Answer:
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.
Answer:
A delta is formed when the river deposits its material faster than the sea can remove it. ... Cuspate - the land around the mouth of the river juts out arrow-like into the sea. The Ebro Delta. Bird's foot - the river splits on the way to the sea, each part of the river juts out into the sea, rather like a bird's foot.
I hope it's helpful!
Answer:
B. To change from a liquid state to a solid state is called Freezing
Answer:
The molarity of urea in this solution is 6.39 M.
Explanation:
Molarity (M) is <em>the number of moles of solute in 1 L of solution</em>; that is
To calculate the molality, we need to know the number of moles of urea and the volume of solution in liters. We assume 100 grams of solution.
Our first step is to calculate the moles of urea in 100 grams of the solution,
using the molar mass a conversion factor. The total moles of 100g of a 37.2 percent by mass solution is
60.06 g/mol ÷ 37.2 g = 0.619 mol
Now we need to calculate the volume of 100 grams of solution, and we use density as a conversion factor.
1.032 g/mL ÷ 100 g = 96.9 mL
This solution contains 0.619 moles of urea in 96.9 mL of solution. To express it in molarity, we need to calculate the moles present in 1000 mL (1 L) of the solution.
0.619 mol/96.9 mL × 1000 mL= 6.39 M
Therefore, the molarity of the solution is 6.39 M.
Answer:
Explanation:
A 12.48 g sample of an unknown metal, heated to 99.0 °C was then plunged into 50.0 mL of 25.0 °C water. The temperature of the water rose to 28.1 Go to calculating final temperature when mixing two samples of water ... Problem #1: A 610. g piece of copper tubing is heated to 95.3 °C and placed in an ... The two rings are heated to 65.4 °C and dropped into 12.4 mL of water at 22.3 °C. ... Problem #4: A 5.00 g sample of aluminum (specific heat capacity = 0.89 J g¯1
