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vampirchik [111]
2 years ago
11

A basic solution is 1. 35×10−5m in calcium hydroxide, ca(oh)2. what is the ph of the solution at 25. 0∘c?

Chemistry
1 answer:
Andreyy892 years ago
8 0

Slaked lime, often known as calcium hydroxide, is an inorganic substance having the chemical formula Ca(OH)2. When water is combined with quicklime (calcium oxide), a colorless crystal or white powder is created.

Step:-1,(a (OH)₂ →) (9²+ (aq) +201 (az)-5 of C₂₂ (01) ₂ = 1.35 x 10²5 m0

OH = 2 X Concentration of Ca(OH)₂

Concentration of = 2 X 1.35 X 105 m

Step 2- pOH = -log [OH-]=2= 2.7x 10 log [2.7 x 105]- log 3.7 - (-5) log 2.7 + 5×1

0.9313 +5

5.4313

stepз- PH +poH =14

рн = 14-рон

= 14-5.4313

=8.5687

Learn more about basic solution brainly.com/question/18720954

#SPJ4

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Answer: Theoretical Yield = 0.2952 g

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Explanation:

Calculation of limiting reactant:

n-trans-cinnamic acid moles = (142mg/1000) / 148.16 = 9.584*10⁻⁴ mol

pyridium tribromide moles = (412mg/1000) / 319.82= 1.288*10⁻³ mol

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The molar ratio according to the equation mentioned is equals to 1:1

The brominated product moles is also = 9.584*10⁻⁴ mol

Theoretical yield = (9.584*10⁻⁴ mol) * (Mr of brominated product)

                             =  (9.584*10⁻⁴ mol) * (307.97) = 0.2952 g

Percentage Yield is : Actual Yield / Theoretical Yield = 0.2223/0.2952

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Yet a third pair of compounds of manganese and oxygen is 50.48% and 36.81% oxygen respectively. In what small whole number ratio
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Answer:

The number ratio is 4:7

Explanation:

Step 1: Data given

Compound 1 has 50.48 % oxygen

Compound 2 has 36.81 % oxygen

Molar mass oxygen = 16 g/mol

Molar mass manganese = 54.94 g/mol

Step 2: Calculate % manganes

Compound 1: 100 - 50.48 = 49.52 %

Compound 2: 100 - 36.81 = 63.19 %

Step 3: Calculate mass

Suppose mass of compounds = 100 grams

Compound 1:

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 49.52 % Mn = 49.52 grams

Compound 2:

36.81 % O = 36.81 grams

63.19 % Mn = 63.19 grams

Step 4: Calculate moles

Compound 1

Moles O = 50.48 grams / 16.0 g/mol = 3.155 moles

Moles Mn = 49.52 grams / 54.94 g/mol = 0.9013 moles

Compound 2

Moles O = 36.81 grams / 16.0 g/mol = 2.301 moles

Moles Mn = 63.19 grams / 54.94 g/mol = 1.150 moles

Step 5: calculate mol ratio

We will divide by the smallest amount of moles

Compound 1

O: 3.155/0.9013 = 3.5

Mn: 0.9013 / 0.9013 = 1

Mn2O7

Compound 2

O: 2.301 / 1.150 = 2

Mn: 1.150 / 1.150 = 1

MnO2

The number ratio is 2:3.5 or 4:7

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