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garik1379 [7]
3 years ago
7

Tritium is an isotope of hydrogen (H) that has:

Chemistry
2 answers:
ivann1987 [24]3 years ago
8 0

Answer:

D.1 proton and 1 neutron is the correct answer    

Explanation:

Anika [276]3 years ago
7 0
D. 1 proton and 1 neutron
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Aleks04 [339]

Answer:

less metallic

Explanation:

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3 years ago
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Which structure is not an isomer of pentane and state why?
Musya8 [376]

Answer:

2,2-dimethylpropane : C5H12

becoz formula of propane is

C3H8

5 0
2 years ago
CH3CN major species present when dissolved in water
Margaret [11]

Answer: CH₃CN and H₂O.

Explanation:

1) The spieces present in a solution may be either the molecules, in case of covalent compounds, or ions, in case of ionic compounds that dissociate (ionize).

2) Both, CH₃CN and H₂O are covalent (polar covalent) substances, so they do not ionize and the spieces in the solution are the molecules per se.

3) In solution, the molecules of H₂O will solvate the molecules of CH₃CN, meaning that H₂O molecules are able to separate the molecules of CH₃N from each other, and so every molecule of CH₃CN will end surrounded by many molecules of H₂O.

This happens because the interaction between the polar molecules of the two different compounds is strong enough to overcome the intermolecular forces between the molecules of the same compound.

3 0
3 years ago
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Help please!!
frez [133]

Answer:

6.25%

Explanation:

Given data:

Half life of lutetium-117 = 6.75 days

Percentage remaining after 27 days = ?

Solution;

Number of half lives = Time elapsed / half life

Number of half lives = 27 days / 6.75 days

Number of half lives = 4

At time zero = 100%

At first half life = 100%/2 = 50%

At second half life = 50%/2 = 25%

At 3rd half life = 25%/2 = 12.5%

At 4th half life = 12.5%/2 = 6.25%

7 0
2 years ago
Conduct metric Titration of H_2(SO_4) and Ba(OH)_2 Write an equation (including states of matter) for the reaction between H_2(S
meriva

Answer:

a) H₂SO₄ + Ba(OH)₂ ⇄ BaSO₄(s) + 2 H₂O(l)

b) H₂SO₄, H⁺, HSO₄⁻, SO₄²⁻. H₂O, H⁺, OH⁻.

c) H⁺, HSO₄⁻, SO₄²⁻

d) As the titration takes place, reaction [1] proceeds to the right. The conductivity of the solution decreases because the amount of H⁺, HSO₄⁻, SO₄²⁻ decreases. The formed solid is barium sulfate BaSO₄. Since BaSO₄ is very insoluble, the main responsible for conductivity are still H⁺, HSO₄⁻ and SO₄²⁻,

e) At the equivalence point equivalent amounts of H₂SO₄ and Ba(OH)₂ react. The conducting species are Ba²⁺, SO₄²⁻, H⁺ and OH⁻.

f) After the equivalence point there is an excess of Ba(OH)₂. The ions Ba²⁺ and OH⁻ are responsible for the increase in the conductivity, being the major conducting species.

Explanation:

a) Write an equation (including states of matter) for the reaction between H₂SO₄ and Ba(OH)₂.

The <em>balanced equation</em> is:

H₂SO₄ + Ba(OH)₂ ⇄ BaSO₄(s) + 2 H₂O(l)   [1]

b) At the very start of the titration, before any titrant has been added to the beaker, what is present in the solution?

In the beginning there is H₂SO₄ and the ions that come from its <em>dissociation reactions</em>: H⁺, HSO₄⁻, SO₄²⁻. There is also H₂O and a very small amount of H⁺ and OH⁻ coming from its <em>ionization</em>.

H₂SO₄(aq) ⇄ H⁺(aq) + HSO₄⁻(aq)

HSO₄⁻(aq) ⇄ H⁺(aq) + SO₄²⁻(aq)

H₂O(l)  ⇄ H⁺(aq) + OH⁻(aq)

c) What is the conducting species in this initial solution?

The main responsible for conductivity are the <em>ions</em> coming from H₂SO₄: H⁺, HSO₄⁻, SO₄²⁻.

d) Describe what happens as titrant is added to the beaker. Why does the conductivity of the solution decrease? What is the identity of the solid formed? What is the conducting species present in the beaker?

As the titration takes place, reaction [1] proceeds to the right. The conductivity of the solution decreases because the amount of H⁺, HSO₄⁻, SO₄²⁻ decreases. The formed solid is barium sulfate BaSO₄. Since BaSO₄ is very insoluble, the main responsible for conductivity are still H⁺, HSO₄⁻ and SO₄²⁻,

e) What happens when the conductivity value reaches its minimum value (which is designated as the equivalence point for this type of titration)? What is the conducting species in the beaker?

At the <em>equivalence point</em> equivalent amounts of H₂SO₄ and Ba(OH)₂ react. Only BaSO₄ and H₂O are present, and since they are <em>weak electrolytes</em>, there is a small amount of ions to conduct electricity. The conducting species are Ba²⁺ and SO₄²⁻ coming from BaSO₄ and H⁺ and OH⁻ coming from H₂O.

f) Describe what happens at additional titrant is added past the equivalence point. Why does the conductivity of the solution increase? What is the conducting species present in the beaker?

After the equivalence point there is an excess of Ba(OH)₂. The ions Ba²⁺ and OH⁻ are responsible for the increase in the conductivity, being the major conducting species.

7 0
3 years ago
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