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3 years ago

Which of the following is a legume?

Chemistry

2 answers:

nirvana33 [79]3 years ago

5 0

Answer:

The answer to your question is Clover

Explanation:

Legumes are plants or the seed of plants. Legumes are harvested for human consumption, for livestock forage and silage.

These plants are also important during the Nitrogen cycle due to they fix nitrogen.

Examples of legumes are:

Beans, alfalfa, clover, lentils, peas, mesquite, carob, tamarind, peanuts, soybeans, etc.

ololo11 [35]3 years ago

3 0

Answer:

Clover because legume is a plant in the family Fabaceae, or the fruit or seed of such a plant.

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diamong [38]

Answer:

The attractive force between them decreases

Explanation:

This is because they become localised.

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3 years ago

An astronaut weighs 104 newtons on the moon,where the strength of gravity is 1.6 newtons per kilogram

belka [17]

I'd say he ways about 35 kilograms, but I'm probably wrong, xD

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3 years ago

Read 2 more answers

Many double-displacement reactions are enzyme-catalyzed via the "ping pong" mechanism, so called because the reactants appear to

zhenek [66]

Answer:

D. It will decrease by a factor of 4

Explanation:

According to the question , the equation follows :

$A+B\rightarrow C+D$

Rate law : This states the rate of reaction is directly proportional to concentration of reactants with each reactant raised to some power which may or may not be equal to the stoichiometeric coefficient.

$Rate\ \alpha [A]^{a}[B]^{b}$

$r=[A]^{a}[B]^{b}$ .................(1)

STEP": First, find out the power "a" and "b"

a+b = 3 (because it is given that the reaction follow 3rd order-kinetics)

According to question, doubling the concentration of the first reactant causes the rate to increase by a factor of 2 means,

r' = 2r if [A'] = 2[A]

Here [B] is uneffected means [B']=[B]

hence new rate =

$r'=[A']^{a}[B']^{b}$

Put the value of [A'] , [B'] and r' in the above equation:

$2r=[2A]^{a}[B]^{b}$ ...........(2)

Divide equation (1) by (2) we , get

$\frac{2r}{r}=\frac{[2A]^{2}[B]^{b}}{[A]^{a}[B]^{b}}$

$2= 2(\frac{A}{A})^{a}\times (\frac{B}{B})^{b}$

Here A and A cancel each other

B and B cancel each other

We get,

$2= 2^{a}\times 1^{b}$

1^b = 1 ( power of 1 = 1)

$2= 2^{a}$

This is possible only when a = 1

We know that : a + b = 3

1 + b = 3

b =3 -1 = 2

b = 2

Hence the rate law becomes :

$r=[A]^{a}[B]^{b}$

$r=[A]^{1}[B]^{2}$ .............(3)

Look in the question now, it is asked to calculate the concentration of [B],if cut in half

Hence

[B']=1/2[B]

Insert the value of [B'] in equation (3)

$r'=[A]^{1}[B']^{2}$

$r'=[A]^{1}(\frac{1}{2}[B])^{2}$

$r'=\frac{1}{4}[A]^{1}[B]^{2}$ ............(a)

But

$r=[A]^{a}[B]^{b}$ ..............(b)

Compare equation (a) and (b) , we get

new rate r' =

r' = 1/4 r

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3 years ago

If you have a heartburn, how you could you safely help the situation

Ket [755]

The answer is:

:Avoid lying down after eating

That should help

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3 years ago

Calculate the Density of the substances below to determine if they will float or sink in water.

Reil [10]

Explanation:

If the density of the object is more than that of water, it will sink. Otherwise it will float. The density of water is 1 g/mL.

Substance 1,

Mass, m = 450 g, Volume, V = 90 mL

Density = mass/volume

So,

$d_1=\dfrac{450}{90}\\\\=5\ g/mL$

It will sink.

Substance 2,

Mass, m = 35 g, Volume, V = 70 mL

Density = mass/volume

So,

$d_2=\dfrac{35}{70}\\\\=0.5\ g/mL$

It will float.

Substance 3,

Mass, m = 24 g, Volume, V = 12 mL

Density = mass/volume

So,

$d_3=\dfrac{24}{12}\\\\=2\ g/mL$

It will sink.

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3 years ago