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Ahat [919]
3 years ago
11

Difference between vegetative and reproductive part​

Chemistry
2 answers:
andrezito [222]3 years ago
8 0

Answer:

The vegetative parts includes the roots, stems,leaves and the reproductive parts include flowers. ... The vegetative parts can reproduce new plants by the process of vegetative propagation but the reproductive parts cannot produce new plants directly instead they produce precursor of new plants that are seeds or fruits.

gavmur [86]3 years ago
6 0
The vegetative parts includes the roots, stems,leaves and the reproductive parts include flowers. ... The vegetative parts can reproduce new plants by the process of vegetative propagation but the reproductive parts cannot produce new plants directly instead they produce precursor of new plants that are seeds or fruits.

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Prokaryote cells are single celled organisms
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Express the result of the following calculation in scientific notation: 0.0263 cm2÷ 88.2 cm
VikaD [51]

<u>Answer:</u> The result of the given calculation is 2.98\times 10^{-4}cm

<u>Explanation:</u>

Scientific notation is defined as the notation in which a number is expressed in the decimal form. This means that the number is always written in the power of 10 form.

We are given an expression:

\Rightarrow \frac{0.0263cm^2}{88.2cm}

The operation carried out in the above expression is division operation. The units are also treated in the same way.

The solution for the above expression is:

\Rightarrow \frac{0.0263cm^2}{88.2cm}=0.000298cm

The value in scientific notation is 2.98\times 10^{-4}cm

Hence, the result of the given calculation is 2.98\times 10^{-4}cm

3 0
4 years ago
Now we need to find the amount of NF3 that can be formed by the complete reactions of each of the reactants. If all of the N2 wa
Alex777 [14]

The question is incomplete, the complete question is:

Nitrogen and fluorine react to form nitrogen fluoride according to the chemical equation:

N_2(g)+3F_2(g)\rightarrow 2NF_3(g)

A sample contains 19.3 g of N_2 is reacted with 19.3 g of F_2. Now we need to find the amount of NF_3 that can be formed by the complete reactions of each of the reactants.

If all of the N_2 was used up in the reaction, how many moles of NF_3 would be produced?

<u>Answer:</u> 1.378 moles of NF_3 are produced in the reaction.

<u>Explanation:</u>

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}       ......(1)

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

In the given chemical reaction, N_2 is considered as a limiting reagent because it limits the formation of the product and it was completely consumed in the reaction.

We are given:

Mass of N_2 = 19.3 g

Molar mass of N_2 = 28.02 g/mol

Putting values in equation 1:

\text{Moles of }N_2=\frac{19.3g}{28.02g/mol}=0.689mol

For the given chemical reaction:

N_2(g)+3F_2(g)\rightarrow 2NF_3(g)

By the stoichiometry of the reaction:

1 mole of N_2 produces 2 moles of NF_3

So, 0.689 moles of N_2 will produce = \frac{2}{1}\times 0.689=1.378mol of NF_3

Hence, 1.378 moles of NF_3 are produced in the reaction.

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Which is a a characteristic of a strong acid ?
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Answer:

A strong acid ionises COMPLETELY in water to produce HIGH conc. of hydrogen ion

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Hope helped you!!
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