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3 years ago

Convert 34.59cm3 to cubic inches (in3) given that 1 inch = 2.54cm

Chemistry

1 answer:

JulsSmile [24]3 years ago

8 0

Answer:

$2.111in^3$

Explanation:

Hello,

In this case, we are able to perform this unit conversion by cubing the relationship between inches and centimetres as shown below:

$=34.59cm^3*(\frac{1in}{2.54cm})^3 \\\\=34.59cm^3*(\frac{1in^3}{16.39cm^3} )\\\\=2.111in^3$

Best regards.

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For the following reaction at 600. K, the equilibrium constant, Kp, is 11.5. PCl5(g) equilibrium reaction arrow PCl3(g) + Cl2(g)

Lubov Fominskaja [6]

Answer:

a) pPCl5 = 0.856 atm

b)pPCl5 = 0.0557 atm

pCl2 = pCl3 = 0.800 atm

c) Ptotal = 1.66 atm

d) 93.5

Explanation:

Step 1: Data given

Temperature = 600 K

Kp = 11.5

Mass of PCl5 = 2.010 grams

Volume of the bulb = 555 mL = 0.555 L

The bulb is heated to 600 K

Step 2: The balanced equation

PCl5(g) ⇄ PCl3(g) + Cl2(g)

Step 3:

a) pv = nrt

⇒with p = the pressure = TO BE DETERMINED

⇒with V = the volume = 0.555 L

⇒ with n =the number of moles PCl5 = 2.010 grams / 208.24 g/mol = 0.00965 moles

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 600K

p = (0.00965 *0.08206*600)/0.555

pPCl5 = 0.856 atm

The initial pressures

pPCl5 = 0.856 atm

pCl2 = pCl3 = 0 atm

For 1 mol PCl5 we'll have PCl3 and 1 mol Cl2

The pressure at the equilibrium

pPCl5 = (0.856 -x) atm

pCl2 = pCl3 = x atm

Kp = pPCl3 * pCl2/pPCl5

11.5 = x*x / (0.856 - x)

11.5 = x²/(0.856- x)

x = 0.8003

pPCl5 = (0.856 -x) atm = 0.0557 atm

pCl2 = pCl3 = x atm = 0.800 atm

c) Since x = 0.8003 and PCl3 and PCl2 are x

Ptotal = 0.8003 + 0.8003 +0.0557 = 1.66 atm

The degree of dissociation = (x / initial pressure PCl5)

(0.8003/0.856) * 100 = 93.5

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4 years ago

What is the resultant pressure if 1.7 mol of ideal gas at 273 K and 2.79 atm in a closed container of constant volume is heated

dedylja [7]

Answer: The resultant pressure is 3.22 atm

Explanation:

Gay-Lussac's Law: This law states that pressure is directly proportional to the temperature of the gas at constant volume and number of moles.

$P\propto T$ (At constant volume and number of moles)

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 2.79 atm

$P_2$ = final pressure of gas = ?

$T_1$ = initial temperature of gas = 273K

$T_2$ = final temperature of gas = 315 K

$\frac{2.79}{273}=\frac{P_2}{315}$

$P_2=3.22atm$

Thus the resultant pressure is 3.22 atm

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3 years ago

A 1.00 g sample of octane (C8H18) is burned in a bomb calorimeter with a heat capacity of 837J∘C that holds 1200. g of water at

lubasha [3.4K]

Answer:

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

Explanation:

Step 1: Data given

Mass of octane = 1.00 grams

Heat capacity of calorimeter = 837 J/°C

Mass of water = 1200 grams

Temperature of water = 25.0°C

Final temperature : 33.2 °C

Step 2: Calculate heat absorbed by the calorimeter

q = c*ΔT

⇒ with c = the heat capacity of the calorimeter = 837 J/°C

⇒ with ΔT = The change of temperature = T2 - T1 = 33.2 - 25.0 : 8.2 °C

q = 837 * 8.2 = 6863.4 J

Step 3: Calculate heat absorbed by the water

q = m*c*ΔT

⇒ m = the mass of the water = 1200 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = The change in temperature = T2 - T1 = 33.2 - 25 = 8.2 °C

q = 1200 * 4.184 * 8.2 = 41170.56 J

Step 4: Calculate the total heat

qcalorimeter + qwater = 6863.4 + 41170. 56 = 48033.96 J = 48 kJ

Since this is an exothermic reaction, there is heat released. q is positive but ΔH is negative.

Step 5: Calculate moles of octane

Moles octane = 1.00 gram / 114.23 g/mol

Moles octane = 0.00875 moles

Step 6: Calculate heat combustion for 1.00 mol of octane

ΔH = -48 kJ / 0.00875 moles

ΔH = -5485.7 kJ/mol

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

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