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mart [117]
3 years ago
7

Which type of carbon fixation stores carbon dioxide in acid form?. a. C3. b. C4. c. CAM. d. all of the above

Chemistry
2 answers:
Tcecarenko [31]3 years ago
4 0

Answer is: c) CAM.

CAM (crassulacean acid metabolism) is a carbon fixation pathway.

In plants with CAM carbon fixation, carbon dioxide is stored as the malic acid malate in vacuoles at night.

During the daytime, the malic acid is transported to chloroplasts where it is converted back to carbon dioxide, which is then used during photosynthesis.

Malic acid is an organic dicarboxylic acid with the molecular formula C₄H₆O₅.

7nadin3 [17]3 years ago
3 0
I think its C. CAM

hope that helps

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An atom of gold has a mass of 3.271 X 10-22 g. How many atoms of gold are in 5.00 g of gold? (Give your answer in scientific not
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Answer:

1.53 × 10²² atoms Ag

Explanation:

Step 1: Define conversions

3.271 × 10⁻²² g = 1 atom

Step 2: Use Dimensional Analysis

5.00 \hspace{3} g \hspace{3} Ag(\frac{1 \hspace{3} atom \hspace{3} Ag}{3.271(10)^{-22} \hspace{3} g \hspace{3} Ag} ) = 1.52858 × 10²² atoms Ag

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<u>Answer:</u> 0.0237 g of calcium carbonate would be required to neutralize the given amount of HCl

<u>Explanation:</u>

pH is defined as the negative logarithm of hydrogen ion concentration present in the solution

pH=-\log [H^+]      .....(1)

Given value of pH = 1.5

Putting values in equation 1:

1.5=-\log[H^+]

[H^+]=10^{(-1.5)}=0.0316M

Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:

\text{Molarity of solution}=\frac{\text{Number of moles of solute}\times 1000}{\text{Volume of solution (mL)}}       .....(2)

We are given:

Volume of solution = 15.0 mL

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Putting values in equation 2:

0.0316=\frac{\text{Moles of HCl}\times 1000}{15.0}\\\\\text{Moles of HCl}=\frac{0.0316\times 15.0}{1000}=4.74\times 10^{-4}mol

The chemical equation for the reaction of HCl and calcium carbonate follows:

2HCl+CaCO_3\rightarrow H_2CO_3+CaCl_2

By the stoichiometry of the reaction:

2 moles of HCl reacts with 1 mole of calcium carbonate

So, 4.74\times 10^{-4}mol of HCl will react with = \frac{1}{2}\times 4.74\times 10^{-4}=2.37\times 10^{-4}mol of calcium carbonate

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of calcium carbonate = 2.37\times 10^{-4}mol

Molar mass of calcium carbonate = 100.01 g/mol

Putting values in the above equation:

\text{Mass of }CaCO_3=(2.37\times 10^{-4}mol)\times 100.01g/mol\\\\\text{Mass of }CaCO_3=0.0237g

Hence, 0.0237 g of calcium carbonate would be required to neutralize the given amount of HCl

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