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3 years ago

What type of base is defined as a substance that forms hydroxide ions (OH-) in water?

Chemistry

2 answers:

jok3333 [9.3K]3 years ago

7 0

Answer:

B:Arrhenius base

Explanation:

Acccording to Svante Arrhenius a base is a substance the dissociates in water to form hydroxide ion.In other words a base is that substance which when dissolved in water,increases the concentration of hydroxide ion.The examples of Arrhenius bases are given below;

NaOH ,KOH,LiOH etc

Dissociation of NaOH in aqueous solution is given as;

NaOH=>Na+ + OH-

skad [1K]3 years ago

3 0

Answer:

A.Brønsted-Lowry base

Explanation:

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kherson [118]

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8 0

2 years ago

A blacksmith heated an iron bar to 1445 °C. The blacksmith then tempered the metal by dropping it into 42,800 mL of

Wittaler [7]

Answer:

6626 g

Explanation:

Given that:

Density of water = 1.00 g/ml, volume of water = 42800 ml.

Since density = mass/ volume

mass of water = volume of water * density of water = 42800 ml * 1 g/ml = 42800 g

Initial temperature of water = 22°C and final temperature of water = 45°C.

specific heat capacity for water = 4.184 J/g°C

ΔT water = 45 - 22 = 23°C

For iron:

mass = m,

specific heat capacity for iron = 0.444 J/g°C

Initial temperature of iron = 1445°C and final temperature of water = 45°C.

ΔT iron = 45 - 1445 = -1400°C

Quantity of heat (Q) to raised the temperature of a body is given as:

Q = mCΔT

The quantity of heat required to raise the temperature of water is equal to the temperature loss by the iron.

Q water (gain) + Q iron (loss) = 0

Q water = - Q iron

42800 g × 4.184 J/g°C × 23°C = -m × 0.444 J/g°C × -1400°C

m = 4118729.6/621.6

m = 6626 g

8 0

3 years ago

What is the pH of a 3.40 mM of acetic acid, 500 mL in which 50 mL of 1.00 M NaOAc has been added? Ka HOAc is 1.8 x 10-5? What pe

Deffense [45]

Answer:

a) pH = 4.213

b) % dis = 2 %

Explanation:

Ch3COONa → CH3COO- + Na+

CH3COOH ↔ CH3COO- + H3O+

∴ Ka = 1.8 E-5 = ([ CH3COO- ] * [ H3O+ ]) / [ CH3COOH ]

mass balance:

⇒ C CH3COOH + C CH3COONa = [ CH3COOH ] + [ CH3COO- ]

∴ C CH3COOH = 3.40 mM = 3.4 mmol/mL * ( mol/1000mmol)*(1000mL/L)

∴ C CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL

⇒ [ CH3COOH ] = 4.4 - [ CH3COO- ]

charge balance:

⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH- ]....is negligible [ OH-], comes from water

⇒ [ CH3COO- ] = [ H3O+ ] + 1.00

⇒ Ka = (( [ H3O+ ] + 1 )* [ H3O+ ]) / ( 3.4 - [ H3O+])) = 1.8 E-5

⇒ [ H3O+ ]² + [ H3O+ ] = 6.12 E-5 - 1.8 E-5 [ H3O+ ]

⇒ [ H3O+ ]² + [ H3O+ ] - 6.12 E-5 = 0

⇒ [ H3O+ ] = 6.12 E-5 M

⇒ pH = - Log [ H3O+ ] = 4.213

b) (% dis)* mol acid = C CH3COOH = 3.4

∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol

⇒ % dis = 3.4 / 1.7 = 2 %

7 0

3 years ago

What is the mass of a 5.00 cm^3 piece of copper having a density of 8.96 g/cm^3

erik [133]

Answer:

44.8 g

Explanation:

Density = mass / Volume

Mass = density x Volume = 8.96x 5 = 44.8 g

5 0

3 years ago

Calculate the ph of a 0.20 m solution of kcn at 25.0 ∘c. express the ph numerically using two decimal places

storchak [24]

The pH of the solution is basically the negative logarithm of the concentration of hydrogen ions or H+. In equation form, pH = -log[H+]. It could also be in terms of the concentration of hydroxide ions or OH- as pOH, where pOH = -log[OH-]. The sum of pH and pOH is 14. These are the important equations to know when it comes to equilibrium pH problems.

KCN is a basic salt coming from the reaction of a weak acid, HCN, and a strong base, KOH. In the hydrolysis of KCN, only the strong conjugate base (SCB) is involved. Since HCN is the weak acid, the SCB is CN-. The reaction would be

CN- + H2O ⇔ HCN + OH-

The important data is the equilibrium constant of acidity of the weak acid. Ka for HCN is 6.2×10^-10. Then, let's do the ICE(Initial-Change-Equilibrium) analysis.

CN- + H2O ⇔ HCN + OH-

I 0.2 m ∞ 0 0
C -x ∞ +x +x
-----------------------------------------------------------
E 0.2-x +x +x

The value x denotes the number of moles CN- reacted. There is no value for H2O because the solution is dilute such that H2O>>>CN-. Then, we apply the ratio:

$K_{H} = \frac{ K_{W} }{ K_{A} } = \frac{[HCN][OH-]}{[CN-]}$

where K,H is the equilibrium constant of hydrolysis and Kw is equilibrium constant for water solvation which is equal to 1×10^-14. Therefore,

$K_{H} = \frac{ 1x10^-14}{ 6.2x10^-10 } = \frac{[X][X]}{[0.2-X]}$

x = 0.001788 m
Since the value of OH- is also x, then OH-=0.001788 m. Consequently,

pOH = -log(0.001788) = 2.75
pH = 14 - pOH = 14 - 2.75
pH = 11.25

4 0

3 years ago

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