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mezya [45]
2 years ago
13

Be sure to answer all parts. Select the anti periplanar geometry for the E2 reaction of (CH3)2CHCH2Br with base. mc2x mc2xd1 mc2

xans mc2xd2 Then draw the product that results after elimination of HBr. draw structure ...
Chemistry
1 answer:
sleet_krkn [62]2 years ago
6 0

The anti periplanar geometry for the E2 reaction of (CH3)2CHCH2Br with base is shown in the image attached as well as the structure of the product formed in the reaction.

In organic chemistry, an antiperiplanar conformation is one in which the groups point up and down at a dihedral angle of 180° away from one another. In the image attached, the antiperiplanar conformation of (CH3)2CHCH2Br is shown.

Recall that an E2 reaction is a synchronous elimination reaction where to atoms leave at the same time. The product of this reaction is also shown in the image attached.

Learn more: brainly.com/question/2510654

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How many grams of F2 gas are there in a 5.00-L cylinder at 4.00 × 10^3 mm Hg and 23°C?
GuDViN [60]

Answer:

41.17g

Explanation:

We are given the following parameters for Flourine gas(F2).

Volume = 5.00L

Pressure = 4.00× 10³mmHG

Temperature =23°c

The formula we would be applying is Ideal gas law

PV = nRT

Step 1

We find the number of moles of Flourine gas present.

T = 23°C

Converting to Kelvin

= °C + 273k

= 23°C + 273k

= 296k

V = Volume = 5.00L

R = 0.08206L.atm/mol.K

P = Pressure (in atm)

In the question, the pressure is given as 4.00 × 10³mmHg

Converting to atm(atmosphere)

1 mmHg = 0.00131579atm

4.00 × 10³ =

Cross Multiply

4.00 × 10³ × 0.00131579atm

= 5.263159 atm

The formula for number of moles =

n = PV/RT

n = 5.263159 atm × 5.00L/0.08206L.atm/mol.K × 296K

n = 1.0834112811moles

Step 2

We calculate the mass of Flourine gas

The molar mass of Flourine gas =

F2 = 19 × 2

= 38 g/mol

Mass of Flourine gas = Molar mass of Flourine gas × No of moles

Mass = 38g/mol × 1.0834112811moles

41.169628682grams

Approximately = 41.17 grams.

3 0
2 years ago
Oxygen gas can be prepared by heating potassium chlorate according to the following equation: 2KClO3(s)2KCl(s) + 3O2(g) The prod
Ipatiy [6.2K]

Answer:

The number of moles of KClO₃ reacted was 0,15 mol

Explanation:

For the reaction:

2KClO₃(s) → 2KCl(s) + 2O₂(g)

The only gas product is O₂.

Total pressure is the sum of vapor pressure of water with O₂ gas formed. Thus, pressure of O₂ is:

749mmHg - 23,8mmHg = 725,2mmHg

Using gas law:

PV/RT = n

Where:

P is pressure (725,2mmHg ≡ <em>0,9542atm</em>)

V is volume (<em>5,76L</em>)

R is gas constant (<em>0,082 atmL/molK</em>)

And T is temperature (25°C ≡ <em>298,15K</em>)

Replacing, number of moles of O₂ are <em>0,2248 moles</em>

As 2 moles of KClO₃ react with 3 moles of O₂ the moles of KClO₃ that reacted was:

0,2248 mol O₂×\frac{2 mol KClO_{3}}{3 mol O_{2}} = <em>0,15 mol of KClO₃</em>

I hope it helps!

8 0
3 years ago
Which of the following is a source of water pollution?
Mashutka [201]
What are " of the following"? 

4 0
2 years ago
Read 2 more answers
Someone answer the limited regent...
OleMash [197]

Answer:

20 g Ag

General Formulas and Concepts:

<u>Chemistry - Stoichiometry</u>

  • Using Dimensional Analysis

<u>Chemistry - Atomic Structure</u>

  • Reading a Periodic Table

Explanation:

<u>Step 1: Define</u>

[RxN]   Cu (s) + AgNO₃ (aq) → CuNO₃ (aq) + Ag (s)

[Given]   10 g Cu

<u>Step 2: Identify Conversions</u>

[RxN]   1 mol Cu = 1 mol Ag

Molar Mass of Cu - 63.55 g/mol

Molar Mass of Ag - 197.87 g/mol

<u>Step 3: Stoichiometry</u>

<u />10 \ g \ Cu(\frac{1 \ mol \ Cu}{63.55 \ g \ Cu})(\frac{1 \ mol \ Ag}{1 \ mol \ Cu} )(\frac{197.87 \ g \ Ag}{1 \ mol \ Ag} ) = 16.974 g Ag

<u>Step 4: Check</u>

<em>We are given 1 sig fig. Follow sig fig rules and round.</em>

16.974 g Ag ≈ 20 g Ag

6 0
3 years ago
I need an answer now please asap and you will me marked brainiest please it's missing I need it now.
valentina_108 [34]

Answer:

C. NaHCO3 → Na2CO3 + H2O + CO2

Explanation:

» For a balanced equation, the number of atoms of reactants and products must be equal.

» In equation C, the reactant side has one sodium atom while ptoduct side has 2 sodium atoms.

» The balanced equations mus be;

{ \rm{ 2NaHCO _{3(aq)} → Na _{2}CO _{3(s)}+ H _{2} O _{(l)} + }CO _{2(g)} }

7 0
2 years ago
Read 2 more answers
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