What is <br><br> 114.1 kPa to atm

The statement is true because , double bonded oxygen atom of carboxylic group interacts with hydrogen atom of water molecule and oxygen atom of water molecule is in same interaction with hydrogen atom of carboxylic acid.

B. The long alkane ''tail" $CH_3(CH_2)_{16} -$ , is hydrophobic.

The statement is true because , region which do not interacts with water molecule is termed as hydrophobic region and region which do interacts with water molecule is termed as hydrophilic region. Generally ,alkanes are are hydrophobic in nature so is the alkane chain in stearic acid.

C. Stearic acid and water are probably miscible.

The statement is false because of higher hydrophobic region the interaction between the acid molecules and water molecules results in immiscibility.

D. Stearic acid would probably dissolve in non polar solvents such as hexane $C_6H_{14}$ .

The statement is true because of hydrophobic part in stearic acid is larger so that is why it get easily dissolve in non polar solvents.

E. Stearic acid contains hydrophobic regions

The statement is true because of higher hydrophobic region due to which they easily get dissolved in organic and non polar solvents.

6 0

3 years ago

Which is an example of plasmas in nature? solar cells plasma balls auroras clouds

Anna [14]

Answer:

plasma balls

Explanation:

Plasma balls because of the name and the fact that plasma has the highest heat signature.

6 0

3 years ago

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What mass of potassium hypochlorite (FW-90.6 g/mol) must be added to 4.50 x 10 mL of water to give a solution with pH 10.20? [Ka

marta [7]

Answer : The mass of potassium hypochlorite is, 4.1 grams.

Explanation : Given,

pH = 10.20

Volume of water = $4.50\times 10^2ml=0.45L$

The decomposition of KClO will be :

$KClO\rightarrow K^++ClO^-$

Now the further reaction with water $(H_2O)$ to give,

$ClO^-+H_2O\rightarrow HClO+OH^-$

First we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-10.20=3.8$

Now we have to calculate the $OH^-$ concentration.

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

$3.8=-\log [OH^-]$

$[OH^-]=1.58\times 10^{-4}M$

Now we have to calculate the base dissociation constant.

Formula used : $K_b=\frac{K_w}{K_a}$

Now put all the given values in this formula, we get :

$K_b=\frac{1.0\times 10^{-14}}{4.0\times 10^{-8}}=2.5\times 10^{-7}$

Now we have to calculate the concentration of $ClO^-$ .

The equilibrium constant expression of the reaction is:

$K_b=\frac{[OH^-][HClO]}{[ClO^-]}$

As we know that, $[OH^-]=[HClO]=1.58\times 10^{-4}M$

$2.5\times 10^{-7}=\frac{(1.58\times 10^{-4})^2}{[ClO^-]}$

$[ClO^-]=0.0999M$

Now we have to calculate the moles of $ClO^-$ .

$\text{Moles of }ClO^-=\text{Molarity of }ClO^-\times \text{Volume of solution}$

$\text{Moles of }ClO^-=0.0999mole/L\times 0.45L=0.0449mole$

As we know that, the number of moles of $ClO^-$ are equal to the number of moles of KClO.

So, the number of moles of KClO = 0.0449 mole

Now we have to calculate the mass of KClO.

$\text{Mass of }KClO=\text{Moles of }KClO\times \text{Molar mass of }KClO$

$\text{Mass of }KClO=0.0449mole\times 90.6g/mole=4.07g\approx 4.1g$

Therefore, the mass of potassium hypochlorite is, 4.1 grams.

6 0

3 years ago

What is 114.1 kPa to atm