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muminat
2 years ago
9

Define an Arrhenius base and describe properties of bases. Use an example to explain how an Arrhenius base will behave in water.

Chemistry
1 answer:
frozen [14]2 years ago
8 0

An Arrhenius base is a molecule that when dissolved in water will break down to yield an OH^-or hydroxide ion in solution.

<h3>What is Arrhenius base?</h3>

An Arrhenius base is a compound that increases the OH^- ion concentration in aqueous solution.

An Arrhenius base is a substance that, when dissolved in an aqueous solution, increases the concentration of hydroxide, or , OH^- ions in the solution.

Bases Properties

Arrhenius bases that are soluble in water can conduct electricity.

Bases often have a bitter taste and are found in foods less frequently than acids. Many bases, like soaps, are slippery to the touch.

Bases also change the colour of indicators. Red litmus turns blue in the presence of a base (see figure below), while phenolphthalein turns pink.

Some bases react with metals to produce hydrogen gas.

Acids (pH < 7.0) react with bases (pH > 7.0) to produce a salt and water. When equal moles of an acid and a base are combined, the acid is neutralized by the base. The resulting mixture will have a more neutral pH.

An Arrhenius acid is a substance that dissociates in water to form hydrogen ions or protons. In other words, it increases the number of H^+ ions in the water. In contrast, an Arrhenius base dissociates in water to form hydroxide ions OH^-.

Example, sodium hydroxide, is added to an aqueous solution. NaOH dissociates into sodium, Na^+, and hydroxide, OH^- ions.

Learn more about the Arrhenius bases here:

https://brainly.in/question/8273595

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Answer:

Erosion/Weathering

Explanation:

Because the rock is exposed to the ocean, the constant push and pull of the water against the rock erodes the material away and moves it elsewhere. The wind likely aided in this process as well. This caused the big gap in the center of the rock.

6 0
3 years ago
Consider the reaction 2CO(g) + O2(g)2CO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr
Marizza181 [45]

<u>Answer:</u> The value of \Delta S^o for the surrounding when given amount of CO is reacted is 432.52 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]

For the given chemical reaction:

2CO(g)+O_2(g)\rightarrow 2CO_2(g)

The equation for the entropy change of the above reaction is:  

\Delta S^o_{rxn}=[(2\times \Delta S^o_{(CO_2(g))})]-[(1\times \Delta S^o_{(O_2(g))})+(2\times \Delta S^o_{(CO(g))})]

We are given:

\Delta S^o_{(CO_2(g))}=213.74J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(CO)}=197.674J/K.mol

Putting values in above equation, we get:

\Delta S^o_{rxn}=[(2\times (213.74))]-[(1\times (205.14))+(2\times (197.674))]\\\\\Delta S^o_{rxn}=-173.008J/K

Entropy change of the surrounding = - (Entropy change of the system) = -(-173.008) J/K = 173.008 J/K

We are given:

Moles of CO gas reacted = 2.25 moles

By Stoichiometry of the reaction:

When 2 moles of CO is reacted, the entropy change of the surrounding will be 173.008 J/K

So, when 2.25 moles of CO is reacted, the entropy change of the surrounding will be = \frac{173.008}{1}\times 2.25=432.52J/K

Hence, the value of \Delta S^o for the surrounding when given amount of CO is reacted is 432.52 J/K

3 0
3 years ago
A sample of hydrated magnesium sulfate (MgSO4)
yaroslaw [1]

Answer:

MgSO4.7H2O

Explanation:

Let the formula for the hydrated magnesium sulphate be MgSO4.xH2O

Mass of the hydrated salt (MgSO4.xH2O) = 12.845g

Mass of anhydrous salt (MgSO4) = 6.273g

Mass of water molecule(xH2O) = Mass of the hydrated salt — Mass of anhydrous salt = 12.845 — 6.273 = 6.572g

Now,we can obtain the number of mole of water molecule present in the hydrated salt as follows:

Molar Mass of hydrated salt (MgSO4.xH2O) = 24 + 32 + (16x4) + x(2 + 16) = 24 + 32 + 64 + x(18) = 120 + 18x

Mass of xH2O/ Molar Mass of MgSO4.xH2O = Mass of water / mass of hydrated salt

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Cross multiply to express in linear form

18x x 12.845 = 6.572(120 + 18x)

231.21x = 788.64 + 118.296x

Collect like terms

231.21x — 118.296x = 788.64

112.914x = 788.64

Divide both side by 112.914

x = 788.64 /112.914

x = 7

Therefore the formula for the hydrated salt (MgSO4.xH2O) is MgSO4.7H2O

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