A solution

Give the reason for the evaporation of the solution to dryness with 6 m hcl in the first procedure of the copper-arsenic group a

SCORPION-xisa [38]

Evaporation occurred between the reaction of the hydrochloric acid and the arsenic group because there is a formation of gas. A liquid is cooled if condensation exceeds evaporation. This is because the cooling of the liquid decreases the kinetic energy of the molecules. Their movement is being restricted by the colder temperature. The molecules tend to be closer with each other. Also, because of their restricted movement, the liquid may turn into solid due the colder temperature. An example of this is the cooling of liquid water by placing it into the refrigerator into solid water (ice).

4 0

3 years ago

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

For the following reaction, KcKc = 255 at 1000 KK.

bonufazy [111]

Answer :

The equilibrium concentration of CO is, 0.016 M

The equilibrium concentration of Cl₂ is, 0.034 M

The equilibrium concentration of COCl₂ is, 0.139 M

Explanation :

The given chemical reaction is:

$CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)$

Initial conc. 0.1550 0.173 0

At eqm. (0.1550-x) (0.173-x) x

As we are given:

$K_c=255$

The expression for equilibrium constant is:

$K_c=\frac{[COCl_2]}{[CO][Cl_2]}$

Now put all the given values in this expression, we get:

$255=\frac{(x)}{(0.1550-x)\times (0.173-x)}$

x = 0.139 and x = 0.193

We are neglecting value of x = 0.193 because equilibrium concentration can not be more than initial concentration.

Thus, we are taking value of x = 0.139

The equilibrium concentration of CO = (0.1550-x) = (0.1550-0.139) = 0.016 M

The equilibrium concentration of Cl₂ = (0.173-x) = (0.173-0.139) = 0.034 M

The equilibrium concentration of COCl₂ = x = 0.139 M

5 0

3 years ago

33 protons 36 electrons 45 neutrons

Artist 52 [7]

As has an atomic number of 33, so it has 33 protons.

It has a charge of 3-, so there are three more electrons than protons. Thus, there are 36 electrons.

It has a mass of 75, which is the sum of neutrons and protons.

33+n=75 ---> n = 75 - 33 = 42 neutrons

The answer is e) 33 protons, 42 neutrons and 36 electrons.

6 0

3 years ago

How many orbitals are in 2s sublevel?

kykrilka [37]

4 orbitals. P sublevel has 3 orbitals. 2nd level has 4 orbitals. An f sublevel has 7 orbitals.

7 0

2 years ago