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At equal concentrations would a nonelectrolyte (e.g. glucose) or electrolyte (e.g. nacl) containing solution have a lower freezi

Anestetic [448]

Answer is: solution of electrolyte will have lower freezing point than solution of nonelectrolyte.
This is because salt solution has more particles in of sodium chloride (sodium and chlorine ions) than in same concentration of glucose. Electrolytes better separates into particles in water because of their ionic bond.

6 0

3 years ago

Calculate the molarity of the two solutions.

zmey [24]

Answer:

a) 0.100 M

b) 0.395 M

Explanation:

a) Calculate the molarity of a solution that contains 0.200 moles of NaOH (solute) in 2.00 L of solution

We will use the following expression for molarity.

[NaOH] = moles of solute / liters of solution

[NaOH] = 0.200 mol/2.00 L = 0.100 M

b) Calculate the molarity for a solution that contains 15.5 g of NaCl (solute, 58.44 g/mol) in 671 mL of solution

We will use the following expression for molarity.

[NaCl] = mass of solute / molar mass of solute × liters of solution

[NaCl] = 15.5 g / 58.44 g/mol × 0.671 L = 0.395 M

4 0

2 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 )

Step 1 : To find pka

H₂PO₄⁻ <=> HPO₄²⁻

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

Step 2 : Plug values in Hasselbalch- Henderson equation .

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)

Step 3 : To find molarity of H₂PO₄⁻ and HPO₄²⁻

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

Step 6 : To find mass of Na₂HPO₄ and NaH₂PO₄

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

What is the empirical formula? A compound is used to treat iron deficiency in people. It contains 36.76% iron, 21.11% sulfur, an

Mandarinka [93]

Answer: The empirical formula of the compound is $Fe_1S_1O_4$

Explanation:

Empirical formula is defined formula which is simplest integer ratio of number of atoms of different elements present in the compound.

Percentage of iron in a compound = 36.76 %

Percentage of sulfur in a compound = 21.11 %

Percentage of oxygen in a compound = 42.13 %

Consider in 100 g of the compound:

Mass of iron in 100 g of compound = 36.76 g

Mass of iron in 100 g of compound = 21.11 g

Mass of iron in 100 g of compound = 42.13 g

Now calculate the number of moles each element:

Moles of iron= $\frac{36.76 g}{55.84 g/mol}=0.658 mol$

Moles of sulfur= $\frac{21.11 g}{32.06 g/mol}=0.658 mol$

Moles of oxygen= $\frac{42.13 g}{16 g/mol}=2.633 mol$

Divide the moles of each element by the smallest number of moles to calculated the ratio of the elements to each other

For Iron element = $\frac{0.658 mol}{0.658 mol}=1$

For sulfur element = $\frac{0.658 mol}{0.658 mol}=1$

For oxygen element = $\frac{2.633 mol}{0.658 mol}=4.001\approx 4$

So, the empirical formula of the compound is $Fe_1S_1O_4$

4 0

2 years ago

Read 2 more answers

PLEASE ANSWER QUICKLY!!!in an ionic compound , which ions are attracted to each other?

Sauron [17]

Answer:

Oppositely charged ions attract each others i.e, positive and negative ions

Explanation:

Ionic Compound:

Ionic compounds consist have ionic bonds. It is the bond which is formed by the transfer of electron from one atom to the atom of another element.

Both bonded atoms have very large electronegativity difference.

The atom with large electronegativity value accept the electron from other with smaller value of electronegativity.

For example:

Sodium chloride is ionic compound. The electronegativity of chlorine is 3.16 and for sodium is 0.93. There is large difference is present. That's why electron from sodium is transfer to the chlorine. Sodium becomes positive and chlorine becomes negative ion. Sodium have one valance electron while chlorine have 7 valance electrons. In order to complete the octet chlorine require one electron while sodium need to lose its one electrons. That's why when both atom combine sodium lose its electron and becomes positive ion i.e cation while chlorine accept its electron and becomes negative ion called anion and bond between them is ionic bond. The electrostatic forces created between both atoms.

6 0

3 years ago