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3 years ago

What is the difference between

Chemistry

1 answer:

Nutka1998 [239]3 years ago

4 0

Answer:

A physical property is a characteristic of a substance that can be observed or measured without changing the identity of the substance. Physical properties can be color, density, hardness, melting or boiling points, while a chemical property describes the ability of a substance to undergo a specific chemical change.

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Entropy of a system decreases with decrease in Temperature T/F

Svet_ta [14]

Answer: The given statement is true.

Explanation:

Entropy means the measure of randomness present in a substance. That is, an increase in temperature will lead cause more motion in the particles of a substance more will be their kinetic energy.

As a result, there will occur more collisions due to which randomness of molecules will increase. Hence, there will be increase in entropy.

So, when we decrease the temperature then there will be decrease in motion of particles. As a result, lesser number of collisions will take place between them. Hence, degree of randomness will also decrease.

Thus, we can conclude the statement entropy of a system decreases with decrease in temperature, is true.

3 0

3 years ago

What is thermal equilibrium? equal temperature equal heat equal thermal energy equal conductivity

Margarita [4]

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I believe the answer would be A. equal temperature

Have A Nice Day ❤

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6 0

3 years ago

Read 2 more answers

Which of the following possess the greatest concentration of hydroxide ions?

jek_recluse [69]

Answer : The correct option is (d) a solution of 0.10 M NaOH

Explanation :

(a) a solution of pH 3.0

First we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-3.0=11$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$11=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-11}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-11}M$

(b) a solution of 0.10 M $NH_3$

As we know that 1 mole of $NH_3$ is a weak base. So, in a solution it will not dissociates completely.

So, the $OH^-$ concentration will be less than 0.10 M

(c) a solution with a pOH of 12.

We have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$12=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-12}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-12}M$

(d) a solution of 0.10 M NaOH

As we know that $NaOH$ is a strong base. So, it dissociates to give $Na^+$ ion and $OH^-$ ion.

So, 0.10 M of $NaOH$ in a solution dissociates to give 0.10 M of $Na^+$ ion and 0.10 M of $OH^-$ ion.

Thus, the $OH^-$ concentration is, 0.10 M

(e) a $1\times 10^{-4}M$ solution of $HNO_2$

As we know that 1 mole of $HNO_2$ in a solution dissociates to give 1 mole of $H^+$ ion and 1 mole of $NO_2^-$ ion.

So, $1\times 10^{-4}M$ of $HNO_2$ in a solution dissociates to give $1\times 10^{-4}M$ of $H^+$ ion and $1\times 10^{-4}M$ of $NO_2^-$ ion.

The concentration of $H^+$ ion is $1\times 10^{-4}M$

First we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (1.0\times 10^{-4})$

$pH=4$

Now we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-4=10$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$10=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-10}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-10}M$

From this we conclude that, a solution of 0.10 M NaOH possess the greatest concentration of hydroxide ions.

Hence, the correct option is (d)

3 0

4 years ago

Write the oxidation and reduction half reactions;

luda_lava [24]

Answer:

$Fe^{2+}$ ⇒ $Fe^{3+}+e^-$

$Br_2+2e^-$ ⇒ $2Br^-$

$Mg$ ⇒ $Mg^{2+}+2e^-$

$Cr^{3+}+e^-$ ⇒ $Cr^{3+}$

Explanation:

Remember that positive number superscripts mean electrons lack and negative numbers mean electrons 'excess' (if we compare it with the neutral element). So, for the case of Fe2+ which is converted to Fe3+, we know that in Fe2+ there is a two electrons lack, while in Fe3+ there is a 3 electrons lack; it means that Fe2+ was converted to Fe3+ but releasing one electron:

$Fe^{2+}$ ⇒ $Fe^{3+}+e^-$

The same analysis is applied to Br2; Br2 is a molecule which is said to have a zero superscript because it is an apolar covalent bond; and it is converted to Br-, which, according to what I wrote above, means that there is a one electron excess. So, Br2 must have received an electron in order to change to Br-; but Br2 can't change to Br- as simple as that because Br2 is a molecule, not an atom; it is a molecule that has two Br atoms, so, Br2 must give two Br- ions as products, but receiving one electron for each one:

$Br_2+2e^-$ ⇒ $2Br^-$

Applying the same, in Mg2+ there is a 2 electrons lack, and in Mg is not electron lack (its superscript is zero), so Mg must have released two electrons in order to change to Mg2+:

$Mg$ ⇒ $Mg^{2+}+2e^-$

Cr3+ has a 3 electrons lack, and Cr2+ a two electrons one, so, Cr3+ must receive an electron to convert to Cr2+:

$Cr^{3+}+e^-$ ⇒ $Cr^{3+}$

3 0

4 years ago

Read 2 more answers

What is Oxidation state of sulphur in Na2S4O6?Help me..

quester [9]

Answer:

oxidation state of sulphur=x

Explanation:

Na2S4O6=2[+1]+4x+6[-2]=0

+2+4x-12=0

4x-10=0

4x=10

x=10/4=2.5

6 0

3 years ago