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2 years ago

5

Draw the structure of the major organic product(s for the following reaction between an acetylenic anion and an alkyl halide. (t

he reaction stoichiometry is 1:1.

1 answer:

Alina [70]2 years ago

7 0

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What is the molarity of a potassium triiodide solution, KI3(aq), if 30.00 mL of the solution is required to completely react wit

vfiekz [6]

Answer:

The molarity is 0.203 M

Explanation:

Using the formula C(oxi) x V(oxi) / [C(red) x V(red)] = N(oxi) / N(red)

Where oxi and red means reducing agent and oxidising agent respectively.

C = Concentration, V = Volume and N = number of moles.

C(oxi) = 0.5 M

V(oxi) = mL

C(red) = ?

V(red) = 30mL

Equation of reaction = 2K2S2O3 + KI3 = K2S4O6 + 3KI

so N(red) = 1 , N(oxi) = 2

from the equation above,

C(red) = 0.5 x 25 x 1 / (2 x 30)

= 0.203 M.

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2 years ago

If 35 ml of 6.0 m h2so4 was spilled, calculate the minimum mass of nahco3 that must be added to the spill to neutralize the acid

belka [17]

The balanced equation between the $H_2SO_4$ and $NaHCO_3$ is:

$H_2SO_4+2NaHCO_3\rightarrow Na_2SO_4+2CO_2+2H_2O$

Formula of molarity is:

$Molarity = \frac{Moles of solute}{Volume of solution in Liters}$

$Molarity = 6.0 M, Volume = 35 mL = 0.035 L$

Substituting the values,

$6 = \frac{Moles of solute}{0.035}$

$Moles of solute = 0.035 L\times 6 mol/L = 0.21 mole$

So, number of moles of $H_2SO_4$ is $0.21 mole$ .

From the balanced equation it is clear that for 1 mole of $H_2SO_4$ , 2 moles of $NaHCO_3$ are required.

Hence, 0.21 mole of $H_2SO_4$ = $2\times 0.21 mole = 0.42 mole of NaHCO_3$

Molar mass of $NaHCO_3$ = $84.007 g/mol$

So, the mass of $NaHCO_3$ = $84.007 g/mol \times 0.42 mol = 35.283 g$

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2 years ago

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dalvyx [7]

Answer:

I think the answer is Toledo

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Ray Of Light [21]

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vaieri [72.5K]

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