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Vitek1552 [10]
3 years ago
10

You are given stocks of 4 M NaCl, 40% Glucose, and 1M Tris-HCl (pH 8.5). You need to make 400 ml of a buffer containing 0.5 M Na

Cl, 8% Glucose and 50 mM Tris-HCl (pH 8.5). How would you make the buffer?
Chemistry
1 answer:
OLga [1]3 years ago
8 0

Answer:

50mL of 4M NaCl, 80mL of 40% glucose, 20mL of 1M Tris-HCl (pH 8.5) and 250mL of water.

Explanation:

To make 400mL containing 0.5M NaCl you need to add:

4M / 0.5M = 8 (dilution 1/8). 400mL / 8 = <em>50 mL of 4M NaCl.</em>

Glucose 8% you need to add:

40% / 8% = 5 (dilution 1/5). 400mL / 5 = <em>80 mL of 40% glucose </em>

Buffer 50mM you need to add:

1000mM / 50mM = 20 (dilution 1/20). 400mL / 20 = <em>20mL of 1M Tris-HCl (pH 8.5)</em>

<em></em>

The resting volume: 400mL - 50mL of 4M NaCl - 80mL of 40% glucose - 20mL of 1M Tris-HCl (pH 8.5) = 250 mL must be completed with water.

Thus, to make the solution you need: <em>50mL of 4M NaCl, 80mL of 40% glucose, 20mL of 1M Tris-HCl (pH 8.5) and 250mL of water.</em>

<em></em>

I hope it helps!

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3 years ago
Vanadium (V) and oxygen (O) form a series of compounds with the following compositions: Mass % V 76.10 67.98 61.42 56.02 Mass %
11Alexandr11 [23.1K]

Answer:

For every given mass of Vanadium, the relative number of oxygen atoms present or the mole ratio of Oxygen to Vanadium is:

A. 1:1

B. 3:2

C. 2:1

D. 5:2

<em>Note: The question is stated more clearly below:</em>

<em>Vanadium (V) and oxygen (O) form a series of compounds with the following compositions: Mass % V 76.10 67.98 61.42 56.02 Mass % O 23.90 32.02 38.58 43.98 Compound Mass % N 1 33.28 2 39.94.</em>

<em>What are the relative numbers of atoms of oxygen in the compounds for a given mass of vanadium?</em>

Explanation:

Number of moles in 100 g mass = % mass / molar mass

Molar mass of Vanadium, V = 51 g/mol

Molar mass of oxygen atom, O = 16 g/mol

1. Percentage mass of V and O is 76.10% and 23.90% respectively.

Number of moles of each atom;

V = 76.10/51.0 = 1.5 moles

O = 23.9/16 = 1.5 moles

Mole ratio of oxygen to vanadium = 1.5/1.5 = 1 : 1

2. Percentage mass of V and O is 67.98% and 32.02% respectively

Number of moles of each atom:

V = 67.98/51 = 1.33

O = 32.02/16 = 2

Mole ratio of oxygen to vanadium = 2/1.33 = 1.5 : 1 = 3 : 2

3. Percentage mass of V and O is 61.42% and 38.58% respectively

Number of moles of each atom:

V = 61.42/51 = 1.2

O = 38.58/16 = 2.4

Mole ratio of oxygen to vanadium = 2.4/1.2 = 2 : 1

4. Percentage mass of V and O is 56.02% and 43.98% respectively

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V = 56.02/51 = 1.10

O = 43.98/16 = 2.75

Mole ratio of oxygen to vanadium = 2.75/1.10 = 2.5 : 1 = 5 : 2

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Answer:

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Answer:

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4. Decrease in the number of electron shells in the cation.

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I hope I helped you^_^

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x 0,025
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1 1


c=1M
V=25ml=0,025l ( you need the volume to be in l)
c = n/V
n=c*V
n=1*0,025=0,025 m KOH


x=0,025m
c=1 M
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