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3 years ago

What is the molarity of a NaCl solution containing 9.0 moles of NaCl in 3.0 L of solution?

Chemistry

1 answer:

ss7ja [257]3 years ago

4 0

Answer:

3 M

Explanation:

Molarity equation: M = n/v

n = moles of solute

v = liters of solution

9 moles of NaCl / 3 L

9/3 = 3 M

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Picture a neutral P atom. This neutral atom will have

aleksandr82 [10.1K]

Answer:

It will have 5 valence electrons as its in group 5.

The atom will gain as its closer to the full configuration 8.

The charge will be 3- as it will gain 3 electrons.

3 0

2 years ago

What are the half equations that exist in thus reaction: 4Na + O2 --->2Na2O .

givi [52]

4Na +2e- ---->2Na2
O2 +2e- ------>2O

8 0

3 years ago

The enthalpy of combustion of hard coal averages -35 kJ/g, that of gasoline, 1.28 x 105 kJ/gal. How many kilograms of hard coal

Ganezh [65]

Answer:

3.657 kg

Explanation:

Given:

Enthalpy of combustion of hard coal = -35 kJ/g

Enthalpy of combustion of gasoline = 1.28 × 10⁵ kJ/gal

Density of gasoline = 0.692 g/mL

now,

The heat provide 1 gallon of gasoline provide = 1.28 × 10⁵ kJ

and,

heat provided by the 1 gram of coal = 35 kJ

1 kJ of heat is provided by (1/35) gram of hard coal

therefore,

For 1.28 × 10⁵ kJ of heat, mass of hard coal = 1.28 × 10⁵ kJ × (1 / 35)

For 1.28 × 10⁵ kJ of heat, mass of hard coal = 3657.14 grams = 3.657 kg

3 0

3 years ago

In the reaction: CH3COO-+NH4+----------CH3COOH+NH3 what is the reactant acid and its conjugate base?

WINSTONCH [101]

Answer : The reactant acid and conjugate base in this reaction is, $NH_4^$ and $NH_3$ .

Explanation :

Conjugate acid : A species that is formed by receiving of a proton $(H^+)$ by a base is known as conjugate acid.

Conjugate base : A species that is formed by donating of a proton by an acid is known as conjugate base.

The given chemical reaction is,

$CH_3COO^-+NH_4^+\rightleftharpoons CH_3COOH+NH_3$

In this reaction, $CH_3COO^-$ (base) react with $NH_4^$ (acid) to give $CH_3COOH$ (conjugate acid) and $NH_3$ (conjugate base).

Therefore, the reactant acid and conjugate base in this reaction is, $NH_4^$ and $NH_3$ .

8 0

3 years ago

Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.150 M acetic acid (Ka = 1.75x10-5) with 0.150

mylen [45]

Answer:

a) pH = 2.793

b) pH = 4.280

c) pH = 4.933

d) pH = 8.816

e) pH = 8.861

f) pH = 8.891

Explanation:

a) VNaOH = 0 mL

∴ CH3COOH ↔ CHECOO- + H3O+

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

mass balance:

⇒ C CH3COOH = [ CH3COO- ] + [ CH3COOH ] = 0.150 M

charge balance:

⇒ [ H3O+ ] = [ CH3COO- ]

⇒ Ka = 1.75 E-5 = [ H3O+ ]² / ( 0.150 M - [ H3O+ ] )

⇒ [ H3O+ ]² + 1.75 E-5 [ H3O+ ] - 2.625 E-6 = 0

⇒ [ H3O+ ] = 1.61146 E-3 M

⇒ pH = - Log [ H3O+ ] = 2.793

b) after 5.0 mL NaOH:

∴ CH3COOH + NaOH ↔ CH3COONa + H2O

⇒ C NaOH = (5 E-3 L * 0.150 mol/L) / (0.025+0.01 ) = 0.02143 M

⇒ C CH3COOH = ((0.025*0.150) - (0.01*0.150)) / (0.025 + 0.01) = 0.0643 M

mass balance:

⇒ 0.02143 + 0.0643 = [ CH3COOH ] + [ CH3COO- ] = 0.086 M

charge balance:

⇒ [ H3O+ ] + [Na ] = [ CH3COO- ]

⇒ [ H3O+ ] + 0.02143 = [ CH3COO- ]

⇒ Ka = [ H3O+ ] * ( [ H3O+ ] + 0.150 ) / (0.086 - 0.02143 - [ H3O+ ]) = 1.75 E-5

⇒ [ H3O+ ]² + 0.02143 [ H3O+ ] = 1.13 E-6 - 1.75 E-5 [ H3O+ ]

⇒ [ H3O+ ]² + 0.02144 [ H3O+ ] - 1.13 E-6 = 0

⇒ [ H3O+ ] = 5.26 E-5 M

⇒ pH = 4.28

c) after 15 mL NaOH:

⇒ C CH3COOH = 0.0375 M

⇒ C NaOH = 0.05625 M

mass balance:

⇒ 0.09375 M = [ CH3COO- ] +[ CH3COOH ]

charge balance:

⇒ [ H3O+ ] + 0.05625 = [ CH3COO- ]

⇒ Ka = 1.75 E-5 = [ H3O+ ] * ([ H3O+ ] + 0.05625) / (0.09375 - 0.05625 - [H3O+])

⇒ [H3O+]² + 0.05625[H3O+] = 6.5625 E-7 - 1.75 E-5 [H3O+]

⇒ [ H3O+]² + 0.05626[H3O+] - 6.5625 E-7 = 0

⇒ [ H3O+ ] = 1.1662 E-5 M

⇒ pH = 4.933

d) after 25 mL NaOH:

⇒ C NaOH = 0.075 M

⇒ C CH3COOH = 0 M....equiv. point

⇒Kh = Kw/Ka = 1 E-14 / 1.75 E-8 = 5.7143 E-10 = [ OH-]² / ( 0.075 - [OH-])

⇒ [OH-]² + 5.7143 E-10[OH-] - 4.286 E-11 = 0

⇒ [ OH- ] = 6.5463 E-6 M

⇒ pOH = 5.184

⇒ pH = 8.816

e) after 40 mL NaOH:

⇒ C NaOH = 0.0923 M

⇒ [OH-]² + 5.7143 E-10 [OH-] - 5.275 E-11 = 0

⇒ [OH-] = 7.2624 E-6 M

⇒ pOH = 5.139

⇒ pH = 8.861

f) after 60 mL NaOH:

⇒ C NaOH = 0.106 M

⇒ [OH-]² + 5.7143 E-10 [OH-] - 6.05 E-11 = 0

⇒ [OH-] = 7.7782 E-6 M

⇒ pOH = 5.11

⇒ pH = 8.891

5 0

3 years ago