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Vlad [161]
3 years ago
15

Using the 5M NaCI and 10% glucose stock solutions you generated, how would you prepare 100 ml of a solution that is both 150mM N

aCl and 1 % glucose?
Chemistry
1 answer:
fredd [130]3 years ago
4 0

Answer:

Take 3 mL of the 5 M NaCl solution, 10 mL of the 10% glucose solution, and add water for a final volume of 100 mL.

Explanation:

  • In order to calculate the required volume of the 5 M NaCl solution, we calculated the moles contained in a 100 mL solution that has a concentration of 150 mM:

0.1 L * 0.150 M = 0.015 moles of NaCl

With those moles we can calculated the required volume, using the concentration of the stock solution:

0.015 mol / 5 M = 0.003 L = 3 mL.

  • To make a solution that has a 1 % concentration of glucose, from a 10 % glucose solution, is the same as to make it ten times less concentrated. Thus, with a final volume of 100 mL, you would need to take 10 mL of the 10% glucose solution, because 100mL * 10/100 = 10.

So in order to prepare the solution, you would need to take 3 mL of the 5 M NaCl solution, 10 mL of the 10% glucose solution, and add water for a final volume of 100 mL.

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H e l p,,, ill give brainliest :)
mote1985 [20]

Hi there! :)

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6 0
2 years ago
Human use of groundwater has _______ over time.
Harman [31]
Increased is the answer
7 0
3 years ago
A transformer has 90 turns in the primary coil and 9 turns in the secondary coil If the output voltage is 6 V, what is the input
Afina-wow [57]

Answer:

60 V

Explanation:

From;

Vs/Vp = Ns/Np

Where;

Vs = voltage in the secondary coil = 6V

Vp = voltage in the primary coil= ??

Ns = number of turns in the secondary coil = 9

Np= number of turns in the primary coil = 90

6/Vp = 9/90

Vp= 90 * 6/9

Vp=  60 V

6 0
3 years ago
Suppose that Daniel has a 3.00 3.00 L bottle that contains a mixture of O 2 O2 , N 2 N2 , and CO 2 CO2 under a total pressure of
Alenkinab [10]

Answer:

Partial pressure O₂ → 2.74 atm

Explanation:

Let's analyse the data given:

Volume → 3L

In the bottle there is a mixture of gases that contains, O₂, N₂ and CO₂.

Total pressure is 4.80 atm

Let's apply the Ideal Gases Law to determine the total moles of the mixture

P . V = n .  R. T

4.80 atm . 3L = n . 0.082 . 273K

n = 4.80 atm . 3L / 0.082 . 273K → 0.643 moles

We apply the concept of mole fraction:

Mole fraction of a gas X = moles of gas X / Total moles

Mole fraction of a gas X = Partial pressure X / Total pressure

In a mixture, sum of mole fraction of each gas = 1

We determine mole fraction of N₂ → 0.230 / 0.643 = 0.357

We determine mole fraction of CO₂ → 0.350 atm / 4.80 atm = 0.0729

1 - mole fraction N₂ - mole fraction CO₂ = mole fraction O₂

1 - 0.357 - 0.0729 = 0.5701 → mole fraction O₂

We replace in the formula: Mole fraction O₂ = Partial pressure O₂ / 4.80 atm

0.5701 . 4.80 atm = Partial pressure O₂ → 2.74 atm

5 0
3 years ago
Read 2 more answers
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Artist 52 [7]

Answer:

603 mL

Explanation:

A milliliter is a unit of volume equal to 1/1000th of a liter. It is the same as a cubic centimeter.

6 0
3 years ago
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