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2 years ago

In each reaction box, place the best reagent and conditions from the list below.

Chemistry

1 answer:

Serggg [28]2 years ago

8 0

This is a two step step reaction as shown below,

Step 1:
Epoxidation of Alkene:
In first step Cyclohexene is treated with peroxybenzoic acid in the presence of Dichloromethane which results in the formation of epoxide moiety.

Step 2:
Nucleophillic Substitution Reaction:
In second step the epoxide formed is treated with Sodium Methoxide which acts as a nucleophile and attacks the less hindered carbon of epoxide resulting in the formation of final product as shown below.

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A sample of neon has a volume of 40.81 m3 at 23.5C. At what temperature, in Kelvins, would the gas occupy 50.00 cubic meters? As

mezya [45]

At $\fbox{\begin \\363 K \end{minispace}}$ temperature, a sample of neon gas will occupy $50.00 \text{ m}^{3}$ volume.

Further Explanation:

The given problem is based on the concept of Charles’ law. Charles’ law states that “at constant pressure and fixed mass the volume occupied an ideal gas is directly proportional to the Kelvin temperature.”

Mathematically the law can be expressed as,

$\fbox{ \begin \\ V \propto T \end{minispace}}$

Or,

$\frac{V}{T}=k$

Here, V is the volume of the gas, T is Kelvin temperature, and k is proportionality constant.

Given information:

The initial volume of neon gas is $40.81 \text{ m}^{3}$ .

The final volume of neon gas is $50.00 \text{ m}^{3}$ .

The initial temperature value is $23.5 \text{ } ^{\circ} \text{C}$ .

To calculate:

The final temperature

Given Condition:

The pressure is constant.

Mass of gas is fixed.

Solution:

Step 1: Modify the mathematical expression for Charles’ law for two different temperature and volume values as follows:

$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$

Here,

$V_{1}$ is the initial volume of the gas.

$V_{2}$ is the final volume of the gas.

$T_{1}$ is the initial temperature of the gas.

$T_{2}$ is the final temperature of the gas.

Step 2: Rearrange equation (2) for .

$\fbox {\begin \\T_{2}=\frac{(V_{2}) \times (T_{1})}{V_{1}}\\\end{minispace}}$ …… (2)

Step 3: Convert the given temperature from degree Celsius to Kelvin.

The conversion factor to convert degree Celsius to Kelvin is,

$T(\text{K}) = T(^{\circ}\text{C}) + 273.15$ …… (3)

Substitute $23.5\text{ }^{\circ} \text{C}$ for $T(^{\circ}\text{C})$ in equation (3) to convert temperature from degree Celsius to Kelvin.

$T(\text{K}) = 23.5 \text{ } ^{\circ} \text{C} + 273.15\\T(\text{K})= 296.65 \text{ K}$

Step 4: Substitute $40.81 \text{ m}^{3}$ for $V_{1}$ , $50.00 \text{ m}^{3}$ for $V_{2}$ and $296.65 \text{ K}$ for $T_{1}$ in equation (2) and calculate the value of $T_{2}$ .

$T_{2}=\frac{(50.00 \text{ m}^{3}) \times (296.65 \text{ K})}{40.81 \text{ m}^{3}}\\T_{2}=363.45 \text{ K}\\T_{2} \approx 363 \text{ K}$

Important note:

The temperature must be in Kelvin.

The condition of fixed mass and fixed pressure must be fulfilled in order to apply Charles’ law.

Learn More:

1. Gas laws brainly.com/question/1403211

2. Application of Charles’ law brainly.com/question/7434588

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: States of matter

Keywords: neon, volume, occupies, temperature, Kelvin, degree Celsius, Charle’s law, constant pressure, fixed mass, 40.81 m^3 , 50.00 m^3 , 23.5 degree C , celsius , 363 K , sates of matter, initial volume, final volume, initial temperature, final temperature, V1 , V2 , T1 , T2 .

5 0

2 years ago

Read 2 more answers

Chem help what's the answer to 8 and why will give brainliest

laiz [17]

avogadros law states that under constant temp n pressure, volume of a gas is directly proportional to amount of gas

amount of gas1 = 352millimoles

amount of gas2 = 352+100 = 452millimoles

new vol = old vol x 452/352 = 25.2 x 452/352

= 32.4 mL

4 0

3 years ago

Read 2 more answers

A 0.43g samle of KHP required 24.11cm of NaOH for neutralization. Calculate the molarity of NaOH

expeople1 [14]

Answer:

0.083 M

Explanation:

We'll begin by calculating the number of mole in 0.43 g of KHP (potassium hydrogen phthalate, C₈H₅O₄K). This is can be obtained as follow:

Mass of C₈H₅O₄K = 0.43 g

Molar mass of C₈H₅O₄K = (8×12) + (5×1) + (16× 4) + 39

= 96 + 5 + 64 + 39 = 204 g/mol

Mole of C₈H₅O₄K =?

Mole = mass / molar mass

Mole of C₈H₅O₄K = 0.43 / 204

Mole of C₈H₅O₄K = 0.002 mole

Next, we shall determine the number of mole of NaOH required to react with 0.43 g (i.e 0.002 mole) of KHP. This can be obtained as follow:

C₈H₅O₄K + NaOH → C₈H₄O₄KNa + H₂O

From the balanced equation above,

1 mole of KHP reacted with 1 mole of NaOH.

Therefore, 0.002 mole of KHP will also react with 0.002 mole of NaOH.

Next, we shall convert 24.11 cm³ to L. This can be obtained as follow:

1000 cm³ = 1 L

Therefore,

24.11 cm³ = 24.11 cm³ × 1 L / 1000 cm³

24.11 cm³ = 0.02411 L

Finally, we shall determine the molarity of NaOH. This can be obtained as follow:

Mole of NaOH = 0.00 2 mole

Volume = 0.02411 L

Molarity of NaOH =?

Molarity = mole /Volume

Molarity of NaOH = 0.002 / 0.02411

Molarity of NaOH = 0.083 M

3 0

2 years ago

If 12.5 grams of strontium hydroxide is reacted with 150 mL of 3.5 M carbonic acid, identify the limiting reactant.

kiruha [24]

Answer:

Sr(OH)₂ will be the limiting reagent.

Explanation:

First of all, you should know the following balanced chemical equation:

2 H₂CO₃ + 2 Sr(OH)₂ → 4 H₂O + Sr₂(CO₃)₂

The balanced equation is based on the Law of Conservation of Mass, which says that matter cannot be created or destroyed. Therefore, the number of each type of atom on each side of a chemical equation must be the same.

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use the reaction stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction). By stoichiometry the following amounts in moles react:

strontium hydroxide: 2 moles
carbonic acid: 2 moles

Now, you know the following masses of the elements:

Sr: 87.62 g/mole
O: 16 g/mole
H: 1 g/mole

So the molar mass of strontium hydroxide is:

Sr(OH)₂= 87.62 g/mole + 2*(16 g/mole + 1 g/mole)= 121.62 g/mole

You apply the following rule of three, if 121.62 grams of hydroxide are present in 1 mole, 12.5 grams in how many moles are they?

$moles of strontium hydroxide=\frac{12.5 grams*1 mole}{121.62 grams}$

moles of hydroxide= 0.103 moles

On the other hand, you have 150 ml of 3.5 M carbonic acid. Since molarity is the concentration of a solution expressed in the number of moles dissolved per liter of solution, you can apply the following rule of three: if in 1 L there are 3.5 moles of carbonic acid, in 0.150 L (being 1 L = 1000 mL, 0.150 L = 150 mL) how many moles of acid are there?

$molesofcarbonicacid=\frac{0.150 L*3.5 moles}{1 L}$

moles of carbonic acid= 0.525 moles

Finally, to calculate the limiting reagent, you can use a simple rule of three as follows: if by stoichiometry 2 mole of strontium hydroxide reacts with , how much moles of carbonic acid will be needed if 0.103 moles of strontium hydroxide react?

$molesofcarbonicacid=\frac{0.103 moles of strontium hydroxide*2 moles of carbonic acid}{2 moles of strontium hydroxide}$

moles of carbonic acid= 0.103 moles

But 0.525 moles are available. Since more moles are available than you need to react with 0.103 moles of strontium hydroxide, Sr(OH)₂ will be the limiting reagent.

7 0

3 years ago

Explain why we use 1/12 in finding atomic mass unit ?

Nat2105 [25]

Answer:

The u (amu is the old unit name) is 1/12 of the weight of an 12C atom. The way the u is chosen ensures that all core and atom masses are multiples of 1(±0.1) u.

Explanation:

Further explanation if needed...

Carbon 12 was chosen because the chemical atomic weights based on C12 are almost identical to the chemical atomic weights based on the natural mix of oxygen. Simply because the atomic mass is defined as 1/12 of the mass of 12C. Others isotopes of carbon (13C mostly, with an abundance of 1.1% approximately) account for an average atomic mass slightly above 12.

7 0

2 years ago