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hammer [34]
2 years ago
9

What type of intermediate is present in the sn2 reaction of cyanide with bromoethane?

Chemistry
1 answer:
elixir [45]2 years ago
8 0

Answer:

in said reaction, there is a reaction by substitution, to form said compound. the product between both chemical compounds is propanonitrile.

 

Explanation:

cyanide is usually associated with potassium to form a salt, and as a product more stable chemical situations occur.

It is important to note that both products are very toxic.

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An archeologist finds a 1.62 kg goblet that she believes to be made of pure gold. She determines that the volume of the goblet i
katovenus [111]
The  answer is

<span>The density (D) is quotient of mass (m) and volume (V):
</span>D= \frac{m}{V}
The unit is g/cm³

It is given:
m = 1.62 kg = 1620 g
V = 205 mL = 205 cm³
D = ?

Thus:
D= \frac{m}{V} = \frac{1620g}{205cm ^{3} } = 7.90 g/cm ^{3}

The density of the goblet is 7.90 g/cm³.
5 0
2 years ago
A 0.3870-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.7191Â g of co
olchik [2.2K]
The chemical formula for the compound can be written as,

    CxHyOz 

where x is the number of C atoms, y is the number of H atoms, and z is the number of O atoms. The combustion reaction for this compound is,
   
    CxHyOz + O2 --> CO2 + H2O 

number of moles of C:
     (0.7191 g)(1 mol CO2/44 g of CO2) = 0.0163 mol CO2
 This signifies that 0.0163 mole of C and the mass of carbon in the compound,
        (0.0163 mols C)(12 g C/ 1 mol C) = 0.196 g C

number of moles H:
      (0.1472 g H2O)(1 mol H2O/18 g H2O) = 0.00818 mol H2O

This signifies that there are 0.01635 atoms of H in the compound.
      mass of H in the compound = (0.01635 mols H)(1 g of H) = 0.01635 g H

Mass of oxygen in the compound,
   0.3870 - (0.196 g C + 0.01635 g H) = 0.1746 g

Moles O in the compound = (0.1746 g O)(1 mol O/16 g O) = 0.0109 mols O

The formula of the compound is,
      C0.0163H0.01635O0.0109

Dividing the numbers by the least number,
    C3/2H3/2O

The empirical formula of the compound is therefore,
    <em>  C₃H₃O₂</em>
5 0
2 years ago
Predict the charge on the monatomic ions formed from the following atoms in binary ionic compounds:(a) |(b) Sr(c) K(d) N(e) S(f)
Pavel [41]

Answer:

(a) I⁻ (charge 1-)

(b) Sr²⁺ (charge 2+)

(c) K⁺ (charge 1+)

(d) N³⁻ (charge 3-)

(e) S²⁻ (charge 2-)

(f) In³⁺ (charge 3+)

Explanation:

To predict the charge on a monoatomic ion we need to consider the octet rule: atoms will gain, lose or share electrons to complete their valence shell with 8 electrons.

(a) |

I has 7 valence electrons so it gains 1 electron to form I⁻ (charge 1-).

(b) Sr

Sr has 2 valence electrons so it loses 2 electrons to form Sr²⁺ (charge 2+).

(c) K

K has 1 valence electron so it loses 1 electron to form K⁺ (charge 1+).

(d) N

N has 5 valence electrons so it gains 3 electrons to form N³⁻ (charge 3-).

(e) S

S has 6 valence electrons so it gains 2 electrons to form S²⁻ (charge 2-).

(f) In

In has 3 valence electrons so it loses 3 electrons to form In³⁺ (charge 3+).

3 0
2 years ago
Molecular bromine is 24 percent dissociated at 1600 k and 1.00 bar in the equilibrium br2 (
dlinn [17]
Br2 == 2Br

24% dissociated => n total moles, 0.24 mol*n of Br, and 0.76*n mol of Br2

=> partial pressure of Br, P Br = 0.24 bar, and
     partical pressure of Br2, P Br2 = 0.76 bar

kp = (P Br)^2 / P Br2 = (0.24)^2 / 0.76 = 0.0758


3 0
2 years ago
Use average bond energies to calculate ΔHrxn for the following hydrogenation reaction: H2C=CH2(g)+H2(g)→H3C−CH3(g)
marissa [1.9K]

Answer:

The\Delta H_{rxn} of the given reaction is -129.6 kJ

Explanation:

The given chemical reaction is as follows.

H_{2}C=CH_{2}(g)+H_{2}(g)\rightarrow H_{3}C-CH_{3}(g)

Enthalpy of each reactant and products are as follows.

\Delta H_{C=C}\,=615.0\,kJ\,mol^{-1}

\Delta H _{H-H}\,=435.1\,kJ\,mol^{-1}

\Delta H _{C-C}\,=347.3\,kJ\,mol^{-1}

\Delta H _{C-H}\,=416.2\,kJ\,mol^{-1}

In the given chemical reaction involved two C-H bonds in the reactant side and one C-C bond in the product side therefore, the enthalpy of formation will be the negative.

\Delta H_{rxn}=-\Delta H_{C-C}-2\Delta H_{C-H}+\Delta H_{C=C}+\Delta H_{H-H}

=-347.4-2\times416.2+615.0+435.1

=-129.6 \,kJ

Therefore, The\Delta H_{rxn} of the given reaction is -129.6 kJ

4 0
1 year ago
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