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uranmaximum [27]
2 years ago
10

Arrange the following in order of increasing boiling point: RbCl, CH3Cl, CH3OH, CH4. A. CH3OH < CH4 < CH3Cl < RbCl B. R

bCl < CH3Cl < CH3OH < CH4 C. CH4 < CH3OH < CH3Cl < RbCl D. CH3OH < CH3Cl < RbCl < CH4 E. CH4 < CH3Cl < CH3OH < RbCl
Chemistry
1 answer:
rosijanka [135]2 years ago
4 0

Answer:

E. CH₄ < CH₃Cl < CH₃OH < RbCl

Explanation:

The molecule with the stronger intermolecular forces will have the higher boiling point.

The order of strength of intermolecular forces (strongest first) is

  • Ion-Ion
  • Hydrogen bonding
  • Dipole-dipole
  • London dispersion

RbCl is a compound of a metal and a nonmetal. It is an ionic compound, so it has the highest boiling point.

CH₃Cl has a C-Cl polar covalent bond. It has dipole-dipole forces, so it has the second lowest boiling point.

CH₃OH has an O-H bond. It has hydrogen bonding, so it has the second highest boiling point.

CH₄ has nonpolar covalent C-H bonds. It has only nonpolar bonds, so the only attractive forces are London dispersion forces. It has the lowest boiling point.

Thus, the order of increasing boiling points is

CH₄ < CH₃Cl < CH₃OH < RbCl

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By mistake, a quart of oil was dumped into a swimming pool that measures 25.0 m by 30.0 m. The density of the oil was 0.750 g/cm
kondor19780726 [428]

The oil slick thick = 1.256 x 10⁻⁴ cm

<h3>Further explanation</h3>

Volume is a derivative quantity derived from the length of the principal

The unit of volume can be expressed in liters or milliliters or cubic meters

The conversion is

1 cc = 1 cm3

1 dm = 1 Liter

1 L = 1.06 quart

<em>so for 1 quart = 0.943 L</em>

\tt 0.943~L=0.943\times 10^{-3}m^3

Volume of oil dumped = volume of swimming pool

\tt 0.943\times 10^{-3}~m^3=25\times 30\times h(h=thick)\\\\h=\dfrac{0.943\times 10^{-3}}{750~m^2}=1.257\times 10^{-6}~m=\boxed{\bold{1.256\times 10^{-4}~cm}}

3 0
2 years ago
What is the number of moles in 15.0 g AsH3?
Grace [21]

Answer:

0.192 mol.

Explanation:

  • To calculate the no. of moles of a substance (n), we use the relation:

<em>n = mass / molar mass.</em>

mass of AsH₃ = 15.0 g.

molar mass of AsH₃ = 77.95 g/mol.

∴ The number of moles in 15.0 g AsH₃ = mass / molar mass = (15.0 g) / (77.95 g/mol) = 0.192 mol.

7 0
2 years ago
A 10.0-ml sample of 0.200 m hydrocyanic acid (hcn) is titrated with 0.0998 m naoh. what is the ph at the equivalence point? for
tatyana61 [14]
When the titration of HCN with NaOH is:

HCN (aq) + OH- (aq) → CN-(aq) + H2O(l)

So we can see that the molar ratio between HCN: OH-: CN- is 1:1 :1

we need to get number of mmol of HCN = molarity * volume 

                      = 0.2 mmol / mL* 10 mL = 2 mmol

so the number of mmol of NaOH = 2 mmol according to the molar ratio

so, the volume of NaOH = moles/molarity

                                          = 2 mmol / 0.0998mL

                                          = 20 mL

and according to the molar ratio so, moles of CN- = 2 mmol

∴the molarity of CN- =  moles / total volume 

                                   = 2 mmol / (10mL + 20mL ) = 0.0662 M

when we have the value of PKa = 9.31 and we need to get Pkb

so, Pkb= 14 - Pka

            = 14 - 9.31 = 4.69 

when Pkb = -㏒Kb

         4.69 = -㏒ Kb 

∴ Kb = 2 x 10^-5

and when the dissociation reaction of CN- is:

CN-(aq) + H2O(l) ↔ HCN(aq) + OH- (aq) 

by using the ICE table:

∴ the initials concentration are:

[CN-] = 0.0662 M

and [HCN] = [OH]- = 0 M

and the equilibrium concentrations are:

[CN-] = (0.0662- X)

[HCN] = [OH-]= X

when Kb expression = [HCN][OH-] /[CN-]

by substitution:

2 x 10^-5 = X^2 / (0.0662 - X)

X = 0.00114 

∴[OH-] = X = 0.00114

when POH = -㏒[OH]

                    = -㏒ 0.00114

POH = 2.94

∴PH = 14 - 2.94 = 11.06



 

6 0
2 years ago
Estimate ΔG°rxn for the following reaction at 387 K. HCN (g) + 2 H2 (g) → CH3NH2 (g) ΔH° = −158.0 kJ; ΔS° = −219.9
Lina20 [59]

Answer:

ΔG°rxn = -72.9 kJ

Explanation:

Let's consider the following reaction.

HCN(g) + 2 H₂(g) → CH₃NH₂(g)

We can calculate the standard Gibbs free energy of the reaction (ΔG°rxn) using the following expression:

ΔG°rxn = ΔH° - T.ΔS°

where,

ΔH° is the standard enthalpy of the reaction

T is the absolute temperature

ΔS° is the standard entropy of the reaction

ΔG°rxn = -158.0 KJ - 387 K × (-219.9 × 10⁻³ J/K)

ΔG°rxn = -72.9 kJ

4 0
2 years ago
Air tends to move from _____.
beks73 [17]
A: The Equator To The Poles

4 0
2 years ago
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