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1 year ago

In a fixed cylinder are 3moles of oxygen gas at 300Kelvin and 1.25atm. What is the volume of the container?

Chemistry

1 answer:

vladimir2022 [97]1 year ago

6 0

Answer:

The volume of the container is 59.112 L

Explanation:

Given that,

Number of moles of Oxygen, n = 3

Temperature of the gas, T = 300 K

Pressure of the gas, P = 1.25 atm

We need to find the volume of the container. For a gas, we know that,

PV = nRT

V is volume

R is gas constant, R = 0.0821 atm-L/mol-K

So,

$V=\dfrac{nRT}{P}\\\\V=\dfrac{3\ mol\times 0.0821\ L-atm/mol-K \times 300\ K}{1.25\ atm}\\\\V=59.112\ L$

So, the volume of the container is 59.112 L

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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

1 year ago

Each answer below describes the connectivity around a single atom.

MrRissso [65]

Answer:

d. One single bond and two double bonds.

Explanation:

The octate rule is a chemical rule in which the atoms prefer to have eight electrons in the valence shell. Where a single bond provide two electrons and a double bond provide 4 electrons. Thus:

a. Two double bonds . Two double bonds provide 8 electrons. Octate rule is not violated

b. Three single bonds and one pair of electrons . Three single bonds provide 6 electrons and one pair of electrons provide two electrons. Thus, you have eight electrons and octate rule is not violated

c. Two single bonds and one double bond . Two single bonds provide four electrons and one double bond 4. Thus, you have eight electrons and octate rule is not violated.

d. One single bond and two double bonds. One single bond provides two electrons and two double bonds 8. Thus, you have 10 electrons and octate rule is violated.

e. Four single bonds. Four single bonds provide 8 electrons. Octate rule is not violated.

I hope it helps!

4 0

1 year ago

Calculate the magnitude of the current in amperes if the number of moles of hydrogen gas produced is 5.13×10−4 mol (from part A)

Marat540 [252]

Answer:

1.65 A

Explanation:

The hydrogen gas dissociates following the equation:

H₂(g) → 2H⁺ + 2e⁻

So, for 1 mol of H₂, 2 moles of electrons are given. The current is formed because of the presence of the electrons. Thus, the number of moles of electrons is the double of the moles of hydrogen gas: 1.026x10⁻³ mol.

By the Faraday law, 1 mol of electrons produces 96500 C of charge, thus:

1 mol --------------- 96500 C

1.026x10⁻³ mol ---- x

By a simple direct three rule:

x = 99.009 C

The current is the charge divided by the time. So, for 1 min = 60 s,

i = 99.009/60

i = 1.65 A

5 0

1 year ago

Draw the Lewis structure for CH3CH2O2H?

kotegsom [21]

Answer:

hydroperoxyethane

Explanation:

tomsFor the Lewis structure we have to remember that all the atoms must have 8 electrons (except for hydrogen). In this structure, we have three types of atoms, Carbon, Hydrogen and Oxygen. So, we have to remember the valence electrons for each atom:

-) Carbon : 4 electrons

-) Hydrogen: 1 electron

-) Oxygen: 6 electrons

We can start with the " $CH_3$ " part. We can put 3 hydrogen bond arroun the carbon. We can use this same logic with " $CH_2$ ". Finally for oxygens, we can put it one bond with $CH_2$ and a bond between oxygens with a final bond with hydrogen to obtain hydroperoxyethane.

See figure 1 for further explanations.

6 0

1 year ago

At equilibrium, the concentrations in this system were found to be [ N 2 ] = [ O 2 ] = 0.200 M and [ NO ] = 0.400 M . N 2 ( g )

stiks02 [169]

Answer: The equilibrium concentration of NO after it is re-established is 0.55 M

Explanation:

For the given chemical equation:

$N_2(g)+O_(g)\rightleftharpoons 2NO(g)$

The expression of $K_{eq}$ for above equation follows:

$K_{eq}=\frac{[NO]^2}{[N_2][O_2]}$ .....(1)

We are given:

$[NO]_{eq}=0.400M$

$[N_2]_{eq}=0.200M$

$[O_2]_{eq}=0.200M$

Putting values in expression 1, we get:

$K_{eq}=\frac{(0.400)^2}{0.200\times 0.200}\\\\K_{eq}=4$

Now, the concentration of NO is added and is made to 0.700 M

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle. This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

The equilibrium will shift in backward direction.

$N_2(g)+O_(g)\rightleftharpoons 2NO(g)$

Initial: 0.200 0.200 0.700

At eqllm: 0.200+x 0.200+x 0.700-2x

Putting values in expression 1, we get:

$4=\frac{(0.700-2x)^2}{(0.200+x)\times (0.200+x)}\\\\x=0.075$

So, equilibrium concentration of NO after it is re-established = (0.700 - 2x) = [0.700 - 2(0.075)] = 0.55 M

Hence, the equilibrium concentration of NO after it is re-established is 0.55 M

4 0

1 year ago