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3 years ago

How much is one liter

Chemistry

1 answer:

Bas_tet [7]3 years ago

8 0

Answer:

1 litre = 1000 millilitres

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Fire extinguishers that spray carbon dioxide on the fire, work very effectively because it forms a blanket around the burning ma

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Answer:

I think it will option B it will retain enough heat

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3 years ago

Which one is an interjection? boo! running people?

12345 [234]

Boo! Is the interjection

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3 years ago

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 you were asked to convert 4.2 × 1022 atoms of aluminum to moles, which of the following should you use for the conversion?

DanielleElmas [232]

4.2×1022NA N_{A} is the avagadro number

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3 years ago

If a pork roast must absorb

Black_prince [1.1K]

Let us assume propane was the fuel
C3H8(g) + 5O2(g) ---> 3CO2(g) + 4H2O(g) = 2217kJ
1 mole ofpropane produces 3 moles of CO2
heat absorbed by pork = 0.11 x 2217
= 243.87 kJ/mol
number of moles of propane = 1700kJ / 243.87 kJ/mol
= 6.971 moles
1 mole of C3H8 = 3 moles ofCO2
6.971 moles of C3H8 = ?
3 x 6.971 = 20.913 moles of CO2
Convert to grams
mass = MW x mole
= 44 x 20.913
= 920.172g of CO2 emitted

7 0

3 years ago