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mamaluj [8]
3 years ago
9

The Periodic Table List 3 elements that belong to the same Group/Family.

Chemistry
2 answers:
gizmo_the_mogwai [7]3 years ago
4 0

Answer:lithium(LI),sodium(NA) and rubidium(RB)

Explanation:

these 3 elements are part of the same family(alkali metals) and group(1)

hichkok12 [17]3 years ago
3 0

Answer:

Lithium,Sodium,Potassium all belong to transition metals

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If the molar mass of NaCl is 58.44 g/mol, what amount of NaCl (in mol) is contained in 250.0 mg of NaCl?
4vir4ik [10]

Answer: 0.004 moles

Explanation:

Amount of substance= mass/ molar mass

Mass= 250mg or 0.25g

Molar mass= 58.44gmol-1

Amount= 0.25/58.44

= 0.004 moles

5 0
2 years ago
Answer the following questions for H2CrO4:
Sholpan [36]

If this molecule is one half of a buffer, then the formula of the second half of the buffer is M2CrO4 where M is a univalent metal.

<h3>What is a strong acid?</h3>

A weak acid is one that is able to ionize completely in solution. The acid called chromic acid H2CrO4 is not  able to ionize completely in solution.

We know that a buffer is composed of a weak acid and its salt or a weak base and its salt hence if the acid H2CrO4 is present in a buffer then the other half must be  salt of the acid.

If this molecule is one half of a buffer, then the formula of the second half of the buffer is M2CrO4 where M is a univalent metal.

Learn more about buffer:brainly.com/question/22821585

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7 0
2 years ago
Write a chemical equation for the reaction that occurs in the following cell: Cu|Cu2+(aq)||Ag+(aq)|Ag Express your answer as a b
Alinara [238K]

Answer:

Explanation:

The cell reaction properly written is shown below:

              Cu|Cu²⁺_{aq} || Ag⁺_{aq} | Ag

From this cell reaction, to get the net ionic equation, we have to split the reaction into their proper oxidation and reduction halves. This way, we can know that is happening at the electrodes and derive the overall net equation.

  Oxidation half:

                  Cu_{s}  ⇄ Cu²⁺_{aq} + 2e⁻

At the anode, oxidation occurs.

  Reduction half:

                  Ag⁺_{aq} + 2e⁻ ⇄ Ag_{s}

At the cathode, reduction occurs.

To derive the overall reaction, we must balance the atoms and charges:

             Cu_{s}  ⇄ Cu²⁺_{aq} + 2e⁻

              Ag⁺_{aq} + e⁻ ⇄ Ag_{s}

  we multiply the second reaction by 2 to balance up:

         2Ag⁺_{aq} + 2e⁻ ⇄ 2Ag_{s}

The net reaction equation:

Cu_{s} + 2Ag⁺_{aq} + 2e⁻⇄ Cu²⁺_{aq} + 2e⁻ + 2Ag_{s}

We then cancel out the electrons from both sides since they appear on both the reactant and product side:

  Cu_{s} + 2Ag⁺_{aq} ⇄ Cu²⁺_{aq} + 2Ag_{s}

6 0
2 years ago
Why did you get that wrong?
victus00 [196]

Answer:

Got what wrong? A question?

5 0
2 years ago
Conclusion Questions:
Lerok [7]
That’s the question?
5 0
2 years ago
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