Convection currents are caused by differences in what 2 things?

Consider the reactionI2(g) + Cl2(g)2ICl(g)Using standard thermodynamic data at 298K, calculate the entropy change for the surrou

Georgia [21]

We know,

$\Delta H_{I_2(g)}=62.438\ KJ/mol\\\\\Delta H_{Cl_2(g)}= 0.0\ KJ/mol\\\\\Delta H_{ICl(g)}=17.78\ KJ/mol$

For given reaction, $I_2(g)+Cl_2(g)\ -->\ 2ICl(g)$

$\Delta H_{rxn}=2\Delta H_{ICl(g)}-\Delta H_{I_2(g)}-\Delta H_{Cl_2(g)}\\\\\Delta H_{rxn}=2(17.78)-0-62.438\ KJ/mol\\\\\Delta H_{rxn}=-26.878\ KJ/mol$

For , 2.41 moles of $I_2$ :

$\Delta H_{rxn}=2.41\times (-26.878)\ KJ\\\\\Delta H_{rxn}=-64.78\ KJ$

We know :

$\Delta S = -\dfrac{\Delta H_{rxn}}{T}\\\\\Delta S = -\dfrac{-64.78}{298}\ KJ/K\\\\\Delta S =-0.21738 \ KJ/K\\\\\Delta S=-217.38\ J/K$

Hence, this is the required solution.

7 0

4 years ago

In a test of an automobile engine 1.00 L of octane (702 g) is burned, but only 1.84 kg of carbon dioxide is produced. What is th

Reptile [31]

Answer:

The % yield of CO2 is 85.05 %

Explanation:

Step 1: Data given

Mass of octane = 702 grams

Molar mass octane = 114.23 g/mol

Mass CO2 =1.84 kg = 1840 grams

Molar mass of CO2

Step 2: The balanced equation

2C8H18 + 25O2 → 16CO2 + 18H2O

Step 3: Calculate moles of octane

Moles octane = mass octane / molar mass octane

Moles octane = 702.0 grams / 114.23 g/mol

Moles octane = 6.145 moles

Step 4: Calculate moles of CO2

For 2 moles octane we need 25 moles O2 to produce 16 moles CO2 and 18 moles H2O

For 6.145 moles octane we'll have 8*6.145 moles =49.16 moles

Step 5: Calculate mass of CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 49.16 moles * 44.01 g/mol

Mass CO2 = 2163.5 grams

Step 6: Calculate % yield of carbon dioxide

% yield = (actual yield / theoretical yield)*100%

% yield = (1840/2163.5)*100%

% yield = 85.05 %

The % yield of CO2 is 85.05 %

4 0

3 years ago

Which element has the strongest attraction for electrons?

NemiM [27]

Answer:

The answer is Flourine

3 0

3 years ago

Calculate the number of kilojoules to warm 125 g of iron from 23.5 °C to 78.0 °C.

insens350 [35]

Given:
Iron, 125 grams T
1 = 23.5 degrees Celsius, T2 = 78 degrees Celsius.

Required:
Heat produced in kilojoules

Solution:
The molar mass of iron is 55.8 grams per mole. SO we need to change the given mass of iron into moles.

Number of moles of iron = 125 g/(55.8 g/mol) = 2.24 moles

Q (heat) = nRT = nR(T2 = T1)
Q (heat) = 2.24 moles (8.314 Joules per mol degrees Celsius) (78.0 degrees Celsius – 23.5 degrees Celsius)
Q (heat) = 1014.97 Joules or 1.015 kilojoules This is the amount of heat produced in warming 125 g f iron.

7 0

3 years ago

I have a few questions from my homework and I was wondering if I can get some help

sashaice [31]

Well it breaks down into small parts

3 0

3 years ago