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Makovka662 [10]
3 years ago
11

1. Define the term disproportionation reaction. 2. Write the balanced equation for the conversion of 4-chlorobezaldehye into 4-c

hlorobenzyl alcohol and 4-chlorobenzoic acid. 3. What is the oxidation number of the aldehyde carbon in each product of the Cannizzaro reaction? Show all work for credit.
Chemistry
1 answer:
sergejj [24]3 years ago
7 0

Answer:

3 is the answer

Explanation:

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I really have no idea what to do here sorry. Help please.
horrorfan [7]

1. An ion is a charged atom. A molecule is a neutrally-charged combination of atoms.

2. A molecule is a combination of atoms. It can consist of atoms from one or more elements. For example, an oxygen molecule comprises two oxygen atoms. A compound is a substance made up of a combination of atoms of different elements. For example, water is a compound of hydrogen and oxygen.

3. An electron dot diagram is a simple way of representing the bond and electronic structure of molecules. A formula is a written representation of the types and numbers of atoms in a molecule.

4. As above...a formula denotes which atoms are in a molecule and how many. For example, H2SO4 tells us there are two hydrogen atoms, one sulfur atom and four oxygen atoms in each molecule of sulfuric acid.

5. An ionic bond is a type of chemical bond that stems from electrostatic attraction between ions with opposite charges. A covalent bond is another type of chemical bond that involves sharing of electrons between atoms in order to achieve a stable electronic structure for the molecule as a whole.

5 0
3 years ago
Be sure to answer all parts. Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after
Fantom [35]

Answer:

(a) pH = 12.73

(b) pH = 10.52

(c) pH = 1.93

Explanation:

The net balanced reaction equation is:

KOH + HBr ⇒ H₂O + KBr

The amount of KOH present is:

n = CV = (0.1000 molL⁻¹)(30.00 mL) = 3.000 mmol

(a) The amount of HBr added in 9.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(9.00 mL) = 0.900 mmol

This amount of HBr will neutralize an equivalent amount of KOH (0.900 mmol), leaving the following amount of KOH:

(3.000 mmol) - (0.900 mmol) = 2.100 mmol KOH

After the addition of HBr, the volume of the KOH solution is 39.00 mL. The concentration of KOH is calculated as follows:

C = n/V = (2.100 mmol) / (39.00 mL) = 0.0538461 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0538461) = 1.2688

pH = 14 - pOH = 14 - 1.2688 = 12.73

(b) The amount of HBr added in 29.80 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(29.80 mL) = 2.980 mmol

This amount of HBr will neutralize an equivalent amount of KOH, leaving the following amount of KOH:

(3.000 mmol) - (2.980 mmol) = 0.0200 mmol KOH

After the addition of HBr, the volume of the KOH solution is 59.80 mL. The concentration of KOH is calculated as follows:

C = n/V = (0.0200 mmol) / (59.80 mL) = 0.0003344 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0003344) = 3.476

pH = 14 - pOH = 14 - 3.476 = 10.52

(c) The amount of HBr added in 38.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(38.00 mL) = 3.800 mmol

This amount of HBr will neutralize all of the KOH present. The amount of HBr in excess is:

(3.800 mmol) - (3.000 mmol) = 0.800 mmol HBr

After the addition of HBr, the volume of the analyte solution is 68.00 mL. The concentration of HBr is calculated as follows:

C = n/V = (0.800 mmol) / (68.00 mL) = 0.01176 M HBr

The pH of the solution can then be calculated:

pH = -log[H⁺] = -log(0.01176) = 1.93

4 0
3 years ago
1. 2.0 mL/s x 4.0 hrs = ? L<br><br> How do you do this conversion problem
Phantasy [73]

Answer:

29 L.

Explanation:

Hello!

In this case, considering that we are performing a conversion by which the time should be cancelled out to obtain liters, we first need to convert the seconds on bottom to hours and then the volume on top to liters, just a shown down below:

2.0\frac{mL}{s} *\frac{1L}{1000mL} *\frac{3600s}{1hr}*4.0hr\\\\=28.8L

Which turns out 29 L with 2 significant figures.

Best regards!

6 0
2 years ago
Us
MrRissso [65]

Answer:

24.9%

Explanation:

According to this question, mole fraction of NaCl in an aqueous solution is 0.0927. This means that the mole percent of NaCl in the solution is:

0.0927 × 100 = 9.27%

Let's assume that the solution contains water (solvent) + NaCl (solute), hence, the mole fraction of water will be;

100% - 9.27% = 90.73%

THEREFORE, it can be said that, NaCl contains 0.0927moles while H2O contains 9.073moles

N.B: mole = mass/molar mass

Given the Molar Mass

NaCl: 58.44 g/mol

H2O: 18.016 g/mol

For NaCl;

0.0927 = mass/58.44

mass = 0.0927 × 58.44

5.42g

For H2O;

9.073 = mass/18.016

mass = 9.073 × 18.016

= 16.35g

Total mass of solution = 16.35g + 5.42g = 21.77g

Mass percent of NaCl = mass of NaCl/total mass × 100

% mass of NaCl = 5.42g/21.77g × 100

= 0.249 × 100

= 24.9%

5 0
3 years ago
Read 2 more answers
Convert 850 mm Hg to kPa. *
Feliz [49]

Answer:

113.324 KPA

Explanation:

4 0
3 years ago
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