true or false. a less reactive halogen will displace a more reactive halogen from an aqueous solution of its salt.

Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O,

WITCHER [35]

Answer: Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

$2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]$

We are given:

$\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ$

Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

6 0

2 years ago

How is percent by mass and concentration related?

Julli [10]

You can have a solution of hydrogen peroxide that might say 10% that means that 10% per mass of the hydrogen peroxide solution is the hydrogen peroxide the rest is water.

concentration is the amount of mass in the solution eg 5gdm-3

hope that helps

6 0

2 years ago

The name given to the substance formed when a liquid is miscible in another liquid.

m_a_m_a [10]

Answer:

Solute

Explanation:

5 0

3 years ago

The atmospheric gases in the thermosphere are heated _____. Pick the best one.

lozanna [386]

by radiation from the Earth's surface

4 0

3 years ago

Read 2 more answers

THIS IS A THREE PART QUESTION IF YOU CAN HELP IT WOULD BE REALLY APPRECIATED SO I DONT FAIL.

Nostrana [21]

Answer:

1. B = 1.13M

2. A. 8.46%

3. D = 0.0199

Explanation:

1. Molarity of a a solution = number of moles of solute/ volume of solution in L

Number of moles of solute = mass of solute/molar mass of solute

Molar mass of NaC₂H₃O₂ = 82 g/mol, mass of NaC₂H₃O₂ = 245 g

Number of moles of NaC₂H₃O₂ = 245 g / 82 g/mol = 2.988 moles

Molarity of solution = 2.988 mols/ 2.65 L = 1.13 M

2. Percentage by mass of a substance = mass of substance /mass of solution × 100%

Mass of 2.65 L of water = density × volume

Density of water = 1 Kg/L = 1000 g/L; volume of waterb= 2.65 L

Mass of water = 1000 g/L × 2.65 L = 2650 g

Mass of solution = mass of water + mass of solute = 245 + 2650 =2895 g

Percentage by mass of NaC₂H₃O₂ = 245/2895 × 100% = 8.46%

3. Mole fraction of NaC₂H₃O₂ = moles of NaC₂H₃O₂/moles of solution

Moles of water = mass /molar mass

Mass of water = 2650 g; molar mass of water = 18 g/mol

Moles of water = 2650 g / 18 g/mol = 147.222 moles

Moles of solution = moles of solute + moles of water = 147.222 + 2.988 = 150.21

Moles of NaC₂H₃O₂ =2.988 moles

Moles fraction of NaC₂H₃O₂ =2.988/150.21 = 0.0199

6 0

3 years ago

true or false. a less reactive halogen will displace a more reactive halogen from an aqueous solution of its salt.​

true or false. a less reactive halogen will displace a more reactive halogen from an aqueous solution of its salt.