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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.

7 0

4 years ago

Why helium is duplet ??

Llana [10]

Answer:

The outermost shell of helium for example is filled with only two electrons. Hydrogen and lithium become stable by acquiring the electronic configuration of the nearest inert gas, which is helium. This is the duplet (or duet) rule.

Explanation:

6 0

4 years ago

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What volume of the stock solution (Part A) would contain the number of moles present in the diluted solution (Part B)?

vova2212 [387]

Idk what is is i am sorry

6 0

3 years ago

What is the volume of 33.25g of butane gas at 293 C and 10.934 kPa?

PSYCHO15rus [73]

Answer:

V = 240.79 L

Explanation:

Given data:

Volume of butane = ?

Temperature = 293°C

Pressure = 10.934 Kpa

Mass of butane = 33.25 g

Solution:

Number of moles of butane:

Number of moles = mass/ molar mass

Number of moles = 33.25 g/ 58.12 g/mol

Number of mole s= 0.57 mol

Now we will convert the temperature and pressure units.

293 +273 = 566 K

Pressure = 10.934/101 = 0.11 atm

Volume of butane:

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

V = nRT/P

V = 0.57 mol × 0.0821 atm.L/ mol.K ×566 K / 0.11 atm

V = 26.49 L/0.11

V = 240.79 L

6 0

3 years ago

Will an empty balloon have precisely the same apparent weight on a scale as a balloon filled with air? Explain.

hram777 [196]

The notion <span>an empty balloon have precisely the same apparent weight on a scale as a balloon filled with air depends on the diameter of the balloon. The weight of the balloon filled with air is equal to the mass of the balloon and the mass of the air inside. The mass of air inside is equal to the density of air multiplied by the volume of the balloon. If the balloon is large, then the two masses are equal whereas if not, the mass of air inside the inflation is neglible</span>

4 0

4 years ago

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