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2 years ago

When an alpha particle is emitted from an unstable nucleus, the atomic mass numberof the nucleus

Chemistry

1 answer:

barxatty [35]2 years ago

3 0

Answer:

The correct option is D.

Explanation:

Radioactive substances usually emit different types of particles when they are decaying. Such particles include alpha particles, beta particles and gamma ray. When an alpha particle is emitted from an unstable radioactive nucleus such nucleus usually lost an atomic mass that correspond to that of helium atom. Note that an alpha particle is made up of two protons and two neutrons, which result in mass number of 4. Thus, a nucleus that emit an alpha particle will have its mass number (atomic mass) reduce by 4 and atomic number that is reduced by 2.

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Determine whether each description applies to electrophilic aromatic substitution or nucleophilic aromatic substitution.

Alborosie

Answer:

a. electrophilic aromatic substitution

b. nucleophilic aromatic substitution

c. nucleophilic aromatic substitution

d. electrophilic aromatic substitution

e. nucleophilic aromatic substitution

f. electrophilic aromatic substitution

Explanation:

Electrophilic aromatic substitution is a type of chemical reaction where a hydrogen atom or a functional group that is attached to the aromatic ring is replaced by an electrophile. Electrophilic aromatic substitutions can be classified into five classes: 1-Halogenation: is the replacement of one or more hydrogen (H) atoms in an organic compound by a halogen such as, for example, bromine (bromination), chlorine (chlorination), etc; 2- Nitration: the replacement of H with a nitrate group (NO2); 3-Sulfonation: the replacement of H with a bisulfite (SO3H); 4-Friedel-CraftsAlkylation: the replacement of H with an alkyl group (R), and 5-Friedel-Crafts Acylation: the replacement of H with an acyl group (RCO). For example, the Benzene undergoes electrophilic substitution to produce a wide range of chemical compounds (chlorobenzene, nitrobenzene, benzene sulfonic acid, etc).

A nucleophilic aromatic substitution is a type of chemical reaction where an electron-rich nucleophile displaces a leaving group (for example, a halide on the aromatic ring). There are six types of nucleophilic substitution mechanisms: 1-the SNAr (addition-elimination) mechanism, whose name is due to the Hughes-Ingold symbol ''SN' and a unimolecular mechanism; 2-the SN1 reaction that produces diazonium salts 3-the benzyne mechanism that produce highly reactive species (including benzyne) derived from the aromatic ring by the replacement of two substituents; 4-the free radical SRN1 mechanism where a substituent on the aromatic ring is displaced by a nucleophile with the formation of intermediary free radical species; 5-the ANRORC (Addition of the Nucleophile, Ring Opening, and Ring Closure) mechanism, involved in reactions of metal amide nucleophiles and substituted pyrimidines; and 6-the Vicarious nucleophilic substitution, where a nucleophile displaces an H atom on the aromatic ring but without leaving groups (such as, for example, halogen substituents).

3 0

2 years ago

A beach has a supply of sand grains composed of calcite, ferromagnesian silicate minerals, and non-ferromagnesian silicate miner

bezimeni [28]

Ferromagnesian silicate minerals (i looked it up)

4 0

1 year ago

How many moles of n are in 0.187 g of n2o?

Ainat [17]

We know that the molar mass of N is 14 and O is 16, therefore the molar mass of N2O is:

molar mass N2O = 14 * 2 + 16 = 44 g/mol

The number of moles:

moles N2O = 0.187 / 44

moles N2O = 0.00425 mol

There are 2 moles of N per 1 mole of N2O hence:

moles N = 0.00425mol * 2

<span>moles N = 0.0085 mol</span>

8 0

1 year ago

Read 2 more answers

A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor

AlexFokin [52]

Answer : The partial pressure of $C_2HBrClF_3$ and $O_2$ are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of $C_2HBrClF_3$ = 15.0 g

Mass of $O_2$ = 22.6 g

Molar mass of $C_2HBrClF_3$ = 197.4 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $C_2HBrClF_3$ and $O_2$ .

$\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole$

and,

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole$

Now we have to calculate the mole fraction of $C_2HBrClF_3$ and $O_2$ .

$\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971$

and,

$\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.706}{0.0759+0.706}=0.903$

Now we have to partial pressure of $C_2HBrClF_3$ and $O_2$ .

According to the Raoult's law,

$p^o=X\times p_T$

where,

$p^o$ = partial pressure of gas

$p_T$ = total pressure of gas

$X$ = mole fraction of gas

$p_{C_2HBrClF_3}=X_{C_2HBrClF_3}\times p_T$

$p_{C_2HBrClF_3}=0.0971\times 862torr=84torr$

and,

$p_{O_2}=X_{O_2}\times p_T$

$p_{O_2}=0.903\times 862torr=778torr$

Therefore, the partial pressure of $C_2HBrClF_3$ and $O_2$ are, 84 torr and 778 torr respectively.

6 0

1 year ago

When a 375 mL sample of nitrogen is kept at constant temperature, it has a pressure of 1.2 atmospheres. What pressure does it ex

kolezko [41]

Answer:

It exerts a pressure of 3.6 atm

Explanation:

This is a gas law problem. We are looking at volume and pressure with temperature being kept constant, thus, the gas law to use is Boyle’s law. It states that at a given constant temperature, the volume of a given mass of gas is inversely proportional to the pressure of the gas.

Mathematically; P1V1 = P2V2

Let’s identify the parameters according to the question.

P1 = 1.2 atm

V1 = 375mL

P2 = ?

v2 = 125mL

We arrange the equation to make room for P2 and this can be written as:

P2 = P1V1/V2

P2 = (1.2 * 375)/125

P2 = 3.6 atm

3 0

1 year ago

Read 2 more answers