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3 years ago

Li(OH) --> Li2O + H20

Chemistry

1 answer:

Zigmanuir [339]3 years ago

5 0

Answer:

2Li(OH)-->Li2O+H20

Explanation:

There are 2Li, 2H and 2O on each side

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explain how energy changes from one form to another in a exothermic reaction. in an endothermic reaction.

Alexxx [7]

Answer:

Exothermic reaction: In exothermic reaction, energy is transferred to the surroundings, and the surrounding temperature increases, this is known as exothermic reaction. In other words energy exits in exothermic reaction. Some example of exothermic reactions are:

1) Neutralisation reaction.

2) Combustion reaction.

3) Some oxidation reaction.

Endothermic reaction: In endothermic reaction, energy is taken in from the surrounding, and the surrounding temperature decreases, this is known as endothermic reaction. In other words energy enters in endothermic reaction. Some example of exothermic reactions are:

1) Thermal decomposition.

2) Reaction between citric acid and sodium hydrogen carbonate.

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3 years ago

13. As heat is added to a lúguid that is boiling at constant

11Alexandr11 [23.1K]

Tbh I would pick A because the liquid* decreases. B doesn’t increase

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3 years ago

Question.2

fiasKO [112]

Answer:i dont know the answer

3 0

3 years ago

Suppose 0.795 g of sodium iodide is dissolved in 100. mL of a 39.0 m M aqueous solution of silver nitrate.

lubasha [3.4K]

Answer:

The final molarity of iodide anion is 0.053 M

Explanation:

Step 1: Data given

Mass of sodium iodide (NaI) = 0.795 grams

Volume of the solution = 100 mL = 0.1 L

Molarity of aqueous solution of silver nitrate (AgNO3) = 39 mM = 0.039M

The molecular mass of sodium iodide is 149.89 g/mol.

Step 2: The balanced equation

AgNO3(aq) + NaI(aq) → AgI(s) + NaNO3(aq)

Step 3: Calculate number of moles of sodium iodide

Moles NaI = mass NaI / Molar mass NaI

Moles NaI = 0.795 grams / 149.89 g/mol

Moles NaI = 0.0053 moles

For 1 mole AgNO3 consumed, we need 1 mole NaI to produce 1 mole AgI and 1 mole NaNO3

The sodium iodide will dissociate as followed:

NaI(aq) → Na+(aq) + I-(aq)

Step 4: Calculate iodide ions

For 1 mole NaI, we have 1 mole of I-

For 0.0053 moles of NaI we'll have 0.0053 moles I-

Step 5: Calculate molarity of iodide ion

Molarity = moles I- / volume

Molarity I- = 0.0053 moles / 0.1 L

Molarity I- = 0.053 M

The final molarity of iodide anion is 0.053 M

5 0

3 years ago

How many molecules of ethanol (c2h5oh) (the alcohol in alcoholic beverages) are present in 155 ml of ethanol? the density of eth

ozzi

First convert volume to mass:

mass = 0.789 g/mL * 155 mL = 122.295 g

Then convert mass to number of moles:

number of moles = 122.295 g * (1 mole / 46 g) = 2.66 moles

Using avogadros number, we get the molecules:

number of molecules = 2.66 moles * 6.022 x 10^23 molecules / mole = 1.6 x 10^24 molecules

4 0

3 years ago