8 groups, 1 to 7 and most people call group 8 group 0 because the noble gases in group 0 have the outer electron shell full, so for example argon has 18 electrons, so it’s structure is 2.8.8 with its outer shell full. Also the number of the last shell defies what group it’s in, for example sodium is in group 1 so it’s structure is 2.8.1. Hope this helps.
Answer:
The roles of decomposers organisms are to degrade organic matter and contribute to the nutrition of the soil, which involves recycling the chemical elements needed by living organisms.
Explanation:
In nature, decomposers -such as bacteria and fungi- are heterotrophic organisms that require nourishment from organic matter in order to survive, being the last level of the food chains.
<u>The role of decomposers in an ecosystem can be summarized in two main functions</u>:
- <em>Degrading organic matter to convert complex substances into simple substances, which contributes to their own nutrition.</em>
- <em>To provide nutrients to the soil, which will be used by producers, plants, for their growth and development.</em>
An example of this is the degradation of proteins by decomposers, providing nitrogen to the soil that will subsequently be absorbed by plants to synthesize their nutrients.
Learn more:
Decomposers in food chains brainly.com/question/2634998
Answer:
Name Molecular Formula Structural Formula
ethane C2H6 CH3CH3
propane C3H8 CH3CH2CH3
butane C4H10 CH3CH2CH2CH3
pentane C5H12 CH3(CH2)3CH
The volume of ethane gas, C₂H₆ in cm³ that will react to produce 40 cm³ of CO₂ is 20 cm³
<h3>Balanced equation </h3>
2C₂H₆ + 7O₂ —> 4CO₂ + 6H₂O
From the balanced equation above,
4 cm³ of CO₂ were obtained from 2 cm³ of C₂H₆
<h3>How to determine the volume of C₂H₆ needed </h3>
From the balanced equation above,
4 cm³ of CO₂ were obtained from 2 cm³ of C₂H₆
Therefore,
40 cm³ of CO₂ will be obtain from = (40 × 2) / 4 = 20 cm³ of C₂H₆
Thus, 20 cm³ of C₂H₆ are required for the reaction.
Learn more about stoichiometry:
brainly.com/question/14735801
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<h3>
Answer:</h3>
Empirical formula is CrO
<h3>
Explanation:</h3>
<u>We are given;</u>
- Mass of sample of Chromium as 7.337 gram
- Mass of the metal oxide formed as 9.595 g
We are required to determine the empirical formula of the metal oxide.
<h3>Step 1 ; Determine the mass of oxygen used </h3>
Mass of oxygen = Mass of the metal oxide - mass of the metal
= 9.595 g - 7.337 g
= 2.258 g
<h3>Step 2: Determine the moles of chromium and oxygen</h3>
Moles of chromium metal
Molar mass of chromium = 51.996 g/mol
Moles of Chromium = 7.337 g ÷ 51.996 g/mol
= 0.141 moles
Moles of oxygen
Molar mass of oxygen = 16.0 g/mol
Moles of Oxygen = 2.258 g ÷ 16.0 g/mol
= 0.141 moles
<h3>Step 3: Determine the simplest mole number ratio of Chromium to Oxygen</h3>
Mole ratio of Chromium to Oxygen
Cr : O
0.141 mol : 0.141 mol
1 : 1
Empirical formula is the simplest whole number ratio of elements in a compound.
Thus the empirical formula of the metal oxide is CrO