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DedPeter [7]
4 years ago
13

A synthesis reaction can be identified because it only has pure elements on the left side of the arrow and compounds on the righ

t side of the arrow.
True
False
Chemistry
2 answers:
masha68 [24]4 years ago
6 0

Actually, this would be false.

Alik [6]4 years ago
5 0

Answer: False

Explanation:

Element is a pure substance which is composed of atoms of similar elements.It can not be decomposed into simpler constituents using chemical reactions. Example: Hydrogen H_2

Compound is a pure substance which is made from atoms of different elements combined together in a fixed ratio by mass.It can be decomposed into simpler constituents using chemical reactions. Example: water H_2O

Synthesis reaction is a type of reaction in which multiple reactants combine to form a single product.Compounds can also combine to give other compounds as in the given reaction: BaO+CO_2\rightarrow BaCO_3

Thus the statement is false.

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Hot material near Earth's surface is more dense and sinks, and when it cools, it becomes less dense and rises.

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Convection is a process by which less dense material rises and more dense material sinks near Earth's surface. Water, air, rocks and other materials expands when temperature rises and they become less dense. When the water is heated it becomes vapors and carries away thermal energy from the water.

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Explanation:

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The standard free energy of activation of a reaction A is 81.9 kJ mol–1 (19.6 kcal mol–1) at 298 K. Reaction B is one million ti
klasskru [66]

Answer:

54.9 kJ/mol

Explanation:

The relation between the activation energy (Ea) and the rate constant (k) is given by the Arrhenius equation.

k=A.e^{-Ea/RT}

where,

A is a collision factor

R is the ideal gas constant

T is the absolute temperature

Reaction B is one million times faster than reaction A at the same temperature. So k_{B}=10^{6} k_{A}.

Then,

k_{B}=10^{6} k_{A}\\A.e^{-Ea_{B}/RT}=10^{6}A.e^{-Ea_{A}/RT}\\e^{-Ea_{B}/RT}=10^{6}e^{-Ea_{A}/RT}\\ln(e^{-Ea_{B}/RT})=ln(10^{6}e^{-Ea_{A}/RT})\\\frac{-Ea_{B}}{RT} =ln10^{6} -\frac{Ea_{A}}{RT} \\Ea_{B}=(ln10^{6} -\frac{Ea_{A}}{RT}).(-RT)=(ln10^{6}-\frac{89.1kJ/mol}{(8.314\times 10^{-3} kJ/mol.K).298K} ).(-8.314\times 10^{-3} \frac{kJ}{mol.K}.298K )=54.9kJ/mol

8 0
3 years ago
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