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

Answers: 3

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Answers

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

nacl is added to the aqueous of what? ? in order to get a proper answer you have to specify the exact composition of the aqueous solution.

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.