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Gnom [1K]
3 years ago
15

Which of the following aqueous solution will have the highest boiling point

Chemistry
1 answer:
Sliva [168]3 years ago
4 0

Ionic molecules have higher boiling point than covalent molecules.

So that leaves Na2SO4 and NaCl.

The ionic molecule with higher charge will have higher boiling point.

Na2SO4 have ions with charge +-2.

NaCl have ions with charge +-1.

So the answer is C.

You might be interested in
What is the name of the compound with the formula NaOH ?
vladimir2022 [97]

Answer:

Sodium hydroxide

Explanation:

Sodium hydroxide ( NaOH ) -

Sodium hydroxide is an inorganic compound , and is also called caustic soda and lye .

It is an ionic compound , which is white in color and is in solid state .

The cation and anion of this salt are the sodium cation Na⁺  and the hydroxide anions OH⁻ .

<u>It is highly basic in nature and is soluble in water , and when left open in air it can readily absorb moisture from the air , to form a hydrated sodium hydroxide .</u>

4 0
3 years ago
If a balloon containing 3000 L of gas at 39 C and 99 k Pa rises to an altitude where the pressure is 45.5 kPa and the temperatur
mihalych1998 [28]

Answer:

The new volume of the balloon will be 6046.28 L

Explanation:

Initial pressure (P1) = 99 kpa

initial volume (V1) = 3000 L

Initial temperature = 39 C = 39 + 273 = 312 K

Final pressure (P2) = 45.5 kpa

Final temperature = 16 C = 16 +273 = 289K

Final volume = ????

To calculate the final volume using the general gas equation

                  P1 V1 / T1  = P2 V2 / T2

make V2 the subject of the formular

                   V2 = 99000 ×3000× 289 / 45500×312

                   V2 = 85833000 /14196

                     V2 = 6046.28 litres

5 0
3 years ago
A car travels 33.52 miles in 28 minutes. how fast is the car moving
N76 [4]
The answer is 69.69 miles per hour
3 0
3 years ago
How many grams of sodium phosphate monobasic would we add to a liter and how many grams of sodium phosphate dibasic would we add
Nostrana [21]

Answer :

The correct answer   for Mass of Na₂HPO₄ = 4.457 g and mass of  NaH₂PO₄  = 8.23 g

Given :  pH = 6.86

Total concentration of Phosphate buffer = 0.1 M

Asked : Mass of  Sodium phosphate monobasic (NaH₂PO₄) = ?

Mass of  Sodium phosphate dibasic(Na₂HPO₄)= ?

Following steps can be done to find the masses of NaH₂PO₄ and Na₂HPO₄ :

(In phosphate buffer , Na+ ion from  NaH₂PO₄ and Na₂HPO₄ acts as spectator ion , so only H₂PO₄⁻ and HPO₄²⁻ will be considered )

<u>Step 1 : To find pka </u>

H₂PO₄⁻  <=> HPO₄²⁻  

The above reaction has pka = 7.2 ( from image shown )

<u>Step 2 : Plug values in Hasselbalch- Henderson equation </u>.

Hasselbalch -Henderson equation is to find pH  for buffer solution which is as follows :

pH = pka + log\frac{[A^-]}{[HA]}

pH = 6.86         pKa = 7.2

6.86 = 7.2 + log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}

Subtracting  both side by 7.2

6.86-7.2 = 7.2 -7.2+ log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}

-0.34 =  log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}

Removing log

10^-^0^.^3^4 =   \frac{[HPO_4^2^-]}{[H_2PO_4^-]}

\frac{[HPO_4^2^-]}{[ H_2PO_4^-]} = 0.457 ---------------- equation (1)

<u>Step 3 : To find  molarity of H₂PO₄⁻ and HPO₄²⁻</u>

Total concentration of buffer = [H₂PO₄⁻] + [HPO₄²⁻] = 0.1 M

Hence,  [H₂PO₄⁻ ] + [ HPO₄²⁻ ] =  0.1 M

Assume [H₂PO₄⁻ ] = x

So ,  [x ] + [ HPO₄²⁻ ] =  0.1 M

[ HPO₄²⁻ ] =  0.1 - x

Step 4 : Plugging value of [H₂PO₄⁻ ]  and  [ HPO₄²⁻ ]

[H₂PO₄⁻ ]  = x

 [ HPO₄²⁻ ] = 0.1 - x

Equation (1) = >\frac{[HPO_4^2^-]}{[ H_2PO_4^-]} = 0.457

Plug value of [H₂PO₄⁻ ]  and  [ HPO₄²⁻ ] ( from step 3 ) into equation (1)  as :

\frac{[0.1 - x ]}{[ x]} = 0.457

Cross multiplying

0.1 - x  = 0.457 x

Adding x on both side

0.1 -x + x = 0.457 x + x

0.1  = 1.457 x

Dividing both side by 1.457

\frac{0.1}{1.457} = \frac{1.457 x }{1.457}

x = 0.0686 M

Hence , [H₂PO₄⁻ ]  = x  = 0.0686 M

 [ HPO₄²⁻ ] = 0.1 - x

 [ HPO₄²⁻ ]  =   0.1 - 0.0686  

[ HPO₄²⁻ ] = 0.0314 M

Step 5 : To find moles of  H₂PO₄⁻ ( NaH₂PO₄) and HPO₄²⁻ (Na₂HPO₄ ) .

Molarity is defined as mole of solute per 1 L volume of solution .

Molarity of NaH₂PO₄ = 0.0686 M  or 0.0686 mole per 1 L

Molarity of Na₂HPO₄ = 0.0314 M  or 0.0314 mole per 1 L

Since  that volume of buffer solution  is 1 L , so Molarity  = mole

Hence Mole of NaH₂PO₄  = 0.0686 mol

Mole of Na₂HPO₄ = 0.0314 mol

<u>Step 6 : To find mass  of Na₂HPO₄  and NaH₂PO₄ </u>

Moles of  Na₂HPO₄  and NaH₂PO₄  can be converted to their masses using molar mass as follows :

Molar mass of  Na₂HPO₄  = 141.96 \frac{g}{mol}

Molar mass of NaH₂PO₄ = 119.98 \frac{g}{mol}

Mass (g) = mole (mol)* molar mass(\frac{g}{mol})

Mass of Na_2HPO_4 = 0.0314 mol * 141.96 \frac{g}{mol}

Mass of Na₂HPO₄ = 4.457 g

Mass of NaH_2PO_4 = 0.0686 mol * 119.98 \frac{g}{mol}

Mass of  NaH₂PO₄  = 8.23 g

5 0
3 years ago
A property is a characteristic that can be used to identify a substance. Properties can be ______ or _______.
kirill [66]

Answer:

a list of characteristics that can be used to identify a substance include color, odor, taste, density, melting point, boiling point, conductivity, and hardness.

3 0
3 years ago
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