Answer:
1.2×10² mmole of Na₂S₂O₃
Explanation:
From the question given above, the following data were obtained:
Volume = 0.6 L
Molarity = 0.2 mol/L
Mole of Na₂S₂O₃ =?
Molarity is simply defined as the mole of solute per unit litre of water. Mathematically, it is expressed as:
Molarity = mole /Volume
With the above formula, we can obtain the number of mole of Na₂S₂O₃ in the solution as illustrated below:
Volume = 0.6 L
Molarity = 0.2 mol/L
Mole of Na₂S₂O₃ =?
Molarity = mole /Volume
0.2 = Mole of Na₂S₂O₃ / 0.6
Cross multiply
Mole of Na₂S₂O₃ = 0.2 × 0.6
Mole of Na₂S₂O₃ = 0.12 mole
Finally, we shall convert 0.12 mole to millimole (mmol). This can be obtained as follow:
1 mole = 1000 mmol
Therefore,
0.12 mole = 0.12 mole × 1000 mmol / 1 mole
0.12 mole = 120 = 1.2×10² mmole
Thus, the chemist added 1.2×10² mmole of Na₂S₂O₃
Molar mass
C₂H₄O₂ = 60.0 g/mol
n = mass / molar mass
3.00 = mass / 60.0
m = 3.00 * 60.0
m = 180 g of <span>C₂H₄O₂
hope this helps!</span>
<span>2.51 grams
You want to prepare 19.16 g of some solution which will have 13.1% of it's mass being sucrose. So we just need to perform some simple multiplication:
19.16g * 0.131 = 2.50996g
Rounding to 3 significant figures gives 2.51 g.</span>
Answer:
[Cl2] equilibrium = 0.0089 M
Explanation:
<u>Given:</u>
[SbCl5] = 0 M
[SbCl3] = [Cl2] = 0.0546 M
Kc = 1.7*10^-3
<u>To determine:</u>
The equilibrium concentration of Cl2
<u>Calculation:</u>
Set-up an ICE table for the given reaction:

I 0 0.0546 0.0546
C +x -x -x
E x (0.0546-x) (0.0546-x)
![Kc = \frac{[SbCl3][Cl2]}{[SbCl5]}\\\\1.7*10^{-3} =\frac{(0.0546-x)^{2} }{x} \\\\x = 0.0457 M](https://tex.z-dn.net/?f=Kc%20%3D%20%5Cfrac%7B%5BSbCl3%5D%5BCl2%5D%7D%7B%5BSbCl5%5D%7D%5C%5C%5C%5C1.7%2A10%5E%7B-3%7D%20%3D%5Cfrac%7B%280.0546-x%29%5E%7B2%7D%20%7D%7Bx%7D%20%5C%5C%5C%5Cx%20%3D%200.0457%20M)
The equilibrium concentration of Cl2 is:
= 0.0546-x = 0.0546-0.0457 = 0.0089 M
Answer: So if you had 570 cm of ribbon, then 570%2F8.5=67.05 which means that about 67 students can do the experiment (round down to the nearest whole number).
Explanation: If you had 8.5 cm of ribbon, then only 8.5%2F8.5=1 student can do the experiment. If you had 17 cm of ribbon, then 17%2F8.5=2 students can do the experiment.