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3 years ago

What happens to the saturation concentration when a solution’s temperature is increased?

1 answer:

4 0

The saturation concentration of a solution would decrease when the temperature of a solution is increased since the solubility of the solute in the solvent would decrease. This is true for solids and gases. For liquids, it depends on the solute. Hope this helps.

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HELP PLEASEEE !!! ILL MARK YOU BRAINLIEST + 15 POINTS

Shtirlitz [24]

False- water has a high specific heat, so it does not change temperature as easily.

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3 years ago

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At 20 degrees Celsius, the density of air is 1.20 g/L. Nitrogen's density is 1.17 g/L. Oxygen's density is 1.33 g/L. Will balloo

Anna [14]

Balloons filled with oxygen will sink and balloons filled with nitrogen will also sink, this is because oxygen is more dense than air but nitrogen is only slightly less dense than air meaning it is not enough for the nitrogen to rise

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3 years ago

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A concentration cell consists of two sn/sn2+ half-cells. the electrolyte in compartment a is 0.24 m sn(no3)2. the electrolyte in

Vikentia [17]

A:- sn(s) => Sn +2(0.24 M) + 2e-
B:- Sn +2 (0.87 M) +2e- => Sn(s)

solution will become more concentrated and solution B become less concentrated

Sn(s)+ Sn +2(0.87 ) ----> Sn(s) + Sn +2(0.24)
E = Eo - 0.0592 / 2 * log [ (0.24 / 0.87 ) ]
E = 0.0 - 0.0592 / 2 * log ( 0.275)
( n=2 two electrons are transferred)

E = -0.0296 * ( - 0.560)
E = 0.0165 volts

4 0

3 years ago

If 0.400 moles CO and 0.400 moles O2 completely react how many grams of CO2 are produces?

True [87]

If 0.400 moles CO and 0.400 moles O2 completely react, 17.604 grams of CO2 would be produced.

First, let us look at the balanced equation of reaction:

$2CO + O_2 ---> 2CO_2$

According to the equation, the mole ratio of CO and O2 is 2:1. But in reality, the mole ratio supplied is 1:1. Thus, CO is the limiting reactant while O2 is in excess.

Also from the equation, the ratio of CO consumed to that of CO2 produced is 1:1. Thus, 0.400 moles of CO2 would also be produced from 0.400 moles of CO.

Recall that: mole = mass/molar mass

Therefore, the mass in grams of CO2 that would be produced from 0.400 moles can be calculated as:

Mass = mole x molar mass

= 0.400 x 44.01

= 17.604 grams

More on calculating mass from number of moles can be found here: brainly.com/question/12513822

7 0

3 years ago

Ethanol (C2H5OH) melts a - 144 oC and boils at 78 °C. The enthalpy of fusion of ethanol is 5.02 kj/mol, and its enthalpy of vapo

hammer [34]

Answer:

For a: The total heat required is 36621.5 J

For b: The total heat required is 58944.5 J

Explanation:

For a:

To calculate the heat required at different temperature, we use the equation:

$q=mc\Delta T$ .........(1)

where,

q = heat absorbed

m = mass of substance

$c$ = specific heat capacity of substance

$\Delta T$ = change in temperature

To calculate the amount of heat required at same temperature, we use the equation:

$q=m\times \Delta H$ ........(2)

where,

q = heat absorbed

m = mass of substance

$\Delta H$ = enthalpy of the reaction

The processes involved in the given problem are:

$1.)C_2H_5OH(l)(35^oC)\rightarrow C_2H_5OH(l)(78^oC)\\2.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)$

For process 1:

We are given:

Change in temperature remains the same.

$m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=35^oC\\\Delta T=[T_2-T_1]=[78-35]^oC=43^oC=43K$

Putting values in equation 1, we get:

$q_1=42.0g\times 2.3J/g.K\times 43K\\\\q_1=4153.8J$

For process 2:

We are given:

Conversion factor: 1 kJ = 1000 J

Molar mass of ethanol = 46 g/mol

$m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{35.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g$

Putting values in equation 2, we get:

$q_2=42.0g\times 773.04J/g\\\\q_2=32467.7J$

Total heat required = $[q_1+q_2]$

Total heat required = $[4153.8J+32467.7J]=36621.5J$

Hence, the total heat required is 36621.5 J

For b:

The processes involved in the given problem are:

$1.)C_2H_5OH(s)(-155^oC)\rightarrow C_2H_5OH(s)(-144^oC)\\2.)C_2H_5OH(s)(-144^oC)\rightarrow C_2H_5OH(l)(-144^oC)\\3.)C_2H_5OH(l)(-144^oC)\rightarrow C_2H_5OH(l)(78^oC)\\4.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)$