1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Tatiana [17]
3 years ago
8

A compound is 53.31% c, 11.18% h, and 35.51% o by mass. what is its empirical formula?

Chemistry
1 answer:
Alexus [3.1K]3 years ago
3 0
<span>Your final answer would be C4H10O2, which equals 90amu</span>
You might be interested in
Can someone help me?
elena-s [515]
It’s will decrease I have to make it 20 characters but it’s decrease
4 0
2 years ago
Read 2 more answers
Manganese dioxide (MnO2(s), Hf = –520.0 kJ) reacts with aluminum to form aluminum oxide (AI2O3(s), Hf = –1699.8 kJ/mol) and mang
Temka [501]

Answer : The enthalpy of the reaction = -1839.6 KJ

Solution : Given,

\Delta (H_{f})_{MnO_{2}} = -520.0 KJ/mole

\Delta (H_{f})_{Al_{2}O_{3}} = -1699.8 KJ/mole

The balanced chemical reaction is,

3MnO_{2}(s)+4Al(s)\rightarrow 2Al_{2}O_{3}(s)+3Mn(s)

Formula used :

\Delta (H_{f})_{reaction}=\sum n(\Delta H_{f})_{product}-\sum n(\Delta H_{f})_{reactant}

\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))

We know that the standard enthalpy of formation of the element is equal to Zero.

Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.

Now, put all the values in above formula, we get

\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]

                        = (-3399.6) + (1560)

                        = -1839.6 KJ



5 0
3 years ago
An iron chloride compound contains 55.85 grams of iron and 106.5 grams of chlorine. What is the most likely empirical formula fo
mart [117]

Answer:

FeCl_{3}

Explanation:

Moles =\frac {Given\ mass}{Molar\ mass}

mass of Fe = 55.85 g

Molar mass of Fe = 55.85 g/mol

<u>Moles of Fe = 55.85 / 55.85 = 1</u>

mass of Cl = 106.5 g

Molar mass of Cl = 35.5 g/mol

Moles of Cl = 106.5 / 35.5 = 3

Taking the simplest ratio for Fe and Cl as:

1 : 3

The empirical formula is = FeCl_{3}

3 0
3 years ago
The body gets rid of lactic acid in a chemical pathway that requires ___________. A. carbon dioxide . B. oxygen . C. amino acids
Luden [163]
The body gets rid of acid in a chemical pathway that requires oxygen. The correct answer is B, oxygen. 
5 0
2 years ago
At STP, which substance has metallic bonding?
kogti [31]
Silver has metallic bonding.

Silver is a very typical and main metal. The negatively charged electrons distribute themselves throughout the entire piece of metal and form non directional bonds between the positive silver ions, which is metallic bonding, and what silver contributes.
7 0
3 years ago
Read 2 more answers
Other questions:
  • A solution was prepared by dissolving 40.0 g of kcl in 225 g of water. part a calculate the mass percent of kcl in the solution.
    5·2 answers
  • What does the root word for "atom" mean and why is that not an accurate name?
    10·1 answer
  • How does the snake's skull allow it to eat large prey?
    14·1 answer
  • You will learn that matter is neither created nor destroyed in any chemical change. Is this statement a theory or a law?
    9·1 answer
  • (SCIENCE)Which conditions are needed for a hurricane to form?
    14·1 answer
  • When an acetic acid solution is titrated with sodium hydroxide, the slope of the titration curve (pH versus volume of NaOH added
    13·1 answer
  • How many number of orbitals are in Potassium (K)?<br><br> 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1
    13·1 answer
  • 1. What would be the molarity of the sodium ion in solution.​
    15·1 answer
  • Can someone help me pls i understand if you don't know too DONT SEND IN LINKS THO
    6·2 answers
  • 0 clue what is going on. pls help
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!