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aleksandr82 [10.1K]
3 years ago
8

Does data from mass spectrometry indicate that modern scientists have made modifications to Dalton''s model? Justify.

Chemistry
1 answer:
Gelneren [198K]3 years ago
5 0
John Dalton was a scientist who proposed that all matter consists of atoms. At this stage, no one had yet discovered neutrons and the nucleus. As a result, Dalton's model consisted of a single atom i.e. the atom was the smallest object. 
A mass spectrometer is an instrument that is able to see what is inside an atom. Scientists have been able to prove that the item is not the smallest object in the world. Atoms are made up of smaller objects called protons, neutrons and electrons. 
We can, therefore, safely conclude that data  from mass spectrometry has helped modern scientists to make modifications to Dalton's model. <span>
</span>
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Answer:

   true

Explanation:

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5.36 litars of nitrogen gas are at :25°C and 733 mm Hg
ArbitrLikvidat [17]

Answer:

4.6L

Explanation:

Use the equation (P1*V1)/(T1)=(P2*V2)/(T2)

P= pressure

V= volume

T= temperature in kelvins (remember K= C + 273)

Convert atm to mmHg or vise versa

1.5atm*(760mmhg/1atm)= 1140mmHg

(733mmHg * 5.36L)/(298K)=(1140mmHg * V)/(402K)

V= 4.6 or 4.65L (depending on sig figs)

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2 years ago
Webb has calculated the percent composition of a compound. How can he check his result?
goldenfox [79]

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3 years ago
In addition to mass balance, oxidation-reduction reactions must be balanced such that the number of electrons lost in the oxidat
Fiesta28 [93]

Answer:

Part A: (1, 1, 4, 1, 1, 1)

Part B: (2, 6, 4, 2, 3, 8)

Explanation:

Redox reactions can be balanced using the half-reaction method. It has the following steps:

  1. We write both half-reactions (reduction and oxidation)
  2. We balance the masses using H⁺ and H₂O in acidic media or OH⁻ and H₂O in basic media.
  3. We add electrons to balance electrically the half-reaction
  4. We multiply the half-reaction by numbers to make sure the number of electrons gained and lost are the same.
  5. We add both half-reactions and take the numbers to the general equation.

<em>Acidic solution</em>

SO₄²⁻(aq) + Sn²⁺(aq) + X ⇄ H₂SO₃(aq) + Sn⁴⁺(aq) + Y

1.

Reduction: SO₄²⁻ ⇒ SO₃²⁻

Oxidation: Sn²⁺ ⇒ Sn⁴⁺

2.

2 H⁺ + SO₄²⁻ ⇒ SO₃²⁻ + H₂O

Sn²⁺ ⇒ Sn⁴⁺

3.

2 H⁺ + SO₄²⁻ + 2 e⁻ ⇒ SO₃²⁻ + H₂O

Sn²⁺ ⇒ Sn⁴⁺ + 2 e⁻

4.

1 x [2 H⁺ + SO₄²⁻ + 2 e⁻ ⇒ SO₃²⁻ + H₂O]

1 x [Sn²⁺ ⇒ Sn⁴⁺ + 2 e⁻]

5.

2 H⁺ + SO₄²⁻ + 2 e⁻ + Sn²⁺ ⇄ SO₃²⁻ + H₂O + Sn⁴⁺ + 2 e⁻

2 H⁺ + SO₄²⁻ + Sn²⁺ ⇄ SO₃²⁻ + H₂O + Sn⁴⁺

Taking this to the general equation:

SO₄²⁻(aq) + Sn²⁺(aq) + 2 H⁺(aq) ⇄ H₂SO₃(aq) + Sn⁴⁺(aq) + H₂O(l)

Since H⁺ are spectator ions, they are not balanced automatically through this method and we have to balance them manually. In this case, we need to add 2 more H⁺ to the left.

SO₄²⁻(aq) + Sn²⁺(aq) + 4 H⁺(aq) ⇄ H₂SO₃(aq) + Sn⁴⁺(aq) + H₂O(l)

<em>Basic solution</em>

MnO₄⁻(aq) + F⁻(aq) + X ⇄ MnO₂(s) + F₂(aq) + Y

1.

Reduction: MnO₄⁻ ⇒ MnO₂

Oxidation: F⁻ ⇒ F₂

2.

2 H₂O + MnO₄⁻ ⇒ MnO₂ + 4 OH⁻

2 F⁻ ⇒ F₂

3.

2 H₂O + MnO₄⁻ + 3 e⁻ ⇒ MnO₂ + 4 OH⁻

2 F⁻ ⇒ F₂ + 2 e⁻

4.

2 × (2 H₂O + MnO₄⁻ + 3 e⁻ ⇒ MnO₂ + 4 OH⁻)

3 × (2 F⁻ ⇒ F₂ + 2 e⁻)

5.

4 H₂O + 2 MnO₄⁻ + 6 e⁻ + 6 F⁻ ⇄ 2 MnO₂ + 8 OH⁻ + 3 F₂ + 6 e⁻

4 H₂O + 2 MnO₄⁻ + 6 F⁻ ⇄ 2 MnO₂ + 8 OH⁻ + 3 F₂

Taking this to the general equation:

2 MnO₄⁻(aq) + 6 F⁻(aq) + 4 H₂O ⇄ 2 MnO₂(s) + 3 F₂(aq) + 8 OH⁻

This equation is balanced.

6 0
2 years ago
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