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almond37 [142]
3 years ago
12

Consider the solubilities of a particular solute at two different temperatures. Temperature ( ∘ C ) Solubility ( g / 100 g H 2 O

) 20.0 32.2 30.0 70.2 Suppose a saturated solution of this solute was made using 56.0 g H 2 O at 20.0 °C. How much more solute can be added if the temperature is increased to 30.0 ∘ C? mass:
Chemistry
1 answer:
grin007 [14]3 years ago
8 0

Answer:

21.28 grams solute can be added if the temperature is increased to 30.0°C.

Explanation:

Solubility of solute at 20°C = 32.2 g/100 grams of water

Solute soluble in 1 gram of water = \frac{32.2}{100}g=0.322 g

Mass of solute in soluble in 56.0 grams of water:

0.322\times 56.0=18.032 g

Solubility of solute at 30°C = 70.2g/100 grams of water

Solute soluble in 1 gram of water = \frac{70.2}{100}g=0.702 g

Mass of solute in soluble in 56.0 grams of water:

0.702 \times 56.0=39.312 g

If the temperature of saturated solution of this solute using 56.0 g of water at 20.0 °C raised to 30.0°C

Mass of solute in soluble in 56.0 grams of water 20.0°C = 18.032 g

Mass of solute in soluble in 56.0 grams of water at 30.0°C = 39.312 g

Mass of of solute added If the temperature of the saturated solution increased to 30.0°C:

39.312 g - 18.032 g = 21.28 g

21.28 grams solute can be added if the temperature is increased to 30.0°C.

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By the transfer of one or more electrons from one atom to another

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2Na + Cl₂ → 2Na⁺Cl⁻

Na - e⁻ → Na⁺
Cl₂ + 2e⁻ → 2Cl⁻
6 0
3 years ago
Read 2 more answers
A state property X has a value 89.6 units. It undergoes the following changes, first increase by
yanalaym [24]

Answer:

B) -4.1 units

Explanation:

According to this question, a state property X has a value 89.6 units. It undergoes the certain changes as follows:

- first increase by 3.6 units

- then increase by another 18.7 units

- then decrease by 12.2 units

- and finally attains a value of 85.5 units

This can be mathematically represented by 89.6 - {3.6 + 18.7 - 12.2 - x) = 85.5

To get x, we say;

89.6 + 3.6 = 93.2

93.2 + 18.7 = 111.9

111.9 - 12.2 = 99.7

99.7 - 85.5 = 14.2units.

The changes that occured is represented as follows:

= (3.6 + 18.7) - (12.2 + 14.2)

= 22.3 - 26.4

= -4.1 units

7 0
3 years ago
If water were a non-polar<br> Molecule
JulijaS [17]

Answer:

Water would not be able to transport nutrients -‐-‐ in plants, or in our bodies -‐-‐ nor to dissolve and transport waste products out of our bodies. ... Cohesiveness, adhesiveness, and surface tension: would decrease because without the +/-‐ polarity, water would not form hydrogen bonds between H20 molecules.

8 0
3 years ago
For a process Arightwards harpoon over leftwards harpoonB, at 25 °C there is 10% of A at equilibrium while at 75 °C, there is 80
Lostsunrise [7]

This question is describing the following chemical reaction at equilibrium:

A\rightleftharpoons B

And provides the relative amounts of both A and B at 25 °C and 75 °C, this means the equilibrium expressions and equilibrium constants can be written as:

K_1=\frac{90\%}{10\%}=9\\\\K_2=\frac{20\%}{80\%}  =0.25

Thus, by recalling the Van't Hoff's equation, we can write:

ln(K_2/K_1)=-\frac{\Delta H}{R}(\frac{1}{T_2} -\frac{1}{T_1} )

Hence, we solve for the enthalpy change as follows:

\Delta H=\frac{-R*ln(K_2/K_1)}{(\frac{1}{T_2} -\frac{1}{T_1} ) }

Finally, we plug in the numbers to obtain:

\Delta H=\frac{-8.314\frac{J}{mol*K} *ln(0.25/9)}{[\frac{1}{(75+273.15)K} -\frac{1}{(25+273.15)K} ] } \\\\\\\Delta H=4,785.1\frac{J}{mol}

Learn more:

  • brainly.com/question/10038290
  • brainly.com/question/19671384
5 0
3 years ago
Determine the molarity of a 6.0 mole% sulfuric acid solution with SG-a 1.07 Note: Atomic Weight: S (32), O 16); H (O)
marin [14]

Answer:

The molarity of a 6.0 mole% sulfuric acid solution is 2.8157 Molar.

Explanation:

Suppose there are 100 moles in solution:

Moles of sulfuric acid = 6% of 100 moles = 6 moles

Mass of 6 moles of sulfuric acid = 6 mol × 98 g/mol=588 g

Moles of water = 100%- 6% = 94%= 94 moles

Mass of water = 94 mol × 18 g/mol = 1692 g

Specific gravity of the solution ,S.G= 1.07

Density of solution = D

S.G=\frac{D}{d_w}

d_w = density of water = 1 g/mL

D=S.G\times d_w=1.07\times 1 g/mL=1.07 g/mL

Mass of the solution = 588 g + 1692 g = 2280 g

Volume of the solution = V

Volume = \frac{Mass}{Density}

=\frac{2280 g}{1.07 g/mL}=2130.84 mL=2.13084 L

1 mL = 0.001 L

Molarity = \frac{n}{V(L)}

n = number of moles of compound

V = volume of the solution in L

here we have ,n = 6 moles of sulfuric acid

V = 2.13084 L

So, the molarity of the solution is :

Molarity=\frac{6 mol}{2.13084 L}=2.8157 mol/L

5 0
3 years ago
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