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katrin [286]
3 years ago
9

Net ionic equation for Li+(aq) + OH-(aq) + H+(aq) + Cl-(aq) ---> ?

Chemistry
1 answer:
hram777 [196]3 years ago
5 0
Learn how to write, balance net ionic equations<span> with examples. ... </span>A<span> chemical equation shows the reactant molecules which involve in reaction and products which are formed during the reaction as well as .... Na</span>+(aq<span>) + </span>OH-(aq) + H+(aq) + NO3-(aq) ? ... For example: Aqueous Sodium chloride is to be written as Na+<span> and </span>Cl-<span>.

</span>
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Elements in Group 18 are known
o-na [289]

Answer:

C the number and placement of the electrons

Explanation:

Elements react with one another because they want to achieve a full valence shell of electrons. The noble gases already have a full valence shell of electrons, so they are unlikely to react with other elements.

6 0
3 years ago
Which description best characterizes the motion of particles in a solid?
Masja [62]

Answer:

D. Vibrating around fixed positions

Explanation:

It's right on edge

6 0
3 years ago
Which characteristic is typical of metals? A. They are brittle. B. They do not conduct electricity. C. They are highly soluble i
shutvik [7]

D. They are malleable.


7 0
3 years ago
Read 2 more answers
What is the molar concentration of chloride ions in a
Yanka [14]

Answer:

The concentration of chloride ions in the final solution is 3 M.

Explanation:

The number of moles present in a solution can be calculated as follows:

number of moles = concentration in molarity * volume

In 100 ml of a 2 M KCl solution, there will be (0.1 l * 2mol/l) 0.2 mol Cl⁻

For every mol of CaCl₂, there are 2 moles of Cl⁻, then, the number of moles of Cl⁻ in 50 l of a 1.5 M solution will be:

number of moles of Cl⁻ = 2 * number of moles of  CaCl₂

number of moles of Cl⁻ = 2 ( 50 l * 1.5 mol / l ) = 150 mol Cl⁻

The total number of moles of Cl⁻ present in the solution will be (150 mol + 0.2 mol ) 150.2 mol.

Assuming ideal behavior, the volume of the final solution will be ( 50 l + 0.1 l) 50.1 l. The molar concentration of chloride ions will be:

Concentration = number of moles of Cl⁻ / volume

Concentration = 150.2 mol / 50.1 l = 3.0 M

8 0
3 years ago
Consider a galvanic cell consisting of the following two redox couplesAlB+ 3e Al , Mga + 2e Mg, (ag. E® = 1.676 V ) E° = 2.356 V
malfutka [58]

Answer:

0.68 V

Explanation:

For anode;

3Mg(s) ---->3Mg^2+(aq) + 6e

For cathode;

2Al^3+(aq) + 6e -----> 2Al(s)

Overall balanced reaction equation;

3Mg(s) + 2Al^3+(aq) ----> 3Mg^2+(aq) + 2Al(s)

Since

E°anode = -2.356 V

E°cathode = -1.676 V

E°cell=-1.676 -(-2.356)

E°cell= 0.68 V

4 0
3 years ago
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