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Leya [2.2K]
3 years ago
12

Explain the difference between qualitative and quantitative properties.

Chemistry
1 answer:
SVETLANKA909090 [29]3 years ago
3 0
Qualitative properties are properties that are observed and can generally not be measured with a numerical result. They are contrasted to quantitative properties which have numerical characteristics.
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For each of the following compounds, identify what type of bonding holds them together.
svet-max [94.6K]

Explanation

NaCl:  Ionic crystal lattice forces

Hg:   Metallic bonding

CO₂:  London dispersion forces

CH₄:  London dispersion forces

Li₂O:  Ionic crystal lattice forces

Ag:  Metallic bonds

Ionic crystal lattice forces are strong electrostatic force of attraction between oppositely charged ions arranged into a crystal lattice of ionic compound. NaCl and Li₂O are ionic compounds

London dispersion forces holds the molecules of carbon dioxide and methane. They are weak attractions found between non-polar (and polar) molecules.

Metallic bonds exists between Mercury and Gold atoms. This is due to sea of electrons present.

4 0
3 years ago
Calculate the following quantity: molarity of a solution prepared by diluting 45.45 mL of 0.0404 M ammonium sulfate to 550.00 mL
dybincka [34]

Answer:

M_2=3.34x10^{-3}M

Explanation:

Hello!

In this case, since a dilution process implies that the moles of the solute remain the same before and after the addition of diluting water, we can write:

M_1V_1=M_2V_2

Thus, since we know the volume and concentration of the initial sample, we compute the resulting concentration as shown below:

M_2=\frac{M_2V_2}{V_1} =\frac{45.45mL*0.0404M}{550.00mL}\\\\M_2=3.34x10^{-3}M

Best regards!

5 0
2 years ago
How do you work out question 1a?
Sliva [168]

Answer:

-125 kJ

Explanation:

You calculate the energy required to break all the bonds in the reactants. Then you subtract the energy to break all the bonds in the products.

                     H₂C=CH₂   +    H₂ ⟶    H₃C-CH₃

Bonds:       4C-H + 1C=C     1H-H     6C-H + 1C-C

D/kJ·mol⁻¹:  413       612        436       413      347

The formula relating ΔHrxn and bond dissociation energies (D) is

ΔHrxn = Σ(Dreactants) – Σ(Dproducts)

(Note: This is an exception to the rule. All other thermochemical reactions are “products – reactants”. With bond energies, it’s “reactants – products”. The reason comes from the way we define bond energies.)

<em>For the reactant</em>s:

Σ(Dreactants) = 4 × 413 + 1 × 612 + 1 × 436 = 2700 kJ

<em>For the products:</em>

Σ(Dproducts) = 6 × 413 + 1 × 347 = 2825 kJ

<em>For the system</em> :

ΔHrxn = 2700 - 2825 = -125 kJ

4 0
3 years ago
What is the formula for dinitrogen pentoxide?<br> A)NO5<br> B)N5O2<br> C)N2O5<br> D)N2O7
Juliette [100K]

Answer:

nick gur

Explanation:

5 0
3 years ago
Read 2 more answers
A balloon containing helium gas expands from 230
Anit [1.1K]

The answer for the following problem is mentioned below.

  • <u><em>Therefore the final  moles of the gas is 14.2 × </em></u>10^{-4}<u><em> moles.</em></u>

Explanation:

Given:

Initial volume (V_{1}) = 230 ml

Final volume (V_{2}) = 860 ml

Initial moles (n_{1}) = 3.8 ×10^{-4} moles

To find:

Final moles (n_{2})

We know;

According to the ideal gas equation;

    P × V = n × R × T

where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of the moles of the gas

R represents the universal gas constant

T represents the temperature of the gas

So;

    V ∝ n

\frac{V_{1} }{V_{2} } = \frac{n_{1} }{n_{2} }

where,

(V_{1}) represents the initial volume of the gas

(V_{2}) represents the final volume of the gas

(n_{1}) represents the initial  moles of the gas

(n_{2}) represents the final moles of the gas

Substituting the above values;

   \frac{230}{860} = \frac{3.8 * 10^-4}{n_{2} }

  n_{2} = 14.2 × 10^{-4} moles

<u><em>Therefore the final  moles of the gas is 14.2 × </em></u>10^{-4}<u><em> moles.</em></u>

7 0
3 years ago
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