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3 years ago

When co2 levels are low and o2 levels are high, rubisco adds an o2 molecule to rubp. what are the consequences of this reaction?

1 answer:

4 0

Rubisco is an important enzyme that helps in making lifeless carbon of carbon dioxide into organic molecules. Rubisco takes carbon dioxide and attaches it to ribulose bisphosphate, a short sugar chain with five carbon atoms that has rubp as its shortcut. Rubisco then clips the lengthened chain into to polyglycerate pices, which are pretty flexible molecules and are also used in the feeding of the plant. Most of it is used in the photosynthesis pathway, but some of it is used to make sucrose (table sugar) to feed the rest of the plant, or stored away in the form of starch for later use. Hence, rubisco is crucial in the storing of the energy that is created from photosynthesis.

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A cube of steel has dimension 0.2m x 02.m x 0.2m. What is the volume of the cube in cubic centimeters ?

drek231 [11]

Answer:

8000

Explanation:

0.2 x 0.2 x 0.2

then m to cm^3

7 0

3 years ago

A weather balloon contains 1.10X10 to the power of 5 mol of helium and has a volume of 2.70x10 to the power of 6 L at 1.00 atm p

olchik [2.2K]

Answer:

299.14 K or 26°C

Explanation:

The ideal gas law, also called the general gas equation, is the equation of state of a hypothetical ideal gas.

The ideal gas law is often written as

PV = nRT

where P ,V and T are the pressure, volume and absolute temperature;

n is the number of moles of gas and R is the ideal gas constant.

n=1.10 x 10^5 mol

V= 2.70 x 10^6 L

P= 1.00 atm= 101.325 kPa

R= 8.314 kPa*L/ mol*K

when the formula is rearranged, T=PV/ nR

T = (101.325kPa * 2.70 x 10^6 L)/ (1.10 x 10^5 mol * 8.314 kPa*L/ mol*K)

T = 299.1421917 K

T = 299.14 - 273.15 = 25.99 = 26°C

4 0

3 years ago

Mara and Ivy are asked to identify two minerals in science class. To do this, they decide to study the physical properties of th

zmey [24]

Answer: determine how they break

Explanation:

hope this helps!

3 0

3 years ago

Read 2 more answers

In reality, a hydrate of iron(III) nitrate had to be used, not the anhydrous salt. As you may guess, some of the hydrate’s mass

liq [111]

<u>Answer:</u> The mass of nonahydrate iron (III) nitrate is 16.2 g

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of $Fe(NO_3)_3$ = 0.0020 M

Volume of solution = 2 L

Putting values in above equation, we get:

$0.0200M=\frac{\text{Moles of }Fe(NO_3)_3}{2L}\\\\\text{Moles of }Fe(NO_3)_3=(0.0200mol/L\times 2L)=0.04mol$

The chemical equation for the decomposition of hydrated iron (III) nitrate follows:

$Fe(NO_3)_3.9H_2O\rightarrow Fe(NO_3)_3+9H_2O$

By Stoichiometry of the reaction:

1 mole of iron (III) nitrate is produced from 1 mole of hydrated iron (III) nitrate

So, 0.04 moles of iron (III) nitrate will be produced from = $\frac{1}{1}\times 0.04=0.04mol$ of hydrated iron (III) nitrate

To calculate the mass from given number of moles, we use the equation:

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

Molar mass of nonahydrate iron (III) nitrate = 404.0 g/mol

Moles of nonahydrate iron (III) nitrate = 0.04 moles

Putting values in above equation, we get:

$0.04mol=\frac{\text{Mass of nonahydrate iron (III) nitrate}}{404.0g/mol}\\\\\text{Mass of nonahydrate iron (III) nitrate}=(0.04mol\times 404.0g/mol)=16.2g$

Hence, the mass of nonahydrate iron (III) nitrate is 16.2 g

7 0

3 years ago

If more sugar were added to the solution on the left side of the tube, what would happen to the water level on the right side of

Vesna [10]

The given question is incomplete. The complete question is as follows.

If more sugar were added to the solution on the left side of the tube, what would happen to the water level on the right side of the tube?

There is more solute in the left arm of the tube than in the right arm of the tube.

There is less free water in the right arm of the tube than in the left arm of the tube.

Water is tightly clustered around the hydrophilic solute molecules on both sides of the membrane.
Solutes can pass through the selectively permeable membrane from right to left, but free water cannot.

Explanation:

It is known that when water is able to diffuse through a semi-permeable membrane from a region of high water concentration to low water concentration is called osmosis.

Since, more amount of solute is present on the right side of the tube so, there will be less amount of free water present over there as compared to the left side of the tube.

Also, water will be clustered around the hydrophilic solute molecules which is present on both sides of the membrane.

Thus, we can conclude that the true statements are as follows.

There is less free water in the right arm of the tube than in the left arm of the tube.

Water is tightly clustered around the hydrophilic solute molecules on both sides of the membrane.

5 0

3 years ago