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weeeeeb [17]
2 years ago
15

A trend in the periodic table means?

Chemistry
1 answer:
Delvig [45]2 years ago
4 0

Answer:

A periodic trend is a regular variation the properties of an element with increasing atomic number.

Explanation:

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Websites that end in .com or .org should be avoided.
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3 years ago
Consider the following voltaic cell:(j) In which direction do anions within the salt bridge move to maintain charge neutrality?
aniked [119]

In order to maintain neutrality, the negatively charged ions in the salt bridge will migrate into the anodic half-cell. A similar (but reversed) situation is found in the cathodic cell.

<h3>What purpose does a salt bridge serve in an oxidation process?</h3>

Anions (negatively charged particles) are added to the solution of the oxidation half of the cell by the salt bridge, and cations (positively charged particles) are added to the solution of the reduction half of the reaction. 

<h3>What purpose does the salt bridge serve in a galvanic cell?</h3>

For instance, KCl, AgNO3, etc. In a galvanic cell, such as a voltaic cell or Daniel cell, salt bridges are typically used. A salt bridge's primary job is to assist in preserving the electrical neutrality of the internal circuit. Additionally, it aids in keeping the cell's response from reaching equilibrium.

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How do geologists use seismic waves to locate an earthquakes epicenter
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3 years ago
Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3)2(aq)+2
Setler79 [48]

Answer:

a) volume of ammonium iodide required =349 mL

b) the moles of lead iodide formed = 0.0436 mol

Explanation:

The reaction is:

Pb(NO_{3})_{2}+2NH_{4}I -->PbI_{2}+2NH_{4}NO_{3}

It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.

Let us calculate the moles of lead nitrate taken in the solution.

Moles=molarityX volume (L)

Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol

the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol

The volume of ammonium iodide required will be:

volume=\frac{moles}{molarity}=\frac{0.0872}{0.250}=0.349L=349mL

the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol

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