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3 years ago

How do you know whether or not an element has high electronegativity?

Chemistry

1 answer:

ale4655 [162]3 years ago

5 0

All elements have either a positive charge or a negative charge. And if I'm not mistaken, I do know that Chlorine has a negative charge of 1.

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Consider this reaction:

aleksklad [387]

Answer: 0.0345 sec

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$Rate=k[H_3PO_4]^2$

k= rate constant = $46.6s^{-1}$

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process

for completion of 20 % of reaction

$t=\frac{2.303}{46.6}\log\frac{0.660}{\frac{20}{100}\times 0.660}$

$t=\frac{2.303}{46.6}\log\frac{0.660}{0.132}$

$t=0.0345sec$

The time taken for the concentration of $H_3PO_4$ to decrease to 20% to its natural value is 0.0345 sec

6 0

3 years ago

I need half reactions, for the following equation.

zhuklara [117]

Answer : Half reaction is defined as the reaction which will be either the oxidation or reduction reaction of particular species in a given redox reaction.

So as per the definition in the reaction of,

 $Ba + ZnSO_{4} ----\ \textgreater \ BaSO_{4} + Zn$

 $Ba_{(s)} + Zn ^{+2} + SO_{4}^{-2} = Ba^{+2} + SO_{4}^{-2} + Zn_{(s)}$

here on left and right side $SO_{4}^{-2}$ will get cancelled.

and we will get, $Ba_{(s)} + Zn ^{+2} = Ba^{+2} + Zn_{(s)}$

Here,
$Ba_{(s)} ----\ \textgreater \ Ba^{+2} - 2e^{-}$ ...this is oxidation reaction as Ba is getting reduced to $Ba^{+2}$ by losing 2 electrons.

whereas,on the other hand, $Zn ^{+2} ----\ \textgreater \ Zn_{(s)} + 2e ^{-}$ ...this is reduction reaction as $Zn ^{+2}$ is getting reduced to Zn by gaining 2 electrons.

8 0

3 years ago

Given that E°red = -0.40 V for Cd2+/Cd at 25°C, find E° and E for the concentration cell expressed using shorthand notation belo

tensa zangetsu [6.8K]

From the information in the question, the E° and E for the cell is 0.00 V and 0.12 V.

Using the Nernst equation;

Ecell = E°cell - 0.0592/n log Q

We know that E°cell = 0.00 V since the anode and cathode are both made up of cadmium.

Substituting the given values into the Nernst equation;

Ecell = 0.00 V - 0.0592/2 log (1.0 × 10-5 M/0.100 M)

Ecell = 0.00 V - 0.0296 log(1 × 10^-4)

Ecell = 0.12 V