Answer:
c) the study of matter and the changes it undergoes
Explanation:
Chemistry -
It is the sub topic of science , dealing with the study of matter , the properties of matter , the type of interaction between the particles of matter , the reason for the particle of matter to combine and separate in order to form new substance .
The basic concepts of chemistry are applicable on the day to day activities.
Hence, from the given options , the correct statement is c) the study of matter and the changes it undergoes.
The Law of conservation of mass states that option C: matter is neither created nor destroyed.
<h3>What is the law of conservation of matter?</h3>
Physical and chemical changes can cause matter to transform into different forms, but no matter what happens, matter is always conserved. There is no creation or destruction of matter; the amount of matter is the same before and after the transformation.
The principle of matter conservation. argues that matter cannot be generated or destroyed during a chemical reaction. The same number of atoms exist before and after the alterations even though the matter may shift from one form to another. reactant.
Therefore, According to the principle of mass conservation, neither chemical processes nor physical changes can create or destroy mass in an isolated system. The mass of the products and reactants of a chemical reaction must be equal, in accordance with the law of conservation of mass.
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See full question below
1. Multiple-choice
Q.
Conservation of matter article questions
Law of conservation of mass states that
answer choices
matter is created
matter is destroyed
matter is neither created nor destroyed
matter does not change
Answer:
There are 2 types of forces, contact forces and act at a distance force. Every day you are using forces. Force is basically push and pull. When you push and pull you are applying a force to an object.
Explanation:
Answer:
<h2>
= (
1.08 /
2.2
) 100% = 49%</h2>
Explanation:
Balanced Equation: 2CH₃CH₃(g) + 5O₂(g) → 2CO₂(g) + 6H₂O(g)
Calculate moles of CH₃CH₃ and O₂
1.2 ₃₃ (
1 ₃₃/
30.0694 ₃₃
) = 0.040 ₃₃
8.6 ₂ (
1 2/
31.998 ₂
) = 0.27 ₃₃
Find limiting reagent 0.040 ₃₃ (
5 ₂/
2 ₃₃
) = 0.10 ₂
CH₃CH₃ is the limiting Reagent
CH₃CH₃ (L.R.) O₂ CO₂ H₂O
Initial (mol) 0.040 0.27 0 0
Change
(mol)
-2x=-0 -5x=
-0.10 +2x=+0.040 +6x=+0.12
Final (mol) 0 0.117 0.040 0.12
0.040 − 2 = 0 = 0.020
Determine percent yield
0.12 ₂ (
18.0148 ₂
/1 ₂
) = 2.2 ₂
= (
1.08 /
2.2
) 100% = 49%
