Where am I missing an electron?

What heat is liberated in the formation of 10.0 grams of sulfur hexafluoride, SF6, from the elements sulfur and fluorine?

tensa zangetsu [6.8K]

Answer:

$Q=-76.7kJ$

Explanation:

Hello,

In this case, for such formation of sulfur hexafluoride, the standard enthalpy of formation is -1220.47 kJ/mol (data extracted from NIST database). Next, we compute the moles in 10.0 grams of sulfur hexafluoride as shown below:

$n_{SF_6}=10.0gSF_6*\frac{1molSF_6}{146.06 gSF_6}=0.0685mol$

Next, for the given energy, we compute the total heat that is liberated:

$Q=-1220.47\frac{kJ}{mol}*0.0685 mol\\\\Q=-76.7kJ$

Finally, we conclude such symbol has sense since negative heat is related with liberated heat.

Best regards.

8 0

3 years ago

A 25.0 mL solution of 0.100 M CH3COOH is titrated with a 0.200 M KOH solution. Calculate the pH after the following additions of

zaharov [31]

Answer:

a) pH = 2.88

b) pH = 4.598

c) pH = 5.503

d) pH = 8.788

e) pH = 12.097

Explanation:

CH3COOH ↔ CH3COO- + H3O+

∴ Ka = 1.75 E-5 = [H3O+]*[CH3COO-] / [CH3COOH]

a) 0.0 mL KOH:

mass balance:

⇒ C CH3COOH = [CH3COOH] + [CH3COO-] = 0.100 M

charge balance:

⇒ [H3O+] = [CH3COO-]

⇒ 1.75 E-5 = [H3O+]²/(0.100 - [H3O+])

⇒ [H3O+]² + 1.75 E-5[H3O+] - 1.75 E-6 = 0

⇒ [H3O+] = 1.314 E.3 M

∴ pH = - Log [H3O+]

⇒ pH = 2.88

b) 5.0 mL KOH:

CH3COOH + KOH ↔ CH3COONa + H2O

∴ C CH3COOH = ((0.025)(0.100) - (5 E-3)(0.200))/(0.025+5 E-3)

⇒ C CH3COOH = 0.05 M

∴ C KOH = ((5 E-3)(0,200))/(0.025+5 E-3) = 0.033 M

mass balance:

⇒ C CH3COOH + C KOH = [CH3COOH] + [CH3COO-] = 0.05 + 0.033 = 0.083 M

charge balance:

⇒ [H3O+] + [K+] = [CH3COO-]

⇒ [CH3COO-] = [H3O+] + 0.033

⇒ 1.75 E-5 = ([H3O+]*([H3O+] + 0.033))/(0.083 - ([H3O+] + 0.033))

⇒ 1.75 E-3 = ([H3O+]² + 0.033[H3O+])/(0.05 - [H3O+])

⇒ 8.75 E-7 - 1.75 E-5[H3O+] = [H3O+]² + 0.033[H3O+]

⇒ [H3O+]² +0.03302[H3O+] - 8.75 E-7 = 0

⇒ [H3O+] = 2.523 E-5 M

⇒ pH = 4.598

equivalent point:

(C*V)acid = (C*V)base

⇒ (0.100 M)*(0.025 L) = (0.200 M)( Vbase)

⇒ Vbase = 0.0125L = 12.5 mL

c) 10.0 mL KOH:

∴ C CH3COOH = 0.0143 M

∴ C KOH = 0.057 M

as in the previous point, starting from the mass and charge balances, we obtain:

⇒ [H3O+] = 3.1386 E-6 M

⇒ pH = 5.503

d) 12.5 mL KOH:

at the equivalence point, there is complete salt formation, then the pH is calculated through the salt:

CH3COO- + H2O ↔ CH3COOH - OH-

∴ Kw/Ka = 1 E-14/1.75 E-5 = 5.714 E-10 = [CH3COOH]*[OH-]/[CH3COO-]

∴ [CH3COO-] = (0.025)(0.100))/(0.025+0.0125) = 0.066 M

mass balance:

⇒ 0.066 = [CH3COOH] + [CH3COO-]..........(1)

charge balance:

⇒ [K+] = [OH-] + [CH3COO-] = 0.066 M.........(2)

∴ [K+] = C CH3COO- = 0.066 M

(1) = (2):

⇒ [OH-] = [CH3COOH].......(3)

⇒ 5.714 E-10 = [OH-]² / (0.066 - [OH-])

⇒ [OH-]² + 5.714 E-10[OH-] - 3.7712 E-11 = 0

⇒ [OH-] = 6.1408 e-6 m

⇒ pOH = 5.212

⇒ pH = 14 - pOH = 8.788

d) 15.0 mL KOH:

after the equivalence point there is salt and excess base (OH-); ph is calculated from excess base:

⇒ C KOH = ((0.015)(0.200) - (0.025)(0.100)) / (0.025 + 0.015) = 0.0125 M

⇒ [OH-] ≅ C KOH = 0.0125 M

⇒ pOH = 1.903

⇒ pH = 12.097

8 0

2 years ago

What are the charges of families 1, 2, & 13 (boron family) if/when they become Ions

Natalka [10]

Answer:

Explanation:

Group one elements are alkali metals. All alkali metal have one valance electron. They loses their one valance electron and from cation with charge of +1.

Charges on group one.

Hydrogen = +1

Lithium = +1

Sodium = +1

Potassium = +1

Rubidium = +1

Cesium = +1

Francium = +1

Group two elements are alkaline earth metals. All alkaline earth metal have two valance electron. They loses their two valance electron and from cation with charge of +2.

Charges on group two.

Beryllium = +2

Magnesium = +2

Calcium = +2

Strontium = +2

Barium= +2

Radium = +2

Group 13 elements are boron family. All elements have three valance electrons. They loses their three valance electron and from cation with charge of +3.

Charges on group 13.

Boron = +3

Aluminium = +3

Gallium = +3

Indium = +3

Thallium= +3

Group 13 elements are also shows +1 charge by losing one valance electron.

8 0

3 years ago

The Ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. In t

Sloan [31]

Answer:

Mass = 42.8g

Explanation:

4 NH 3 ( g ) + 5 O 2 ( g ) ⟶ 4 NO ( g ) + 6 H 2 O ( g )

Observe that every 4 mole of ammonia requires 5 moles of oxygen to obtain 4 moles of Nitrogen oxide and 6 moles of water.

Step 1: Determine the balanced chemical equation for the chemical reaction.

The balanced chemical equation is already given.

Step 2: Convert all given information into moles (through the use of molar mass as a conversion factor).

Ammonia = 63.4g × 1mol / 17.031 g = 3.7226mol

Oxygen = 63.4g × 1mol / 32g = 1.9813mol

Step 3: Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.

If all of the 1.9831 moles of oxygen were to be used up, there would need to be 1.9831 × 4 / 5 or 1.5865 moles of Ammonia. We have 3.72226 moles of ammonia - Far excess. Because there is an excess of Ammonia, the Oxygen amount is used to calculate the amount of the products in the reaction.

Step 4: Use the amount of limiting reactant to calculate the amount of H2O produced.

5 moles of O2 = 6 moles of H2O

1.9831 moles = x

x = (1.9831 * 6 ) / 5

x = 2.37972 moles

Mass of H2O = Molar mass * Molar mass

Mass = 2.7972 * 18

Mass = 42.8g

6 0

3 years ago

An object that is already moving can . . .

Yuri [45]

An object that is not already moving will begin to move in the direction of the larger force. An object that is already moving will change its speed and/or its direction.

4 0

2 years ago