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Andru [333]
3 years ago
14

How many grams are in 0.35 moles of C2H4? please show dimensional analysis

Chemistry
1 answer:
fredd [130]3 years ago
5 0

Answer:

9.8g

Explanation:

Using periodic table find molar mass of C and H:

C=12.01g/mol

H=1.008g/mol

Molar mass of C2H4=(12.01)2+(1.008)4=28.03g/mol

Using molar mass times moles of the chemical to find the mass in 0.35 moles of c2h4:

28.03 x 0.35=9.8grams

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Using the principles of VSEPR theory, you can predict the geometry around any atom in any molecule, no matter how complex. Enant
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Explanation:

The O atom is sp3 in a water molecule, with two sigma bonds and two lone pairs of electrons like that in water. The steric integer is thus 4, and its structure is tetrahedral.

The C atom is sp hybridised into two identical bonds and two identical bonds in acetylene.

The steric integer is therefore 2 because only sigma bonds are engaged in deciding hybridization, and its structure is linear.

The C atom is sp2 hybridised in ethene with single pi bond and three sigma identical bonds.

Thus the steric integer is 3, and its structure is planar trigonal.

The C atom is sp2 hybridized in ethene, with one pi bond and three sigma identical bonds.

The steric integer would therefore be 3 and its structure is planar trigonal.

The O atom is sp3 in a water molecule with two bond pairs and two lone pairs of electrons like that. The steric integer is thus 4, and its structure is tetrahedral.

The C atom is sp3 in a methane ring, with 4 bond pairs and no solitary pairs of electrons like that. The steric integer is thus 4, and its structure is tetrahedral.

6 0
3 years ago
In the unbalanced equation given below, what is the element that is gaining electrons?
viktelen [127]

Hey there!:

HCl + MnO2 → MnCl2 + H2O + Cl2

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* on the product side the oxidation state is 0 .

* therefore Cl gains electrons .

* in  MnO2 the oxidation state of Mn is +4

* in MnCl2 the oxidation state of Mn is +2

Therefore Mn loses electrons

Answer A

Hope That helps!

4 0
2 years ago
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What is the approximate pressure of a storage cylinder of recovered R-404A that does not contain any non-condensable impurities
gregori [183]

Answer:

288 psig

Explanation:

4 0
3 years ago
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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

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What experiments did Lavoisier do?
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He did experiments with combustion and gas's
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