a mixture of two amino acids, glycine and alanine, what is the best separating technique to separate them

The specific heat of water is 4.18 J/g•°C. How much heat does 225.0 g of water release when it cools from 85.5°C to 50.0°C? Ente

Sergio [31]

The answer is 33400J

4 0

2 years ago

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A volumetric pipette has an uncertainty of 0.01cm3. What are the lowest and highest possible volumes for a measurement of 0.20cm

Anarel [89]

Answer:

Possible lowest volume = 0.19 cm

Possible highest volume = 0.21 cm

Explanation:

given data

volumetric pipette uncertainty = 0.01 cm³

total volume = 0.20 cm³

solution

we will get here Possible lowest and highest volume that is express as

Possible lowest volume = total volume - uncertainty .....................1

Possible highest volume = total volume + uncertainty ....................2

put here value in both equation and we get

Possible lowest volume = 0.20 cm - 0.01 cm

Possible lowest volume = 0.19 cm

and

Possible highest volume = 0.20 cm + 0.01 cm

Possible highest volume = 0.21 cm

3 0

2 years ago

When an electron of an atom falls from a higher energy level to the ground state, the atom loses 9.4145 x 10-25 joules of energy

Ann [662]

$E = \frac{hc}{wavelength} 
$
given E = 9.4145E-25
h = 6.626E-34
c = 2.998E8
sub values into the equation above, and solve for wavelength.
You will get 0.211m

3 0

2 years ago

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Describe how to prepare the solution using a 250.0 mL volumetric flask by placing the steps in the correct order. Not all of the

padilas [110]

Answer:

The correct order will be

a. Transfer the measured amount of NaCl to the volumetric flask.

e. Dissolve the NaCl in less than 250 mL of water and mix well.

b. Dilute the solution with water to the 250.0 mL mark.

Explanation:

Preparation of NaCl solution in 250.0 ml volumetric flask:

Add the weighed NaCl directly to volumetric flask and add small amount of water to it and mix it will until all NaCl gets dissolved( if not add small water amount of water more)

After dissolving NaCl add the water upto the mark.

The correct order will be

a. Transfer the measured amount of NaCl to the volumetric flask.

e. Dissolve the NaCl in less than 250 mL of water and mix well.

b. Dilute the solution with water to the 250.0 mL mark.

6 0

2 years ago

A precipitate of zinc hydroxide can be formed using the reaction below.

worty [1.4K]

Answer:

Option B is correct. KOH is the limiting reagent, and 0.27 mole of Zn(OH)2 precipitate is produced.

Explanation:

Step 1: Data given

Number of moles ZnCl2 = 0.36 moles

Number of moles KOH = 0.54 moles

Step 2: The balanced equations

ZnCl2(aq) + 2 KOH(aq) → Zn(OH)2(s) + 2 KCl(aq)

For 1 mol ZnCl2 we need 2 moles KOH to produce 1 mol Zn(OH)2 and 2 moles KCl

Step 3: Calculate the limiting reactant.

KOH is the limiting reactant. It will completely be consumed (0.54 moles). ZnCl2 is in excess. There will react 0.54/2 = 0.27 moles

There will remain 0.36 - 0.27 = 0.09 moles.

Step 4: The products

There will be produced 2 moles KCl and 1 mol Zn(OH)2. Zn(OH)2 is the precipitate produced.

For 1 mol ZnCl2 we need 2 moles KOH to produce 1 mol Zn(OH)2 and 2 moles KCl

For 0.54 moles KOH, we will produce 0.27 moles precipitate (Zn(OH)2)

Option A is not correct because 0.27 mol of Zn(OH)2 will precipitate, not 0.54 mol

Option B is correct

Option C is not correct because ZnCl2 is not the limiting reactant, but the excess reactant

Option D is not correct because ZnCl2 is not the limiting reactant, but the excess reactant

7 0

2 years ago