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4 years ago

How to compare and contrast a solution of hydrogen chloriide gas in water and toluene?

1 answer:

4 0

<span>The solute is the substance that is being dissolved while the solvent is the base that the solute is bring dissolved in. For example, in salt water, salt would be the solute that dissolves into the water, and the water is the solvent that the salt is being dissolved in.</span>

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What is the name of this alkane?

Otrada [13]

It's a cyclohexane ring with an ethyl group at 1 and a methyl group at 3. The Ethyl group is bigger and more important group get's the first position.

1-ethyl-3-methylcyclohexane

3 0

4 years ago

Read 2 more answers

Calculate ΔH∘f (in kilojoules per mole) for benzene, C6H6, from the following data: 2C6H6(l) + 15O2(g)→12CO2(g)+6H2O(l) ΔH∘=−653

Pachacha [2.7K]

Answer: 48.6 kJ/mol

Explanation:

The balanced chemical reaction is,

$2C_6H_6(l)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

$\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_6H_6}\times \Delta H_{C_6H_6})]$

where,

n = number of moles

$\Delta H_{O_2}=0$ (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

$-6534.0=[(12\times -393.5)+(6\times -285.8)]-[(15\times 0)+(2\times \Delta H_{C_6H_6})]$

$\Delta H_{C_6H_6}=48.6kJ/mol$

Therefore, the enthalpy change for benzene is 48.6 kJ/mol

4 0

3 years ago

Heat is being removed from a substance, but its temperature does not change. What is

ololo11 [35]

So its temperature will not rise, since kinetic energy of molecules remains the same. The quantity of heat absorbed or released when a substance changes its physical phase at constant temperature (e g. From solid to liquid at melting point or from liquid to gas at boiling point) is termed as its latent heat.

4 0

2 years ago

The following reaction shows the products when sulfuric acid and aluminum hydroxide react.

SSSSS [86.1K]

Answer:

There will remain 11.47 grams of Al(OH)3

Explanation:

Step 1: Data given

Mass of sulfuric acid = 35.0 grams

Molar mass sulfuric acid = 98.08 g/mol

Mass of aluminium hydroxide = 30.0 grams

Molar mass of aluminium hydroxide = 78.0 g/mol

Step 2: The balanced equation

2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles Al(OH)3 = 30.0 grams / 78.0 g/mol

Moles Al(OH)3 = 0.385 moles

Moles H2SO4 = 35.0 grams / 98.08 g/mol

Moles H2SO4 = 0.357 moles

Step 4: Calculate the limiting reactant

For 2 moles Al(OH)3 we need 3 moles H2SO4 to produce 1 mol Al(SO4)3 and 6 moles H20

H2SO4 is the limiting reactant. It will completely be consumed ( 0.357 moles). Al(OH)3 is in excess. There will be react 2/3 * 0.357 = 0.238 moles

There will remain 0.385 - 0.238 = 0.147 moles

Mass of Al(OH)3 remaining = 0.147 moles* 78.0 g/mol = 11.47 grams

There will remain 11.47 grams of Al(OH)3

6 0

3 years ago

Which process begins the formation of sedimentary rock? the movement of sediment the cementation of rock sediment the breakdown

max2010maxim [7]

Answer:

C- the breakdown of rock into sediment

Explanation:

Because sedimentary rocks have layers and are made of of those layers which are made up of the breakdown of rock into sediment.

hope this helps:)

5 0

3 years ago