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3 years ago

One method of favoring the formation of an ester is to add excess acetic acid.

Chemistry

1 answer:

AleksandrR [38]3 years ago

4 0

Answer: True

Explanation:

An ester refers to the class of organic compounds which react with water in order to produce organic acids, inorganic acids or alcohol. They are usually gotten from carboxylic acids.

A method of favoring the formation of an ester is to add excess acetic acid. Therefore, the above statement is true.

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The density of lead is 11.3g/cm^3 and the density of chromium is 7.14g/cm^3. If we have one gram of each metal, what is the rati

Lady bird [3.3K]

The ratio of the volume of lead to the volume of chromium is 22 : 35

Density is simply defined as the mass of a substance per unit volume of the substance.

<h3>Density = mass / volume </h3>

To solve the question given above, we'll begin by calculating the volume of lead and chromium. This can be obtained as follow:

Mass of lead = 1 g

Density of lead = 11.3 g/cm³

<h3>Volume of lead =? </h3>

Density = mass /volume

11.3 = 1 / Volume

Cross multiply

11.3 × Volume = 1

Divide both side by 11.3

Volume = 1 / 11.3

<h3>Volume of lead = 0.088 cm³</h3>

<h3>For Chromium:</h3>

Mass of chromium = 1 g

Density of chromium = 7.14 g/cm³

<h3>Volume of chromium =? </h3>

Density = mass /volume

7.14 = 1 / Volume

Cross multiply

7.14 × Volume = 1

Divide both side by 7.14

Volume = 1 / 7.14

<h3>Volume of chromium = 0.14 cm³</h3>

Finally, we shall determine the ratio of the volume of lead to the volume chromium. This can be obtained as follow:

Volume of lead = 0.088 cm³

Volume of chromium = 0.14 cm³

<h3>Ratio of the volume of lead to chromium =? </h3>

Ratio = Volume of lead / Volume of chromium

Ratio = 0.088 / 0.14

Ratio = 22 / 35

<h3>Ratio = 22 : 35</h3>

Therefore, the ratio of the volume of lead to the volume chromium is 22 : 35

Learn more: brainly.com/question/17262276

5 0

3 years ago

How many moles of oxygen gas, o2, are in a storage tank with a volume of 1.000×105 l at stp?

Mumz [18]

STP means standard temperature and pressure which is equivalent to 273 K and 1 atm, respectively. Assuming ideal gas behavior, the solution for this problem is as follows:

PV = nRT
Solve for n,
n = RT/PV
n = (0.0821 L-atm/mol-K)(273 K)/(1 atm)(1×10⁵ L)
<em>n = 2.24×10⁻⁴ moles</em>

6 0

3 years ago

Read 2 more answers

How does the polarity of water make water more cohesive than other liquids?

MatroZZZ [7]

Cohesion holds hydrogen bonds together to create surface tension on water. Since water is attracted to other molecules, adhesive forces pull the water toward other molecules.

3 0

3 years ago

Be sure to answer all parts. calculate δg ocell for the reaction between cr(s) and cu2+(aq). e ocell = 1.08 j/c. enter your answ

Mademuasel [1]

Answer:

$\boxed{-6.29 \times10^{5}\text{ J}}$

Explanation:

Step 1. Determine the cell potential

2×[Cr ⟶ Cr³⁺ + 3e⁻] 0.744 V

2Cr + 3Cu²⁺ ⟶ 3Cu + 2Cr³⁺ 1.086 V

Step 2. Calculate ΔG°

$\Delta G^{\circ} = -nFE_{\text{cell}}^{^{\circ}} = -6 \times 96 485 \times 1.086 = \text{-629 000 J}\\\\= \boxed{-6.29 \times10^{5}\text{ J}}$

6 0

3 years ago

For the following reaction, 7.53 grams of benzene (C6H6) are allowed to react with 8.33 grams of oxygen gas. benzene (C6H6) (l)

juin [17]

Answer:

The maximum amount of CO2 that can be formed is 9.15 grams CO2

O2 is the limiting reactant

There will remain 4.82 grams of benzene

Explanation:

Step 1: Data given

Mass of benzene = 7.53 grams

Mass of oxygen gas = 8.33 grams

Molar mass of benzene = 78.11 g/mol

Molar mass oxygen gas = 32.00 g/mol

Step 2: The balanced equation

2C6H6 + 15O2 → 12CO2 + 6H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles C6H6 = 7.53 grams / 78.11 g/mol

Moles C6H6 = 0.0964 moles

Moles O2 = 8.33 grams / 32.00 g/mol

Moles O2 = 0.2603 moles

Step 4: Calculate the limiting reactant

For 2 moles benzene we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed ( 0.2603 moles). Benzene is in excess. there will react 2/15 * 0.2603 = 0.0347 moles

There will remain 0.0964 - 0.0347 = 0.0617 moles benzene

This is 0.0617 moles * 78.11 g/mol = <u>4.82 grams benzene</u>

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Step 5: Calculate moles CO2

For 2 moles benzene we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O

For 0.2603 moles O2 we'll have 12/15 * 0.2603 = 0.208 moles CO2

Step 6: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.208 moles * 44.01 g/mol

<u>Mass CO2 = 9.15 grams</u>

4 0

4 years ago