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3 years ago

8

If more solvent is added to solution, the value of heat of solution

1 answer:

TiliK225 [7]3 years ago

8 0

if more solvent is added to the solution the heat of solution[enthalpy of solution] is decreased because by adding more solution the heat is lost to the surroundings that is why it is decreased

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Please answer, with explanation. Thanks!

nadya68 [22]

Answer:

Explanation:

a = 40.1 g of Ca

Number of moles = mass / molar mass

Number of moles = 40.1 g/ 40.1 g/mol

Number of moles = 1 mol

b = 11.5 g Na

Number of moles = mass / molar mass

Number of moles = 11.5 g/ 23 g/mol

Number of moles = 0.5 mol

c = 5.87 g Ni

Number of moles = mass / molar mass

Number of moles = 5.87 g/ 58.7 g/mol

Number of moles = 0.1 mol

d = 150 g of S

Number of moles = mass / molar mass

Number of moles = 150 g/ 32 g/mol

Number of moles = 4.7 mol

e = 2.65 g Fe

Number of moles = mass / molar mass

Number of moles = 2.65 g/ 55.85 g/mol

Number of moles = 0.05 mol

f = 0.00750 g Ag

Number of moles = mass / molar mass

Number of moles = 0.00750 g/ 107.9 g/mol

Number of moles = 6.95 × 10⁻⁵ mol

g = 2.25 × 10²⁵ atoms Zn

1 mole = 6.022 × 10²³ atoms

1 mol / 6.022 × 10²³ atoms × 2.25 × 10²⁵ atoms

0.17 × 2.25 × 10²⁵ moles

38.25 moles

h = 50 atoms of Ba

1 mole = 6.022 × 10²³ atoms

1 mol / 6.022 × 10²³ atoms ×50 atoms

0.17 × 10²³ × 50 moles

8.5 × 10²³ moles

6 0

3 years ago

13. Fill in the following table

Mandarinka [93]

Answer:

Potassium Bromide = KBr

Nitrogen dioxide = NO₂

Lithium oxide = Li₂O

Explanation:

Potassium Bromide (KBr)

4 0

3 years ago

1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2) The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3)The given statement is false.

Explanation:

1)

Solubility of lead chloride = $S=3.1\times 10^-2M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

S 2S

The solubility product of the lead(II) chloride = $K_{sp}$

$K_{sp}=[Pb^{2+}][Cl^-]^2$

$K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}$

The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2)

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion = $1\timed 0.000010 M=0.000010 M$

Solubility of aluminium hydroxide in aluminum nitrate solution = $S$

$Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)$

S 3S

The solubility product of the aluminium nitrate = $K_{sp}=1.0\times 10^{-33}$

$K_{sp}=[Al^{3+}][OH^-]^3$

$1.0\times 10^{-33}=(0.000010+S)\times (3S)^3$

$S=1.6\times 10^{-10} M$

The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3.

$Molarity=\frac{Moles}{Volume (L)}$

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = $\frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol$

Volume of the solution = 0.250 L

$[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M$

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

$[Cl^-]=[NaCl]=0.00024 M$

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

$n=0.12 M]\times 0.250 L=0.030 mol$

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

$[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

Solubility of lead(II) chloride = $K_{sp}=1.2\times 10^{-4}$

Ionic product of the lead chloride in solution :

$Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}$

$Q_i$ ( no precipitation)

The given statement is false.

3 0

4 years ago

What is the advantage of having valves in the

fenix001 [56]

The valves stop the blood from flowing backwards

6 0

3 years ago

3. Calculate the volume, in mL, of 100.0 g of alcohol, if the density is 0.79 g/mL.

Naya [18.7K]

Answer:

The answer is

<h2>126.58 mL</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

<h3> $volume = \frac{mass}{density}$ </h3>

From the question

mass of alcohol = 100 g

density = 0.79 g/mL

The volume is

$volume = \frac{100}{0.79} \\ = 126.58227848...$

We have the final answer as

<h3>126.58 mL</h3>

Hope this helps you

4 0

3 years ago