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2 years ago

How many carbon atoms are represented by the model below A) 0 B) 3 C) 2 D) 1

Chemistry

1 answer:

Bogdan [553]2 years ago

5 0

Answer:

Option (B) 3.

Explanation:

In the model represented above, the two extreme represent carbon atoms since no other group are attached to it. The joint at the middle also represent carbon atom.

Thus, we can write a more simplify illustration for the model above as

C—C—C

From the above illustration, we can see that the model contains 3 carbon atom.

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During studies of the following reaction (i), a chemical engineer measured a less-than-expected yield of N2 and discovered that

bonufazy [111]

Answer:

Maximum expected yield = 87.2%

Explanation:

Equations of reactions:

Main reaction: N₂O₄(l) + 2N₂H₄(l) ---> 3N₂(g) + 4H₂O(g)

Side reaction: 2N₂O₄(l) + N₂H₄(l) ----> 6NO(g) + 2H₂O(g)

Molar mass of N₂O₄ = 92 g/mol; molar mass of N₂H₄ = 32 g/mol; molar mass of N₂ = 28 g/mol; molar mass of of NO = 30 g/mol; molar mass of water = 18 g/mol

In the main reaction, 92 g of N₂O₄ reacts with 2 * 32 g of N₂H₄ to produce 3 * 14 g of N₂.

101.1 g of N₂O₄ will react with 2 * 32 * 101.1 / 92 g of N₂H₄ = 70.33 g of N₂H₄

<em>N₂O₄ is the limiting reactant</em>

101.1 g of N₂O₄ will react to produce 3 * 14 * 101.1 / 92 g of N₂ = 46.15 g of N₂

In the side reaction, (6 * 30 g) of NO is produced from (2 * 92 g) of N₂O₄ and 32 g of N₂H₄

12.7 g of N₂O₄ will be produced from ( 2 * 92 * 12.7/180 g) of N₂O₄ and (32 * 12.7/180) g of N₂H₄ to produce

mass of N₂O₄ used = 12.98 g

mass of N₂H₄ used = 2.26 g

mass of N₂O₄ left for main reaction = 101.1 - 12.98 = 88.12 g

mass of N₂H₄ left for main reaction = 101.1 - 2.26 = 98.84 g

In the main reaction, 92 g of N₂O₄ reacts with 2 * 32 g of N₂H₄ to produce 3 * 14 g of N₂

88.12 g of N₂O₄ will react with 2 * 32 * 88.12 / 92 g of N₂H₄ = 61.30 g of N₂H₄

N₂O₄ is the limiting reactant.

88.12 g of N₂O₄ will to react produce 3 * 14 * 88.12 / 92 g of N₂ = 40.23 g of N₂

Percentage yield = (theoretical yield/actual yield) * 100%

Percentage yield = (40.23/46.15) * 100% = 87.2%

Therefore, maximum expected yield = 87.2%

4 0

2 years ago

The gas cyclobutane, C4H8(g), can be used in welding. When cyclobutane is burned in oxygen, the reaction is: C4H8(g) + 6 O2(g)4

Snowcat [4.5K]

Answer:

$\Delta H^o _{rxn} = -2568.9 \ kJ$

$H = 350 JK^{-1}$

$T_{max} = 32.4 ^o C$

Explanation:

From the question we are told that

The reaction of cyclobutane and oxygen is

$C_4H_8_{(g)} + 6 O_2_{(g)} \to 4 CO_2_{(g)} + 4 H_2O_{(g)}$

ΔH°f (kJ mol-1) : C4H8(g) = 27.7 ; CO2(g) = -393.5 ; H2O(g) = -241.8 ΔH° = kJ

Generally ΔH° for this reaction is mathematically represented as

$\Delta H^o _{rxn} = [[4 * \Delta H^o_f (CO_2_{(g)} ) + 4 * \Delta H^o_f(H_2O_{(g)} ] -[\Delta H^o_f (C_2H_6_{(g)} + 6 * \Delta H^o_f (O_2_{(g)}) ] ]$

=> $\Delta H^o _{rxn} = [[4 * (-393.5) + 4 * (-241.8) ] -[ 27.7 + 6 * 0]$

=> $\Delta H^o _{rxn} = -2568.9 \ kJ$

Generally the total heat capacity of 4 mol of CO2(g) and 4 mol of H2O(g), using CCO2(g) = 37.1 J K-1 mol-1 and CH2O(g) = 33.6 J K-1 mol-1. C = J K-1 is mathematically represented as

$H = [ 4 * C_{CO_2_{(g)}} + 6* C_{CH_2O_{(g)}}]$

=> $H = [ 4 * 37.1 + 6* 33.6 ]$

=> $H = 350 JK^{-1}$

From the question the initial temperature of reactant is $T_i = 25^oC$

Generally the enthalpy change( $\Delta H^o _{rxn}$ ) of the reaction is mathematically represented as

$|\Delta H^o _{rxn} |= H * (T_{max} -T_i)$

$2568.9 = 350 * (T_{max} -25)$

=> $\frac{2568.9 }{350} = T_{max} - 25$

=> $T_{max} = 32.4 ^o C$

4 0

2 years ago

Which is true of protons and neutrons?

Reptile [31]

Answer:

the answet is no.3.by The have different mass and different charge.

O sette

5 0

2 years ago

Read 2 more answers

Balance the following redox reaction if it occurs in acidic solution. What are the coefficients in front of h⁺ and fe3 in the ba

Nana76 [90]

The coefficients in front of H⁺ and Fe³⁺ in the balanced reaction is 8 and 5.

<h3>What is a redox reaction ?</h3>

An oxidation-reduction (redox) reaction is a type of chemical reaction that involves a transfer of electrons between two species.

An oxidation-reduction reaction is any chemical reaction in which the oxidation number of a molecule, atom, or ion changes by gaining or losing an electron.

Redox reactions are common and vital to some of the basic functions of life, including photosynthesis, respiration, combustion, and corrosion or rusting.

Fe²⁺ +MnO₄⁻ ⇄ Fe³⁺ + Mn²⁺

Assigning Oxidation numbers

Fe²⁺ : +2

+Mn : + 7 O₄⁻ : -2

Fe³⁺ : +3

Mn²⁺ : +2

Oxidation half reaction Fe²⁺ -----> Fe³⁺ +e-

Reduction half reaction MnO₄⁻ + 5 e⁻ ----> Mn²⁺

After balancing the charge & oxygen atoms

5Fe²⁺ -----> 5Fe³⁺ +5e-

MnO₄⁻ + 5 e⁻ +8H⁺ ----> Mn²⁺ +4H₂O

After adding half reaction

5Fe +MnO₄⁻+8H⁺ ⇄5Fe³⁺ +Mn²⁺ +4H₂O

This is the final balanced reaction.

The coefficients in front of H⁺ and Fe³⁺ in the balanced reaction is 8 and 5.

To know more about Redox Reaction

brainly.com/question/13293425

#SPJ4

3 0

2 years ago

Compound A reacts with alcoholic KOH to yield compound B which on ozonolyzises followed by the reaction with Zn/H2O gives methan

Simora [160]

Answer:

1-chlorobutane

Explanation:

Given that :

-- The compound A is made to react with alcoholic KOH and produce compound B.

-- The compound B on Ozonolysis produces methanol as well as propanol

The compound A should be a haloalkanes as treatment of haloalkanes with the alcoholic KOH, it will give alkene.

$CH_3CH_2CH_2CH_2Cl+KOH( \text{alcohol}) \rightarrow CH_3CH_2CH=CH_2+KCl+H_2O$

So, the compound B should be Butene

Now the ozonolysis of the compound B or butene will give methanol and propanal as shown :

$CH_3CH_2CH=CH_2 \rightarrow CH_2O+CH_3CH_2CHO$

Therefore, the compound A is 1-chlorobutane.

5 0

2 years ago