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pantera1 [17]
3 years ago
7

What are the advantages and diadvantages of preparing a standard solution?​

Chemistry
1 answer:
Ghella [55]3 years ago
3 0

Answer:

advantages:

. The error due to weighing the solute can be minimised by using concentrated stock solution that requires large quantities of solute. 2. We can prepare working standards of different concentrations by diluting the stock solution which is more efficient since consistency is maintained

3. Some of the concentrated solutions are more stable and are less likely to support microbial growth than working standards used in the experiments.

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Given the following equation,, how many moles of water will be produced when 12 moles of ethane (C2H6) are burned?
VARVARA [1.3K]

Answer: 36 mol H2O (l)

Explanation:

Let's use stoichiometry to solve this problem.

12 mol C2H6 * (6 mol H2O / 2 mol C2H6) = 36 mol H2O

Hope this helps!

6 0
3 years ago
At stp what element is a good conductor of electricity?
Vlad1618 [11]
Copper is a good conductor
3 0
3 years ago
A. Convert the mass of the Salt weighed out to fg.
omeli [17]

Answer:

6.564×10¹⁶ fg.

Explanation:

The following data were obtained from the question:

Mass of beaker = 76.9 g

Mass of beaker + salt = 142.54 g

Mass of salt in fg =?

Next, we shall determine the mass of the salt in grams (g). This can be obtained as follow:

Mass of beaker = 76.9 g

Mass of beaker + salt = 142.54 g

Mass of salt =?

Mass of salt = (Mass of beaker + salt) – (Mass of beaker)

Mass of salt = 142.54 – 76.9

Mass of salt = 65.64 g

Finally, we shall convert 65.64 g to femtograms (fg) as illustrated below:

Recall:

1 g = 1×10¹⁵ fg

Therefore,

65.64 g = 65.64 g × 1×10¹⁵ fg / 1g

65.64 g = 6.564×10¹⁶ fg

Therefore, the mass of the salt is 6.564×10¹⁶ fg.

3 0
3 years ago
One mole of potassium permanganate, KMnO4, contains how many<br> grams of oxygen.
Nutka1998 [239]
36 grams of oxygen is the answer
6 0
3 years ago
A patient receives 3.3 L of glucose solution intravenously (IV). If 100. mL of the solution
artcher [175]

Answer:

660kcal

Explanation:

The question is missing the concentration of the glucose solution. Standard glucose concentration for IV solution is 5% or 5g of glucose every 100mL of solution.  

We need to determine how many grams of glucose are there inside the solution. The number of glucose in 3.3L solution will be:  

3.3L * (1000mL / L) * (5g/100mL)= 165 g.

If glucose will give 4kcal/ g, then the total calories 165g glucose give will be: 165g * 4kcal/ g= 660kcal.

6 0
4 years ago
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