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Tema [17]
3 years ago
14

Liquid nitrogen has a density of 0.807 g/ml at –195.8 °c. if 1.00 l of n2(l) is allowed to warm to 25°c at a pressure of 1.00 at

m, what volume will the gas occupy? (r = 0.08206 l×atm/k×mol)
Chemistry
1 answer:
Flauer [41]3 years ago
8 0
Step 1: Change density from g/mL to g/L;

                                  0.807 g/mL  =  807 g/L

Step 2: Find Moles of N₂;
As,
             Density  =  Mass / Volume
Or,
             Mass  =  Density × Volume

Putting Values,

            Mass  =  807 g/L × 1 L

            Mass  =  807 g
Also,
            Moles  =  Mass / M.mass

Putting values,

            Moles  =  807 g / 28 g.mol⁻¹

            Moles  =  28.82 moles

Step 3: Apply Ideal Gas Equation to Find Volume of gas occupied,

As,
                        P V  =  n R T

                            V  =  n R T / P
Putting Values, remember! don't forget to change temperatue into Kelvin (25 °C + 273 = 298 K)

                 V  =  (28.82 mol × 0.08206 atm.L.mol⁻¹.K⁻¹ × 298 K) ÷ 1 atm

                 V  =  704.76 L
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8 0
3 years ago
A 59.1g sample of aluminum is put into a calorimeter (see sketch at right) that contains 250.0g of water. The aluminum sample st
Rainbow [258]

Answer:

The specific heat capacity of aluminum according to this experiment is 0.863 J/g°C

Explanation:

Step 1: Data given

Mass of aluminium = 59.1 grams

Mass of water = 250.0 grams

Initial temperature of aluminium = 91.3 °C

Initial temperature of water = 16.0 °C

Final temperature = 19.5 °C

Pressure remains constant

Specific heat capacity of water = 4.186 J/g°C

Step 2: Calculate specific heat of aluminium

Heat lost = heat gained

Qlost = -Q heat

Q = m*c*ΔT

heat aluminium = - heat water

m(aluminium) * c(aluminium) * ΔT(aluminium) = -m(water) * c(water) * ΔT(water)

⇒m(aluminium) = mass of aluminium = 59.1 grams

⇒c(aluminium) = the specific heat of aluminium = TO BE DETERMINED

⇒ΔT = the change in temperature = T2 -T2 = 19.5 - 91.3 = -71.8 °C

⇒ m(water) = 250.0 grams

⇒c(water) = the specific heat of water = 4.186 J/g°C

⇒ΔT = the change in temperature = T2 -T2 = 19.5 - 16.0 = 3.5 °C

59.1 * c(aluminium) * -71.8 °C = 250.0 * 4.186 J/g°C * 3.5 °C

c(aluminium) = 0.863 J/g°C

The specific heat capacity of aluminum according to this experiment is 0.863 J/g°C

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what is the molecular formula for a compound with the empirical formula: K2SO4 and a molecular mass of 696g​
LekaFEV [45]

Answer:

K8S4O16 or K8(SO4)4 depending on if the SO4 is supposed to represent sulfate or not

Explanation:

Find the molar mass of K2SO4 first:

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Divide the goal molar mass of 696 by the molar mass of the empirical formula:

696 / 174 = 4

This means you need to multiply everything in the empirical formula by 4:

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Estimate the enthalpy change for the combustion of one mole of acetylene, C2H2, to form carbon dioxide and water vapor. BE(C?H)
tia_tia [17]

Answer:

ΔH rxn =  -1010 kJ/molC₂H₂

Explanation:

To obtain the enthalpy change for a reaction from bond energies what we do is to make an inventory of the bonds broken and formed for the balanced chemical reaction:

C₂H₂ + 5/2O₂   ⇒   2CO₂ + H₂O

Bond Broken                                    Bonds Formed

2 C-H + 1 C≡C + 5/2 O=O                4C=O + 2 H-O

Enthalpy bonds broken:

2 mol (456 kJ/mol)+ 1 mol (962 kJ/mol) + 5/2 mol (499 kJ/mol)  =  3121.5 kJ

Enthalpy bond formed:

4 mol (802 kJ/mol) + 2 mol (462 kJ/mol) = 4132.0 kJ

ΔH rxn = H broken - H formed =  3121.5 kJ - 4132.0 kJ = - 1010  kJ (per mol C₂H₂ )

4 0
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