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Over [174]
3 years ago
9

The organic compounds that are structural elements in the cell walls of plants and bacteria are known as

Chemistry
1 answer:
ohaa [14]3 years ago
3 0
An organic compound that contains a carbonyl group with a cell wall. They are only found in plants and bacteria
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Peter is 5 ft, 11 inches tall. Paul is 176 cm tall. Who is taller?
arlik [135]
Peter is taller, convert cm to in.
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3 years ago
A 100.0 mL solution containing 0.864 g of maleic acid (MW=116.072 g/mol) is titrated with 0.276 M KOH. Calculate the pH of the s
Lilit [14]

Answer:

pH = 1.32

Explanation:

                 H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺

This problem involves a weak diprotic acid which we can solve by realizing they amount  to buffer solutions.  In the first  deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:

So first calculate the moles reacted and produced:

n H₂M = 0.864 g/mol x 1 mol/ 116.072 g  =  0.074 mol H₂M

54 mL x  1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH

it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.

moles H₂M left = 0.074 - 0.015 = 0.059

moles HM⁻ produced = 0.015

Using the Henderson - Hasselbach equation to solve for pH:

ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325

Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.

For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.

           

3 0
3 years ago
The half-life period for a first order reaction is independent of
maria [59]
I believe the answer is concentration

Sorry if I’m wrong hope this helps
4 0
2 years ago
Read 2 more answers
Into which two smaller groups are plants divided.
vova2212 [387]
It’s B

Because that is the smallest way scientists divide plants your welcome
5 0
3 years ago
A 2.5 g sample of french fries is placed in a calorimeter with 500.0 g of water at an initial temperature of 21 °C. After combus
SIZIF [17.4K]
Q=m°C<span>ΔT
=(500g) x (1 cal/g.</span>°C) x (48°C-21°C) = 13500 cal
13500 cal / 1000 = 13.5 kcal

<span>"What is the caloric value (kcal/g) of the french fries?"
13.5 kcal/ 2.5 g = 5.4 kcal/g</span>
8 0
3 years ago
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