All categories

2 years ago

Which is the best example of tropism in plants?

Physics

1 answer:

seraphim [82]2 years ago

3 0

Answer:

Here are some examples of tropism in plants : Sunflowers are one of the best examples of positive phototropism , as they always grow facing the sun.

Explanation:

Although some experts consider that perhaps such a clear growth movement should not be considered tropism.

You might be interested in

archer shoots his arrow towards a target at a distance of 90 m, and hits ‘bullseye’ . Calculate the acceleration and time taken

Cerrena [4.2K]

Answer:

Acceleration = 45m/s^2
Time = 2 s

Explanation:

6 0

3 years ago

In science, a hypothesis must be true.

Anon25 [30]

I would believe that this is false.

6 0

3 years ago

it is defined as the intersection of two lines;describes a position in space and has no length, width or height

Ksivusya [100]

Answer:

Could you explain more?

Explanation:

5 0

2 years ago

A tornado warning means that the current weather conditions could produce a tornado. True or False

WARRIOR [948]

true because it means that the cloud formation of a tornado has been already spotted.

4 0

3 years ago

Read 2 more answers

A 12.5843-gram sample of metal bromide, MBr4, was dissolved and, after reaction with silver nitrate, AgNO3, all of the bromide w

barxatty [35]

Answer:

zirconium

Explanation:

Given, Mass of AgBr(s) = 23.0052 g

Molar mass of AgBr(s) = 187.77 g/mol

The formula for the calculation of moles is shown below:

$Moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles\ of\ AgBr= \frac{23.0052\ g}{187.77\ g/mol}$

$Moles\ of\ AgBr= 0.1225\ mol$

The reaction taking place is:

$MBr_4+4AgNO_3\rightarrow 4AgBr+M(NO_3)__4$

From the reaction,

4 moles of AgBr is produced when 1 mole of $MBr_4$ undergoes reaction

1 mole of AgBr is produced when 1 / 4 mole of $MBr_4$ undergoes reaction

0.1225 mole of AgBr is produced when $\frac {1}{4}\times 0.1225$ mole of $MBr_4$ undergoes reaction

Moles of $MBr_4$ got reacted = 0.030625 moles

Mass of the sample taken = 12.5843 g

Let the molar mass of the metal = x g/mol

So, Molar mass of $MBr_4$ = x + 4 × 79.904 g/mol = 319.616 + x g/mol

Thus,

$0.030625 = \frac{12.5843}{319.616 + x}$

Solve for x,

we get, x = 91.2999 g/mol

<u>The metal shows +4 oxidation state and has mass of 91.2999 g/mol . The metals is zirconium.</u>

8 0

2 years ago