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3 years ago

How do chaetoceros contribute to global atmospheric oxygen and carbon dioxide levels?

1 answer:

3 0

Chaetoceros is planktonic diatoms and known to be the largest genus with 400 described species. It has a cell wall made up of silica containing long thin spins. These spines connect the cell walls together to form a colony of cells.

Due to its high rate of growth and high lipid concentration, they can be used as bio fuel. Colonies of chaetoceros act as an important food source inside the water and they are major contributor of carbon.

In north water, they contribute 91% of the total phytoplankton cells thus acting as an important producer in the area. This contributes to the production of oxygen in north water. Total phytoplankton contributes to half of the oxygen production of Earth.

In this way, chaetoceros contribute to global atmospheric oxygen and carbon dioxide levels.

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What does not add co2 to the atmosphere

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Answer:

Abiotic objects

Ex. bed, table, lamp, chair, blanket, etc.

Abiotic means Non-living.

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2 years ago

How many branches of chemistry are there .

faltersainse [42]

Answer:

Explanation:

Analytical chemistry

Organic chemistry

Inorganic chemistry

physical chemistry

Biochemistry

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2 years ago

Read 2 more answers

This is all the water; the liquid portion of the planet.

Dominik [7]

The oceanic part of the earth is the liquid portion Of the planet

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3 years ago

What is the mass of the solid NH4Cl formed when 75.5 g of NH3 is mixed with an equal mass of HCl? What is the volume of the gas

Gekata [30.6K]

Answer : The volume of the gas remaining is 56.5 liters.

The gas is hydrochloric acid and the formula of the gas is HCl.

The mass of $NH_4Cl$ produced is, 110.7 grams.

Explanation :

The balanced chemical reaction will be:

$NH_3+HCl\rightarrow NH_4Cl$

First we have to calculate the moles of $NH_3$ and HCl

$\text{Moles of }NH_3=\frac{\text{Mass of }NH_3}{\text{Molar mass of }NH_3}$

Molar mass of NH_3 = 17 g/mole

$\text{Moles of }NH_3=\frac{75.5g}{17g/mole}=4.44mole$

and,

$\text{Moles of }HCl=\frac{\text{Mass of }HCl}{\text{Molar mass of }HCl}$

Molar mass of HCl = 36.5 g/mole

$\text{Moles of }HCl=\frac{75.5g}{36.5g/mole}=2.07mole$

Now we have to calculate the limiting and excess reagent.

From the balanced reaction we conclude that

As, 1 mole of HCl react with 1 mole of $NH_3$

So, 2.07 mole of HCl react with 2.07 mole of $NH_3$

From this we conclude that, $NH_3$ is an excess reagent because the given moles are greater than the required moles and HCl is a limiting reagent and it limits the formation of product.

The remaining moles of HCl gas = 4.44 - 2.07 = 2.37 moles

Now we have to calculate the volume of the gas remaining.

Using ideal gas equation :

PV = nRT

where,

P = Pressure of gas = 752 mmHg = 0.989 atm (1 atm = 760 mmHg)

V = Volume of gas = ?

n = number of moles of gas = 2.37 moles

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = $14.0^oC=273+14.0=287K$

Putting values in above equation, we get:

$0.989atm\times V=2.37mole\times (0.0821L.atm/mol.K)\times 287K$

V = 56.5 L

Now we have to calculate the moles of $NH_4Cl$

As, 1 mole of HCl react with 1 mole of $NH_4Cl$

So, 2.07 mole of HCl react with 2.07 mole of $NH_4Cl$

Now we have to calculate the mass of $NH_4Cl$

$\text{ Mass of }NH_4Cl=\text{ Moles of }NH_4Cl\times \text{ Molar mass of }NH_4Cl$

Molar mass of $NH_4Cl$ = 53.5 g/mole

$\text{ Mass of }NH_4Cl=(2.07moles)\times (53.5g/mole)=110.7g$

Thus, the volume of the gas remaining is 56.5 liters.

The gas is hydrochloric acid and the formula of the gas is HCl.

The mass of $NH_4Cl$ produced is, 110.7 grams.

3 0

3 years ago

Combustion of 6.90g of this compound produced 13.8g of CO2 and 5.64g of H2O. What is the empirical formula of the unknown compou

Alla [95]

The empirical formula for the unknown compound would be: C2H4O (2 molecules of Carbon, 4 molecules of Hydrogen, and 1 molecule of Oxygen)

4 0

3 years ago