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Wittaler [7]
2 years ago
8

The mole fraction of hcl in a solution prepared by dissolving 5.5 g of hcl in 200 g of c2h6o is ________. the density of the sol

ution is 0.79 g/ml.
Chemistry
1 answer:
Thepotemich [5.8K]2 years ago
5 0
The answer:
all that we search for is the number of mole of HCl and the number of mole of C2H6O

M(HCl) = 5.5g/ mole of HCl , so mole of HCl = 5.5/M(HCl), where M(HCl) is the molar mass.
M(HCl) = 1+ 36.5= 37.5 

moles of HCl = 5.5/37.5=0.14 

M(C2H6O) = 200g / moles of C2H6O, so moles of C2H6O=200g / M(C2H6O)

M(C2H6O)= 2x12+ 6 + 16=46,

moles of C2H6O=200g / 46 =<span>4.35 </span><span> moles 
</span>
the sum of the moles is    0.14 + <span>4.35 </span> = 4.501 moles

finally,  <span>The mole fraction of hcl in a solution prepared by dissolving 5.5 g of hcl in 200 g of c2h6o is 0.031
</span>
because it can be found by  0.14 / 4.501= 0.031

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The combustion of propane (c3h8) produces co2 and h2o: c3h8 (g) + 5o2 (g) → 3co2 (g) + 4h2o (g) the reaction of 2.5 mol of o2 wi
Andrews [41]
The answer is 2 mol of H₂O will be produced.
The balanced equation for the chemical reaction is:

<span>c3h8 (g) + 5o2 (g) → 3co2 (g) + 4h2o (g) 
5 moles of O</span>₂ produces 4 moles of H₂O, and when there is 2.5 mol of O₂, moles of H₂O will be:
2.5 x 4/5 = 2 mol of H₂O
8 0
2 years ago
What name is used to describe the H-N-H group?
mamaluj [8]

Answer:

an amine group

Explanation:

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5 0
2 years ago
A transition in the balmer series for hydrogen has an observed wavelength of 434 nm. Use the Rydberg equation below to find the
DanielleElmas [232]

Answer:

i. n = 5

ii. ΔE = 7.61 × 10^{-46} KJ/mole

Explanation:

1. ΔE = (1/λ) = -2.178 × 10^{-18}(\frac{1}{n^{2}_{final} } - \frac{1}{n^{2}_{initial}  })

    (1/434 × 10^{-9}) = -2.178 × 10^{-18} (\frac{n^{2}_{initial} - n^{2}_{final}  }{n^{2}_{final} n^{2}_{initial}   })

⇒ 434 × 10^{-9} = (1/-2.178 × 10^{-18})\frac{n^{2}_{final} *n^{2}_{initial}   }{n^{2}_{initial} - n^{2}_{final}    }

But, n_{final} = 2

434 × 10^{-9} = (1/2.178 × 10^{-18})\frac{2^{2} n^{2}_{initial}  }{n^{2}_{initial} - 2^{2}  }

434 × 10^{-9}  × 2.178 × 10^{-18} = (\frac{4n^{2}_{initial}  }{n^{2}_{initial} - 4 })

⇒ n_{initial} = 5

Therefore, the initial energy level where transition occurred is from 5.

2. ΔE = hf

     = (hc) ÷ λ

    = (6.626 × 10−34 × 3.0 × 10^{8} ) ÷ (434 × 10^{-9})

    = (1.9878 × 10^{-25}) ÷ (434 × 10^{-9})

    = 4.58 × 10^{-19} J

    = 4.58 × 10^{-22} KJ

But 1 mole = 6.02×10^{23}, then;

energy in KJ/mole = (4.58 × 10^{-22} KJ) ÷ (6.02×10^{23})

         = 7.61 × 10^{-46} KJ/mole

7 0
2 years ago
Consider the skeletal structure of naphthalein (C10H8), the active ingredient in mothballs. How many double bonds must be added
Greeley [361]

Answer:

four (4)

Explanation:

Naphthalein is an organic compound with formula C 10H 8. It is the simplest polycyclic aromatic hydrocarbon, and is a white crystalline solid with a characteristic odor that is detectable at concentrations as low as 0.08 ppm by mass. As an aromatic hydrocarbon, naphthalene's structure consists of a fused pair of benzene rings. It is best known as the main ingredient of traditional mothballs.

The molecule is planar, like benzene. Unlike benzene, the carbon–carbon bonds in naphthalene are not of the same length. The bonds C1−C2, C3−C4, C5−C6 and C7−C8 are about 1.37 Å (137 pm) in length, whereas the other carbon–carbon bonds are about 1.42 Å (142 pm) long. This difference, established by X-ray diffraction is consistent with the valence bond model in naphthalene and in particular, with the theorem of cross-conjugation. This theorem would describe naphthalene as an aromatic benzene unit bonded to a diene but not extensively conjugated to it (at least in the ground state), which is consistent with two of its three resonance structures.

Because of this resonance, the molecule has bilateral symmetry across the plane of the shared carbon pair, as well as across the plane that bisects bonds C2-C3 and C6-C7, and across the plane of the carbon atoms. Thus there are two sets of equivalent hydrogen atoms: the alpha positions, numbered 1, 4, 5, and 8, and the beta positions, 2, 3, 6, and 7. Two isomers are then possible for mono-substituted naphthalenes, corresponding to substitution at an alpha or beta position. Bicyclo[6.2.0]decapentaene is a structural isomer with a fused 4–8 ring system.

Therefore four (4) double bonds will be added to give each carbon atom an octet structure.

8 0
2 years ago
What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig
Mashutka [201]

Answer:

P_H =2.86

c=1.4\times 10^{-4}

Explanation:

first write the equilibrium equaion ,

C_3H_6O_3  ⇄ C_3H_5O_3^{-}  +H^{+}

assuming degree of dissociation \alpha =1/10;

and initial concentraion of C_3H_6O_3 =c;

At equlibrium ;

concentration of C_3H_6O_3 = c-c\alpha

[C_3H_5O_3^{-}  ]= c\alpha

[H^{+}] = c\alpha

K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}

\alpha is very small so 1-\alpha can be neglected

and equation is;

K_a = {c\alpha \times \alpha}

[H^{+}] = c\alpha = \frac{K_a}{\alpha}

P_H =- log[H^{+} ]

P_H =-logK_a + log\alpha

K_a =1.38\times10^{-4}

\alpha = \frac{1}{10}

P_H= 3.86-1

P_H =2.86

composiion ;

c=\frac{1}{\alpha} \times [H^{+}]

[H^{+}] =antilog(-P_H)

[H^{+} ] =0.0014

c=0.0014\times \frac{1}{10}

c=1.4\times 10^{-4}

6 0
2 years ago
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