The tsunami walls at Miyako were high enough to hold back the water. True False

What ions exist in acid solutions

Iteru [2.4K]

Hydrogen Ions exist in acid solutions.

5 0

3 years ago

You have dissolved 10 g sodium oxide in 200 ml water.calculate concentration of the solution

e-lub [12.9K]

Answer:

0.85 Molar Na2O

Explanation:

Determine the moles of sodium oxide, Na2O, in 10 grams by dividing by the molar mass of Na2O (61.98 g/mole).

(10 g Na2O)/(61.98 g/mole) = 0.161 moles Na2O.

Molar is a measure of concentration. It is defined as moles/liter. A 1 M solution contains 1 mole of solute per liter of solvent. [200 ml water = 0.2 Liters water.]

In this case, we have 0.161 moles Na2O in 0.200 L of solvent.

(0.161 moles Na2O)/(0.200 L) = 0.85 Molar Na2O

8 0

1 year ago

What example show the transformation for electrical change to radiant?

zlopas [31]

When you plug in an electrical heater, it turns the electrical energy into radiant energy.

3 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

Study this chemical reaction: (aq)(s)(s)(aq) Then, write balanced half-reactions describing the oxidation and reduction that hap

disa [49]

The question is incomplete, complete question is:

Study this chemical reaction:

$FeSO_4 (aq) + Zn (s)\rightarrow Fe (s) + ZnSO_4 (aq)$

Then, write balanced half-reactions describing the oxidation and reduction that happen in this reaction.

Oxidation:

Reduction:

Answer:

Oxidation taking place in given reaction :

$Zn(s)\rightarrow Zn^{2+}+2e^-$

Reduction taking place in given reaction;

$Fe^{2+}(aq)+2e^-\rightarrow Fe(s)$

Explanation:

Redox reaction is defined as the reaction in which oxidation and reduction reaction occur side by side.

Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

$X\rightarrow X^{n+}+ne^-$

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

$X^{n+}+ne^-\rightarrow X$

$FeSO_4 (aq) + Zn (s) \rightarrow Fe (s) + ZnSO_4 (aq)$

In the given reaction, iron(II) ions are getting reduced and zinc metal is getting oxidized to zinc(II) ions.

Oxidation :

$Zn(s)\rightarrow Zn^{2+}+2e^-$

Reduction ;

$Fe^{2+}(aq)+2e^-\rightarrow Fe(s)$

6 0

2 years ago