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3 years ago

13

What are the terms used to describe a location on a sanddun?

1 answer:

maria [59]3 years ago

4 0

Means exact place on a location

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What is the molarity of a solution of 17.0 g of nh4br in enough h2o to make 158 ml of solution? answer in units of m?

vivado [14]

Unit of M is also mole/L, where mole is the moles of solute and L is the volume of the solution. The latter is given: 158 mL or 0.158 L. So we need to find out the moles of NH4Br.

Moles of NH4Br = Mass of NH4Br/molar mass of NH4Br = 17.0g/(14+1*4+79.9)g/mol = 0.1736 mole.

So, the molarity of the solution = 0.1736mole/0.158L = 1.10 mole/L = 1.10 M

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3 years ago

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Answer:

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3 years ago

Match the terms with their correct definition:

geniusboy [140]

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3 years ago

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2 years ago

Calculate the pH of the following simple solutions:

IceJOKER [234]

Answer :

(1) pH = 1.27

(2) pH = 13.35

(3) The given solution is not a buffer.

Explanation :

<u>(1) 53.1 mM HCl</u>

Concentration of HCl = $53.1mM=53.1\times 10^{-3}M$

As HCl is a strong acid. So, it dissociates completely to give hydrogen ion and chloride ion.

So, Concentration of hydrogen ion= $53.1\times 10^{-3}M$

pH : It is defined as the negative logarithm of hydrogen ion concentration.

$pH=-\log [H^+]$

$pH=-\log (53.1\times 10^{-3})$

$pH=1.27$

<u>(2) 0.223 M KOH</u>

Concentration of KOH = 0.223 M

As KOH is a strong base. So, it dissociates completely to give hydroxide ion and potassium ion.

So, Concentration of hydroxide ion= 0.223 M

Now we have to calculate the pOH.

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

$pOH=-\log (0.223)$

$pOH=0.65$

Now we have to calculate the pH.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-0.65=13.35$

<u>(3) 53.1 mM HCl + 0.223 M KOH</u>

Buffer : It is defined as a solution that maintain the pH of the solution by adding the small amount of acid or a base.

It is not a buffer because HCl is a strong acid and KOH is a strong base. Both dissociates completely.

As we know that the pH of strong acid and strong base solution is always 7.

So, the given solution is not a buffer.

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3 years ago