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3 years ago

How much lead (pb) is in 5.07 × 1012 atoms of lead? answer in units of mol?

Chemistry

1 answer:

Butoxors [25]3 years ago

4 0

Answer:

2.49⋅10−12moles Pb

Explanation:

Before doing any calculations, it's worth noting that atoms do not contain moles, it's the other way around.

A mole is simply a collection of atoms. More specifically, you need to have exactly 6.022⋅1023 atoms of an element in order to have one mole of that element - this is known as Avogadro's number.

In your case, you must determine how many moles of lead would contain 1.50⋅1012atoms of lead.

Well, if you know that one mole of lead must contain 6.022⋅1023 atoms of lead, it follows that you get 1.50⋅1012 atoms of lead in

1.50⋅1012atoms of Pb⋅1 mole Pb6.022⋅1023atoms of Pb=2.49⋅10−12moles Pb

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Answer:

See explaination

Explanation:

Since X is more reactive than Y

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3 years ago

The solubility of an ionic compound can be expressed as the number of moles of the compound that will dissolve per liter of solu

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Answer:

Number of moles of sodium dissolved = 6.0 *10^23

Explanation:

The image for the question is attached

Solution

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3 years ago

How many moles of chromium III nitrate are produced When chromium reacts with 0.85 moles of lead for nitrate to produce chromium

Pepsi [2]

0.85 moles formula units of lead nitrate will produce 0.57 moles formula units of chromium (III) nitrate.

<h3>Explanation</h3>

Typically, the oxidation state of Pb in lead nitrate tend to be +2. In other words, Pb in lead nitrate tends to exist as $\text{Pb}^{2+}$ ions. The formula for a nitrate ion is ${\text{NO}_3}^{-}$ . The charge on each of the nitrate ion is -1. The charge on the two ions should balance. As a result, each $\text{Pb}^{2+}$ ion in lead nitrate would pair up with two ${\text{NO}_3}^{-}$ ions. The formula for lead nitrate will be $\text{Pb}({\text{NO}_3})_2$ . Each formula unit of lead nitrate will contain one $\text{Pb}^{2+}$ ion and two ${\text{NO}_3}^{-}$ ions.

The "III" in the name "chromium (III) nitrate" is a Roman Numeral. It indicates that the oxidation state of Cr in chromium (III) nitrate is +3. The Cr in that compound will exist as $\text{Cr}^{3+}$ . Similarly, each $\text{Cr}^{3+}$ will pair up with three ${\text{NO}_3}^{-}$ ions. The formula for chromium (III) nitrate will be $\text{Cr}(\text{NO}_3})_3$ . Each formula unit of chromium (III) nitrate will contain one ${\text{NO}_3}^{-}$ ion and three ${\text{NO}_3}^{-}$ ions.

0.85 moles formula units of lead nitrate will contain 0.85 × 2 = 1.7 moles of ${\text{NO}_3}^{-}$ ions. Those nitrate ions will end up in 1.7 / 3 = 0.57 moles formula units of chromium (III) nitrate. As a result, the reaction will produce 0.57 moles formula units of chromium (III) nitrate.

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