is the process in which green plants and other organism convert light energy to chemical energy which is used in activities within the organism. Water that is exposed to the plant reacts with carbon dioxide to produce sugar and oxygen. The radioactive O will hence show up in the oxygen molecule and the radioactive H shows up in the sugar/ glucose molecule.
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What is the % composition of Carbon in Chromium (iii) Carbonate
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Step 1 - Discovering the ionic formula of Chromium (III) Carbonate

Chromium (III) Carbonate is formed by the ionic bonding between Chromium (III) (Cr(3+)) and Carbonate (CO3(2-)):

Cr^{3+}+CO^{2-}_3\rightarrow Cr_2(CO_3)_3

Step 2 - Finding the molar mass of the substance

To find the molar mass, we need to multiply the molar mass of each element by the number of times it appears in the formula of the substance and, finally, sum it all up.

The molar masses are 12 g/mol for C; 16 g/mol for O and 52 g/mol for Cr. We have thus:

\begin{gathered} C\rightarrow3\times12=36 \\  \\ O\rightarrow9\times16=144 \\  \\ Cr\rightarrow2\times52=104 \end{gathered}

The molar mass will be thus:

M=36+104+144=284\text{ g/mol}

Step 3 - Finding the percent composition of carbon

As we saw in the previous step, the molar mass of Cr2(CO3)3 is 284 g/mol. From this molar mass, 36 g/mol come from C. We can set the following proportion:

\begin{gathered} 284\text{ g/mol ---- 100\%} \\ 36\text{ g/mol ----- x} \\  \\ x=\frac{36\times100}{284}=\frac{3600}{284}=12.7\text{ \%} \end{gathered}

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The transuranium synthesis process involves creating a transuranium element through the transmutation process . The transmutation process  is the process of creating heavy elements from light elements. Hence the process is the transmutation of light elements. There are two types: artificial and natural transmutation.

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6 0
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Henry finds an element that is light blue, breaks easily in his hand, and does not reflect light. How should Henry classify the
pshichka [43]

Answer:

The options are not given, here are the options.

metal

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b. a metalloid

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Explanation:

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