Answer:
In order to prepare 200.0 mL of an aqueous solution of iron (III) chloride, at a concentration of 1.25 x 10⁻² M, you need to weight 0.4055 g of FeCl₃ and add to 200.0 mL of water.
Explanation:
Concentration: 1.25 x 10⁻² M
1,25 x 10⁻² mol FeCl₃ ___ 1000 mL
x ___ 200.0 mL
x = 2.5 x 10⁻³ mol FeCl₃
Mass of FeCl₃:
1 mol FeCl₃ _____________ 162.2 g
2.5 x 10⁻³ mol FeCl₃ _______ y
y = 0.4055 g FeCl₃
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.
The template that was specified as described would be the nucleic acid synthesis. The nucleic acid synthesis is a process in which DNA and RNA are "synthesised" so that the information from the genes would be successfully transferred to the proteins.
Answer:
The answer to your question is: letter D 10 l
Explanation:
Data
V1 = 5.8 l
P1= 760 mmHg
T 0 constant
P2 = 430 mmHg
V2 = ?
Formula
V1P1 = V2P2
and we clear V2 from the equation
V2 = V1P1/P2
V2 = (5.8)(760)/430)
V2 = 10.25 l
