3Na2O(at) + 2Al(NO3)3(aq) —> 6NaNO3(aq) + Al2O3(s)
This is a double replacement reaction and NaNO3 is aqueous because Na is an alkali metal, plus nitrate is in the solution. Both of these are soluble. Al2O3 is not soluble because it does not contain any element that is soluble and is hence the precipitate.
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Answer:
Flammability.
Explanation:
Chemical properties of a substance are properties of a substance that occur in the chemical reactions in which it is involved. These are the basis for the classification of substances, such as metals, halogens, acids, alcohols, alkenes, arenes, etc. A class of substances is characterized by chemical properties common to that group. Differences in the chemical and physical properties of substances allow the substance to be identified, and the components of a mixture of substances to be separated and purified.
The chemical properties of a substance are its chemical stability, reactivity, oxidizability, acidity, bromine number, polymerisability, flammability, toxicity, etc. These can be determined by chemical exposure.
According to octet rule element has eight electrons in its valence shell.This give it the same electronic configuration as a group 18 elements.Atom with electronic configuration 1s2 2s2 2p1 require five electrons to achieve octet. This is because the valence electrons are 3 in 2s2 and 2p1 and hence require 5 electrons to be filled.
Answer:
45.8 mL
Explanation:
If all variables are held constant, the new volume can be found using the Boyle's Law equation. The equation looks like this:
P₁V₁ = P₂V₂
In this equation, "P₁" and "V₁" represent the initial pressure and volume. "P₂" and "V₂" represent the final pressure and volume. You can find the new volume by plugging the given values into the equation and simplifying.
P₁ = 3.1 atm P₂ = 10.5 atm
V₁ = 155 mL V₂ = ? mL
P₁V₁ = P₂V₂ <----- Boyle's Law equation
(3.1 atm)(155 mL) = (10.5 atm)V₂ <----- Insert values
480.5 = (10.5 atm)V₂ <----- Multiply 3.1 and 155
45.8 = V₂ <----- Divide both sides by 10.5
Exothermic gives off heat/energy and endothermic takes in heat/energy. Exothermic example: a candle flame
Endothermic example: baking bread
In Exothermic, you can expect the surrounding temp. to rise, and in Endothermic you can expect the surrounding temperature to fall.
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