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svetlana [45]
3 years ago
7

Which molecule will undergo only London dispersion forces when interacting with other molecules of the same kind? a.HCl b.C2H5OH

c.NH3 d.CH4
Chemistry
1 answer:
ohaa [14]3 years ago
5 0

Answer:

CH4

Explanation:

London dispersion forces are a type of force acting between atoms and molecules that are normally electrically symmetric; that is, the electrons are symmetrically distributed with respect to the nucleus. They are part of the van der Waals forces(Wikipedia).  It is a temporary attractive force that results when the electrons in two adjacent atoms occupy positions that make the atoms form temporary dipoles. This force is sometimes called an induced dipole-induced dipole attraction(Chem Purdue).

If we consider HCl, C2H5OH  and NH3, we will discover that they all have polar bonds. Consequently, dipole - dipole interaction as well as hydrogen bonds exist between their molecules.

CH4 is nonpolar. Being a nonpolar molecule, the only possible intermolecular interaction forces between its molecules is the London dispersion forces.

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6.) 50.0 mol H2O<br> ? molecules
8_murik_8 [283]
<h3>Answer:</h3>

3.01 × 10²⁵ molecules H₂O

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>Explanation:</h3>

<u>Step 1: Define</u>

50.0 mol H₂O

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

<u />\displaystyle 50.0 \ mol \ H_2O(\frac{6.022 \cdot 10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O} ) = 3.011 × 10²⁵ molecules H₂O

<u>Step 4: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

3.011 × 10²⁵ molecules H₂O ≈ 3.01 × 10²⁵ molecules H₂O

5 0
3 years ago
Ann and Carol each lift the timber with a force of 80 kg/m/s2. If they lift the timber a distance of 1.25 meters, how much work
Novosadov [1.4K]
Work is defined energy transferred from one to another. 
The formula for work done is work done = force x distance

So in our problem, force is equal to 80 kg/ m / s^2 and distance is equal to 1.25 meters. So plugging in our values will give us:

work done = 80 kg/ m/ s^2 * 1.25 m
= 100.00 J is the answer.
4 0
3 years ago
How many molecules are in 50 grams of water?
blondinia [14]
The Molar mass of 50g of water is (18.015 g/mol). Hope this helps
3 0
3 years ago
Read 2 more answers
During lab, a student used a Mohr pipet to add the following solutions into a 25 mL volumetric flask. They calculated the final
kompoz [17]

Answer:

(FeSCN⁺²) = 0.11 mM

Explanation:

Fe ( NO3)3 (aq) [0.200M] + KSCN (aq) [ 0.002M] ⇒ FeSCN+2

M (Fe(NO₃)₃  = 0.200 M

V (Fe(NO₃)₃ =  10.63 mL

n (Fe(NO₃)₃ = 0.200*10.63 = 2.126 mmol

M (KSCN) =  0.00200 M

V (KSCN) = 1.42 mL

n (KSCN) =  0.00200 * 1.42 = 0.00284 mmol

Total volume = V (Fe(NO₃)₃  + V (KSCN)

                       = 10.63 + 1.42

                       = 12.05 mL

Limiting reactant = KSCN

So,

FeSCN⁺² = 0.00284 mmol

M (FeSCN⁺²) = 0.00284/12.05

                     = 0.000236 M

Excess reactant = (Fe(NO₃)₃

n(Fe(NO₃)₃ =  2.126 mmol -  0.00284 mmol

                  =2.123 mmol

For standard 2:

n (FeSCN⁺²) = 0.000236 * 4.63

                    =0.00109

V(standard 2) = 4.63 + 5.17

                       = 9.8 mL

M (FeSCN⁺²)  = 0.00109/9.8

                      = 0.000111 M = 0.11 mM

Therefore, (FeSCN⁺²) = 0.11 mM

7 0
3 years ago
How did the elements begin to stratify?
Bogdan [553]

Answer:

The process of elemental stratification relies on the diffusion velocity, which causes the migration of the different chemical elements within stars.

Explanation:

8 0
3 years ago
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