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Vlad [161]
4 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]4 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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3 H2 (g) + N2 (g) 2 NH3 (g)
ale4655 [162]

Answer:

Mass = 0.697 g

Explanation:

Given data:

Volume of hydrogen = 1.36 L

Mass of ammonia produced = ?

Temperature = standard = 273.15 K

Pressure = standard = 1 atm

Solution:

Chemical equation:

3H₂ + N₂       →      2NH₃

First of all we will calculate the number of moles of hydrogen:

PV  = nRT

R = general gas constant = 0.0821 atm.L/mol.K

1atm ×1.36 L = n × 0.0821 atm.L/mol.K × 273.15 K

1.36 atm.L = n × 22.43 atm.L/mol

n = 1.36 atm.L / 22.43 atm.L/mol

n = 0.061 mol

Now we will compare the moles of hydrogen and ammonia:

                 H₂         :          NH₃

                  3          :            2

                0.061     :         2/3×0.061 = 0.041

Mass of ammonia:

Mass = number of moles × molar mass

Mass = 0.041 mol × 17 g/mol

Mass = 0.697 g

4 0
3 years ago
Colorimetric methods of analysis are based on having a reagent that changes _____________ as a function of the concentration of
GREYUIT [131]

Colorimetric methods of analysis make use of Color changes in reagents to decipher the concentration of solutions.

Therefore, color completes the sentence

<h3>Colourimetric analysis</h3>

Generally, Colorimetric analysis is used to know the concentration of a chemical element in a solution while using colour indicators or reagents.

Therefore

Colorimetric methods of analysis are based on having a reagent that changes Color as a function of the concentration of the analyte.

Color

For more information on  Compound

brainly.com/question/704297

3 0
2 years ago
Great Sand Dunes National Park in Colorado is famous for its giant sand dunes. Sand dunes are landforms that are found in desert
Anit [1.1K]
Either b or c :), i’m not quite sure
5 0
3 years ago
Structural formula for:<br>5 - methyl hept - 3 - enal​
leonid [27]

Answer:

draw the carbon chain is containing 6 carbon then attach the aldehyde group with edge carbon in chain then put the put double bond at 3 no. carbon

6 0
3 years ago
If 0.0106 g of a gas dissolves in 0.792 L of water at 0.321 atm, what quantity of this gas (in grams) will dissolve at 5.73 atm?
Alex73 [517]

Answer:

0.189 g.

Explanation:

  • This problem is an application on <em>Henry's law.</em>
  • Henry's law states that the solubility of a gas in a liquid is directly proportional to its partial pressure of the gas above the liquid.
  • Solubility of the gas ∝ partial pressure
  • If we have different solubility at different pressures, we can express Henry's law as:

<em>S₁/P₁ = S₂/P₂,</em>

S₁ = 0.0106/0.792 = 0.0134 g/L and P₁ = 0.321 atm

S₂ = ??? g/L and P₂ = 5.73 atm

  • So, The solubility of the gas at 5.73 atm (S₂) = S₁.P₂/P₁ = (0.0134 g/L x 5.73 atm) / (0.321 atm) = 0.239 g/L.

<em>The quantity in (g) = S₂ x V = (0.239 g/L)(0.792 L) = 0.189 g.</em>

<em></em>

8 0
3 years ago
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