Answer:
Explanation:
If we look at the structure of 1-Bromopropane; we will see that it is a derivative of alkane family by the the substitution of an alkyl group. The position of the Bromine in the propane is 1, making 1-Bromopropane a primary alkyl-halide.
Primary alkyl - halide undergo SN2 mechanism. This nucleophilic reaction needs to be a strong alkyl halide , such as 1-Bromopropane used otherwise it will result to a reactive mechanism if a weak electrophile is used.
However, the critical and the main objective here is to Draw the major substitution product if the reaction proceeds in good yield. If no reaction is expected or yields will be poor, draw the starting material in the box. If a charged product is formed, be sure to draw the counterion.
The attached diagrams portraying this notions is shown in the attached file below.
Answer:
pH = 2.69
Explanation:
The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>
<em />
The reaction of HNO₂ with KOH is:
HNO₂ + KOH → NO₂⁻ + H₂O + K⁺
Moles of HNO₂ and KOH that react are:
HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>
KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>
That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:
NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻
HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂
It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:
pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]
pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]
<h3>pH = 2.69</h3>
<h3>Further explanation</h3>
The basic oxide is an oxide-forming a base solution.
These oxides are mainly from group 1 alkaline and group 2-alkaline earth
If this oxide is dissolved in water it will form an alkaline solution
LO + H₂O --> L(OH)₂ ---> alkaline earth
L₂O + H₂O --> LOH --> alkaline
So the basic oxides : Na₂O and MgO
Na₂O + H₂O --> NaOH (sodium hydroxide, strong base)
MgO + H₂O --> Mg(OH)₂ (magnesium hydroxide, strong base)
The aqueous solution of CO₂ , obtained by dissolving CO₂ in water
CO₂ + H₂O --> H₂CO₃ (carbonic acid)
In general, basic oxide is obtained from metal oxide, while acid oxide is obtained from non-metal oxide
1) Formulas:
a) mole fraction of component 1, X1
X1 = number of moles of compoent 1 / total number of moles
b) Molar mass = number grams / number of moles => number of moles = number of grams / molar mass
2) Application
Number of moles of CaI2 = 0.400
Molar mass of water = 18.0 g/mol
Number of moles of water: 850.0 g / 18.0 g/mol = 47.22 mol
Total number of moles = 0.400 + 47.22 =47.62
Molar fraction of CaI2 = 0.400 / 47.62 = 0.00840
Answer:
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