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nika2105 [10]
3 years ago
8

What is basicity in acids?​

Chemistry
1 answer:
timama [110]3 years ago
6 0
Basicity of an acid is the number of hydrogen ions which can be produced by one molecule of an acid
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What will be the new volume if 125 mL of He gas at 100 degree Celsius and .979 atm is cooled to 26 degree Celsius and the pressu
ycow [4]
Use Boyle's Law of Pressure: P1 x V1 = P2 x V2. Givens: P1=0.9 atm V1=                4 P2= 0.9 atm Find: V2 Equation: 0.9 atm x 4 x 4 L = 0.20 atm x V2Solve: 36 atmL= 0.20 atm x V2 18 : = V2 Short answer: The new volume is 104 ml. 
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3 years ago
Mannoheptulose is a sugar found in avocados. If each C-C bond contains 76 kcal of energy and each C-H bond contains 91 kcal, how
nadya68 [22]

Answer:

Total Kcal energy produced in the catabolism of mannoheptulose = 1184 Kcal

Explanation:

The molecular formula of mannoheptulose is C₇H₁₄O₇.

The structure is as shown in the attachment below.

Number of C-C bonds present in mannoheptulose = 6

Number of C-H bonds present in mannoheptulose = 8

Since the each C-C bond contains 76 Kcal of energy,

Amount of energy present in six C-C bonds = 6 * 76 = 456 Kcal

Also, since each C-H bond contains 91 Kcal of energy;

amount of energy present in eight C-H bonds = 8 * 91 = 728 Kcal

Total Kcal energy produced in the catabolism of mannoheptulose = 456 + 728 = 1184 Kcal

6 0
3 years ago
A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 23.0 mL of
Georgia [21]

NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa 

NaOH + CH3COOH → CH3COONa + H2O 

Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH 

Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH 

These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L 

Molarity of CH3COOH = 0.0106/0.071 = 0.1493M 

CH3COONa = 0.0076 / 0.071 = 0.1070M 

pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74. 

pH using Henderson - Hasselbalch equation: 

pH = pKa + log ([salt]/[acid]) 

pH = 4.74 + log ( 0.1070/0.1493) 

pH = 4.74 + log 0.717 

pH = 4.74 + (-0.14) 

pH = 4.60.

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When HBr is added to pure water, HBr molecules lose protons, while water molecules gain protons. In this reaction, HBr is a(n)
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Acid Sorry if I am wrong but I am pretty positive it’s acid
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