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3 years ago

How do I calculate the standard enthalpy change for the following reaction at 25 °C ?

1 answer:

4 0

MgCl2(s) + H2O(l) → MgO(s) + 2 HCl(g)

Using the standard enthalpies of formation given in the source below:

(−601.24 kJ) + (2 x −92.30 kJ) − (−641.8 kJ) − (−285.8 kJ) = +141.76 kJ
So:

MgCl2(s) + H2O(l) → MgO(s) + 2 HCl(g), ΔH = +141.76 kJ

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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

3 years ago

A certain gas at 2oC and 1.00 atm pressure fills a 4.0 container. What volume will the gas occupy at 100oC and 780 torr pressure

artcher [175]

The final volume V₂=4.962 L

<h3>Further explanation</h3>

Given

T₁=20 + 273 = 293 K

P₁= 1 atm

V₁ = 4 L

T₂=100+273 = 373 K

P₂=780 torr=1,02632 atm

Required

The final volume

Solution

Combined gas law :

P₁V₁/T₁=P₂V₂/T₂

Input the value :

V₂=(P₁V₁T₂)/(P₂T₁)

V₂=(1 x 4 x 373)/(1.02632 x 293)

V₂=4.962 L

6 0

3 years ago

What is a ‘control’ in an experiment?

tensa zangetsu [6.8K]

Answer:

The answer fo this is D because of the person controlling the outcome for both variables

3 0

3 years ago

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What are sme important facts/info I need to know about Aluminum? Like, is it malleable? I need to know abot this for a science p

grigory [225]

Aluminum is light and is quite unreactive it is malleable and strong too.

6 0

3 years ago

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A chemist mixes 500 g of lead at 500°c with 1,200 g of water at 20°c. she then mixes 500 g of copper at 500°c with 1,200 g of wa

Artemon [7]

The final temperature of the lead-water system will be lower than the final temperature of the copper-water system.

8 0

3 years ago