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andrezito [222]
3 years ago
6

Calculate the mass of arsenic(III) chloride produced when 0.150 g of arsenic reacts with excess chlorine according to the equati

on:
2As+3Cl2 → 2AsCl3
Chemistry
1 answer:
adell [148]3 years ago
6 0

Answer:

0.363 g

Explanation:

2As + 3Cl₂      →   2AsCl₃

calculate the number of moles of arsenic

Number of moles = mass/ molar mass

Number of moles =  0.150 g/ 75 g/mol

Number of moles = 0.002 mol

compare the moles of arsenic with arsenic chloride.

                         As             :             AsCl₃

                          2              :                2

                        0.002        :              0.002

Mass of arsenic chloride produced:

Mass = number of  moles × molar mass

Mass = 0.002 × 181.28 g/mol

Mass = 0.363 g

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Answer:The correct names for the structure are:

--> 1. 1-bromo-3-chlorobenzene.

--> 3. meta-bromochlorobenzene.

--> 6. m-bromochlorobenzene.

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Benzene is the simplest member of the aromatic hydrocarbons. It has a ring structure consisting of six carbon and six hydrogen atoms. This equally means that a benzene can have up to six substituents. One of the chemical properties is that benzene and other members of its series undergo substitution reaction whereby one or more of its six hydrogen atoms is replaced by monoatomic reagents.

Disubstituted benzene consists of two substituents which are described based on either numerical locants or specific words for the three possible forms.

The numerical locant method are used the same naming substitutes of other hydrocarbons. From the question, the numerical locant method was derived through using the following steps:

--> the functional group is benzene

--> there are two substituents which includes bromine( written as bromo) and chlorine ( written as chloro)

--> while placing the number, it's done alphabetically ('1-bromo' comes before '3-chloro') in a clockwise manner. This is to give chorine the lowest locant number.

The second naming method for a disubstituted benzene is the the ortho-, meta-, para- (or their singel letter equivalent) nomenclature method. This is only used for benzene structures.

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8 0
3 years ago
1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium
lilavasa [31]

Answer:

This question is incomplete, here's the complete question:

1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium ion present in 20.0 mL of each buffer.

Buffer A

Mass of sodium acetate used: 0.3730 g

Actual ph of the buffer 5.27

volume of the buffer used in buffer capacity titration 20.0 mL

Concentration of standardized NaOH 0.100M

moles of Naoh needed to change the ph by 1 unit for the buffer 0.00095mol

the buffer capacity 0.0475 M

Buffer B

Mass of sodium acetate used 1.12 g

Actual pH of the buffer 5.34

Volume of the buffer used in buffer capacity titration 20.0 mL

Concentration if standardized NaOH 0.100 M

moles of Naoh needed to change the ph by 1 unit 0.0019 mol

the buffer capacity 0.095 M

2.) A change of pH by 1 unit means a change in hydronium ion concentration by a factor of 10. Calculate the number of moles of NaOH that would theoretically be needed to decrease the moles of hydronium you calculated in #1 by a factor of 10 for each buffer. Are there any differences between your experimental results and the theoretical calculation?

3.) which buffer had a higher buffer capacity? Why?

Explanation:

Formula,

moles = grams/molar mass

molarity = moles/L of solution

1. Buffer A

molarity of NaC2H3O2 = 0.3731 g/82.03 g/mol x 0.02 L = 0.23 M

molarity of HC2H3O2 = 0. 1 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.23/0.1)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

Buffer B

molarity of NaC2H3O2 = 1.12 g/82.03 g/mol x 0.02 L = 0.68 M

molarity of HC2H3O2 = 0.3 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.68/0.3)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

2. let x moles of NaOH is added,

Buffer A,

pH = 5.10

[H3O+] = 7.91 x 10^-6 M

new pH = 4.10

new [H3O+] = 7.91 x 10^-5 M

moles of NaOH to be added = (7.91 x 10^-5 - 7.91 x 10^-6) x 0.02 L

= 1.42 x 10^-6 mol

3. Buffer B with greater concentration of NaC2H3O2 and HC2H3O2 has higher buffer capacity as it resists pH change to a wider range due to addition of acid or base to the system as compared to low concentration of Buffer A

5 0
3 years ago
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