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LUCKY_DIMON [66]
3 years ago
6

Calculate the pHpH of a 0.10 MM solution of barium hydroxide, Ba(OH)2Ba(OH)2. Express your answer numerically using two decimal

places.
Chemistry
2 answers:
Temka [501]3 years ago
8 0

Answer:

The pH of this barium hydroxide solution is 13.30

Explanation:

Step 1: Data given

Concentration Ba(OH)2 = 0.10 M

Step 2: Calculate [OH-]

Ba(OH)2 ⇒ Ba^2+ + 2OH-

[OH-] = 2*0.10 M

[OH-] = 0.20 M

Step 3: Calculate pOH

pOH = -log[OH-]

pOH = -log(0.20)

pOH = 0.70

Step 4: Calculate pH

pH + pOH = 14

pH = 14 -pOH

pH = 14 - 0.70

pH = 13.30

The pH of this barium hydroxide solution is 13.30

Maurinko [17]3 years ago
6 0

Answer:

13.301

Explanation:

To calculate the pH of the solution, we must obtain the pOH of the solution as illustrated below:

The dissociation equation is given below

Ba(OH)2 <==> Ba^2+ + 2OH^-

Since Ba(OH)2 dissociate to produce 2moles of OH^-, the concentration of OH^- = 2x0.1 = 0.2M

pOH = - Log[OH^-]

pOH = - Log 0.2

pOH = 0.699

But

pH + pOH = 14

pH = 14 — pOH

pH = 14 — 0.699

pH = 13.301

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GarryVolchara [31]

Answer: 0.35 m NaCl > 0.20 m MgCl_2 >0.10 m Na3PO_4 > 0.15 m CH_3COOH > 0.15 m C_6H_{12}O_6

Explanation:

Depression in freezing point:

T_f^0-T_f=i\times k_b\times m

where,

T_f= freezing point of solution

T^o_f = freezing point of solvent

k_f = freezing point constant

m = molality

1. For 0.10 m Na3PO_4

Na_3PO_4\rightarrow 3Na^++PO_4^{3-}

, i= 4 as it is a electrolyte and dissociate to give 4 ions. and concentration of ions will be 4\times 0.10=0.40

2. For 0.35 m NaCl

NaCl\rightarrow Na^++Cl^-

, i= 2 as it is a electrolyte and dissociate to give 2 ions. and concentration of ions will be 2\times 0.35=0.70 

3. For 0.20 m MgCl_2

MgCl_2\rightarrow Mg^{2+}+2Cl^-

, i= 3 as it is a electrolyte and dissociate to give 3 ions. and concentration of ions will be 3\times 0.20=0.60 

4. For 0.15 m C_6H_{12}O_6

, i= 1 as it is a non electrolyte and does not dissociate to give ions.

5. For 0.15 m CH_3COOH

CH_3COOH\rightarrow CH_3COO^-+H^+

, i= 2 as it is a electrolyte and dissociate to give 2 ions and thus have 2\times 0.15=0.30 

As concentration is highest for 0.35 m NaCl , freezing point depression will be highest and thus has lowest freezing point. As concentration is lowest for 0.15 m C_6H_{12}O_6 , freezing point depression will be lowest and thus has highest freezing point

8 0
4 years ago
Formaldehyde (h2co) reacts with oxygen to form co2 and h2o. how many moles of co2 will be produced from reacting 2 moles of h2co
bogdanovich [222]
Answer:
            2 mol of CO₂

Solution:

The reaction is as follow,

                              H₂CO  +  O₂     →     CO₂  +  H₂O

According to this equation,

                         1 mole of H₂CO produces  =  1 mole of CO₂
So,
                   2 moles of H₂CO will produce  =  X moles of CO₂

Solving for X,
                      X  =  (2 mol × 1 mol) ÷ 1 mol

                      X  =  2 mol of CO₂
7 0
3 years ago
Read 2 more answers
N2 (g) + 2O2(g) = 2NO2 (g) ΔH = 66.4 kJ 2NO (g) + O2 (g) = 2NO2 (g) ΔH = -114.2 kJ the enthalpy of the reaction of the nitrogen
RSB [31]

Answer:

ΔH  = 180.6 kJ

Explanation:

Given that:

N2 (g) + 2O2(g) = 2NO2 (g)           ΔH = 66.4 kJ

<u>2NO (g) + O2 (g) = 2NO2 (g)         ΔH = -114.2 kJ                     </u>

N2 (g) + O2 (g) = 2NO (g)              ΔH  = ????

The subtraction of both equations would yield the unknown ΔH , therefore:

ΔH = 66.4 - ( - 114.2 kJ)

ΔH  = 180.6 kJ

3 0
3 years ago
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1a) consider the mixed aldol condensation reaction of 1-methylcyclopentane-1-carbaldehyde (shown below) and 3,3-dimethyl-2-butan
zubka84 [21]
Aldehydes and ketones having α-hydrogen atoms, undergoes aldol condensation, in present of base (NaOH). 

The initial product formed during this reaction is β-hydroxy alcohol, which then undergoes dehydration to form α,β-unsaturated aldehyde or ketone.

In present case, 3,3-dimethyl-2-butanone  has 2α-hyrogen atom, while methylcyclopentane-1-carbaldehyde has 1α-hydrogen atom. So the major product formed during cross aldol condensation reaction of these reactants is:
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The complete reaction product formed is shown below. 

7 0
4 years ago
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tigry1 [53]

Answer:

The Percentage concentration of acetic acid = 1.39 %

Explanation:

Density of acetic acid solution = 1.05 g/mL

Volume of acetic acid solution = 0.1 L = 100 mL

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Molar mass of acetic acid, CH₃COOH = (12 × 2 + 1 ×4 + 16 ×2) = 60 g/mol

From the formula, mass concentration = molar concentration × molar mass

Mass concentration of acetic acid, CH₃COOH = 0.243 mol/L × 60 g/mol = 14.58 g/L

In one liter of acetic acid solution, there are 14.58 g of acetic acid. Therefore, in 0.1 L, there will be 14.58 × 0.1 = 1.458 g of acetic acid.

Percentage concentration of acetic acid = mass of acetic acid / mass of acetic acid solution × 100%

Percentage concentration of acetic acid = (1.458 / 105) × 100% = 1.39 %

The Percentage concentration of acetic acid = 1.39 %

8 0
3 years ago
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