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ollegr [7]
3 years ago
9

What happens to glucose in the Calvin cycle? Plz help

Chemistry
1 answer:
Leona [35]3 years ago
7 0

The reactions of the Calvin cycle add carbon from carbon dioxide in the atmosphere to a five-carbon molecule known as RuBP. These reactions use chemical energy that were produced in the light reactions, from NADPH and ATP. The final product of the Calvin cycle is glucose.

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The mass of 0.10 mole of methane
saveliy_v [14]

Answer:

A methane molecule is made from one carbon atom and four hydrogen atoms. Carbon has a mass of 12.011 u and hydrogen has a mass of 1.008 u. This means that the mass of one methane molecule is 12.011 u + (4 × 1.008u), or 16.043 u. This means that one mole of methane has a mass of 16.043 grams.

メタン分子は、1つの炭素原子と4つの水素原子から作られています。炭素の質量は12.011uで、水素の質量は1.008uです。これは、1つのメタン分子の質量が12.011 u +(4×1.008u)、つまり16.043uであることを意味します。これは、1モルのメタンの質量が16.043グラムであることを意味します。^>^

5 0
3 years ago
One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K
mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy (s_{2} - s_{1}), in joules per gram-Kelvin, by the following model:

s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}} (1)

Where:

m - Mass, in kilograms.

c_{w} - Specific heat of water, in joules per kilogram-Kelvin.

T_{1}, T_{2} - Initial and final temperatures of water, in Kelvin.

If we know that m = 1\,kg, c_{w} = 4190\,\frac{J}{kg\cdot K}, T_{1} = 373.15\,K and T_{2} = 288.15\,K, then the change in entropy for the entire process is:

s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}

s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}

The change in entropy is -1083.112 joules per kilogram-Kelvin.

7 0
3 years ago
Suppose the gas resulting from the sublimation of 1.00 g carbon dioxide is collected over water at 25.0◦c into a 1.00 l containe
jeyben [28]

Answer:

0.55 atm

Explanation:

First of all, we need to calculate the number of moles corresponding to 1.00 g of carbon dioxide. This is given by

n=\frac{m}{M_m}

where

m = 1.00 g is the mass of the gas

Mm = 44.0 g/mol is the molar mass of the gas

Substituting,

n=\frac{1.00 g}{44.0 g/mol}=0.0227 mol

Now we can find the pressure of the gas by using the ideal gas law:

pV=nRT

where

p is the gas pressure

V = 1.00 L is the volume

n = 0.0227 mol is the number of moles

R = 0.082 L/(atm K mol) is the gas constant

T = 25.0 C + 273 = 298 K is the temperature of the gas

Solving the formula for p, we find

p=\frac{nRT}{V}=\frac{(0.0227 mol)(0.082 L/(atm K mol))(298 K)}{1.00 L}=0.55 atm

8 0
3 years ago
How much space does a 246g box take up. Its density is listed at 3.2 g/cm3.
CaHeK987 [17]

Answer:

76.875 cm3

<h2>Explanation:</h2>

Since: Density = \frac{Mass}{volume\\}

Then: Volume = \frac{Mass}{Density\\}

The mass of the box = 246g

and the density = 3.2 g/cm3

Then: The volume = \frac{246}{3.2}

= 76.875 cm3

4 0
3 years ago
Plz help me out!!!!!
Lelu [443]

Answer:

i am not 100% sure but im pretty sure there is.

Explanation:

7 0
2 years ago
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