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3 years ago

Describe some tests you might perform to prove if an unknown solid material is either a element or compound.

1 answer:

6 0

Always remember that a compound can be separated into simpler substances by chemical methods/reactions. While elements cannot be broken down into simpler substances by chemical reactions. You can do a flame test and spectrum analysis to determine whether a solid material is an element or a compound. Check the boiling and/or melting point, color or density. Also check the solid material’s reaction with oxygen, hydrogen, calcium, or various acids. Examine and study its physical chemistry. The element(s) that may be present may be identified by checking the absorption edges from an x-ray spectrum.

For instance, a solid white material (A) is heated in the absence of air, decomposing to form a new white material (B) and gas (C). Gas (C) has the same properties as the products when carbon is burned in an excess of oxygen (CO2). How can we determine whether A, B, and C are elements or compounds? A would be a compound since it was broken down into B and C. So how do we determine B and C?

A compound can be decomposed, for the reason that a compound is composed of 2 or more elements. On the contrary, an element cannot be decomposed.

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Which of the molecules represented below contains carbon with sp2 hybridization?

Darya [45]

Answer:

Explanation:

sp² hybridization is found in those compounds having double bond .

Out of the given compounds only C₂H₂Cl₂ has double bond so this compound contains carbon with sp² hybridization .

Rest have sp³ hybridization because they are saturated compounds .

3 0

3 years ago

What is the mass of 8.90 moles of magnesium chloride, MgCl2?

Mazyrski [523]

(8.90/95.211) =0.09347 moles

3 0

2 years ago

Sulfuryl dichloride is formed when sulfur dioxide reacts with chlorine.

zubka84 [21]

Answer: The value of $\Delta G^o$ of the reaction is 28.38 kJ/mol

Explanation:

For the given chemical reaction:

$SO_2(g)+Cl_2(g)\rightarrow SO_2Cl_2(g)$

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

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

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

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(SO_2Cl_2(g))})]-[(1\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(Cl_2(g))})]$

We are given:

$\Delta H^o_f_{(SO_2Cl_2(g))}=-364kJ/mol\\\Delta H^o_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H^o_f_{(Cl_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-364))]-[(1\times (-296.8))+(1\times 0)]=-67.2kJ/mol=-67200J/mol$

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

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

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

$\Delta S^o_{rxn}=[(1\times \Delta S^o_{(SO_2Cl_2(g))})]-[(1\times \Delta S^o_{(SO_2(g))})+(1\times \Delta S^o_{(Cl_2(g))})]$

We are given:

$\Delta S^o_{(SO_2Cl_2(g))}=311.9J/Kmol\\\Delta S^o_{(SO_2(g))}=248.2J/Kmol\\\Delta S^o_{(Cl_2(g))}=223.0J/Kmol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(1\times 311.9)]-[(1\times 248.2)+(1\times 223.0)]=-159.3J/Kmol$

To calculate the standard Gibbs's free energy of the reaction, we use the equation:

$\Delta G^o_{rxn}=\Delta H^o_{rxn}-T\Delta S^o_{rxn}$

where,

$\Delta H^o_{rxn}$ = standard enthalpy change of the reaction =-67200 J/mol

$\Delta S^o_{rxn}$ = standard entropy change of the reaction =-159.3 J/Kmol

Temperature of the reaction = 600 K

Putting values in above equation, we get:

$\Delta G^o_{rxn}=-67200-(600\times (-159.3))\\\\\Delta G^o_{rxn}=28380J/mol=28.38kJ/mol$

Hence, the value of $\Delta G^o$ of the reaction is 28.38 kJ/mol

7 0

3 years ago

When a person has excess stomach acid, he or she may ingest some sodium bicarbonate. The rationale for doing this is that the so

murzikaleks [220]

A base will neutralize the excess acid

4 0

2 years ago

Read 2 more answers

How many grams of sodium sulfide can be produced when 45.3 g Na react with 105 g S?

RoseWind [281]

The chemical reaction would be as follows:

2Na + S → Na2S

We are given the amount of the reactants to be used in the reaction. We use these to calculate the amount of product. We do as follows:

45.3 g Na ( 1 mol / 22.99 g ) = 1.97 mol Na
105 g S ( 1 mol / 32.06 g ) = 3.28 mol S

The limiting reactant would be Na. We calculate as follows:

1.97 mol Na ( 1 mol Na2S / 2 mol Na ) (78.04 g / mol ) = 76.87 g Na2S produced

8 0

2 years ago