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4 years ago

Why is the electronegativity of an element important?

Chemistry

2 answers:

Vinvika [58]4 years ago

6 0

Answer: Because it let the element to form covalent bonding which is too important for forming compounds........

Explanation:

MArishka [77]4 years ago

3 0

Because it let the element to form covalent bonding which is too important for forming compounds........

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In a closed container with a fixed volume, when the temperature of a gas doubles, the pressure of the gas is?

nevsk [136]

Answer:

The pressure of the gas is:

Doubled

Explanation:

Byole's Law = Pressure of fixed amount of gas is inversely proportional to Volume at constant temperature.

PV = k = constant

P1V1 = P2V2

Charle's Law: The volume of ideal gas at fixed pressure is directly proportional to Pressure.

V/T = constant

V1/T1 = V2/T2

On combinig the charles and byole's law , we get:

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

Now , According to question :

V2 = V1

T2 = 2(T1)

We have to find the relation between the pressure :

insert the value in the equation

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{1}}{2T_{1}}$

V1 and V1 & T1 and T1 cancels each other,

So we get

P2 = 2 P1

So the new pressure is double of the original pressure

3 0

3 years ago

For a solution equimolar in HCN and NaCN, which statement is false? a. [H+] is larger than it would be if only the HCN were in s

balandron [24]

Answer:

option (d) is false.

Explanation:

Acid dissociation equilibrium of HCN is represented as-

$HCN\rightleftharpoons H^{+}+CN^{-}$

Acid dissociation constant, $K_{a}$ , is represented as-

$K_{a}=\frac{[H^{+}][CN^{-}]}{[HCN]}$

where species inside third bracket represents equilibrium concentrations of respective species

So, evidently, presence of excess $CN^{-}$ (or NaCN) in solution will combine with $H^{+}$ to produce HCN. Hence $H^{+}$ will be larger that it would be if only the HCN solution were present.

According to Le-chatlier principle, addition of HCN will shift equilibrium towards right and addition of NaCN will shift equilibrium towards left to keep constant $K_{a}$ value at a particular temperature.

NaOH gives acid-base reaction with HCN to produce NaCN and water. So, addition of NaOH will increase concentration of $CN^{-}$ and decrease concentration of HCN

6 0

4 years ago

You are working in a laboratory, and you are given the task of converting cyclopentene into 1, 5-pentanediol. Your first thought

Helen [10]

Answer:

Reagent A = $H_3O^+$

Reagent B= $Na_2Cr_2O_7~H_2SO_4~H_2O$

Intermediate C= δ-Valerolactone

Explanation:

In the reaction from the alkene to the alcohol, we can use the alkene hydration in which the hydronium ion is added to the double bond followed by the attack of water to produce the alcohol.

Then in the conversion from alcohol to ketone can be produced if an oxidant reactive is used. In this case the Jones reagent ( $Na_2Cr_2O_7~H_2SO_4~H_2O$ ).

The intermediate is a structure produced by a peroxyacid. This reaction would introduce an ester group in the cycle generating the δ-Valerolactone (Figure 1).

7 0

3 years ago

1. The vertical columns on the periodic table are called groups, how many groups are

zheka24 [161]

Answer:

Explanation:

the numbers at the top tell you

6 0

3 years ago

Read 2 more answers

How can I make equilibrium shift to the right in a reversible reaction?

labwork [276]

Answer:

According to Le Chatelier's principle, adding additional reactant to a system will shift the equilibrium to the right, towards the side of the products. By the same logic, reducing the concentration of any product will also shift equilibrium to the right.

Explanation:

8 0

3 years ago