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Flauer [41]
3 years ago
12

A compound is 52.14% C, 13.13% H, and 34.73% O by mass. What is the empirical formula of the compound?

Chemistry
2 answers:
yarga [219]3 years ago
7 0
Give the compound a mass of 100 grams, so that the mass of each element is
C=52.14% of 100g = 52.14g
H=13.13g
O=34.73g

convert the mass to moles
divide the mole value by the smallest number of moles calculated
round to nearest whole number, this is the mole ratio / subscripts
Sindrei [870]3 years ago
3 0
C=52.14% of 100g = 52.14g
H=13.13g
O=34.73g
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Answer:

1, just I) color.

Explanation:

Physical properties are the properties that can be observed without changing the composition of a substance, such as color, temperature, density, and boiling point.

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Chemical properties are the properties that need a reaction to being observed, such as the combustibility, which needs a combustion reaction to being quantified.

When a chemical reaction occurs, and the composition of the substance change, it's a chemical change.

So, the heating copper with carbon is a chemical reaction, and purification by electrolysis is too. Color is the only physical property.

4 0
3 years ago
Which of the following is an example of qualitative observation?Select one of the options below as your answer:. A. Joseph is ob
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<h2>Answer : Option C) Joseph is observing the color of the reaction mixture to see whether proteins are present in the given solution.</h2><h3>Explanation :</h3>

An example of qualitative observation is the one where one uses the five senses to identify the changes in the reaction.

Here, when Joseph is studying a reaction mixture he is trying to observe a color change which will confirm that there is proteins present in the reaction mixture or not If there is a color change observed then it will confirm the presence of proteins.

Usually qualitative observations are those which can be easily predicted by using five senses.


5 0
3 years ago
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What two things must all matter have?
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6 0
3 years ago
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The atomic mass of Cu is 63.5. Find its electrochemical equivalent​
FrozenT [24]

Answer:

The electrochemical equivalent of copper, Cu, is 3.29015544 × 10⁻⁷ g/C

Explanation:

The given parameters are;

The element for which the electrochemical equivalent is sought = Copper

The atomic mass of copper = 63.5

The electrochemical equivalent, 'Z', of an element or a substance is the mass, 'm', of the element or substance deposited by one coulomb of electricity, which is equivalent to a 1 ampere current flowing for a period of 1 second

Mathematically, we have;

m = Z·I·t = Z·Q

We have;

Cu²⁺ (aq) + 2·e⁻ → Cu

Therefore, one mole of Cu, is deposited by 2 moles of electrons

The charge carried one mole of electrons = 1 Faraday = 96500 C

∴ The charge carried two moles of electrons, Q = 2 × 96500 C = 193,000 C

Given that the mass of an atom of Cu = 63.5 a.m.u., the mass of one mole of Cu, m = 63.5 g

Z = \dfrac{m}{Q} = \dfrac{63.5 \ g}{193,000 \ C} = 3.29015544 \times 10^{-4} \, g \cdot C^{-1}

∴ Z = 3.29015544 × 10⁻⁴ g/C = 3.29015544 × 10⁻⁷ g/C

The electrochemical equivalent of copper, Cu, is Z = 3.29015544 × 10⁻⁷ g/C

7 0
3 years ago
A 265-mL flask contains pure helium at a pressure of 751 torrs. A second flask with a volume of 465 mL contains pure argon at a
Nadya [2.5K]

Answer:

Total Pressure = 745.6 torr

Partial Pressure of He = 272.8 torr

Partial Pressure of Ar =  472.8 torr

Explanation:

Step 1: Data given

Volume of the flask helium = 265 mL

Pressure in the helium flask = 751 torr = 751/760 atm

Volume of the flask argon = 465 mL

Pressure in the argon flask = 727 torr = 727/760 atm

The total pressure exerted by a gaseous mixture is equal to the sum of the partial pressures of each individual component in a gas mixture.

Step 2: Calculate total volume

Total volume = 265 mL + 465 mL = 730 mL =  0.730 L

Step 3: Boyle's Law:

P1V1=P2V2

⇒ with P1 = total pressure gas exerts in its own flask

 ⇒ with V1 = volume of flask with stopcock valve closed

 ⇒ with P2 = partial pressure of gas exerts on total volume of both flasks when stopcock valve is opened  

 ⇒ with V2 = total volume of both flasks with stopcock valve opened

Helium using Boyle's Law equation from above:

P1V1=P2V2

⇒ with P1 = Pressure of helium = 751 /760 = 0.98816 atm

 ⇒ with V1 = volume of helium = 0.265 L

 ⇒ with P2 = The new partial pressure of helium

 ⇒ with V2 = total volume = 0.730 L

(0.98816 atm)(0.265L)=P2(0.730L)

P2=0.359 atm

Argon using Boyle's Law equation from above:

P1V1=P2V2

⇒ with P1 = Pressure of argon = 727/760 = 0.95658 atm

 ⇒ with V1 = volume of argon = 0.465 L

 ⇒ with P2 = The new partial pressure of argon

 ⇒ with V2 = total volume = 0.730 L

(0.95658 atm)(0.465L)=P2(0.730L)

P2=0.609 atm

Step 4: Convert pressure in atm to torr

Pressure helium = 0.359 atm = 272.8 torr

Pressure argon = 0.609 atm = 472.8 torr

Step 5: Calculate Total pressure

Ptotal = P(He)+P(Ar)

⇒ Pt  = total pressure of the gas mixture

⇒ P(He) = partial pressure of Helium

 ⇒ P(Ar)  = partial pressure of Argon

Pt = 272.8 torr + 472.8 torr

Pt = 745.6 torr

Total Pressure = 745.6 torr

Partial Pressure of He = 272.8 torr

Partial Pressure of Ar =  472.8 torr

5 0
3 years ago
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