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alukav5142 [94]
4 years ago
5

A chemistry student needs 60.0 ml of ethanolamine for an experiment. By consulting the CRC Handbook of Chemistry and Physics, th

e student discovers that the density of ethanolamine is 1.02 g cm^-3. Calculate the mass of ethanolamine the student should weigh out. Round your answer to 3 significant digits.
....... * 10^ ........
Chemistry
1 answer:
BARSIC [14]4 years ago
6 0

Answer:

The student should weigh out 61.2g of ethanolamine [6.12 * 10]

Explanation:

In this question, we are expected to calculate the mass of ethanolamine needed to make 60.0ml of it given that the density of the ethanolamine in question is 1.02g/cm^3

Mathematically, it has been shown that mass = density * volume

Hence, by multiplying the density by the volume, we get the mass.

Now, from the question we can see that we have the values for the density and the volume. We now need to get the mass.

Since cm^3 is same as ml, we need not perform any conversion.

Hence, the needed mass is:

60 * 1.02 = 61.2g

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If xenon has an atomic number of 54 and a mass number of 108, how many neutrons does it have?
vladimir2022 [97]

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Answer 2: d. Isotopes

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5 0
4 years ago
Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (Ka = 1.54 x 10^-5), with 0.1000 M NaOH so
Vanyuwa [196]

Answer:

pH after the addition of 10 ml NaOH = 4.81

pH after the addition of 20.1 ml NaOH = 8.76

pH after the addition of 25 ml NaOH = 8.78

Explanation:

(1)

Moles of butanoic acid initially present = 0.1 x 20 = 2 m moles  = 2 x 10⁻³ moles,

Moles of NaOH added = 10 x 0.1 = 1 x 10⁻³ moles

                          CH₃CH₂CH₂COOH + NaOH ⇄ CH₃CH₂CH₂COONa + H₂O

Initial conc.            2 x 10⁻³                 1 x 10⁻³           0            

Equilibrium             1 x 10⁻³                   0                  1 x 10⁻³

Final volume = 20 + 10 = 30 ml = 0.03 lit

So final concentration of Acid = \frac{0.001}{0.03} = 0.03mol/lit

Final concentration of conjugate base [CH₃CH₂CH₂COONa]=\frac{0.001}{0.03} = 0.03 mol/lit

Since a buffer solution is formed which contains the weak butanoic acid and conjugate base of that acid .

Using Henderson Hasselbalch equation to find the pH

pH=pK_{a}+log\frac{[conjugate base]}{[acid]}  \\\\=-log(1.54X10^{-5} )+log\frac{0.03}{0.03} \\\\=4.81

5 0
4 years ago
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