Therefore Chlorine is losing electrons and being oxidized. Hope it helps.
Metalloids had properties that fall between those of metals and nonmetals (I believe that to be correct-.-)
Answer:
3.94 L
Explanation:
From the question given above, the following data were obtained:
Mass of O₂ = 5.62 g
Volume of O₂ =?
Next, we shall determine the number of mole present in 5.62 g of O₂. This can be obtained as follow:
Mass of O₂ = 5.62 g
Molar mass of O₂ = 2 × 16 = 32 g/mol
Mole of O₂ =?
Mole = mass / molar mass
Mole of O₂ = 5.62 / 32
Mole of O₂ = 0.176 mole
Finally, we shall determine the volume of 5.62 g (i.e 0.176 mole) of O₂ at STP. This can be obtained as follow:
1 mole of O₂ occupied 22.4 L at STP.
Therefore, 0.176 mole of O₂ will occupy = 0.176 × 22.4 = 3.94 L at STP.
Thus 5.62 g (i.e 0.176 mole) of O₂ occupied 3.94 L at STP
.15 liters of normal saline will drip into the patient's IV per hour.
Answer:
62.586 gram
Explanation:
moles of Al(OH)3 = mass / molar mass = 35.1 / (27+17x3) = 0.45 mol
moles of H2SO4 = mass / molar mass = 53.94 / (2+32+16x4) = 0.55 mol
H2SO4 is the limiting reagent (reacts completely)
⇒ moles of Al2(SO4)3 is worked out by moles of H2SO4
moles of Al2(SO4)3 = moles of H2SO4 / 3 = 0.183 mol
mass of Al2(SO4)3 = mole x molar mass = 0.183 x (27x2 + 96x3) = 62.586 gram
