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hodyreva [135]
3 years ago
12

The c—c—c bond angle in propane, c3h8, is closest to

Chemistry
2 answers:
SVETLANKA909090 [29]3 years ago
5 0
Answer is: bond angle in propane is closest to 109°.

Propane is alkane, organic compound. Carbons in propane have sp3 hybridization (carbon’s 2s and three 2p orbitals combine into four identical sp3 orbitals). Orbitals in sp3 hybridization have <span>a tetrahedral arrangement and form single (sigma) bonds.</span>
Archy [21]3 years ago
3 0
Propane has a molecular formula C3H8.

It is of the form CnH2n+2.
where n = number of C atoms. In present case n = 3.

Each carbon atom in propane is sp3 hydribized. Thus, there are 4 hydrid orbital associated with each carbon atom

The central C atom is sigma bonded to two other carbon atoms and two hydrogen atoms.

Such, sp3 hydridized orbitals are spacial orientated at an  angle of 109^{0}28^{'}, in order to minimize repulsion between the electrons.

Hence, the c—c—c bond angle in propane, c3h8, is closest to 109^{0}28^{'}
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Answer:

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Explanation:

8 0
3 years ago
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Na + CI2 =2NaC is It balanced
scoray [572]

Answer:

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Explanation:

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3 0
3 years ago
What is the empirical formula for a compound that is 31.9% potassium, 28.9% chlorine, and 39.2% oxygen?
notsponge [240]
The  empirical formula  for a compound   is KClO3

     Explanation 
 
find  the  moles of each  element
moles  = % composition/molar  mass

molar  mass of  of  potassium =39g/mol ,chlorine = 35.5 g/mol, oxygen =16 g/mol

moles  of potassium  =  31.9 / 39 = 0.818  moles
moles of  chlorine    = 28.9/35.5 = 0.814 moles
moles of   oxygen  =  39.2/  16 =  2.45  moles


find the  mole ratio  by  dividing   with  the smallest  mole = 0.814  moles

potassium = 0.818/0.814  =1  
chlorine  =  0.814/0.814 = 1
oxygen =  2.45 /0.814 =3 


the empirical  formula   is therefore = KClO3
7 0
3 years ago
A 75.0 g sample of dinitrogen monoxide is confined in a 3.q L vessel. What is the pressure( in atm) at 115 celsius
Nonamiya [84]
Data Given:
                  Pressure  =  P  =  ?

                  Volume  =  V  =  3.0 L

                  Temperature  =  T  =  115 °C + 273  =  388 K

                  Mass  =  m  =  75.0 g

                  M.mass  =  M  =  44 g/mol

Solution:
              Let suppose the Gas is acting Ideally. Then According to Ideal Gas Equation,
                                      P V  =  n R T
Solving for P,
                                      P  =  n R T / V      ------ (1)
Calculating Moles,
                                      n  =  m / M

                                      n  =  75.0 g / 44 g.mol⁻¹

                                      n  =  1.704 mol

Putting Values in Eq. 1,

                    P  =  (1.704 mol × 0.08205 atm.L.mol⁻¹.K⁻¹ × 388 K) ÷ 3.0 L

                    P  =  18.08 atm
7 0
3 years ago
30 its ASAP!!!! What is the volume, in liters, occupied by 0.485 moles of Oxygen gas at 23.0 oC and 0.980 atm?
USPshnik [31]
Use the universal gas formula

PV=nRT
where
P=pressure ( 0.980 atm)
V=volume (L)
T=temperature ( 23 &deg; C = 23+273.15 = 296.15 &deg; K)
n=number of moles of ideal gas (0.485 mol)
R=universal gas constant = <span>0.08205 L atm / (mol·K)

Substitute values,
Volume, V (in litres)
=nRT/P
=0.485*0.08205*296.15/0.980
= 12.0256 L
= 12.0 L (to three significant figures)

</span>
5 0
3 years ago
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