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Serga [27]
3 years ago
14

Draw the product obtained when trans-2-butene is treated first with br2 in ch2cl2, second with nanh2 in nh3, and then finally wi

th h2/lindlar's catalyst.
Chemistry
1 answer:
oee [108]3 years ago
3 0

<span>In order to do this, you have change the alkene into an alkyne. That is the aim of Br2/CH2Cl2 trailed by NaNH2. The Br2 with form a vic dihalide (3,4-dibromo octane). Adding of NaNH2 will execute two E2 reactions. -NH2 will eliminate an H from carbons 3 and 4. This double elimination will make the alkyne. Then handling the alkyne with H2/Lindlar will form the cis alkene. The final product will be CIS-3-octene.</span>

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A gaseous mixture of o2 and n2 contains 35.8 % nitrogen by mass. what is the partial pressure of oxygen in the mixture if the to
lubasha [3.4K]
The partial pressure is the amount of linguistic compound there is which makes the lagitude of the element 64.663
5 0
3 years ago
The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures
QveST [7]

<u>Answer:</u>

<u>For A:</u> The K_p for the given reaction is 4.0\times 10^1

<u>For B:</u> The K_c for the given reaction is 1642.

<u>Explanation:</u>

The given chemical reaction follows:

2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)

  • <u>For A:</u>

The expression of K_p for the above reaction follows:

K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}

We are given:

p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm

Putting values in above equation, we get:

K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1

Hence, the K_p for the given reaction is 4.0\times 10^1

  • <u>For B:</u>

Relation of K_p with K_c is given by the formula:

K_p=K_c(RT)^{\Delta ng}

where,

K_p = equilibrium constant in terms of partial pressure = 4.0\times 10^1

K_c = equilibrium constant in terms of concentration = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature = 500 K

\Delta ng = change in number of moles of gas particles = n_{products}-n_{reactants}=2-3=-1

Putting values in above equation, we get:

4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642

Hence, the K_c for the given reaction is 1642.

7 0
3 years ago
Which element would have the lowest electronegativity? (1 point)
irinina [24]

The element that will have the lowest electronegativity is an element with a small number of valence electrons and a large atomic radius.

Electronegativity of an element is the ability or power of that element in a molecule to attract electrons to its Valence electrons.  The following are the properties of electronegativity:

  • It increases across a period from left to right of the periodic table,
  • It decreases down the periodic table groups
  • Group 1 elements are the least (lowest) electronegative elements. These elements have the lowest valence electrons with a large atomic radius.
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Atomic radius of elements increase down a group because of a progressive increase in the number of shells occupied by electrons which increases the size. But it decreases across a period because electrons are accommodated within the same shell leading to greater attraction by the protons in the nucleus.

Learn more about electronegativity of elements here:

brainly.com/question/20348681

6 0
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Name the element which has twice the number of electrons in its second shell as in the first shell write it's electronic configu
Basile [38]
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4 Which pair of ions is isoelectronic?
Softa [21]

Answer:

B. Na+ and O2-

Explanation:

Na+ plus has 10 electrons and O2- also has 10 electrons

7 0
3 years ago
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