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Leviafan [203]
3 years ago
8

must undergo addition because they have easily broken π bonds. ____________ rule states in the addition of HX to an unsymmetrica

l alkene, the H atom bonds to the less substituted carbon atom. ____________ are unsaturated hydrocarbons because they have fewer than the maximum number of hydrogen atoms per carbon. ____________ have good leaving groups and therefore readily undergo substitution and elimination reactions. In hydroboration, the boron atom bonds to the ____________ substituted carbon.
Chemistry
1 answer:
erastova [34]3 years ago
3 0

Answer:

Alkenes, Markovnikov's, Alkenes, Alkyl halides, and less.

Explanation:

Alkenes must undergo addition because they have easily broken π bonds.

Markovnikov's rule states in the addition of HX to an unsymmetrical alkene, the H atom bonds to the less substituted carbon atom.

Alkenes are unsaturated hydrocarbons because they have fewer than the maximum number of hydrogen atoms per carbon.

Alkyl halides have good leaving groups and therefore readily undergo substitution and elimination reactions.

In hydroboration, the boron atom bonds to the less substituted carbon.

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Alex777 [14]
The Relative Formula Mass of NaH2PO4 is 120 g/mol
Therefore, the number of moles = 6.6/120
                                                   = 0.055 moles of NaH2PO4 which is also equal to the number of moles of H2PO4.
[H2PO4-] = Number of moles oof H2PO4-/Volume of the solution in L
  = 0.055/ ( 355 ×10^-3)
  = 0.155 M
Na2HPO4 undergoes complete dissociation as follows;
Na2HPO4 (aq)= 2Na+ (aq) + HPO4^2- (aq)

1 mole of Na2HPO4 = 142 g/mol
Therefore; number of moles = 8.0/142
                                             = 0.0563 moles
 [HPO4 ^-2] is given by no of moles HPO4^2- /volume of the solution in L
     = 0.0563/(355×10^-3)
     =  0.1586 M
Both H2PO4^2- and HPO4^2- are weak acids the undergoes partial dissociation 
Ka of H2PO4- = 6.20 × 10^-8
 [H+] =Ka*([H2PO4-]/[HPO4(2-)]
        = (6.20 ×10^-8)×(0.155/0.1586)
        = 6.059 ×10^-8 M
pH = - log[H+]
     = - log (6.059×10^-8)
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5 0
4 years ago
A 2.50-l volume of hydrogen measured at â196 °c is warmed to 100 °c. calculate the volume of the gas at the higher temperature
ivann1987 [24]

To solve this we assume that the hydrogen gas is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant pressure and number of moles of the gas the ratio T/V is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

T1 / V1 = T2 / V2

V2 = T2 x V1 / T1

V2 = (100 + 273.15) K x 2.50 L / (-196 + 273.15) K

<span>V2 = 12.09 L</span>

Therefore, the volume would increase to 12.09 L as the temperature is increased to 100 degrees Celsius.

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Answer:

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