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Colt1911 [192]
3 years ago
10

Organic formula for phenol

Chemistry
1 answer:
Mkey [24]3 years ago
8 0
The organic formula for Phenol is C6H5<span>OH</span>
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Every chemical reaction requires a reactant, also called a reagents.

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What is covalent bond
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Covalent bonding is when electrons are shared between atoms. 

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Write the chemical symbols for three different atoms or atomic cations with 18 electrons.
Ugo [173]

<span>Chlorine has 17 electrons in it natural state but it gains an electron and becomes an anion with 18 electrons. </span>

Potassium (K) has 19 electrons in its natural state. It loses an electron and becomes K+ and then is left with 18 <span>electrons .
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The atomic number of calcium is 20 i.e. Ca has 20 electrons.It Loses 2 electrons and becomes Ca++ and  has 18 electrons.

7 0
4 years ago
The mole fraction of CO2 in a certain solution with H2O as the solvent is 3.6 × 10−4. What is the approximate molality of CO2 in
nataly862011 [7]

Answer: C) 0.020 m

Explanation:

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

Molality=\frac{n\times 1000}{W_s}

where,

n = moles of solute  

W_s = weight of solvent in g  

Mole fraction of CO_2 is = 3.6\times 10^{-4} i.e.3.6\times 10^{-4}  moles of CO_2 is present in 1 mole of solution.

Moles of solute (CO_2) = 3.6\times 10^{-4}

moles of solvent (water) = 1 - 3.6\times 10^{-4} = 0.99

weight of solvent =moles\times {\text {Molar mass}}=0.99\times 18=17.82g

Molality =\frac{3.6\times 10^{-4}\times 1000}{17.82g}=0.020

Thus  approximate molality of CO_2 in this solution is 0.020 m

5 0
4 years ago
A 100g of sample of a compound is combusted in excess oxygen and the products are 2.492g of CO2 and 0.6495 of H2O. Determine the
ruslelena [56]

The empirical formula : C₁₁O₁₄O₃

<h3>Further explanation</h3>

The assumption of the compound consists of C, H, and O

mass of C in CO₂ =

\tt \dfrac{12}{44}\times 2.492=0.680~g

mass of H in H₂O =

\tt \dfrac{2.1}{18}\times 0.6495=0.072~g

mass of O :

mass sample-(mass C + mass H)

\tt 1-(0.68+0.072)=0.248`g

mol of  C :

\tt \dfrac{0.68}{12}=0.056

mol of H :

\tt \dfrac{0.072}{1}=0.072

mol of O :

\tt \dfrac{0.248}{16}=0.0155

divide by 0.0155(the lowest ratio)

C : H : O ⇒

\tt \dfrac{0.056}{0.0155}\div \dfrac{0.072}{0.0155}\div \dfrac{0.0155}{0.0155}=3.6\div 4.6\div  1\\\\\dfrac{11}{3}\div \dfrac{14}{3}\div \dfrac{3}{3}=11:14:3

3 0
3 years ago
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