Answer:
(1) 0.10 (2) 17.8 g
Explanation:
Since the reaction ratio is 1:1 what we need is to convert the given masses to moles and you will have the answer:
MW anthracene = 178.23 g/mol
MW maleic anhydride = 98.06 g/mol
a) mass anthracene = 178 mg x 1 g/ 1000 mg = 0.178 g anthracene
Moles anthracene = 0.178 g anthracene/ 178.23 g/mol
= 0.001 mol anthracene
0.001 mol anthracene x 1 mol maleic acid/mol anthracene
= 0.001 mol maleic anhydride
mass maleic anhydride = 0.001 mol x 98.06 g/mol = 0.10 g
b) moles maleic anhydride = 9.8 g/ 98.06 g/mol = 0.099 moles
0.099 moles maleic anhydride x 1 mol anthracene/mol maleic anhydride =
0.099 mol anthracene
g anthracene = 0.10mol x 178 g/mol = 17.8 g
Meta oxides are compounds that are formed by reaction of metals with oxygen. If these compounds are placed in water, the ionic components of this substance will dissociate.
The dissociation of metal oxides in water will likely form,
2M³⁺ + 3O²⁻
Using the combined gas law, where PV/T = constant, we first solve for PV/T for the initial conditions: (4.50 atm)(36.0 mL)/(10.0 + 273.15 K) = 0.57213.
Remember to use absolute temperature.
For the final conditions: (3.50 atm)(85.0 mL)/T = 297.5/T
Since these must equal, 0.57213 = 297.5/T
T = 519.98 K
Subtracting 273.15 gives 246.83 degC.
The pH of a buffer solution : 4.3
<h3>Further explanation</h3>
Given
0.2 mole HCNO
0.8 mole NaCNO
1 L solution
Required
pH buffer
Solution
Acid buffer solutions consist of weak acids HCNO and their salts NaCNO.
![\tt \displaystyle [H^+]=Ka\times\frac{mole\:weak\:acid}{mole\:salt\times valence}](https://tex.z-dn.net/?f=%5Ctt%20%5Cdisplaystyle%20%5BH%5E%2B%5D%3DKa%5Ctimes%5Cfrac%7Bmole%5C%3Aweak%5C%3Aacid%7D%7Bmole%5C%3Asalt%5Ctimes%20valence%7D)
valence according to the amount of salt anion
Input the value :
![\tt \displaystyle [H^+]=2.10^{-4}\times\frac{0.2}{0.8\times 1}\\\\(H^+]=5\times 10^{-5}\\\\pH=5-log~5\\\\pH=4.3](https://tex.z-dn.net/?f=%5Ctt%20%5Cdisplaystyle%20%5BH%5E%2B%5D%3D2.10%5E%7B-4%7D%5Ctimes%5Cfrac%7B0.2%7D%7B0.8%5Ctimes%201%7D%5C%5C%5C%5C%28H%5E%2B%5D%3D5%5Ctimes%2010%5E%7B-5%7D%5C%5C%5C%5CpH%3D5-log~5%5C%5C%5C%5CpH%3D4.3)
Your answer is right.
Important elements to consider:
- to use the balanced equation (which you did)
- divide the masses of each compound by the correspondant molar masses (which you did)
- compare the theoretical proportions with the current proportions
Theoretical: 2 mol of Na OH : 1 mol of CuSO4
Then 4 mol of NaOH need 2 mol of CUSO4.
Given that you have more than 2 mol of of CUSO4 you have plenty of it and the NaOH will consume first, being this the limiting reagent.
