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Harman [31]
3 years ago
6

How does burning affect an object's property? Burning can cause an object to increase in volume. Burning changes the chemical ma

ke up of an object. Burning can create the item to increase in mass. Burning keeps the chemical make up of an object the same.​
Chemistry
1 answer:
Jobisdone [24]3 years ago
3 0

Answer: Burning changes the chemical make up of an object.

Explanation:

A chemical change can be defined as a change in the substance when it combines with other kind of substance to form a new substance. A chemical change can also occur when a substance is broken down into two or more products. These changes cannot be reversed. These changes affect the physical make up of an object. For example, burning as when an object is burned it cannot be transformed into its original form. A wood if burned can be converted into ash, water and carbon dioxide cannot regain its original form after burning so burning brings about chemical change in an object.

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The answers for the following sums is given below.

1.Pd_{2}H_{2}

2.C_{2} H_{6}

3.C_{2} H_{2} O_{2} Cl_{2}

4.C_{3}Cl_{3} N_{3}

5.Tl_{2 } C_{4} H_{4}O_{6}

6.C_{8}H_{8}

7. N_{2}O_{5}

8.P_{4}O_{6}

9.C_{4}H_{8}  O_{2}

Explanation:

1.Given:

        Molar mass=216.8g

Molecular formula=PdH_{2}

       we know;

Molecular formula=n(Empirical formula)

molecular weight of palladium(Pd)=106.4u

molecular weight of hydrogen(H)=1u

Molar mass of PdH_{2}:

Pd=106.4×1=106.4u

H=1×2=2

molar mass of PdH_{2}=106.4+2=108.4

n=\frac{216.8}{106.4}

<u><em>n=2</em></u>

Molecular formula=2(PdH_{2})

Molecular formula=Pd_{2}H_{2}

Therefore the molecular formula of the compound is Pd_{2}H_{2}

2. Given:

        Molar mass=30.0g

Molecular formula=CH_{3}

       we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of hydrogen(H)=1u

Molar mass of CH_{3}:

C=12.01 × 1 = 12.01u

H=1 × 3 = 3u

molar mass of CH_{3}=12.01 + 3 =15.01u

n=\frac{30.0}{15.01}

<u><em>n=2</em></u>

Molecular formula=2(CH_{3})

Molecular formula=C_{2} H_{6}

Therefore the molecular formula of the compound is C_{2} H_{6}

3. Given:

        Molar mass=129g

Molecular formula=CHOCl

       we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of hydrogen(H)=1u

molecular weight of oxygen(O)=16.00u

molecular weight of chlorie(Cl)=35.5u

Molar mass of CHOCl:

C=12.01 × 1 = 12.01u

H=1 × 1 = 1u

O=16.00×1=16.00u

Cl=35.5×1=35.5u

molar mass of CHOCl=12.01+1+16.00+35.5=64.5u

n=\frac{129}{64.5}

<u><em>n=2</em></u>

Molecular formula=2(CHOCl)

Molecular formula=C_{2} H_{2} O_{2} Cl_{2}

Therefore the molecular formula of the compound is C_{2} H_{2} O_{2} Cl_{2}

5. Given:

        Molar mass=577g

Molecular formula=TlC_{2} H_{2}O_{3}

       we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of Thallium(Tl)=204.3u

molecular weight of hydrogen(H)=1u

molecular weight of oxygen(O)=16.00u

Molar mass of TlC_{2} H_{2}O_{3} :

C=12.01 × 2= 24.02u

Tl=204.3×1=204.3u

H=1×2=2u

O=16.00×3=48.00

molar mass of TlC_{2} H_{2}O_{3}=204.3+24.02+1+48.00=278.32u

n=\frac{577}{278.32}

<u><em>n=2</em></u>

Molecular formula=2 (TlC_{2} H_{2}O_{3})

Molecular formula=Tl_{2 } C_{4} H_{4}O_{6}

Therefore the molecular formula of the compound is Tl_{2 } C_{4} H_{4}O_{6}

4. Molar mass=184.5g

Molecular formula=CClN

       we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of Nitrogen(N)=14u

molecular weight of chlorine(Cl)=35.5u

Molar mass of CClN:

C=12.01 × 1 = 12.01u

N=1×14=14U

Cl=35.5×1=35.5u

molar mass of CClN=12.01+14+35.5=61.5u

n=\frac{184.5}{61.5}

<u><em>n=3</em></u>

Molecular formula=3 (CClN)

Molecular formula=C_{3}Cl_{3} N_{3}

Therefore the molecular formula of the compound is C_{3}Cl_{3} N_{3}

6. For the table refer the attached file.

Simplest ratio of elements:

Carbon=8

Hydrogen=8

Empirical formula=C_{8}H_{8}

Molecular formula =C_{8}H_{8}

Molar mass of C_{8}H_{8}:

molecular weight of carbon=12.04u

molecular weight of hydrogen=1u

C=8×12.01=96.08u

H=1×8=8u

molar mass of C_{8}H_{8}=96.08+8=104.08u

n=104.08÷78.0

<em><u>n=1</u></em>

Molecular formula = n(Empirical formula)

Molecular formula = 1(C_{8}H_{8})

Molecular formula =C_{8}H_{8}

Therefore the molecular formula of a compound is C_{8}H_{8}

7. Given:

mass of oxide of nitrogen=108g

mass of nitrogen=4.02g

mass of oxygen=11.48g

moles of nitrogen=\frac{4.04}{14.01} = 0.289 moles

moles of oxygen=\frac{11.46}{15.999} =0.716 moles

We divide through by the lowest molar quantity to give an empirical formula  of N_{2} O{5}.

Now the molecular formula is multiple of the empirical formula.

So,

108 = n × (2×14.01 + 5×15.999)

Clearly,n=1, and the molecular formula is N_{2}O_{5}.

8.For the table refer the attached file.

Simplest ratio of elements:

Phosphorus=2

Oxygen=3

We know;

Empirical formula=P_{2} O_{3}

molecular formula= 2(Empirical formula)

Molecular formula =2(P_{2} O_{3})

Molecular formula =P_{4}O_{6}

Therefore the molecular formula of the compound is P_{4}O_{6}

9. For the table refer the attached file.

Simplest ratio of elements:

Carbon=2

Hydrogen=9

Oxygen=2

We know;

Empirical formula =C_{2} H_{4} O

Molecular formula = 2(Empirical formula)

Molecular formula =2(C_{2} H_{4} O)

Molecular formula =C_{4}H_{8}  O_{2}

Therefore the molecular formula of the compound is C_{4}H_{8}  O_{2}

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Basic substances have a gritty texture.
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<h3>1.<u> Answer;</u></h3>

False

<h3><u>Explanation;</u></h3>

Bases have some of the following properties;

  • They have a bitter taste
  • They have a slimy, or soapy feel on fingers
  • Most bases react with acids and precipitate salts.
  • Strong bases may react violently with acids.
  • Bases turn red litmus paper blue

<h3>2. <u>Answer;</u></h3>

An acid

<h3><u>Explanation;</u></h3>
  • When acids are dissolved in water, the concentration of the acid decreases and it becomes dilute.
  • It dissociates in water to give H+ ions or hydrogen ions.
  • All acidic solutions contain more hydrogen ions than hydroxide ions, therefore when added to water it increases the concentration of H+ ions in water, as water is a neutral substance whose concentration of H+ ions is equal to OH-.
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Write the Keq for the dissolving of salt: NaCl → Na+ + Cl-.
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Keq=[Na+][Cl-]/[NaCl]
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ubstance A is a nonpolar liquid and has only dispersion forces among its constituent particles. Substance B is also a nonpolar l
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Answer:

Their particles exhibit the same type of intermolecular interaction

Explanation:

In chemistry, we commonly say that 'like dissolves like'. This implies that polar solvents dissolves polar solutes while nonpolar solvents dissolve nonpolar solutes.

This phenomenon of 'like dissolves like' is possible because, the dissolution of one substance in another involves intermolecular interaction between the solute and solvent molecules.

If the molecules of solute and solvent are both nonpolar and have about the same magnitude of intermolecular (dispersion) forces, interaction between the both molecules is significant hence the solute dissolves completely in the solvent.

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