Please helpppp!!!!!!!!¡!!!!!¡!!!!!!!!¡

Plz answer quick

SCORPION-xisa [38]

It’s baron I got it right on edg

8 0

3 years ago

Read 2 more answers

Consider the following gas phase reaction:

snow_tiger [21]

Answer:

2NO(g) + O2(g) --> 2NO2(g)

now 400 ml of NO × 2 mol of NO2/2 mol of NO

= 400 ml of NO2

now 500 ml of O2 × 2 mol of NO2/1 mol of O2

= 1000 ml of NO2

now 400 ml of NO2 × 1 mol of O2/2 mol of NO

= 200 ml

subtract that from 500 ml of total i.e. 500-200 =300 ml

The total volume of the reaction mixture is 1000 ml -300ml = 700 ml

6 0

3 years ago

Which element has chemical properties that are most similar to potassium, And why?

bekas [8.4K]

Answer:

Brainliest pls

Explanation:

The components potassium and sodium have comparable substance properties since they have a similar number of valence electrons

8 0

2 years ago

Carbon dioxide, water vapor, and oxygen pass through ---------------

Lelechka [254]

The answer is B) stomata.

5 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

Please helpppp!!!!!!!!¡!!!!!¡!!!!!!!!¡​

Please helpppp!!!!!!!!¡!!!!!¡!!!!!!!!¡