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3 years ago

Select all that apply to physical and chemical properties ?

Chemistry

2 answers:

MaRussiya [10]3 years ago

7 0

Answer:

• Chemical properties can only be observed during a chemical reaction:

Chemical properties can only be established by changing a substance's chemical identity, and chemical properties are different from physical properties, which can be observed by viewing or touching a sample. The internal qualities of a substance must be altered to determine its chemical properties.

Examples:

Some common chemical properties are heat of combustion, enthalpy of formation, toxicity,

and flammability, each of which will be covered in this lesson.

• Rusting of iron is the chemical reaction not physical because:

The rusting of iron is a chemical change because it is two substances reacting together to make a

new substance. When iron rusts, iron molecules react with oxygen molecules to make a compound

called iron oxide. Rusting would only be a physical change if iron molecules remained pure iron

throughout the process.

The equation for this reaction is: 4Fe+ 3O2 → 2Fe2O

• Melting and boiling points are physical properties:

Yes melting and boiling points are the physical properties.

Physical properties can be observed or measured without changing the composition of matter. Physical properties are used to observe and describe matter.

Physical properties include: appearance, texture, color, odor, melting point, boiling point, density, solubility, polarity, and many others.

• Mass and volume are chemical properties:

No mass and volume are the extensive physical properties of matter

All properties of matter are either physical or chemical properties and physical properties are either intensive or extensive.

Extensive properties, such as mass and volume, depend on the amount of matter being measured.

Intensive properties, such as density and color, do not depend on the amount of the substance present.

Physical properties can be measured without changing a substance’s chemical identity.

• Various components of a mixture do not combine chemically:

When various components of a mixture do not combine chemically it is a Combination of substances in which individual components do not combine chemically but retain their individual properties. Mixture in which one or more substances are distributed evenly in another substance.

When various components combine chemically then form chemical compounds that are often divided into two categories. Metals often react with nonmetals to form ionic compounds. These compounds are composed of positive and negative ions formed by adding or subtracting electrons from neutral atoms and molecules

mezya [45]3 years ago

6 0

Answer:

all of the above

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For elements in the third row of the periodic table and beyond, the octet rule is often not obeyed. a friend of yours says this

Rom4ik [11]

The reason that some of the elements of period three and beyond are steady in spite of not sticking to the octet rule is due to the fact of possessing the tendency of forming large size, and a tendency of making more than four bonds. For example, sulfur, it belongs to period 3 and is big enough to hold six fluorine atoms as can be seen in the molecule SF₆, while the second period of an element like nitrogen may not be big to comprise 6 fluorine atoms.

The existence of unoccupied d orbitals are accessible for bonding for period 3 elements and beyond, the size plays a prime function than the tendency to produce more bonds. Hence, the suggestion of the second friend is correct.

4 0

3 years ago

At a certain temperature the rate of this reaction is first order in HI with a rate constant of :0.0632s

Shalnov [3]

Answer : The time taken for the reaction is, 28 s.

Explanation :

Expression for rate law for first order kinetics is given by :

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = 0.0632

t = time taken for the process = ?

$[A_o]$ = initial amount or concentration of the reactant = 1.28 M

$[A]$ = amount or concentration left time 't' = $1.28\times \frac{17}{100}=0.2176M$

Now put all the given values in above equation, we get:

$0.0632=\frac{2.303}{t}\log\frac{1.28}{0.2176}$

$t=28s$

Therefore, the time taken for the reaction is, 28 s.

8 0

3 years ago

amm1812

Solution:

1) Separate out the half-reactions. The only issue is that there are three of them.

Fe2+ ---> Fe3+
S2¯ ---> SO42¯
NO3¯ ---> NO

How did I recognize there there were three equations? The basic answer is "by experience." The detailed answer is that I know the oxidation states of all the elements on EACH side of the original equation. By knowing this, I am able to determine that there were two oxidations (the Fe going +2 to +3 and the S going -2 to +6) with one reduction (the N going +5 to +2).

Notice that I also split the FeS apart rather than write one equation (with FeS on the left side). I did this for simplicity showing the three equations. I know to split the FeS apart because it has two "things" happening to it, in this case it is two oxidations.

Normally, FeS does not ionize, but I can get away with it here because I will recombine the Fe2+ with the S2¯ in the final answer. If I do everything right, I'll get a one-to-one ratio of Fe2+ to S2¯ in the final answer.

2) Balancing all half-reactions in the normal manner.

Fe2+ ---> Fe3+ + e¯
4H2O + S2¯ ---> SO42¯ + 8H+ + 8e¯
3e¯ + 4H+ + NO3¯ ---> NO + 2H2O

3) Equalize the electrons on each side of the half-reactions. Please note that the first two half-reactions (both oxidations) total up to nine electrons. Consequently, a factor of three is needed for the third equation, the only one shown below:

3 [3e¯ + 4H+ + NO3¯ ---> NO + 2H2O]

Adding up the three equations will be left as an exercise for the reader. With the FeS put back together, the sum of all the coefficients (including any that are one) in the correct answer is 15.

Problem #2: CrI3 + Cl2 ---> CrO42¯ + IO4¯ + Cl¯ [basic sol.]

Solution:

Go to this video for the solution

Problem #3: Sb2S3 + Na2CO3 + C ---> Sb + Na2S + CO

Solution:

1) Remove all the spectator ions:

Sb26+ + CO32- + C ---> Sb + CO

Notice that I did not write Sb3+. I did this to keep the correct ratio of Sb as reactant and product. It also turns out that it will have a benefit when I select factors to multiply through some of the half-reactions. I didn't realize that until after the solution was done.

2) Separate into half-reactions:

Sb26+ ---> Sb
CO32- ---> CO
C ---> CO

3) Balance as if in acidic solution:

6e¯ + Sb26+ ---> 2Sb
2e¯ + 4H+ + CO32- ---> CO + 2H2O
H2O + C ---> CO + 2H+ + 2e¯Could you balance in basic? I suppose, but why?

4) Use a factor of three on the second half-reaction and a factor of six on the third.

6e¯ + Sb26+ ---> 2Sb
3 [2e¯ + 4H+ + CO32- ---> CO + 2H2O]
6 [H2O + C ---> CO + 2H+ + 2e¯]The key is to think of 12 and its factors (1, 2, 3, 4, 6). You need to make the electrons equal on both sides (and there are 12 on each side when the half-reactions are added together). You get 12 H+ on each side (3 x 4 in the second and 6 x 2 in the third). You get six waters with 3 x 2 in the second and 6 x 1 in the third.Everything that needs to cancel gets canceled!

5) The answer (with spectator ions added back in):

Sb2S3 + 3Na2CO3 + 6C ---> 2Sb + 3Na2S + 9CO

6) Here's a slightly different take on the solution just presented.

a) Write the net ionic equation:Sb26+ + CO32- + C ---> Sb + COb) Notice that charges must be balanced and that we have zero charge on the right. So, do this:Sb26+ + 3CO32- + C ---> Sb + COc) Now, balance for atoms:Sb26+ + 3CO32- + 6C ---> 2Sb + 9COd) Add back the sodium ions and sulfide ions to recover the molecular equation.Sb2S3 + 3Na2CO3 + 6C ---> 2Sb + 3Na2S + 9CO

7) Here's a discussion of a wrong answer to the above problem.

However, after reading the above wrong answer example, look at problem #10 below for an instance of having to add in a substance not included in the original reaction.

Problem #4: CrI3 + H2O2 ---> CrO42¯ + IO4¯ [basic sol.]

Solution:

1) write the half-reactions:

Cr3+ ---> CrO42¯
I33¯ ---> IO4¯
H2O2 ---> H2O

I wrote the iodide as I33¯ to make it easier to recombine it with the chromium ion at the end of the problem.

2) Balance as if in acidic solution:

4H2O + Cr3+ ---> CrO42¯ + 8H+ + 3e¯
12H2O + I33¯ ---> 3IO4¯ + 24H+ + 24e¯
2e¯ + 2H+ + H2O2 ---> 2H2O

I used water as the product for the hydrogen peroxide half-reaction because that gave me a half-reaction in acid solution. It will all go back to basic at the end of the problem.

3) Recover CrI3 by combining the first two half-reactions from just above:

16H2O + CrI3 ---> 3IO4¯ + CrO42¯ + 32H+ + 27e¯

4) Equalize the electrons:

2 [16H2O + CrI3 ---> 3IO4¯ + CrO42¯ + 32H+ + 27e¯]
27 [2e¯ + 2H+ + H2O2 ---> 2H2O]leads to:32H2O + 2CrI3 ---> 6IO4¯ + 2CrO42¯ + 64H+ + 54e¯
54e¯ + 54H+ + 27H2O2 ---> 54H2O

5) Add the half-reactions together. Strike out (1) electrons, (2) hydrogen ion and (3) water. The result:

2CrI3 + 27H2O2 ---> 2CrO42¯ + 6IO4¯ + 10H+ + 22H2O

6) Add 10 hydroxides to each side. This makes 10 more waters on the right, so combine with the water alreadyon the right-hand side to make 32:

2CrI3 + 27H2O2 + 10OH¯ ---> 2CrO42¯ + 6IO4¯ + 32H2O

3 0

3 years ago

an isotope of cesium (cesium-137) has a half-life of 30 years if 1.0g of cesium-137 disintegrates over a period of 90 years how

Juliette [100K]

The answer is after 3 half lives

4 0

3 years ago

A gas occupies a volume of 31.0 ml at 19.0°C. If the gas temperature rises to 38.0°C at constant pressure. Calculate the new vol

Alik [6]

Answer:

The correct answer is 0.0033 L (33.0 mL)

Explanation:

We uses the Charles's law which describes the changes in the volume (V) of a gas and its temperature in Kelvin (T) at constant pressure. The mathematical expression is the following:

V₁/T₁ = V₂/T₂

We have the following data:

V₁= 31.0 mL = 0.0031 L

T₁= 19.0°C = 292 K

T₂= 38.0°C = 311 K

V₂= ?

We calculate V₂ from the mathematical expression, as follows:

V₂= V₁/T₁ x T₂ = 0.0031 L/(292 K) x 311 K = 0.0033 L

5 0

3 years ago