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marshall27 [118]

3 years ago

5

The net reaction of the calvin cycle is the conversion of co2 into the three-carbon sugar g3p. along the way, reactions rearrang

e carbon atoms among intermediate compounds and use the atp and nadph produced by the light reactions. in this exercise, you will track carbon atoms through the calvin cycle as required for the net production of one molecule of g3p. for each intermediate compound in the calvin cycle, identify the number of molecules of that intermediate and the total number of carbon atoms contained in those molecules. as an example, the output g3p is labeled for you: 1 molecule with a total of 3 carbon atoms. labels may be used once, more than once, or not at all. hints

1 answer:

pychu [463]3 years ago

3 0

a) 3 molecule of carbon dioxide (CO2) - 3 carbon atoms

b) 6 molecule of 3 - PGA - 18 carbon atoms ( 3 cabon atoms in 1 3 PGA molecule)

c) 6 molecule og G3P - 18 carbon atom ( 3 carbon atoms in G3P molecule )

d) 5 molecules of G3P - 15 carbon atom

e) 3 molecule of R5P - 15 carbon atoms ( 5 carbon atom of R5P olecule )

f) 3 molecule of RuBP - 15 carbon atoms ( r carbon atom of RuBp molecule )

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i need to know like now. what is a pure substance that contains atoms of more than one element called?

Galina-37 [17]

Answer:

A compound

Explanation:

A compound is a pure substance that contains atoms of more than one element.

For example, water is a compound.

It is pure because it consists only of water molecules, and each molecule contains atoms of the elements hydrogen and oxygen.

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3 years ago

Is every solution also a mixture?

aleksandrvk [35]

A mixture consists of two or more components. a solution can consists of a solute dissolved in a solvent or two solvents mixed together, for instance. A mixture can also be two gases (CO2 and O2) mixed together or two solids mixed together (sawdust and chalk).

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4 years ago

Plz help I’ll give brainliest!!<br> Describe the temperate coniferous forest?<br> Explain

andriy [413]

Answer:

Temperate coniferous forest is a terrestrial biome defined by the World Wide Fund for Nature. This is because Temperate coniferous forests are found predominantly in areas with warm summers and cool winters, and vary in their kinds of plant life.

Explanation:

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2 years ago

Calculate the activity coefficients for the following conditions:

uysha [10]

<u>Answer:</u>

<u>For a:</u> The activity coefficient of copper ions is 0.676

<u>For b:</u> The activity coefficient of potassium ions is 0.851

<u>For c:</u> The activity coefficient of potassium ions is 0.794

<u>Explanation:</u>

To calculate the activity coefficient of an ion, we use the equation given by Debye and Huckel, which is:

$-\log\gamma_i=\frac{0.51\times Z_i^2\times \sqrt{\mu}}{1+(3.3\times \alpha _i\times \sqrt{\mu})}$ ........(1)

where,

$\gamma_i$ = activity coefficient of ion

$Z_i$ = charge of the ion

$\mu$ = ionic strength of solution

$\alpha _i$ = diameter of the ion in nm

To calculate the ionic strength, we use the equation:

$\mu=\frac{1}{2}\sum_{i=1}^n(C_iZ_i^2)$ ......(2)

where,

$C_i$ = concentration of i-th ions

$Z_i$ = charge of i-th ions

<u>For a:</u>

We are given:

0.01 M NaCl solution:

Calculating the ionic strength by using equation 2:

$C_{Na^+}=0.01M\\Z_{Na^+}=+1\\C_{Cl^-}=0.01M\\Z_{Cl^-}=-1$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.01\times (+1)^2)+(0.01\times (-1)^2)]\\\\\mu=0.01M$

Now, calculating the activity coefficient of $Cu^{2+}$ ion in the solution by using equation 1:

$Z_{Cu^{2+}}=2+\\\alpha_{Cu^{2+}}=0.6\text{ (known)}\\\mu=0.01M$

Putting values in equation 1, we get:

$-\log\gamma_{Cu^{2+}}=\frac{0.51\times (+2)^2\times \sqrt{0.01}}{1+(3.3\times 0.6\times \sqrt{0.01})}\\\\-\log\gamma_{Cu^{2+}}=0.17\\\\\gamma_{Cu^{2+}}=10^{-0.17}\\\\\gamma_{Cu^{2+}}=0.676$

Hence, the activity coefficient of copper ions is 0.676

<u>For b:</u>

We are given:

0.025 M HCl solution:

Calculating the ionic strength by using equation 2:

$C_{H^+}=0.025M\\Z_{H^+}=+1\\C_{Cl^-}=0.025M\\Z_{Cl^-}=-1$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.025\times (+1)^2)+(0.025\times (-1)^2)]\\\\\mu=0.025M$

Now, calculating the activity coefficient of $K^{+}$ ion in the solution by using equation 1:

$Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{ (known)}\\\mu=0.025M$

Putting values in equation 1, we get:

$-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.025}}{1+(3.3\times 0.3\times \sqrt{0.025})}\\\\-\log\gamma_{K^{+}}=0.070\\\\\gamma_{K^{+}}=10^{-0.070}\\\\\gamma_{K^{+}}=0.851$

Hence, the activity coefficient of potassium ions is 0.851

<u>For c:</u>

We are given:

0.02 M $K_2SO_4$ solution:

Calculating the ionic strength by using equation 2:

$C_{K^+}=(2\times 0.02)=0.04M\\Z_{K^+}=+1\\C_{SO_4^{2-}}=0.02M\\Z_{SO_4^{2-}}=-2$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.04\times (+1)^2)+(0.02\times (-2)^2)]\\\\\mu=0.06M$

Now, calculating the activity coefficient of $K^{+}$ ion in the solution by using equation 1:

$Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{ (known)}\\\mu=0.06M$

Putting values in equation 1, we get:

$-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.06}}{1+(3.3\times 0.3\times \sqrt{0.06})}\\\\-\log\gamma_{K^{+}}=0.1\\\\\gamma_{K^{+}}=10^{-0.1}\\\\\gamma_{K^{+}}=0.794$

Hence, the activity coefficient of potassium ions is 0.794

6 0

3 years ago

Draw stearic acid (octadecanoic acid). You do not need to draw hydrogen atoms attached to carbon atoms.

postnew [5]

Answer:

The structure of stearic acid (octadecanoic acid) is shown in the attachment below

Explanation:

Stearic acid is an example of a saturated fatty acid. The structure of stearic acid is shown in the attachment below.

5 0

3 years ago