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djverab [1.8K]
3 years ago
12

Name an alkene that would yield the alcohol above on hydration. (Submit a single name even if there is more than one correct ans

wer. Ignore double bond stereochemistry.) Name: fill in the blank 75a878faaf91fe7_1 2,3-Dimethylpentane . Specify whether you would use hydroboration/oxidation or oxymercuration.
Chemistry
1 answer:
insens350 [35]3 years ago
8 0

Answer:

Enzyme ? ...............

You might be interested in
Gaseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water. Suppose 0.802 g of methane i
Kipish [7]

Answer:

1.07g

Explanation:

Step 1:

We will begin by writing the balanced equation for the reaction. This is given below:

CH4 + 2O2 —> CO2 + 2H2O

Step 2:

Determination of the masses of CH4 and O2 that reacted and the mass of H2O produced from the balanced equation. This is illustrated below:

Molar Mass of CH4 = 12 + (4x1) = 12 + 4 = 16g/mol

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 2 x 32 = 64g

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36g

Summary:

From the balanced equation above,

16g of CH4 reacted with 64g of O2 to produce 36g of H2O.

Step 3:

Determination of the limiting reactant.

We need to know which of the reactant is limiting the reaction in order to obtain the maximum mass of water.

This is illustrated below:

From the balanced equation above,

16g of CH4 reacted with 64g of O2.

Therefore, 0.802g of CH4 will react with = (0.802 x 64)/16 = 3.21g of O2.

From the above calculations, a higher mass of O2 is needed to react with 0.802g of CH4. Therefore, O2 is the limiting reactant.

Step 4:

Determination of the mass of H2O produced from the reaction.

To obtain the maximum mass of H2O produced, the limiting reactant will be used because it will generate the maximum yield of the product.

From the balanced equation above,

64g of O2 produce 36g of H2O.

Therefore, 1.9g of O2 will produce = (1.9 x 36)/64 = 1.07g of H2O.

The maximum mass of water (H2O) produced by the reaction is 1.07g

8 0
3 years ago
What is a successful outcome for using a scientific method?
andrezito [222]
Developing a theory would be a successful outcome<span />
3 0
2 years ago
The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol. Part A If an enzyme increases the
emmasim [6.3K]

Answer:

The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme

Explanation:

From the given information:

The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol.

In this  same concentration for the glucose and fructose; the reaction rate can be calculated by the rate factor which can be illustrated from the Arrhenius equation;

Rate factor in the absence of catalyst:

k_1= A*e^{^{^{ \dfrac {- Ea_1}{RT}}

Rate factor in the presence of catalyst:

k_2= A*e^{^{^{ \dfrac {- Ea_2}{RT}}

Assuming the catalyzed reaction and the uncatalyzed reaction are  taking place at the same temperature :

Then;

the ratio of the rate factors can be expressed as:

\dfrac{k_2}{k_1}={  \dfrac {e^{ \dfrac {- Ea_2}{RT} }} { e^{ \dfrac {- Ea_1}{RT} }}

\dfrac{k_2}{k_1}={  \dfrac {e^{[  Ea_1 - Ea_2 ] }}{RT} }}

Thus;

Ea_1-Ea_2 = RT In \dfrac{k_2}{k_1}

Let say the assumed temperature = 25° C

= (25+ 273)K

= 298 K

Then ;

Ea_1-Ea_2 = 8.314 \  J/mol/K * 298 \ K *  In (10^6)

Ea_1-Ea_2 = 34228.92 \ J/mol

\mathbf{Ea_1-Ea_2 = 34.23 \ kJ/mol}

The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme

8 0
3 years ago
one reaction that produces hydrogen gas can be represented by the unbalanced chemical equation Mg(s)+HCI(aq) -&gt; MgCI(aq)+H2(g
Sonbull [250]
<h3>Answer:</h3>

128 g HCl

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

  • Reaction Mole Ratios
  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Unbalanced] Mg (s) + HCl (aq) → MgCl (aq) + H₂ (g)

↓

[RxN - Balanced] 2Mg (s) + 2HCl (aq) → 2MgCl (aq) + H₂ (g)

[Given] 3.25 mol Mg

[Solve] x g HCl

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

[RxN] 2 mol Mg → 2 mol HCl

[PT] Molar Mass of H - 1.01 g/mol

[PT] Molar Mass of Cl - 35.45 g/mol

Molar Mass of HCl - 1.01 + 35.45 = 36.46 g/mol

<u>Step 3: Stoich</u>

  1. [S - DA] Set up:                                                                                                 \displaystyle 3.25 \ mol \ Mg(\frac{2 \ mol \ HCl}{2 \ mol \ Mg})(\frac{36.46 \ g \ HCl}{1 \ mol \ HCl})
  2. [S - DA] Multiply/Divide [Cancel out units]:                                                    \displaystyle 127.61 \ g \ HCl

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

127.61 g HCl ≈ 128 g HCl

3 0
2 years ago
When naming a binary compound, what ending do you use to represent anions?
spin [16.1K]
I believe You replace the ending of the elements name with -ide. example: magnesium flourine should should be magnesium flouride. 
6 0
2 years ago
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