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3 years ago

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Copper occurs in nature in the form of two isotopes, Cu-63 and Cu-65. Its correctly recorded atomic mass (weighted average) is 6

3.5 amu. Without doing the math, explain why it is safe to say that Cu-63 is more abundant than Cu-65.

1 answer:

OlgaM077 [116]3 years ago

6 0

63.5 is closer to 63 than 65 so you know when they find the average mass there are more Cu-63 isotopes

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How are scented candles made?

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poizon [28]

1. Take 100ml of water as solvent and boil it few minutes.
2. Now add one tea spoon sugar, one tea spoon tea leaves and 50ml of milk. Here sugar, tea leaves and milk are solute.
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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.

$pOH=-\log [OH^-]$

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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)

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eduard

The answer for this is 26.6°c

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2 years ago