Answer:
In the final solution, the concentration of sucrose is 0.126 M
Explanation:
Hi there!
The number of moles of solute in the volume taken from the more concentrated solution will be equal to the number of moles of solute in the diluted solution. Then, the concentration of the first solution can be calculated using the following equation:
Ci · Vi = Cf · Vf
Where:
Ci = concentration of the original solution
Vi = volume of the solution taken to prepare the more diluted solution.
Cf = concentration of the more diluted solution.
Vf = volume of the more diluted solution.
For the first dillution:
26.6 ml · 2.50 M = 50.0 ml · Cf
Cf = 26.6 ml · 2.50 M / 50.0 ml
Cf = 1.33 M
For the second dilution:
16.0 ml · 1.33 M = 45.0 ml · Cf
Cf = 16.0 ml · 1.33 M / 45.0 ml
Cf = 0.473 M
For the third dilution:
20.0 ml · 0.473 M = 75.0 ml · Cf
Cf = 20.0 ml · 0.473 M / 75.0 ml
Cf = 0.126 M
In the final solution, the concentration of sucrose is 0.126 M
<span> iron (Fe), ruthenium (Ru), osmium (Os) and hassium (Hs). They are all transition metals.</span>
Answer:
See explanation
Explanation:
Using the formula
°C = (F-32) × 5/9
Where;
°C = temperature in degrees centigrade
F= temperature in Fahrenheit
F= (9/5 ×°C) +32
F= (9/5 × 110) + 32
F= 230°F
To convert -78°C to Kelvin
-78°C + 273 = 195 K
Answer:
Away from the central sulfur atom.
Explanation:
<u>Answer:</u> The standard potential of the cell is 0.77 V
<u>Explanation:</u>
We know that:
The substance having highest positive 
reduction potential will always get reduced and will undergo reduction reaction.
The half reaction follows:
<u>Oxidation half reaction:</u> 
<u>Reduction half reaction:</u> 
( × 2)
To calculate the 
of the reaction, we use the equation:
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
Putting values in above equation follows:
Hence, the standard potential of the cell is 0.77 V
