Chemistry, 11.02.2020 22:26 aprilreneeclaroxob0c

A group of students used this diagram to classify three organisms into different kingdoms.
Which table correctly identifies the three kingdoms?
A) A
B) B
C) C
D) D

A group of students used this diagram to classify three organisms into different kingdoms. Which tab

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Answer from: binkyv15

The correct classification is option C since plants posses cell walls and chloroplasts while fungi posses cell walls but no chloroplasts. Plant cells do not posses both cell walls and chloroplasts.

In taxonomy, living things are classified into kingdoms, phyla, classes, genius and species. The essence of these classification is for easy reference and also to help scientists to easily study the relationship between organisms.

The classification in table C is correct because plants are multicellular organisms that posses cell walls and chloroplasts. Fungi posses cell walls and no chloroplasts but they are multicellular organisms. Animals are multicellular organisms that possess no chloroplasts and no cell walls.

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Answer from: Blakemiller2020

The correct option is C

Explanation:

Plants contain a cell wall made up of cellulose. They also contain chloroplasts for carrying out photosynthesis which enables plants to make their own food. Hence, Kingdom 1 is plantae.

Fungal cells comprise of a cell wall which is made up of chitin. These organisms do not possess chloroplast and hence cannot make their won food. Hence, Kingdom 2 is fungi.

Animals cells do not possess a cell wall as well as chloroplast. Hence, Kindgom 3 in animalia.

Hence, the correct option is C.

Answer from: Quest

The answer is the first option just took the test

Answer from: Quest

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 : $4x (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) h rxn = -600 kj/mol$

equation 2 : $2x (s) + 3cl_{2} (g) \rightarrow 2xcl_{3} (s) h rxn = -800 kj/mol$

equation 3 : $4xcl_{3} (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) + 6cl_(2} h rxn = -200 kj/mol$

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 : $4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+800) \times 2 kj/mol$

let us add equation 1 & 4

$4x (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) h rxn = -600 kj/mol$

$4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+1600) kj/mol$

$4xcl_{3} )s) + 3o_{2} \rightarrow 2x_{2}o_{3} (s) + 6cl_{2} (g) h rxn = (+1000) kj/mol$

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

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