Answer:
They need to use energy from other tress to survive so that they can do basic functions, like absorbing nutrients from the ground, and growing back leaves. They don't have chlorophyll, so they need that energy to do the functions I mentioned above.
Explanation: Hope this helps!
Answer:
The rock cycle
Explanation:
The three types of rocks are related and connected through the rock cycle. Rocks are not formed independently of one another, the process of their formation are interconnected.
Igneous rocks are made from cooling and solidification of magma. These magma are derived from the melting of rocks that have been subjected to high temperature and pressure.
When igneous rocks are formed, they can follow two path ways. They either get transformed to sedimentary rocks or metamorphic rocks.
When igneous rocks are subjected to metamorphic conditions, their mineralogy is altered to form metamorphic rocks.
If igneous rocks gets weathered and broken down by agents of denudation, they end up getting transformed into sedimentary rocks.
Answer: The percentage by mass of sulphur in
is 9.36%
Explanation:
Mass percent of an element is the ratio of mass of that element by the total mass expressed in terms of percentage.

Given: mass of sulphur = 32 g/mol
mass of
= 342 g/mol
Putting in the values we get:

The percentage by mass of sulphur in
is 9.36%
1.66 M is the concentration of the chemist's working solution.
<h3>What is molarity?</h3>
Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.
In this case, we have a solution of Zn(NO₃)₂.
The chemist wants to prepare a dilute solution of this reactant.
The stock solution of the nitrate has a concentration of 4.93 M, and he wants to prepare 620 mL of a more dilute concentration of the same solution. He adds 210 mL of the stock and completes it with water until it reaches 620 mL.
We want to know the concentration of this diluted solution.
As we are working with the same solution, we can assume that the moles of the stock solution will be conserved in the diluted solution so:
=
(1)
and we also know that:
n = M x 
If we replace this expression in (1) we have:
x
=
x 
Where 1, would be the stock solution and 2, the solution we want to prepare.
So, we already know the concentration and volume used of the stock solution and the desired volume of the diluted one, therefore, all we have to do is replace the given data in (2) and solve for the concentration which is
:
4.93 x 210 = 620 x
= 1.66 M
This is the concentration of the solution prepared.
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The mass will stay the same because of the conservation of mass