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SSSSS [86.1K]
3 years ago
6

Electrophilic addition of HBr to alkenes yields a bromoalkane. The reaction begins with an attack on the hydrogen of the electro

philic HBr by the π electrons of the double bond to give a carbocation. This step follows Markovnikov’s rule with the electrophilic H atom adding to the sp2 carbon containing the most hydrogens, leading to the formation of the most stable carbocation (1°<2°<3°). If possible, a 1,2-shift of either a neighboring hydride or methyl group can occur prior to the last step in order to form a more stable carbocation. In the final step of the reaction, nucleophilic bromide adds to the carbocation to give the neutral product. Draw curved arrows to show the movement of electrons in this step of the mechanism. Arrow-pushing Instructions
Chemistry
1 answer:
Veronika [31]3 years ago
3 0

Answer:

b

Explanation:

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You work for a cutlery manufacturer who wants to electrolytically precipitate 0.500 g of silver onto each piece of a batch of 25
ICE Princess25 [194]

Answer:

373.1 mL of AgCN (aq) must be poured into your electrolysis vat to ensure you have sufficient Ag to plate all of the forks.

Explanation:

Mass of silver to be precipitated on ecah spoon = 0.500 g

Number of silver spoons = 250

Total mass of silver = 250 × 0.500 g = 125 g

Moles (n)=Molarity(M)\times Volume (L)

Moles of AgCN = n  = \frac{125 g}{134 g/mol}=0.9328 mol

Volume of AgCN solution =V

Molarity of the AgCN = 2.50 M

V=\frac{0.9328 mol}{2.50 M}=0.3731 L=373.1 mL

(1 L = 1000 mL)

373.1 mL of AgCN (aq) must be poured into your electrolysis vat to ensure you have sufficient Ag to plate all of the forks.

6 0
4 years ago
What are the terms that descripe these values for the element oxygen?
Serga [27]

Answer:

16 is the mass number. 8 is the atomic number.

5 0
2 years ago
How many grams of calcium phosphate (Ca3(PO4)2) re theoretically produced if we start with 3.40 moles of Ca(NO3)2 and 2.40moles
sattari [20]

1) Balance the chemical equation.

3Ca(NO_3)_2+2Li_3PO_4\rightarrow6LiNO_3+Ca_3(PO_4)_2

2) List the known and unknown quantities.

Reactant 1: Ca(NO3)2.

Amount of substance: 3.40 mol.

Reactant 2: Li3PO4.

Amount of substance: 2.40 mol.

Product: Ca3(PO4)2

Mass: unknown.

3) Which is the limiting reactant?

<em>3.1-How many moles of Li3PO4 do we need to use all of the Ca(NO3)2?</em>

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

mol\text{ }Li_3PO_4=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{2\text{ }mol\text{ }Li_3PO_4}{3\text{ }mol\text{ }Ca(NO_3)_2}=2.2667\text{ }mol\text{ }Li_3PO_4

<em>We need 2.2667 mol Li3PO4 and we have 2.40 mol Li3PO4. We have enough Li3PO4. </em>This is the excess reactant.

<em>3.2-How many moles of Ca(NO3)2 do we need to use all of the Li3PO4?</em>

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

mol\text{ }Ca(NO_3)_2=2.40\text{ }mol\text{ }Li_3PO_4*\frac{3\text{ }mol\text{ }Ca(NO_3)_2}{2\text{ }mol\text{ }Li_3PO_4}=3.60\text{ }mol\text{ }Ca(NO_3)_2

<em>We need 3.60 mol Ca(NO3)2 and we have 3.40 mol Ca(NO3)2. We do not have enough Ca(NO3)2. </em>This is the limiting reactant.

4) Moles of Ca3(PO4)2 produced from the limiting reactant.

We have 3.40 mol Ca(NO3)2 of the limiting reactant.

The molar ratio between Ca(NO3)2 and Ca3(PO4)2 is 3 mol Ca(NO3)2: 1 mol Ca3(PO4)2.

mol\text{ }Ca_3(PO_4)_2=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{1\text{ }mol\text{ }Ca_3(PO_4)_2}{3\text{ }mol\text{ }Ca(NO_3)_2}=1.1313\text{ }mol\text{ }Ca_3(PO_4)_2

5) Mass of Ca3(PO4)2 produced.

The molar mass of Ca3(PO4)2 is 310.1767 g/mol.

g\text{ }Ca_3(PO_4)_2=1.1333\text{ }mol\text{ }Ca_3(PO_4)_2*\frac{310.1767\text{ }g\text{ }Ca_3(PO_4)_2}{1\text{ }mol\text{ }Ca_3(PO_4)_2}g\text{ }Ca_3(PO_4)_2=351.526\text{ }g\text{ }Ca_3(PO_4)_2

<em>The mass of Ca3(PO4)2 produced is</em> 351 g Ca3(PO4)2.

Option D.

.

8 0
1 year ago
An object in space is made of gas and gives off light and heat energy. We can only see it at night. Which object is it? (2 point
strojnjashka [21]

Answer:

stars

earth isn't made of gas. (and you can always see it)

moon us not gas, or give off heat

sun is only visible during the day

7 0
3 years ago
Borax (na2b4o7·10h2o; fw = 381.372 g/mol; density = 1.73 g/ml), a primary standard, was used to standardize a solution of hno3.
SashulF [63]
<span>0.06355391 mol The balanced equation for the reaction is Na2B4O7*10H2O + 2 HNO3 = 2 NaNO3 + 4 H3BO3 + 5 H2O So for each mole of Borax to neutralize, it takes 2 moles of HNO3. Calculate number of moles of Borax 0.2619 g / 381.372 g/mol = 0.0006867 mol Moles of HNO3 used = 0.0006867 mol * 2 = 0.0013734 mol Molarity is defined as moles per liter so divide the number of moles used by the volume in liters. So 0.0013734 / 0.02161 = 0.06355391 mol</span>
3 0
4 years ago
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