Parenchyma is the most simple and specialized tissue which is concerned mainly with the vegetative activities of the plant. The cells are isodiametric with well-developed intercellular spaces, vacuolated cytoplasm and cellulosic cell wall.

Collenchyma is the tissue of the primary body. The cells of the tissue contain protoplasm and are living without intercellular spaces. The cell wall articulate at the corners and are made up of cellulose, hemicellulose, and pectin.

Sclerenchyma is the thick-walled cell tissue. In the beginning, the cell is living and have protoplasm, but due to deposition of impermeable secondary board lignin, they become dead thick and hard.

5 0

3 years ago

Which law states that the volume and absolute temperature of a fixed quantity of gas are directly proportional under constant pr

Delvig [45]

<h2>Hello!</h2>

The answer is: Charle's Law.

The law that states that the volume and absolute temperature of a fixed quantity of gas (ideal gas) are proportional under constant pressure is the Charle's Law, also known as the law of volumes.

The law describes how a gas kept under constant pressure tends to expand when the temperature increases and it's described by the following equation:

$\frac{V}{T}=k$

Where,

$V=Volume\\T=Temperature\\k=constant$

Also, to describe the relationship between two differents volumes at different temperatures, we have:

$\frac{V_{i}}{T_{i}}=\frac{V_{f}}{T_{f}}$

Where,

$V_{i}=InitialVolume\\T_{i}=InitialTemperature\\V_{f}=FinalVolume\\T_{f}=FinalTemperature$

Have a nice day!

8 0

3 years ago

Read 2 more answers

If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi

zlopas [31]

Answer:

$T_{eq}=28.9\°C$

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

$Q_{Cd}+Q_{W}=0$

Thus, we insert mass, specific heat and temperatures to obtain:

$m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0$

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

$T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}$

Now, we plug in to obtain:

$T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C$

NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.

Best regards!

6 0

3 years ago