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Bezzdna [24]
3 years ago
8

Is the entropy change favorable or not, when a nonpolar molecule is transferred from water to a nonpolar solvent?

Chemistry
1 answer:
jonny [76]3 years ago
8 0

Answer:

Entropy change is favorable when a nonpolar molecule is transferred from water to a nonpolar solvent.

Explanation:

A nonpolar molecule is not miscible in water (polar solvent). Therefore, when mixed together, each specie will cluster together and solvation will not happen.

However, when you tranfer the nonpolar molecule to a nonpolar solvent, the solvent molecules will interact with the nonpolar molecule. This will increase entropy as the level of disorder will increase with solvation.

You might be interested in
A 7.0 L sample of gas begins at 2.5 atm and 320. K. What is the new pressure if the temperature is changed to 273 K and the volu
enyata [817]

Answer:

2.1 atm

Explanation:

We are given the following variables to work with:

Initial pressure (P1): 2.5 atm

Initial temperature (T1): 320 K

Final temperature (T2): 273 K

Constant volume: 7.0 L

We are asked to find the final pressure (P2). Since volume is constant, we want to choose a gas law equation that relates initial pressure and temperature to final pressure and temperature. Gay-Lussac's law does this:

\frac{P_{1}}{T_1} =\frac{P_{2}}{T_2} \\

We can rearrange the law algebraically to solve for P_{2}.

{P_{2}} =\frac{(T_2)(P_{1} )}{T_1} \\

Substitute your known variables and solve:

P_2 = \frac{273K(2.5atm)}{320K}  = 2.1 atm

4 0
3 years ago
A party balloon contains 5.50 x 10 ^22 atoms of helium gas. What is the mass, in grams of the
eduard

Answer:

\boxed {\boxed {\sf A. \ 0.365 \ g\ He}}

Explanation:

<u>1. Convert Atoms to Moles</u>

We must use Avogadro's Number: 6.022*10²³. This is the number of particles (atoms, molecules, ions, etc.) in 1 mole of a substance. In this case, the particles are atoms of helium. We can create a ratio.

\frac {6.022*10^{23} \ atoms \ He}{1 \ mol \ He}

Multiply by the given number of helium atoms.

5.50 *10^{22} \ atoms \ He *\frac {6.022*10^{23} \ atoms \ He}{1 \ mol \ He}

Flip the fraction so the atoms of helium cancel.

5.50 *10^{22} \ atoms \ He *\frac {1 \ mol \ He}{6.022*10^{23} \ atoms \ He}

5.50 *10^{22} *\frac {1 \ mol \ He}{6.022*10^{23} }

\frac {5.50 *10^{22} \ mol \ He}{6.022*10^{23} }= 0.09133178346 \ mol \ He

<u>2. Convert Moles to Grams</u>

We must use the molar mass, which is found on the Periodic Table.

  • Helium (He): 4.00 g/mol

Use this as a ratio.

\frac { 4.00 \ g \ He }{ 1 \ mol \ He}

Multiply by the number of moles we calculated. The moles will then cancel.

0.09133178346 \ mol \ He *\frac { 4.00 \ g \ He }{ 1 \ mol \ He}

0.09133178346*\frac { 4.00 \ g \ He }{ 1 }

0.3653271338 \ g\ He

<u>3. Round </u>

The original measurement has 3 significant figures (5, 5, and 0). Our answer must have the same. For the number we calculated, it is thousandth place. The 3 in the ten thousandth place tells us to leave the 5.

0.365 \ g\ He

The mass is <u>0.365 grams of helium</u> so choice A is correct.

7 0
2 years ago
An isotope of yttrium has 39 protons and 59 neutrons. what is the atomic mass of that isotope?
Charra [1.4K]
We can define atomic mass the total of number of protons and number of neutrons in an atom or isotope.
<span>So when an isotope of yttrium has 39 protons and 59 neutrons, its atomic mass is equal to;
number of protons + number of neutrons = 39 + 59 = 98</span>
8 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]

<u>Answer:</u>

<u>For a:</u> The total heat required is 36621.5 J

<u>For b:</u> The total heat required is 58944.5 J

<u>Explanation:</u>

  • <u>For a:</u>

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)

  • <u>For process 1:</u>

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

  • <u>For process 2:</u>

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

  • <u>For b:</u>

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)

  • <u>For process 1:</u>

We are given:

Change in temperature remains the same.

m=42.0g\\c_s=0.97J/g.K\\T_2=-144^oC\\T_1=-155^oC\\\Delta T=[T_2-T_1]=[-144-(-155)]^oC=11^oC=11K

Putting values in equation 1, we get:

q_1=42.0g\times 0.97J/g.K\times 11K\\\\q_1=448.14J

  • <u>For process 2:</u>

We are given:

m=42.0g\\\Delta H_{fusion}=5.02kJ/mol=\frac{5.02kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=109.13J/g

Putting values in equation 2, we get:

q_2=42.0g\times 109.13J/g\\\\q_2=4583.5J

  • <u>For process 3:</u>

We are given:

Change in temperature remains the same.

m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=-144^oC\\\Delta T=[T_2-T_1]=[78-(-144)]^oC=222^oC=222K

Putting values in equation 1, we get:

q_3=42.0g\times 2.3J/g.K\times 222K\\\\q_3=21445.2J

  • <u>For process 4:</u>

We are given:

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

Putting values in equation 2, we get:

q_4=42.0g\times 773.04J/g\\\\q_4=32467.7J

Total heat required = [q_1+q_2+q_3+q_4]

Total heat required = [448.14+4583.5+21445.2+32467.7]J=58944.5J

Hence, the total heat required is 58944.5 J

8 0
3 years ago
. Using the pH scale below what would be the pH value of an Acid? *
earnstyle [38]

Answer : The pH value of an acid is below 7.

Explanation :

pH : It is defined as the negative logarithm of hydrogen ion or hydronium ion concentration.

Mathematically,

pH=-\log [H^+]

When the pH less than 7 then the solution is acidic and the concentration of hydrogen ion is greater than hydroxide ion.

When the pH more than 7 then the solution is basic and the concentration of hydrogen ion is less than hydroxide ion.

When the pH is equal to 7 then the solution is neutral and the concentration of hydrogen ion is equal to the hydroxide ion.

Hence, the pH value of an acid is below 7.

6 0
3 years ago
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