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3 years ago

Select the correct answer.

Chemistry

1 answer:

hichkok12 [17]3 years ago

6 0

Answer:

A. The model could not explain how alpha particles could be deflected at large angles.

Explanation:

The plum pudding model of the atom was proposed by J. J Thomson. He suggested that that an atom is made up of dense particles of electrons surrounded by positive charges.

From the Gold foil experiment carried out by Ernest Rutherford, he noticed that the bulk of the alpha particles targeted at the foil passed through and a little fraction was heavily deflected back.

Rutherford then presented his nuclear model from here. He suggested a massive, dense and tiny nucleus where the protons and neutrons are located. The space outside the mass is dominated by orbiting electrons.

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What must break in order for water to change from solid to liquid to gas? what must break in order for water to change from soli

Step2247 [10]

Answer:

hydrogen bonds between water molecules

Explanation:

The hydrogen bonds between water molecules conditions the bulk of its physical property most especially its relatively high boiling point. The hydrogen bond results from the attraction between the oxygen of a water molecule and the hydrogen of another water molecule. The more electronegative oxygen atom causes a distortion and the attraction leads to a strong intermolecular bond between atoms of the water molecules.

Hydrogen bond is a very strong bond and it is responsible for the physical properties of water.

7 0

3 years ago

How to apply Raoult's law to real solutions Consider mixing a liquid with a vapor pressure of 100 torr with an equimolar amount

garri49 [273]

Answer:

$\boxed{\text{positive deviation}}$

Explanation:

$\text{If the molecules interact less favourably in the solution than in the individual}\\\text{liquids, they can break away more easily than they can in the separate liquids.}\\\text{There will be more molecules in the vapour phase than you would expect.}\\\text{The solution will show a } \boxed{\textbf{positive deviation}} \text{ from Raoult's Law.}$

4 0

3 years ago

Mercury chloride is a commercial fungicide. If the molar mass is 470 g/mol and the percent composition is 85.0% Hg and 15.0% Cl,

ElenaW [278]

Data: molar mass 470 g/mol

Percent composition:

Hg = 85.0%
Cl = 15.0%

Solution:

1) Convert % to molar ratios

A. Base: 100 g

=> Hg = 85.0 g / 200.59 g/mol = 0.4235 mol

Cl = 15.0 g / 35.45 g/mol = 0.4231 mol

B. divide by the higher number and round to whole number

Hg = 0.4325 / 0.4231 = 1.00

Cl = 0.4231 / 0.4231 = 1.00

=> Empirical formula = Hg Cl

2) Find the mass of the empirical formula:

HgCl: 200.59 g/mol + 35.45 g/mol = 236.04

3) Determine how many times is the empirical mass contained in the molecular mass:

470 g/mol / 236.04 = 1.99 ≈ 2

=> Molecular formula = Hg2 Cl2.

Answers:

Empirical formula HgCl
Molecular Formula Hg2Cl2

6 0

3 years ago

Calculation of Molar Ratios of Conjugate Base to Weak Acid from pH For a weak acid with a pKa of 6.0, calculate the ratio of con

Free_Kalibri [48]

Explanation:

According to the Handerson equation,

pH = $pK_{a} + log \frac{\text{salt}}{\text{acid}}$

or, pH = $pK_{a} + log \frac{\text{conjugate base}}{\text{acid}}$

Putting the given values into the above equation as follows.

pH = $pK_{a} + log \frac{\text{conjugate base}}{\text{acid}}$

5.0 = 6.0 + log \frac{\text{conjugate base}}{\text{acid}}[/tex]

$log \frac{\text{conjugate base}}{\text{acid}}$ = -1.0

or, $\frac{\text{conjugate base}}{\text{acid}} = 10^{-1.0}$

= 0.1

Therefore, we can conclude that molar ratios of conjugate base to weak acid for given solution is 0.1.

7 0

3 years ago

Balance each of the following examples of heterogeneous equilibria and write each Kc expression. Then calculate the value of Kc

wolverine [178]

Answer:

(1) 2Al(s) + 2NaOH(aq) + 6H2O(l) ↔2Na[Al(OH)4](aq) + 3H2(g)

∴ Kc = ( PH2³ * [Na[Al(OH)4]² ) / [NaOH]² = 11

(2) H2O(l) + SO3(g) ↔ H2SO4(aq)

∴ Kc = [ H2SO4 ] / PSO3 = 0.0123

(3) 2P4(s) + 6O2(g) ↔ 2P4O6(s)

∴ Kc = Kc = 1 / PO2∧6

Explanation:

(1) 2Al(s) + 2NaOH(aq) + 6H2O(l) ↔ 2Na[Al(OH)4](aq) + 3H2(g)

∴ O / Al: 0 → +2 ≡ 2e-

Na: +1 → +2

∴ R / H: +1 → 0

2 - Al - 2

2 - Na - 1

8 - O - 8

14 - H - 14

⇒ Kc = ( PH2³ * [Na[Al(OH)4]² ) / [NaOH]² = 11

(2) H2O(l) + SO3(g) ↔ H2SO4(aq)

1 - S - 1

4 - O - 4

2 - H - 2

⇒ Kc = [ H2SO4 ] / PSO3 = 0.0123

(3) 2P4(s) + 6O2(g) ↔ 2P4O6(s)

8 - P - 8

12 - O - 12

⇒ Kc = 1 / PO2∧6

6 0

3 years ago