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GrogVix [38]
3 years ago
14

In the process of attempting to characterize a substance, a chemist makes the following observations: (a) The substance is a sil

very white, lustrous metal. (b) The substance melts at 649 ∘C, (c) The substance boils at 1105 ∘C. (d) The density of the substance at 20 ∘C is 1.738 g/cm3. (e) The substance burns in air, producing an intense white light. (f) The substance reacts with chlorine to give a brittle white solid. (g) The substance can be pounded into thin sheets or drawn into wires. (h) The substance is a good conductor of electricity.
Chemistry
1 answer:
Amiraneli [1.4K]3 years ago
7 0

The question is incomplete, this is the complete question;

In the process of attempting to characterize a substance, a chemist makes the following observations. Which are physical properties and which are chemical properties?

(a) The substance is a silvery white, lustrous metal.

(b) The substance melts at 649 ∘C

(c) The substance boils at 1105 ∘C.

(d) The density of the substance at 20 ∘C is 1.738 g/cm³.

(e) The substance burns in air, producing an intense white light.

(f) The substance reacts with chlorine to give a brittle white solid.

(g) The substance can be pounded into thin sheets or drawn into wires.

(h) The substance is a good conductor of electricity.

Answer:

(a) Physical

(b) Physical

(c) Physical

(d) Physical

(e) Chemical

(f) Chemical

(g) Physical

(h) Physical

Explanation:

(a) The substance is a silvery white, lustrous metal - is appearance which is physical.

(b) The substance melts at 649 ∘C metling point is a physical measurement

(c) The substance boils at 1105 ∘C boiling point is a physical measurement

(d) The density of the substance at 20 ∘C is 1.738 g/cm3 how heavy a substance is,is physical.

(e) The substance burns in air, producing an intense white light. Burning is a chemical change. It is a reaction of a substance with oxygen

(f) The substance reacts with chlorine to give a brittle white solid. This is a chemical reaction as the question says.

(g) The substance can be pounded into thin sheets or drawn into wires. Ductility is a physical change, the substance remains chemically unchanged

(h) The substance is a good conductor of electricity. Any conductivity is physical, the substance remains chemically unchanged

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Р<br> +<br> 02<br> =<br> P4O6 <br> balance the equation
sergejj [24]

To balance a chemical equation, both sides should have equal mass, or in other words both sides should have same number of atoms as to follow the conservation of mass rule.

P + O₂ = P₄O₆

LHS:

Number of Phosphorus atoms = 1 atom

Number of Oxygen atoms = 2 atoms

RHS:

Number of Phosphorous atoms = 4 atoms

Number of Oxygen atoms = 6 atoms

Also since P₄O₆ has the most number of atoms we will make the LHS equalize to P₄O₆.

Difference between Phosphorous atoms in LHS to RHS = 3

Since phosphorous is a monatomic we need 3 phosphorous atoms extra

Difference between Oxygen atoms is LHS to RHS = 4

But Oxygen is diatomic, so we need 4/2 = 2 Oxygen molecules

Now lets see if it is balanced

P + 3P + O₂ + 2O₂ -------> P₄O₆

               4P + 3O₂ ------>  P₄O₆

LHS:

Phosphorous atoms = 4 atoms

Oxygen atoms = 3 × 2 = 6 atoms

RHS:

Phosphorous atoms = 4 atoms

Oxygen atoms = 6 atoms

LHS = RHS

Therefore the balanced equation is 4P + 3O₂ =  P₄O₆

Happy to help :)

If you need more explanation or help in any other question, feel free to ask

3 0
2 years ago
What is the balanced form of the following equation? Br 2 + S 2 O 3 2– + H 2 O → Br 1– + SO 4 2– + H +
nikitadnepr [17]

Answer:

4Br₂+ 5H₂O+ S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8Br⁻

Explanation:

Br₂ +  S₂O₃²⁻  + H₂O  → Br⁻ + SO₄²⁻ + H⁺

This is a redox reaction:

Br₂ changes the oxidation state from 0 to -1, so it was reduced

In the S₂O₃⁻² anion S changes the oxidation state from +2 to +6 in sulfate anion. (S₂O₃⁻², it is called thiosulfate)

We have protons in the main equation, so we assume we are in acidic medium:

Br₂ + 2e⁻ → 2Br⁻         Reduction

We balanced the bromide with 2, so the bromine has gained 2 electrons.

<u>5H₂O</u> + S₂O₃²⁻ → 2SO₄²⁻ + <u>10H⁺</u> + <em>8e</em>-  Oxidation

First of all, we add 2 to the sulfate anion in the product side, in order to balance the S.

As we have 8 O in right side, and 3 O in left side, we must add 5 O. We add 5 water in the place where the O are lower (reactant side).

Now, we have 10 H, in the reactant side, so we balance the product side with protons (10 H⁺).

Sulfur changed the oxidation state from +2 to +6, so it released 4 electrons, but, if you see thiosulfate anion you have 2 sulfurs so finally it has released 8 electrons.

Electrons are unbalanced so we multiply reduction x4, and oxidation x1.

(Br₂ + 2e⁻ → 2Br⁻) . 4 = 4Br₂ + 8e⁻ → 8Br⁻

(5H₂O + S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + <em>8e</em>-) . 1 = STAYS THE SAME.

We sum both half reactions, to cancel the elecetrons:

4Br₂ + 8e⁻ + 5H₂O + S₂O₃²⁻  → 2SO₄²⁻ + 10H⁺ + <em>8e</em>- + 8Br⁻

Finally the balanced reaction is: 4Br₂+ 5H₂O+ S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8Br⁻

5 0
3 years ago
How is a pure substance different from a mixture?
QveST [7]
The answer is A. Mixtures can be separated by physical means

A pure substance cannot be separated.
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A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor
AlexFokin [52]

Answer : The partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of C_2HBrClF_3 = 15.0 g

Mass of O_2 = 22.6 g

Molar mass of C_2HBrClF_3 = 197.4 g/mole

Molar mass of O_2 = 32 g/mole

First we have to calculate the moles of C_2HBrClF_3 and O_2.

\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole

and,

\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole

Now we have to calculate the mole fraction of C_2HBrClF_3 and O_2.

\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971

and,

\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.706}{0.0759+0.706}=0.903

Now we have to partial pressure of C_2HBrClF_3 and O_2.

According to the Raoult's law,

p^o=X\times p_T

where,

p^o = partial pressure of gas

p_T = total pressure of gas

X = mole fraction of gas

p_{C_2HBrClF_3}=X_{C_2HBrClF_3}\times p_T

p_{C_2HBrClF_3}=0.0971\times 862torr=84torr

and,

p_{O_2}=X_{O_2}\times p_T

p_{O_2}=0.903\times 862torr=778torr

Therefore, the partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

6 0
3 years ago
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