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Artist 52 [7]
2 years ago
6

What is the empirical formula of a substance that contains 5.28 g of c, 0.887 g of h, and 3.52 g of o?

Chemistry
2 answers:
kicyunya [14]2 years ago
8 0
Not sure but I think it is C2H4O
kifflom [539]2 years ago
8 0

Answer:

C₂H₄O

Explanation:

Given data

  • Mass of C: 5.28 g
  • Mass of H: 0.887 g
  • Mass of O: 3.52 g

The total mass of the compound is 5.28 g + 0.887 g + 3.52 g = 9.69 g

In order to determine the empirical formula of the compound, we must follow a series of steps.

Step 1: Determine the percent composition

C: (5.28g/9.69g) × 100% = 54.5%

H: (0.887g/9.69g) × 100% = 9.15%

O: (3.52g/9.69g) × 100% = 36.3%

Step 2: Divide each percentage by the atomic mass of the element

C: 54.5/12.0 = 4.54

H: 9.15/1.01 = 9.06

O: 36.3/16.0 = 2.27

Step 3: Divide all the numbers by the smallest one

C: 4.54/2.27 = 2

H: 9.06/2.27 ≈ 4

O: 2.27/2.27 = 1

The empirical formula is C₂H₄O.

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A buffer is 0.282 m c6h5cooh(aq and 0.282 m na(c6h5coo(aq. calculate the ph after the addition of 0.150 moles of nitric acid to
Sindrei [870]
You have to use the Henderson-Hasselbalch equation. Keep in mind that because the Pka is given the equation changes form slightly:

PH = Pka + log[acid/base]

Step 1 (Figure out the concentrations):

0.282 M of Acid (C6H5OOH) - 0.150 M = 0.132 M of acid
0.282 M of Base (C6HCOO) + 0.150 M = 0.432 M of bas3

Step 2 (Plug into equation):

PH = Pka + log[acid/base]
PH = 4.20 + log[0.132 M/0.432 M]

PH = 3.69

7 0
3 years ago
Describe the structure of ammonium lauryl sulfate. Refer to the given diagram. Your answer should include the type of bonding, t
Anit [1.1K]

Answer:

The compound contains lauryl sulfate and ammoium ions. Lauryl sulfate contains lauric acid (in black and white) , the fatty acid formed by the covalent bonds between C-C attached to hydrogens, and sulfate ions attached to lauric acid (in red) with C-S covalent bond. Sulfer is attached to oxygen by covalent bonds. In Ammonium ions, N is surrounded by four hydrogen atoms.

6 0
3 years ago
Which equation is correctly balanced?
ludmilkaskok [199]

Answer:

C) 2 H₂ + O₂  →  2 H₂O

Explanation:

4 atoms of hydrogen on reactant side

2 atoms of oxygen on reactant side

4 atoms of hydrogen on product side

2 atoms of oxygen on product side

6 0
2 years ago
How many molecules are there in 79g of Fe2O3?
weqwewe [10]
There are approximately 160 grams in 1 mol of fe203 molecules therefore there would be 79/160 = 0.49375 mols of fe203 molecules in 79 grams therefore 5 atoms in total for each molecule of fe203 therefore 79/160 *5 =79/32=2.46875 mols in atoms
6 0
3 years ago
I want to know which ones are molecular equation, complete ionic equation and net ionic equation
NNADVOKAT [17]

Answer:

The molecular equations are:

1. CuSO₄ (aq) + 2 KOH (aq) ----> Cu(OH)₂ (s) + K₂SO₄ (aq)

2. Ba(NO₃)₂ (aq) + K₂SO₄ (aq) + BaSO₄ (s) + 2 KNO₃ (aq)

The complete ionic equations are:

1. Ag + (aq) + NO₃- (aq) + I- (aq) + Na (aq) ---> AgI (s) + No₃- (aq) + Na+ (aq)

2. Cu²+ + SO₄²- (aq) + 2 K+ (aq) + 2 OH- (aq) ---> Cu(OH)₂ (s) + 2K+ (aq) + SO₄²- (aq)

The net ionic equations are:

1. Ca²+ (aq) + SO₄²- (aq) ---> CaSO₄ (s)

2. Ba²+ (aq) +SO₄²- (aq) ---> BaSO₄ (s)

Explanation:

A molecular equation is a balanced chemical equation which shows the reacting species as molecules rather than as componenet ions in their compounds with subscripts written beside the molecules to indicate the state in which they occur in the chemical reaction.

An ionic equation expresses the reacting species as components ions in a chemical reation. All the ions and molecules reacting are shown.

In a net ionic equation, the ions which remain in the ionic state also known as spectator ions are not written as part of the equation.

From the given attachment;

The molecular equations are:

1. CuSO₄ (aq) + 2 KOH (aq) ----> Cu(OH)₂ (s) + K₂SO₄ (aq)

2. Ba(NO₃)₂ (aq) + K₂SO₄ (aq) + BaSO₄ (s) + 2 KNO₃ (aq)

The complete ionic equations are:

1. Ag + (aq) + NO₃- (aq) + I- (aq) + Na (aq) ---> AgI (s) + No₃- (aq) + Na+ (aq)

2. Cu²+ + SO₄²- (aq) + 2 K+ (aq) + 2 OH- (aq) ---> Cu(OH)₂ (s) + 2K+ (aq) + SO₄²- (aq)

The net ionic equations are:

1. Ca²+ (aq) + SO₄²- (aq) ---> CaSO₄ (s)

2. Ba²+ (aq) +SO₄²- (aq) ---> BaSO₄ (s)

8 0
2 years ago
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