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DENIUS [597]
4 years ago
7

The empirical formula of a group of compounds is CHCl. Lindane, a powerful insecticide, is a member of this group. The molar mas

s of lindane is 290.8 g/mol. How many atoms of carbon does a molecule of lindane contain?
A) 2 B) 3 C) 4 D) 6 E) 8
Chemistry
1 answer:
Sergeeva-Olga [200]4 years ago
3 0

Answer:

<h2>D) 6</h2>

Explanation:

since,   n = molar mass / empirical formula mass

   Empirical formula mass = Total mass of atoms present in empirical formula

                       CHCl = 12+1+35.5

                                = 48.5

 Given, Molar mass = 290.8 g.

           So,        n = 290.8/48.5

                           = 5.995 , that is approx  6.

        So, Molecular formula = n × Empirical formula

                                           = 6 × CHCl

                                           = C_{6}H_{6}Cl_{6}

       So,   Number of C = 6

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When you mix copper sulphate solution and steel wool, what is the chemical property that can be observed.
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Answer:

Explanation:

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6 0
3 years ago
The reaction of sodium peroxide and water produces sodium hydroxide and oxygen gas. The following balanced chemical equation rep
a_sh-v [17]

Answer:

3.925 mol.

Explanation:

  • From the balanced equation:

<em>2 Na₂O₂(s) + 2 H₂O(l) → 4 NaOH(s) + O₂(g) ,</em>

It is clear that 2 moles of Na₂O₂ react with 2 moles of H₂O to produce 4 moles of NaOH and 1 mole of O₂ .

<em>Using cross multiplication:</em>

4 moles of NaOH produced with → 1 mole of O₂ .

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<em>∴ The no. of moles of O₂ made =</em> (1 mole)(15.7 mole)/(4 mole) =  <em>3.925 mol.</em>

8 0
4 years ago
Please help serious answers only
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Answer:

Neptune and Saturn although Neptune isn't on there so just Saturn.

hope it'll help ya out!

3 0
3 years ago
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3 0
3 years ago
Photodissociation of ozone (O3) can be described as producing an oxygen atom with heat of formation 247.5 kJ/mol. However, in re
Lapatulllka [165]

Answer:

0.2193 μm

Explanation:

The reaction showing the Photodissociation of ozone (O3) is given below as:

           O₃         +        hv   -------------------------->      O₂         +       O⁺  

H°        (142.9)                                                         (0)                  (438kJ/mol).

The first thing to do here is to determine the change in the enthalpy of the total reaction, this can be done by subtracting the change in the enthalpy of the reactant from the change in enthalpy in the product. Hence, we have:

ΔH° = [438 kJ/mol + 247.5 kJ/mol] - (142.9) = 542.6 KJ/mol.

This value, that is 542.6 KJ/mol will then be used in the determination of the value for the maximum wavelength that could cause this photodissociation.

Therefore, the maximum wavelength could cause this photodissociation ≤ h × c/ E = [ 1.199 × 10⁻⁴]/ 542.6 = 2.193 × 10⁻⁷ =  0.2193 μm

7 0
3 years ago
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