WILL GIVE BRAINLIEST!!!!!!!!!!!<br><br><br>What are 4 examples of physical properties?<br>

Propane gas, C3H8, is sometimes used as a fuel. In order to measure its energy output as a fuel a 1.860 g sample was combined wi

lisov135 [29]

Answer:

The heat of the reaction is 105.308 kJ/mol.

Explanation:

Let the heat released during reaction be q.

Heat gained by water: Q

Mass of water ,m= 1kg = 1000 g

Heat capacity of water ,c= 4.184 J/g°C

Change in temperature = ΔT = 26.061°C - 25.000°C=1.061 °C

Q=mcΔT

Heat gained by bomb calorimeter =Q'

Heat capacity of bomb calorimeter ,C= 4.643 J/g°C

Change in temperature = ΔT'= ΔT= 26.061°C - 25.000°C=1.061 °C

Q'=CΔT'=CΔT

Total heat released during reaction is equal to total heat gained by water and bomb calorimeter.

q= -(Q+Q')

q = -mcΔT - CΔT=-ΔT(mc+C)

$q=-1.061^oC(1000 g\times 4.184J/g^oC+4.643 J/^oC )=-4,444.15J=-4.444 kJ$

Moles of propane = $\frac{1.860 g}{44 g/mol}=0.0422 mol$

0.0422 moles of propane on reaction with oxygen releases 4.444 kJ of heat.

The heat of the reaction will be:

$\frac{4.444 kJ}{0.0422 mol}=105.308 kJ/mol$

3 0

3 years ago

True or False, Substances have different properties?

4vir4ik [10]

True.....................................

7 0

3 years ago

Read 2 more answers

Calculate the values of ΔU, ΔH, and ΔS for the following process:

ladessa [460]

Answer:

ΔU = 45.814 KJ

ΔH = 46.4375 KJ

ΔS = 18.76 J/K

Explanation:

H2O(l) → H2O(l) → H2O(steam)

298.15K, 1atm ΔHp 373.15K,1atm ΔHv 373.15K,1 atm

∴ ΔHp = Qp = nCpΔT

∴ n H2O = 1 mol

∴ Cp,n = 75.3 J/mol.K

∴ ΔT = 373.25 - 298.15 = 75 K

⇒ Qp = (1 mol)*(75.3 J/mol.K)*(75K) = 5647.5 J

⇒ ΔHp = 5647.5 J = 5.6475 KJ

⇒ ΔH = ΔHp + ΔHv

∴ ΔHv = 40.79 KJ/mol * 1 mol = 40.79 KJ

⇒ ΔH = 5.6475 KJ + 40.79 KJ = 46.4375 KJ

ideal gas:

∴ ΔH = ΔU + PΔV

∴ V1 = nRT1/P1 = ((1)*(0.082)*(298.15))/1 = 24.45 L

∴ V2 = nRT2/P2 = ((1)*(0.082)*(373.15))/ 1 = 30.59 L

⇒ ΔV = V2 - V1 = 6.15 L * (m³/1000L) = 6.15 E-3 m³

∴ P = 1 atm * (Pa/ 9.86923 E-6 atm) = 101325.027 Pa

⇒ ΔU = ΔH - PΔV = 46.4375 KJ - ((101325.027 Pa*6.15 E-3m³)*(KJ/1000J))

⇒ ΔU = 46.4375 KJ - 0.623 KJ

⇒ ΔU = 45.814 KJ

∴ ΔS = Cv,n Ln (T2/T1) + nR Ln (V2/V1)

⇒ ΔS = (75.3) Ln(373.15/298.15) + (1)*(8.314) Ln (30.59/24.45)

⇒ ΔS = 16.896 J/K + 1.863 J/K

⇒ ΔS = 18.76 J/K

3 0

4 years ago

Hat volume of a 0.540 M NaOH solution contains 12.5 g NaOH

amid [387]

Answer:

At 0.58 L of 0.540 M NaOH solution contain 12.5 g NaOH.

Explanation:

Given data:

At volume = ?

Mass of NaOH = 12.5 g

Molarity of solution = 0.540 M

Solution:

First of all we will calculate the number of moles of sodium hydroxide.

Number of moles = mass/molar mass

Number of moles = 12.5 g / 40 g/mol

Number of moles = 0.3125 mol

Volume of NaOH:

Molarity = number of moles / volume in L

Now we will put the values.

0.540 M = 0.3125 mol / volume in L

volume in L = 0.3125 mol / 0.540 mol/L

volume in L = 0.58 L

5 0

3 years ago

How many minutes are in 180,0 days?

iris [78.8K]

Answer:

259,200 minutes

Explanation:

4 0

3 years ago

WILL GIVE BRAINLIEST!!!!!!!!!!!What are 4 examples of physical properties?​

WILL GIVE BRAINLIEST!!!!!!!!!!!What are 4 examples of physical properties?