Answer:
<em><u>V = 25 m/s</u></em>
Explanation:
From first equation of motion, V = U + at
V = 10 + ( 5×3 )
V = 25 m/s
Answer:
ooooooooooooooooooh i dont know this one
Explanation:
im not good a math at all
Answer:
18 g
Explanation:
We'll begin by converting 500 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
500 mL = 500 mL × 1 L / 1000 mL
500 mL = 0.5 L
Next, we shall determine the number of mole of the glucose, C₆H₁₂O₆ in the solution. This can be obtained as follow:
Volume = 0.5 L
Molarity = 0.2 M
Mole of C₆H₁₂O₆ =?
Molarity = mole / Volume
0.2 = Mole of C₆H₁₂O₆ / 0.5
Cross multiply
Mole of C₆H₁₂O₆ = 0.2 × 0.5
Mole of C₆H₁₂O₆ = 0.1 mole
Finally, we shall determine the mass of 0.1 mole of C₆H₁₂O₆. This can be obtained as follow:
Mole of C₆H₁₂O₆ = 0.1 mole
Molar mass of C₆H₁₂O₆ = (12×6) + (1×12) + (16×6)
= 72 + 12 + 96
= 180 g/mol
Mass of C₆H₁₂O₆ =?
Mass = mole × molar mass
Mass of C₆H₁₂O₆ = 0.1 × 180
Mass of C₆H₁₂O₆ = 18 g
Thus, 18 g of glucose, C₆H₁₂O₆ is needed to prepare the solution.
Explanation:
Transpiration is the movement of water through plants using the xylem and also their loss from plants surfaces.
Water is a lost from plants from the leaves and the stomata. This water is usually replaced by the absorption of water by the root.
- Transpiration is an important part of the water cycle in the biosphere.
- Significant amount of water is lost from plant in the process of transpiration.
- Transpiration water condenses in the atmosphere to form rain clouds.
An example of transpiration when water loses water from their leaves.
