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3 years ago

????how many students failed this science class???

Chemistry

1 answer:

deff fn [24]3 years ago

6 0

Answer:

Explanation:

You take the 50-69 and you add up the students and it will allow you to get 7

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imagine that you're able to fully dissolve a powered substance and water what can be said about this creation

Softa [21]

We can say that the water is the solvent, and the powder is the solute. This is also a solution altogether.

Explanation:- A solute is the thing being dissolved into the solvent. While the solvent is what when the solute is being dissolved in. Together, they make a solution.

8 0

3 years ago

Will give brainliest, like, 5 stars and 5 points to best answer

Oliga [24]

evaporation systems allow for an endless source of water. you can grab cups of water straight from the sea or even a lake. the use of evaporation allows for you to drink water thats even healthier than getting it from a cloud and it will leave all of the bad parts that used to be in the water in the first container you pour into. this system is most useful in hot climates such as places near the equator.

6 0

3 years ago

Please help! 35 points for a valid answer!

Furkat [3]

Answer:

Oxidation half reaction is written as follows when using using reduction potential chart

example when using copper it is written as follows

CU2+ +2e- --> c(s) +0.34v

oxidasation is the loos of electron hence copper oxidation potential is as follows

cu (s) --> CU2+ +2e -0.34v

Explanation:

6 0

3 years ago

Read 2 more answers

Which form of energy can be connected closely to climate change?

Art [367]

Answer:

COAL

Explanation:

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