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3 years ago

What are the major parts of the ecosystem

Chemistry

1 answer:

eduard3 years ago

4 0

Answer:

Ecosystems have lots of different living organisms that interact with each other. The living organisms in an ecosystem can be divided into three categories: producers, consumers and decomposers. They are all important parts of an ecosystem. Producers are the green plants.

Explanation:

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the pressure of a sample of argon gas was increased from 2.32 atm2.32 atm to 7.16 atm7.16 atm at constant temperature. if the fi

SOVA2 [1]

initial volume of the argon sample = 5.93L according to Boyle's law

What is Boyle's law ?

Boyle's law, also known as Mariotte's law, is a relationship describing how a gas will compress and expand at a constant temperature. The pressure (p) of a given quantity of gas changes inversely with its volume (v) at constant temperature, according to this empirical connection, which was established by the physicist Robert Boyle in 1662. In equation form, this means that pv = k, a constant.

According to Boyle's law

P1/V1 = P2/V2

P1 = initial pressure

P2 = final pressure

V1 =initial volume

V2= final volume

V1 = P1*V2/P2

V1 = 2.32*18.3/7.16 = 5.93L

initial volume of the argon sample = 5.93L according to Boyle's law

To know about Boyle's law from the link

brainly.com/question/26040104

#SPJ4

6 0

1 year ago

7. There are 7. 0 ml of 0.175 M H2C2O4 , 1 ml of water , 4 ml of 3.5M KMnO4 what is the molar concentration ofH2C2O4 ?

Illusion [34]

Answer:

7. 0.1021 M

8. 1.167 M

10. Increase in volume of water would lower the rate of reaction

Explanation:

7. What is the molar concentration of H₂C₂O₄ ?

Since we have 7.0 ml of 0.175 M H₂C₂O₄, the number of moles of H₂C₂O₄ present n = molarity of H₂C₂O₄ × volume of H₂C₂O₄ = 0.175 mol/L × 7.0 ml = 0.175 mol/L × 7 × 10⁻³ L = 1.225 × 10⁻³ mol.

Also, the total volume present V = volume of H2C2O4 + volume of water + volume of KMnO4 = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L

So, the molar concentration of H₂C₂O₄, M = number of moles of H₂C₂O₄/volume = n/V

= 1.225 × 10⁻³ mol/12 × 10⁻³ L

= 0.1021 mol/L

= 0.1021 M

8. Using the data from question 7 what is the molar concentration of KMnO₄ ?

Since we have 4.0 ml of 3.5 M KMnO₄, the number of moles of KMnO4 present n' = molarity of KMnO₄ × volume of KMnO₄ = 3.5 mol/L × 4.0 ml = 3.5 mol/L × 4 × 10⁻³ L = 14 × 10⁻³ mol.

Also, the total volume present V = volume of KMnO₄ + volume of water + volume of KMnO₄ = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L

So, the molar concentration of KMnO₄, M' = number of moles of KMnO₄/volume = n'/V

= 14 × 10⁻³ mol/12 × 10⁻³ L

= 1.167 mol/L

= 1.167 M

10. From question number 7, what effect increasing the volume of water has on the reaction rate?

Increase in volume of water would lower the rate of reaction because, the particles of both substances would have to travel farther distances to collide with each other, since there are less particles present in the solution and thus, the concentration of the particles would decrease thereby decreasing the rate of reaction.

3 0

3 years ago

how many kilograms of a 35% m/m sodium chlorate solution is needed to react completely with 0.29 l of a 22% m/v aluminum nitrate

Stolb23 [73]

Answer:- 0.273 kg

Solution:- A double replacement reaction takes place. The balanced equation is:

$3NaClO_3+Al(NO_3)_3\rightarrow 3NaNO_3+Al(ClO_3)_3$

We have 0.29 L of 22% m/v aluminum nitrate solution. m/s stands for mass by volume. 22% m/v aluminium nitrate solution means 22 g of it are present in 100 mL solution. With this information, we can calculate the grams of aluminum nitrate present in 0.29 L.

$0.29L(\frac{1000mL}{1L})(\frac{22g}{100mL})$

= 63.8 g aluminum nitrate

From balanced equation, there is 1:3 mol ratio between aluminum nitrate and sodium chlorate. We will convert grams of aluminum nitrate to moles and then on multiplying it by mol ratio we get the moles of sodium chlorate that could further be converted to grams.

We need molar masses for the calculations, Molar mass of sodium chlorate is 106.44 gram per mole and molar mass of aluminum nitrate is 212.99 gram per mole.

$63.8gAl(NO_3)_3(\frac{1mol}{212.99g})(\frac{3molNaClO_3}{1molAl(NO_3)_3})(\frac{106.44g}{1mol})$

= $95.7gNaClO_3$

sodium chlorate solution is 35% m/m. This means 35 g of sodium chlorate are present in 100 g solution. From here, we can calculate the mass of the solution that will contain 95.7 g of sodium chlorate and then the grams are converted to kg.

$95.7gNaClO_3(\frac{100gSolution}{35gNaClO_3})(\frac{1kg}{1000g})$

= 0.273 kg

So, 0.273 kg of 35% m/m sodium chlorate solution are required.

7 0

3 years ago

The table shows a chemical equation and the dimensional analysis used by a student to calculate the number of moles of Ba3N2 req

UNO [17]

This problem is providing the chemical reaction whereby barium nitride reacts with water to produce barium hydroxide and ammonia, so the number of moles of barium nitride are required in order to produce 8.3 moles of ammonia. It asks for us to evaluate the student's setup, so we conclude the answer is C. "1 mol of NH3 should be replaced with 2 mol of NH3", according to:

<h3>Mole ratios:</h3>

In chemistry, stoichiometric calculations are used to figure out the moles or mass of a substance, given information about another one in the reaction. In this case, for the given chemical equation:

$Ba_3N_2 + 6H_2O \rightarrow 3Ba(OH)_2 + 2NH_3$

We evidence a 1:2 mole ratio of barium nitride to ammonia, for that reason, the student's setup:

$8.3 mol NH_3 * \frac{1 mol Ba_3N_2 }{1 mol NH_3}$

Is incorrect, because the ammonia must be accompanied by a 2 rather than the 1 it is given there:

$8.3 mol NH_3 * \frac{1 mol Ba_3N_2 }{2mol NH_3}$

Thereby, the correct answer is C. "1 mol of NH3 should be replaced with 2 mol of NH3"

Learn more about mole ratios: brainly.com/question/15288923

5 0

2 years ago

When 8.21 l of c3h8 (g) burn in oxygen, how many liters of oxygen are consumed? all gas volumes are measured at the same tempera

just olya [345]

<span>8.21 L of C3H8(g) Lets take c as the molar volume at that temperature.
c L <><> 5c L
C3H8 (g) + 5O2 (g) --> 3CO2 + 4H2O + Q
8.21 L <><> x L
x = (8.21 * 5c)/c = 8.21 * 5 = 41.05 L O2 consumed for a 100% yield.</span>

5 0

3 years ago