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3 years ago

7

Which Group 14 element is classified as a metal?

1 answer:

eimsori [14]3 years ago

4 0

Tin is classified as a metal in family 14.

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What are the sub-particles that make up an atom? (select all that apply) *?

Mrac [35]

The answer should be neutrons electrons and protons

7 0

3 years ago

Which of these was NOT listed in the article a a way to stay safe while enjoying winter weather?

lukranit [14]

Answer:

<h2>ignore your body being cold</h2>

Hope it helps

8 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 :

<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

How many moles of copper are there in 6.93 g of copper sample<br>pls be quick

Alex_Xolod [135]

Molar mass of copper=63.5g/mol
Given mass=6.93g

$\\ \sf\longmapsto No\:of\;moles=\dfrac{Given\:mass}{Molar\:mass}$

$\\ \sf\longmapsto No\:of\:moles=\dfrac{6.93}{63.5}$

$\\ \sf\longmapsto No\:of\:moles=0.109\approx 1.11moles$

7 0

2 years ago

Read 2 more answers

At a certain temperature the vapor pressure of pure heptane C7H16 is measured to be 454.mmHg. Suppose a solution is prepared by

OlgaM077 [116]

Answer:

Mass of heptane = 102g

Vapor pressure of heptane = 454mmHg

Molar mass of heptane = 100.21

No of mole of heptane = mass/molar mass = 102/100.21

No of mole of heptane = 1.0179

Therefore the partial pressure of heptane = no of mole heptane *Vapor pressure of heptane

Partial pressure of heptane = 1.0179*454mmHg

Partial pressure of heptane = 462.1096 = 462mmHg

the partial pressure of heptane vapor above this solution = 462mmHg

5 0

3 years ago