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Sidana [21]
3 years ago
15

Lithium diorganocopper (Gilman) reagents are prepared by treatment of an organolithium compound with copper(I) iodide. Decide wh

at lithium diorganocopper (Gilman) reagent is needed to convert 1-bromopropane into 1-pentene. Draw the structure of the organolithium compound that is used to prepare this Gilman reagent.
ether
2 R---- Li + Cul → R2CuLi
or THF
Chemistry
1 answer:
Cloud [144]3 years ago
5 0

Answer: sounds very confusing

Explanation:

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Calculate the mass (in grams) of 250mL of ether at 25 oC. The density of
leonid [27]
  • Volume=250mL
  • Density=0.71g/ml

\boxed{\sf Density=\dfrac{Mass}{Volume}}

\\ \sf{:}\implies Mass=Density(Volume)

\\ \sf{:}\implies Mass=0.71(250)

\\ \sf{:}\implies Mass=177.5g

6 0
2 years ago
An object that is already moving can . . .
Yuri [45]
An object that is not already moving will begin to move in the direction of the larger force. An object that is already moving will change its speed and/or its direction.
4 0
3 years ago
Which gas below has the largest number of moles at STP?
ArbitrLikvidat [17]

Answer:

a.) 22.4 L Ne.

Explanation:

It is known that every 1.0 mol of any gas occupies 22.4 L.

For the options:

  • 22.4 L Ne:

<em>It represents </em><em>1.0 mol of Ne.</em>

<em />

  • 20 L Ar:

using cross multiplication:

1.0 mol occupies → 22.4 L.

??? mol occupies → 20 L.

The no. of moles of (20 L) Ar = (1.0 mol)(20 L)/(22.4 L) = 0.8929 mol.

  • 2.24 L Xe:

using cross multiplication:

1.0 mol occupies → 22.4 L.

??? mol occupies → 2.24 L.

<em>The no. of moles of (2.24 L) Xe </em>= (1.0 mol)(2.24 L)/(22.4 L) = <em>0.1 mol.</em>

  • So, the gas that has the largest number of moles at STP is: a.) 22.4 L Ne.

6 0
3 years ago
A sample of Manganese (II) chloride has a mass of 19.8 grams before heating, and 12.6 grams after heating until all the water is
IRINA_888 [86]

Answer:

no. of water molecules associated to each molecule of MnCl_2 = 4

Explanation:

Mass of MnCl_2 before heating = 19.8 g

Mass of MnCl_2 after heating = 12.6 g

Difference in mass of MnCl_2 before and after heating

                                 = 19.8 - 12.6 = 7.2 g

Difference in mass corresponds to mass of water driven out.

Molar mass of water = 18 g/mol

No. of moles of water = \frac{7.2}{18} = 0.4\ mol

Mass of MnCl_2 obtained after heating is mass of anhydrous MnCl_2.

Mass of anhydrous MnCl_2 = 12.6 g

Molar mass of MnCl_2 = 125.9 g/mol

No. of mol of anhydrous MnCl_2 = \frac{125.9}{125.9} = 0.1\ mol

so,

0.1 mol of MnCl_2 have 0.4 mol of water

1 mol of MnCl_2 will have = \frac{0.4}{0.1} =4\ mol

Hence, no. of water molecules associated to each molecule of MnCl_2 = 4

5 0
3 years ago
If 1.50 L of 0.780 mol/L sodium sulfide is mixed with 1.00 L of a 3.31 mol/L lead(II) nitrate solution, what mass of precipitate
NARA [144]

Answer:

336.1 g of PbS precipitate

Explanation:

The equation of the reaction is given as;

Na2S(aq) + Pb(NO3)2(aq) ----> 2NaNO3(aq) + PbS(s)

Ionically;

Pb^2+(aq) + S^2-(aq) -----> PbS(s)

Number of moles of sodium sulphide= concentration of sodium sulphide × volume of sodium sulphide

Number of moles of sodium sulphide= 0.780 × 1.5 = 1.17 moles

Number of moles of lead II nitrate= concentration of lead II nitrate × volume of lead II nitrate

Number of moles of lead II nitrate= 3.31× 1.00= 3.31 moles

Then we determine the limiting reactant. The limiting reactant yields the least amount of product.

Since 1 moles of sodium sulphide yields 1 mole of lead II sulphide

1.17 moles of sodium sulphide also yields 1.17 moles of lead II sulphide

Hence sodium sulphide is the limiting reactant.

Thus mass of precipitate formed= amount of lead II sulphide × molar mass of sodium sulphide

Molar mass of lead II sulphide= 287.26 g/mol

Mass of lead II sulphide = 1.17 moles × 287.26 g/mol

Mass of lead II sulphide= 336.1 g of PbS precipitate

4 0
3 years ago
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