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3 years ago

Well the awnser is a

Chemistry

2 answers:

Lerok [7]3 years ago

8 0

Answer:

NO ITS C

Explanation:

krok68 [10]3 years ago

5 0

Answer:

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Calculate the number of atoms of sodium in a 4.5-gram sample

allochka39001 [22]

Answer:

1.18×10²³ atoms.

Explanation:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ atoms.

From the above concept, 1 mole of sodium also contains 6.02×10²³ atoms.

1 mole of sodium = 23 g.

Thus,

23 g of sodium contains 6.02×10²³ atoms.

Therefore, 4.5 g of sodium will contain = (4.5 × 6.02×10²³)/23 = 1.18×10²³ atoms.

From the above calculation,

4.5 g of sodium contains 1.18×10²³ atoms.

7 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

Fghgfgfgfftftftft ttcc

Ganezh [65]

Answer:

ybbbbgbg

Explanation:

4 0

3 years ago

Read 2 more answers

Read the article. Use your understanding to answer the questions that follow. What type of source is this article? primary or se

valkas [14]

Answer: option C

Explanation:

3 0

4 years ago

Read 2 more answers

What is the number of molecules of c2h5oh in a 3m solution that contains 4.00kg h2o?

Veronika [31]

The number of C2H5OH in a 3 m solution that contain 4.00kg H2O is calculate as below

M = moles of the solute/Kg of water

that is 3M = moles of solute/ 4 Kg
multiply both side by 4

moles of the solute is therefore = 12 moles

by use of Avogadro law constant

1 mole =6.02 x10^23 molecules

what about 12 moles

=12 moles/1 moles x 6.02 x10^23 = 7.224 x10^24 molecules

3 0

4 years ago