Which one is it (it’s multiple choice too)

The normal boiling point of ethanol (C2H5OH) is 78.3 oC and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the chan

Sloan [31]

Answer:

The change in entropy in the system is -231.5 J/K

Explanation:

Step 1: Data given

The normal boiling point of ethanol (C2H5OH) is 78.3 °C = 351.45 K

The molar enthalpy of vaporization of ethanol is 38.56 kJ/mol = 38560 J/mol

Mass of ethanol = 97.2 grams

Pressure = 1 atm

Step 2: Calculate the entropy change of vaporization

The entropy change of vaporization = molar enthalpy of vaporization of ethanol / temperature

The entropy change of vaporization = 38560 J/mol / 351.45 K

The entropy change of vaporization = -109.72 J/k*mol

Step 3: Calculate moles of ethanol

Moles ethanol = mass / molar mass ethanol

Moles ethanol = 97.2 grams / 46.07 g/mol

Moles ethanol = 2.11 moles

Step 4: Calculate change in entropy

For 1 mol = -109.72 J/K*mol

For 2.11 moles = -231.5 J/K

The change in entropy in the system is -231.5 J/K

7 0

3 years ago

Help help help help help

luda_lava [24]

The sum is 34.688 m.

If m is a variable, then the m would be attached to 34.688.

7 0

3 years ago

Read 2 more answers

A level chem naming isomers

anyanavicka [17]

Answer:

Butan-1-ol or Butanol

2-methylpropan-2-ol

2-methylpropan-1-ol

Butan-2-ol

Explanation:

Data Give:

Three Isomers are given

names of These =?

Details:

Isomers:

Isomers are those chemical compounds that have same molecular formula but different structures.

In the given question the compound have same molecular formula that is C₄H₉OH but the structures are different.

So due to different structures they are named differently.

The 1st and last one are straight Chains and the other 2 are branched. all the structure have four carbon atoms, one OH group and 9 hydrogen atoms.

Names are given in attachment

butan-1-ol or Butanol

As it contain 4 carbon atoms an -OH group is attached at position 1.

________________

2-methylpropan-2-ol

In this -OH and -CH₃ is attached on carbon 2

_____________

2-methylpropan-1-ol

In this -CH₃ attached at position 2 and -OH is attached on Carbon 1.

______________

Butan-2-ol

In this 4 carbons are in straight chain and a -OH group is attached on position 2 of carbon

7 0

4 years ago

Percentage Yield

scoundrel [369]

Q.No. 1:

You have 20.0 g of CaCO₃. You decompose it by heat, and weigh the calcium oxide that remains. You have 10.3 grams. What is the % yield?

Answer:

%age Yield = 92.37 %

Solution:

The balance chemical equation for given decomposition reaction is;

CaCO₃ → CaO + CO₂

Step 1: Calculate Moles of CaCO₃:

Moles = Mass / M.Mass

Moles = 20.0 g / 100.08 g/mol

Moles = 0.199 moles of CaCO₃

Step 2: Calculate theoretical amount of CaO produced;

According to equation,

1 moles of CaCO₃ produced = 1 mole of CaO

So,

0.199 moles of CaCO₃ will produce = X moles of CaO

Solving for X,

X = 0.199 mol × 1 mol / 1 mol

X = 0.199 mol of CaO

Also,

Mass = Moles × M.Mass

Mass = 0.199 mol × 56.07 g/mol

Mass = 11.15 g of CaO

Step 3: Calculate %age Yield as;

%age Yield = Actual Yield / Theoretical Yield × 100

%age Yield = 10.3 g / 11.15 g × 100

%age Yield = 92.37 %

___________________________________________

Q.No. 2:

A student makes sodium chloride by buming 2.3 grams of sodium in chlorine gas. If the yield is 90%, how much sodium chloride is made?

Answer:

Actual Yield = 5.785 g of NaCl

Solution:

The balance chemical equation for given decomposition reaction is;

2 Na + Cl₂ → 2 NaCl

Step 1: Calculate Moles of Na:

Moles = Mass / M.Mass

Moles = 2.3 g / 23 g/mol

Moles = 0.10 moles of Na

Step 2: Calculate theoretical amount of NaCl produced;

According to equation,

2 moles of Na produced = 2 moles of NaCl

So,

0.10 moles of Na will produce = X moles of NaCl

Solving for X,

X = 0.10 mol × 2 mol / 2 mol

X = 0.10 mol of NaCl

Also,

Mass = Moles × M.Mass

Mass = 0.10 mol × 58.44 g/mol

Mass = 5.844 g of NaCl

Step 3: Calculate Actual Yield as;

%age Yield = Actual Yield / Theoretical Yield × 100

Or,

Actual Yield = %age Yield × Theoretical Yield ÷ 100

Actual Yield = 99 × 5.844 g ÷ 100

Actual Yield = 5.785 g of NaCl

___________________________________________

Q.No. 3:

For the chemical reaction Mg(s) + 2 HCl (aq) → H₂ (g) + MgCl₂ (aq) calculate the % yield if 100 grams of magnesium react with excess HCl to produce 310 grams of MgCl₂.

Answer:

%age Yield = 79.13 %

Solution:

The balance chemical equation for given decomposition reaction is;

Mg + 2 HCl → MgCl₂ + H₂

Step 1: Calculate Moles of Mg:

Moles = Mass / M.Mass

Moles = 100 g / 24.30 g/mol

Moles = 4.11 moles of Mg

Step 2: Calculate theoretical amount of MgCl₂ produced;

According to equation,

1 mole of Mg produced = 1 mole of MgCl₂

So,

4.11 moles of Mg will produce = X moles of MgCl₂

Solving for X,

X = 4.11 mol × 1 mol / 1 mol

X = 4.11 mol of MgCl₂

Also,

Mass = Moles × M.Mass

Mass = 4.11 mol × 95.21 g/mol

Mass = 391.73 g of MgCl₂

Step 3: Calculate %age Yield as;

%age Yield = Actual Yield / Theoretical Yield × 100

%age Yield = 310.0 g / 391.73 g × 100

%age Yield = 79.13 %

5 0

4 years ago

Write the formulas of the following compounds:

dem82 [27]

Answer:

a) Li2CO3

b) NaCLO4

c) Ba(OH)2

d) (NH4)2CO3

e) H2SO4

f) Ca(CH3COO)2

g) Mg3(PO4)2

f) Na2SO3

Explanation:

a) 2Li + CO3 ↔ Li2CO3

b) NaOH * HCLO4 ↔ NaCLO4 + H2O

c) Ba + 2H2O ↔ Ba(OH)2 +

d) 2NH4 + H2CO3 ↔ (NH4)2CO3 + H2O

c) SO2 + NO2 +H2O ↔ H2SO4 + NOx

f) 2CH3COOH + CaO ↔ Ca(CH3COOH)2 + H2O

g) 3MgO + 2H3PO4 ↔ Mg3(PO4)2 + H2O

h) NaOH + H2SO3 ↔ Na2SO3 + H2O

6 0

3 years ago