The action that would most likely lead to. Decrease in volume is decrease in temperature.
the mass of aluminum oxide (101.96 g/mol) produced from 1.74 g of manganese(iv) oxide (86.94 g/mol) is 1.36g
The reaction is 3 MnO2 + 4 Al ------ 2Al2o3+ Mn
3 mole of manganese oxide give 2 moles of aluminum oxide so by the reaction n( MnO2)/3 =n(al203)2
the formula is n= mass/M so, now substituting values
m (Al2O3)= m(MnO2) X 2 X M (Al2O3) / M(MnO2 X3
so, by substituting values, 2 X101.96 X1.74g / 3 X 86.94 =1.36g
so mass of aluminum oxide obtained = 1.36g
To learn more about Mass:
brainly.com/question/19694949
#SPJ4
I think you means the KO2 reacts with H2O. The equation of this reaction is 4KO2+2H2O->4KOH +3O2. The ratio of mole number of O2 and KO2 is 3:4. So the mole number of O2 produced is 0.500/4*3=0.375 mol.
Answer:
When 0.250 moles of a gas is placed in a container at 25 °C, it exerts a pressure of 700 mm Hg.
Explanation:
From a solubility curve the solubility of KNO3 at 40°C is about 63 g / 100 g of water.
That means that 100 g of water at 40°C can dissolve 63 g of KNO3.
A saturated solution is one that cannot dissolve more solute, because it already has reached the greatest concentration that it can hold.
Then, the mass of KNO3 that dissolves in 100 g of water is about 63 g.
