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igor_vitrenko [27]
1 year ago
10

What volume (ml) of fluorine gas is required to react with 1. 28 g of calcium bromide to form calcium fluoride and bromine gas a

t stp?
Chemistry
1 answer:
Alexandra [31]1 year ago
4 0

144 mL of fluorine gas is required to react with 1.28 g of calcium bromide to form calcium fluoride and bromine gas at STP.

<h3>What is Ideal Gas Law ? </h3>

The ideal gas law states that the pressure of gas is directly proportional to the volume and temperature of the gas.

PV = nRT

where,

P = Presure

V = Volume in liters

n = number of moles of gas

R = Ideal gas constant

T = temperature in Kelvin

Here,

P = 1 atm  [At STP]

R = 0.0821 atm.L/mol.K

T = 273 K  [At STP]

Now first find the number of moles

F₂  +  CaBr₂  →  CaF₂  +  Br₂

Here 1 mole of F₂ reacts with 1 mole of CaBr₂.

So,  199.89 g CaBr₂ reacts with  = 1 mole of F₂

1.28 g of CaBr₂ will react with = n mole of F₂

n = \frac{1.28\g \times 1\ \text{mole}}{199.89\ g}

n = 0.0064 mole

Now put the value in above equation we get

PV = nRT

1 atm × V = 0.0064 × 0.0821 atm.L/mol.K × 273 K

V = 0.1434 L

V ≈ 144 mL

Thus from the above conclusion we can say that 144 mL of fluorine gas is required to react with 1.28 g of calcium bromide to form calcium fluoride and bromine gas at STP.

Learn more about the Ideal Gas here: brainly.com/question/20348074

#SPJ4

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identify the reagents you would use to convert each of the following compounds into pentanoic acid: (a) 1-pentene (b) 1-bromobut
Morgarella [4.7K]

a)BH3.THF is used to convert 1-pentane to pentanoic acid and b)NaCN is used to convert Bromobutane to pentanoic acid.

a) The conversion of 1-pentane to pentanoic acid using BH3, also known as hydroboration-oxidation, is a two-step reaction involving the reaction of 1-pentane with borane (BH3), followed by oxidation of the resulting 1-pentylborane with hydrogen peroxide or other oxidizing agents.

In the first step, 1-pentane reacts with borane (BH3) to form 1-pentylborane, through a process known as hydroboration. This reaction is catalyzed by a Lewis acid, such as aluminum chloride, and proceeds via a hydride transfer from the borane to the 1-pentane.

In the second step, the 1-pentylborane is oxidized to pentanoic acid using hydrogen peroxide (H₂O₂) or other suitable oxidizing agents. The oxidation is catalyzed by an acid, such as hydrochloric acid (HCl), and proceeds via a proton transfer from the 1-pentylborane to the hydrogen peroxide. The end result is the conversion of 1-pentane to pentanoic acid.

The overall chemical reaction for the conversion of 1-pentane to pentanoic acid using borane (BH₃) and hydrogen peroxide (H₂O₂) is as follows:

1-pentane + BH₃ + H₂O₂ → pentanoic acid + H₂O + BH₂

b)The conversion of 1-Bromo butane to pentanoic acid using sodium cyanide (NaCN) proceeds via a nucleophilic substitution reaction. The reaction mechanism involves the following steps:

1. Attack of the nucleophile, NaCN, on the carbon atom of 1-Bromo butane to form a tetrahedral intermediate.

2. Loss of a proton from the tetrahedral intermediate to form a carbanion.

3. Protonation of the carbanion by water (or another proton source) to form pentanoic acid.

The overall reaction can be represented as follows:

1-Bromo butane + NaCN → Pentanoic Acid + NaBr

To know more about reagents, click below:

brainly.com/question/26283409

#SPJ4

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1 year ago
The spectrochemical series is I &lt; Br&lt; &lt; Cl^- &lt; F
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b. The splitting of the d-orbitals is smaller in the [Ni(Cl)6]4- complex than in the [Ni(en)3]2+ complex.

Explanation:

The spectrochemical series is an arrangement of ligands in increasing order of their magnitude of crystal field splitting.

Ligands that occurs towards the right in the series are called strong field ligands and they tend to cause a greater magnitude of crystal field splitting. Ligands that occur towards the left hand side in the series are called weak field ligands and they tend to cause a lesser magnitude of crystal field splitting.

Since Cl^- is a weak field ligand, it causes a lesser magnitude of d orbital splitting compared to ethylenediammine (en) which causes a greater magnitude of d orbital splitting.

Hence; the splitting of the d-orbitals is smaller in the [Ni(Cl)6]4- complex than in the [Ni(en)3]2+ complex.

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