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Leokris [45]
3 years ago
10

Chemistry students were testing compounds in an aqueous solution to see which one had the strongest electrical current. There ar

e three beakers each with a different compound. The first beaker and the second beaker lit the light bulbs brightly. The third beaker did not light up the bulb. With the information that is provided, what is the best explanation for the first and second beaker being the strongest electrolytes?
A)The first and second beakers contain a solution that partially dissociates the ions. These ions create an electrical current. The third beaker solution completely dissociates the ions. It is a strong electrolyte.

B)The first and second beakers contain a solution that completely dissociates the ions. These ions create an electrical current. The third beaker solution does not dissociate ions. It is a nonelectrolyte.

C)The first and second beakers contain a solution that does not dissociate ions. These ions create an electrical current. The third beaker solution completely dissociates the ions. It is a strong electrolyte.

D)The first and second beakers contain a solution that completely dissociates the ions. These ions do not create an electrical current. The third beaker solution does completely dissociates the ions. It is a strong electrolyte.
Chemistry
2 answers:
jasenka [17]3 years ago
8 0

Answer:

B)The first and second beakers contain a solution that completely dissociates the ions. These ions create an electrical current. The third beaker solution does not dissociate ions. It is a nonelectrolyte.

Explanation:

Electrolytes dissociate ions. They are also conductors. Since the first 2 bulbs light up strongly, we can assume that the ions completely dissociated. Since the 3rd did not light up, it is a nonelectrolyte.

Pavel [41]3 years ago
6 0

Answer:

The first and second beakers contain a solution that completely dissociates the ions. These ions create an electrical current. The third beaker solution does not dissociate ions. It is a nonelectrolyte.

Explanation:

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Explanation:

Ozone, or 03,has two major resonance structures that contribute equally to the overall hybird structure of the molecule. The two structures are equivalent from the stability standpoint, each having a positive and a negative formal charge placed on two of the oxygen atoms.

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4 years ago
What is the molarity of a solution made with 64 grams of sodium hydroxide in 4 liters of water
ehidna [41]

Answer:

The molarity of the solution: 0,27M

Explanation:

First , we calculate the weight of 1 mol of NaCl:

Weight 1mol NaCl= Weight Na + Weight Cl= 23 g+ 35, 5 g= 58, 5 g/mol

58,5 g---1 mol NaCl

64 g--------x= (64 g x1 mol NaCl)/58,5 g= 1, 09 mol NaCl

A solution molar--> moles of solute in 1 L of solution:

4 L-----1,09 mol NaCl

1L----x0( 1L x1,09 mol NaCl)/4L =0,27moles NaCl--->0,27M

7 0
3 years ago
the reaction of aluminum with chlorine gas is shown 2Al + 3Cl2 -> 2AlCl3 based on this equation how many molecules of chlorin
Vanyuwa [196]

45 molecules of chlorine gas (Cl₂) are needed to react with 30 atoms of aluminum (Al)

The balanced equation for the reaction is given below:

2Al + 3Cl₂ —> 2AlCl₃

From the balanced equation above,

2 atoms of Al required 3 molecules of Cl₂.

With the above information, we can determine the number of molecules of Cl₂ needed to react with 30 atoms of Al. This can be obtained as follow:

From the balanced equation above,

2 atoms of Al required 3 molecules of Cl₂.

Therefore,

30 atoms of Al will require = \frac{30 * 3}{2}\\ = 45 molecules of Cl₂.

Thus, 45 molecules of chlorine gas (Cl₂) are needed to react with 30 atoms of aluminum (Al)

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2 years ago
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4 years ago
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How many molecules of ethane are present in 64.28 liters of ethane gas (C2H6) at STP
Rom4ik [11]

Answer : The molecule of ethane present in 64.28 L of ethane gas at STP is, 17.277\times 10^{23} molecule.

Solution :

At STP,

22.4 L volume of ethane present in 1 mole of ethane gas

64.28 L volume of ethane present in \frac{64.28L}{22.4L}\times 1mole=2.869moles of ethane gas

And, as we know that

1 mole of ethane molecule contains 6.022\times 10^{23} molecules of ethane

2.869 moles of ethane molecule contains 2.869\times 6.022\times 10^{23}=17.277\times 10^{23} molecules of ethane

Therefore, the molecule of ethane present in 64.28 L of ethane gas at STP is, 17.277\times 10^{23} molecule.

4 0
3 years ago
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