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2 years ago

2 all substances reacting before the arrow in chemical reactions

Chemistry

1 answer:

Volgvan2 years ago

3 0

Answer:

The substances to the left of the arrow in a chemical equation are called reactants. A reactant is a substance that is present at the start of a chemical reaction. The presence of the arrow also indicates that the reaction goes in one direction under the conditions indicated.

Explanation:

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What is the molar mass of CuCIO3

Genrish500 [490]

<h3>Answer:</h3>

147.05 g/mol

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<h3>Explanation:</h3>

<u>Step 1: Define</u>

CuClO₃

Molar Mass of Cu - 63.55 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CuClO₃ - 63.55 + 35.45 + 3(16.00) = 147.05 g/mol

5 0

2 years ago

During the following chemical reaction, 46.3 grams of C3H6O react with 73.2 grams of O2

ra1l [238]

Answer:

a) O2 is the limiting reactant

b) 75.70 grams CO2 (theoretical yield)

c) There remains 12.81 grams of C3H6O

d) The actual yield CO2 is 34.29 grams

Explanation:

Step 1: Data given

Mass of C3H6O = 46.3 grams

Mass of O2 = 73.2 grams

Molar mass of C3H6O = 58.08 g/mol

Molar mass of O2 = 32 g/mol

Step 2: The balanced equation

C3H6O + 4O2 → 3 CO2 + 3H2O

Step 3: Calculate moles C3H6O

Moles C3H6O = mass C3H6O / molar mass C3H6O

Moles C3H6O = 46.3 grams / 58.08 g/mol

Moles C3H6O = 0.793 moles

Step 4: Calculate moles O2

Moles O2 = 73.2 grams / 32 g/mol

Moles O2 = 2.29 moles

Step 5: Calculate limiting reactant

For 1 mol C3H6O we need 4 moles of O2 to produce 3 moles CO2 and 3 moles H2O

O2 is the limiting reactant. It will completely be consumed. (2.29 moles).

C3H6O is in excess. There will react 2.29/4 = 0.5725 moles C3H6O

There will remain 0.793 - 0.5725 = 0.2205 moles C3H6O

This is 0.2205 moles * 58.08 g/mol =<u> 12.81 grams</u>

Step 6: Calculate moles of CO2

For 1 mol C3H6O we need 4 moles of O2 to produce 3 moles CO2 and 3 moles H2O

For 2.29 moles O2 we need 3/4 * 2.29 = 1.72 moles CO2

This is 1,72 moles * 44.01 g/mol = <u>75.70 grams CO2</u>

Step 7: Calculate actual yield

% yield = 45.3 % = 0.453 = (actual yield / theoretical yield)

actual yield = 0.453 * 75.70 = <u>34.29 grams</u>

3 0

3 years ago

Using phenol, dimethyl sulfate, and naoh, show how you would synthesize methyl phenyl ether.

Jlenok [28]

The synthesis of Methyl Phenyl Ether is shown below,

The synthesis takes place in two steps,

Step 1: Formation of Sodium Phenoxide:
Phenol being more acidic than Alcohols with pKa ≈ 10 can loose its proton attached to oxygen atom, as the resulting phenoxide ion is stabilized by resonance. Therefore, when phenol is treated with NaOH (bBase) it looses proton and forms Sodium Phenoxide.

Step 2: Formation of Methyl Phenyl Ether:
The Phenoxide ion formed in first step when treated with Dimethyl Sulfate produces Methyl Phenyl Ether through a Williamson's Ether synthesis Reaction. Dimethyl sulfate is a well known alkylating agent when treated with phenols, thiols and amines. Dimethyl sulfate readily transfers the methyl group and forms NaSO₄CH₃.

The reaction is as follow,

7 0

3 years ago

how do the components that process information in the ENIAC compare with the components that process in a modern laptop computer

STALIN [3.7K]

For starters, ENIAC was as big as a room and had many parts that don't even exist nowadays. It was operated by humans and broke down often. Information processing was a tedious and a difficult process. Modern components enable much faster data travel, fewer opportunities for the thing to break down, and an easier access for people to process data.

5 0

3 years ago

A 50.0-ml sample of 0.50 m hcl is titrated with 0.50 m naoh. what is the ph of the solution after 28.0 ml of naoh have been adde

hram777 [196]

The pH of the solution after addition of 28 mL of NaOH is added to HCl is $\boxed{{\text{0}}{\text{.85}}}$ .

Further Explanation:

The proportion of substance in the mixture is called concentration. The most commonly used concentration terms are as follows:

1. Molarity (M)

2. Molality (m)

3. Mole fraction (X)

4. Parts per million (ppm)

5. Mass percent ((w/w) %)

6. Volume percent ((v/v) %)

Molarity is a concentration term that is defined as the number of moles of solute dissolved in one litre of the solution. It is denoted by M and its unit is mol/L.

The formula to calculate the molarity of the solution is as follows:

${\text{Molarity of solution}}=\dfrac{{{\text{Moles}}\;{\text{of}}\;{\text{solute}}}}{{{\text{Volume }}\left({\text{L}} \riht){\text{ of solution}}}}$

......(1)

Rearrange equation (1) to calculate the moles of solute.

${\text{Moles}}\;{\text{of}}\;{\text{solute}}=\left( {{\text{Molarity of solution}}}\right)\left({{\text{Volume of solution}}}\right)$ ......(2)

Substitute 0.50 M for the molarity of solution and 50 mL for the volume of solution in equation (2) to calculate the moles of HCl.

$\begin{aligned}{\text{Moles}}\;{\text{of}}\;{\text{HCl}}&= \left({{\text{0}}{\text{.50 M}}}\right)\left( {{\text{50 mL}}} \right)\left( {\frac{{{\text{1}}{{\text{0}}^{ - 3}}{\text{ L}}}}{{{\text{1 mL}}}}} \right)\\&= 0.02{\text{5 mol}}\\\end{aligned}$

Substitute 0.50 M for the molarity of solution and 28 mL for the volume of solution in equation (2) to calculate the moles of NaOH.

$\begin{aligned}{\text{Moles}}\;{\text{of}}\;{\text{NaOH}}&=\left( {{\text{0}}{\text{.50 M}}} \right)\left( {{\text{28 mL}}} \right)\left( {\frac{{{\text{1}}{{\text{0}}^{ - 3}}{\text{ L}}}}{{{\text{1 mL}}}}}\right)\\&= 0.014{\text{ mol}}\\\end{aligned}$

The reaction between HCl and NaOH occurs as follows:

${\text{NaOH}} + {\text{HCl}} \to {\text{NaCl}} + {{\text{H}}_2}{\text{O}}$

The balanced chemical reaction indicates that one mole of NaOH reacts with one mole of HCl. So the amount of remaining HCl can be calculated as follows:

$\begin{aligned}{\text{Amount of HCl remaining}}&= 0.02{\text{5 mol}} - 0.01{\text{4 mol}}\\&= {\text{0}}{\text{.011 mol}} \\\end{aligned}$

The volume after the addition of NaOH can be calculated as follows:

$\begin{aligned}{\text{Volume of solution}} &= {\text{50 mL}} + {\text{28 mL}}\\&= {\text{78 mL}}\\\end{aligned}$

Substitute 0.011 mol for the amount of solute and 78 mL for the volume of solution in equation (1) to calculate the molarity of new HCl solution.

$\begin{aligned}{\text{Molarity of new HCl solution}}&= \left({{\text{0}}{\text{.011 mol}}} \right)\left( {\frac{1}{{{\text{78 mL}}}}}\right)\left( {\frac{{{\text{1 mL}}}}{{{{10}^{ - 3}}\;{\text{L}}}}} \right)\\&= 0.1410{\text{2 M}}\\&\approx {\text{0}}{\text{.141 M}}\\\end{aligned}$

pH:

The acidic strength of an acid can be determined by pH value. The negative logarithm of hydronium ion concentration is defined as pH of the solution. Lower the pH value of an acid, the stronger will be the acid. Acidic solutions are likely to have pH less than 7. Basic or alkaline solutions have pH more than 7. Neutral solutions have pH equal to 7.

The formula to calculate pH of an acid is as follows:

${\text{pH}}=- {\text{log}}\left[ {{{\text{H}}^ + }}\right]$ ......(3)

Here,

$\left[{{{\text{H}}^ + }}\right]$ is hydrogen ion concentration.

HCl is a strong acid so it dissociates completely. So the concentration of also becomes 0.141 M.

Substitute 0.141 M for $\left[{{{\text{H}}^ + }}\right]$ in equation (3).

$\begin{aligned}{\text{pH}}&= - {\text{log}}\left({0.141} \right)\\&=0.85\\\end{aligned}$

So the pH of the solution is 0.85.

Learn more:

1. Which indicator is best for titration between HI and ? brainly.com/question/9236274

2. Why is bromophenol blue used as an indicator for antacid titration? brainly.com/question/9187859

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Acid-base titrations

Keywords: molarity, pH, HCl, NaOH, 0.85, 0.141 M, moles of HCl, moles of NaOH, 50 mL, 0.50 M, 28 mL, 0.025 mol, 0.014 mol, 0.011 mol, 78 mL.

4 0

3 years ago

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