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3 years ago

Why did you wrap test tube 2 and 5 in aluminium foil and not test tube 1 and 6?

1 answer:

5 0

In the dark test tube carbon dioxide will be created, and will bring about the corrosiveness level in the water increment. In the light test tube carbon dioxide will be devoured, making the oxygen level ascent and make the corrosiveness level lessening.

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Calculate the pH during the titration of 20.00 mL of 0.1000 M dimethylamine, (CH3)2NH(aq), with 0.1000 M HCl(aq) after 21.23 mL

Evgesh-ka [11]

The pH value of the solution is mathematically given as

pH=2.35

<h3>What pH value of the solution?</h3>

Question Parameters:

pH during the titration of 20.00 mL of 0.1000 M dimethylamine,

with 0.1000 M HCl(aq) after 21.23 mL of the acid

Generally, the equation for the Chemical Reaction is mathematically given as

(CH3)2NH(aq), +Hcl ---> <---- (CH3)2NH2Cl(aq)

Therefore

$HCl=\frac{0.186mol}{41.86}$

HCL=0.00444M

WHere

HClaq--->H+(aq)+Cl-(aq)

Hence

H+=0.00444M

pH= -log{H+}

pH=log(0.00444)

pH=2.35

For more information on Chemical Reaction

brainly.com/question/11231920

8 0

2 years ago

Can u see water vapour by a microscope?

tigry1 [53]

Answer:

It can be formed either through a process of evaporation or sublimation. Unlike clouds, fog, or mist which are simply suspended particles of liquid water in the air, water vapour itself cannot be seen because it is in gaseous form

Explanation:

hope it help

3 0

3 years ago

Read 2 more answers

What is the predominant intermolecular force in the liquid state of each of these compounds: ammonia (NH3), methane (CH4), and n

MAVERICK [17]

Answer:

The predominant intermolecular force in the liquid state of each of these compounds:

ammonia (NH3)

methane (CH4)

and nitrogen trifluoride (NF3)

Explanation:

The types of intermolecular forces:

1.Hydrogen bonding: It is a weak electrostatic force of attraction that exists between the hydrogen atom and a highly electronegative atom like N,O,F.

2.Dipole-dipole interactions: They exist between the oppositely charged dipoles in a polar covalent molecule.

3. London dispersion forces exist between all the atoms and molecules.

NH3 ammonia consists of intermolecular H-bonding.

Methane has London dispersion forces.

Because both carbon and hydrogen has almost similar electronegativity values.

NF3 has dipole-dipole interactions due to the electronegativity variations between nitrogen and fluorine.

3 0

3 years ago

Can someone help your amiga :(??

bekas [8.4K]

i think the answer is C but don't take my word for it

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