PLEASE HELP!!!! number 9 and 10

I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 mol

galina1969 [7]

Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

Enthalpy of formation of iodine gas = $\Delta H_f_{(I_2)}=62.438 kJ/mol$

Enthalpy of formation of chlorine gas = $\Delta H_f_{(Cl_2)}=0 kJ/mol$

Enthalpy of formation of ICl gas = $\Delta H_f_{(ICl)}=17.78 kJ/mol$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?$

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

$\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]$

$=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol$

Enthaply change when 1.62 moles of iodine gas recast:

$\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ$

Entropy of the surrounding = $\Delta S^o_{surr}=\frac{\Delta H}{T}$

$=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K$

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

4 0

3 years ago

If this is a p1000 micropipette, then this is set to dispense [ Select]ul. If this is a p10

mariarad [96]

Answer:

1000 µL; 10 µL

Explanation:

A p1000 micropipet is set to dispense 1000 µL.

A p10 micropipet set to dispense 10 µL.

3 0

4 years ago

james needs 4.20 moles of nabr for an experiment. how many grams of nabr must he measure out to have 4.20 moles?

kondor19780726 [428]

75.7 grams.
Convert moles to grams using this:

Moles • grams/1 mole = grams

(Grams is over 1 mole)

6 0

3 years ago

Read 2 more answers

Branches of chemistry with definition and examples in real life

nataly862011 [7]

Explanation:The five major branches of chemistry are organic, inorganic, analytical, physical, and biochemistry.

...

Sub-branches of physical chemistry include:

Photochemistry — the study of the chemical changes caused by light.

Surface chemistry — the study of chemical reactions at surfaces of substances

8 0

3 years ago

In Part A, you found the amount of product (1.80 mol P2O5 ) formed from the given amount of phosphorus and excess oxygen. In Par

Finger [1]

First step is to balance the reaction equation. Hence we get P4 + 5 O2 => 2 P2O5

Second, we calculate the amounts we start with

P4: 112 g = 112 g/ 124 g/mol – 0.903 mol

O2: 112 g = 112 g / 32 g/mol = 3.5 mol

Lastly, we calculate the amount of P2O5 produced.

2.5 mol of O2 will react with 0.7 mol of P2O5 to produce 1.4 mol of P2O5.

This is 1.4 * (31*2 + 16*5) = 198.8 g

3 0

4 years ago