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Ksju [112]
3 years ago
14

The empirical formula of styrene is CH; its molar mass is 104.1 g/mol. What is the molecular formula of styrene?

Chemistry
2 answers:
Natalija [7]3 years ago
7 0
You got the answer correct. If you would double check
given:C (12 g/mol) H (1 g/mol). 

12(8) + 1(8) = 104 g/mol

and for C2H4
12(2)+1(4) = 28g/mol

Also, chemical formula of styrene is<span> C6H5CH=CH2.</span>
arlik [135]3 years ago
5 0
1. Find the molecular formula of styrene 

<span>Molecular formula = n x empirical formula </span>

<span>n= 104.14/13.02 = 8 where 13.02 is the empirical mass of CH </span>

<span>Molecular formula = CH x 8 </span>
<span>= C8H8 </span>

<span>2. From step 1, one molecule of styrene contains 8 atoms of hydrogen or one mole of styrene molecules contains 8 moles of hydrogen atoms (remember the formula of styrene is C8H8). </span>

<span>3. Convert 3.49 g styrene to mole styrene </span>

<span>mole styrene = 3.49 g/104.14 g/mol = 4.71 x 10^-3 mole </span>

<span>4. Calculate mole of hydrogens equivalent to 4.71 x 10^-3 mole styrene. Use mole ratio from step 2. </span>

<span>mole H = 4.71 x 10^-3 mole styrene * 8 moles H/1 mole styrene </span>

<span>mole H = 0.0377 mole </span>

<span>5. Find number of H atoms contained in 0.0377 mole of hydrogen atoms. Use Avogadro's number. </span>

<span>no. of H atoms = 0.0377 mole H * 6.02 x 10^23 atoms H/1mole H </span>

<span>no. of H atoms = 2.27 x 10^22 atoms H</span>
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The first step of electrophilic aromatic substitution involves the formation of the arenium ion intermediate.
Anastaziya [24]

Answer:

True

Explanation:

Aromatic compounds undergo substitution rather than addition reactions because the aromatic structure is maintained.

Electrophilic aromatic substitution begins with attack of the electrophile on the aromatic ring to yield a delocalized intermediate called the arenium intermediate. Loss of hydrogen from this intermediate yields the final product.

4 0
3 years ago
The elementary reaction 2 H 2 O ( g ) − ⇀ ↽ − 2 H 2 ( g ) + O 2 ( g ) proceeds at a certain temperature until the partial pressu
daser333 [38]

Answer:

6.75 × 10⁻⁸is the value of the equilibrium constant at this temperature.

Explanation:

2H₂O(g) ⇄ 2H₂(g) + O₂(g)

Partial pressure of H₂O = 0.0500 atm

Partial pressure of H₂ = 0.00150 atm

Partial pressure of O₂ = 0.00150 atm

The expression of Kp for the given chemical equation is:

K_p = \frac{[H_2]^2[O_2]}{H_2O}

= \frac{(0.00150^2)(0.00150)}{(0.0500)} \\= 6.75 * 10^-^8

6.75 × 10⁻⁸is the value of the equilibrium constant at this temperature

7 0
3 years ago
I NEED THIS ASAP TODAY PLEASSSSEEEEEE
alukav5142 [94]

Option C, mass would be same. Only the gravitational pull will be different

7 0
3 years ago
The correct name for P5O2 is
MAVERICK [17]

Answer:

phosphorous(iii) oxide

Explanation:

Hope this helps! :)

7 0
3 years ago
In the summer of 2016, the city of Columbus Dublin Road Water Treatment plan exceeded the regulatory level of nitrate, which is
Verizon [17]

Answer:

a) 10 mg NO_3^-/L

b) 1.61*10^{-4}mol NO_3^-/L

c) 2.26 mg N/L

Carbon: C=26.64 \frac{mg C}{L}

Explanation:

<u>Nitrate</u>

First of all, is important to know that:

1 ppm=1 mg/L

a) 10 ppm of nitrate (NO_3^-) is equal to 10 mg NO_3^-/L

b) The molecular weight of nitrate is 62 g NO_3^-/mol

10 mg NO_3^-/L=0.01 g NO_3^-/L

\frac{0.01 g NO_3^-/L}{62 g NO_3^-/mol}=1.61*10^{-4} mol NO_3^-/L

c) Nitrate has 14 mg of N per 62 mg of NO3

10 mg NO_3^-/L*\frac{14 mg N}{62 mg NO_3^-}=2.26 mg N/L

<u>Carbon</u>

Carbonate has 12 mg of C per 60 mg of CO_3^{-2}

Bicarbonate has 12 mg of C per 61 mg of HCO_3^{-}

C=24 \frac{mg CO_3^{-2}}{L}*\frac{12 mg C}{60 mg CO_3^{-2}}+111 \frac{mg HCO_3^{-}}{L}*\frac{12 mg C}{61 mg HCO_3^{-}}

C=26.64 \frac{mg C}{L}

8 0
3 years ago
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