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2 years ago

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If a sample of water has a mass of 200g and the final solution has a mass of 230g, how much solute dissolved in the water?

1 answer:

Murrr4er [49]2 years ago

6 0

What is the difference between the two numbers? That is your answer

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Consider the following reaction at equilibrium: 2NH3 (g) N2 (g) + 3H2 (g) Le Châtelier's principle predicts that the moles of H2

Lilit [14]

Answer:

A decrease in the total volume of the reaction vessel (T constant)

Explanation:

Le Châtelier's principle predicts that the moles of H2 in the reaction container will increase with a decrease in the total volume of the reaction vessel.
<em><u>According to the Le Chatelier's principle, when a chnage is a applied to a system at equilibrium, then the equilibrium will shift in a way that counteracts the effect causing it.</u></em>
In this case, a decrease in volume means there is an increase in pressure, therefore the equilibrium will shift towards the side with the fewer number of moles of gas.

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3 years ago

Describe the type and nature of the bonding that occurs between reactive metals and nonmetals.

Bingel [31]

It’s ionic bond based on electrons gain/loss.

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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 :

<u>(a) a solution of pH 3.0</u>

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, $1.0\times 10^{-11}M$

<u>(b) a solution of 0.10 M $NH_3$ </u>

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

<u>(c) a solution with a pOH of 12.</u>

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, $1.0\times 10^{-12}M$

<u>(d) a solution of 0.10 M NaOH</u>

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

<u>(e) a $1\times 10^{-4}M$ solution of $HNO_2$ </u>

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

The speed of sound through air at 20°C is 343 meters per second. The speed of sound through fresh water at 20°C is 1,482 meters

AVprozaik [17]

Answer:

C. 5960 meters per second

3 0

2 years ago

Question 5 of 5

rjkz [21]

Probably Fresh vegetables, it can rot out, that’d be my guess, it’s not canned

8 0

2 years ago

Read 2 more answers