Answer:
2HNO3+Ca(OH)2 = Ca(NO3)2+2H2O
Explanation:
The reaction between Nitric acid(HNO3)and Calcium hydroxide(Ca(OH)2) gives Calcium Nitrate( Ca(NO3)2 and Water( H2O)
Answer:
molar mass = 180.833 g/mol
Explanation:
- mass sln = mass solute + mass solvent
∴ solute: unknown molecular (nonelectrolyte)
∴ solvent: water
∴ mass solute = 17.5 g
∴ mass solvent = 100.0 g = 0.1 Kg
⇒ mass sln = 117.5 g
freezing point:
∴ ΔTc = -1.8 °C
∴ Kc H2O = 1.86 °C.Kg/mol
∴ m: molality (mol solute/Kg solvent)
⇒ m = ( - 1.8 °C)/( - 1.86 °C.Kg/mol)
⇒ m = 0.9677 mol solute/Kg solvent
- molar mass (Mw) [=] g/mol
∴ mol solute = ( m )×(Kg solvent)
⇒ mol solute = ( 0.9677 mol/Kg) × ( 0.100 Kg H2O )
⇒ mol solute = 0.09677 mol
⇒ Mw solute = ( 17.5 g ) / ( 0.09677 mol )
⇒ Mw solute = 180.833 g/mol
1) Compund Ir (x) O(y)
2) Mass of iridium = mass of crucible and iridium - mass of crucible = 39.52 g - 38.26 g = 1.26 g
3) Mass of iridium oxide = mass of crucible and iridium oxide - mass of crucible = 39.73g - 38.26g = 1.47g
4) Mass of oxygen = mass of iridum oxide - mass of iridium = 1.47g - 1.26g = 0.21g
5) Convert grams to moles
moles of iridium = mass of iridium / molar mass of iridium = 1.26 g / 192.17 g/mol = 0.00656 moles
moles of oxygen = mass of oxygen / molar mass of oxygen = 0.21 g / 15.999 g/mol = 0.0131
6) Find the proportion of moles
Divide by the least of the number of moles, i.e. 0.00656
Ir: 0.00656 / 0.00656 = 1
O: 0.0131 / 0.00656 = 2
=> Empirical formula = Ir O2 (where 2 is the superscript for O)
Answer: Ir O2
The answer would be 0.25 g/mL.
I determined the density by dividing the mass by the volume which gives you the density. D = mass/volume.
<span>6 g / 24 mL = 0.25 g/mL
</span>
Answer
is: 0.375 moles are present in 8.4 liters of nitrous oxide at stp.
V(N₂O) = 8.4 L.
V(N₂O) =
n(N₂O) · Vm.
Vm = 22,4 L/mol.<span>
n</span>(N₂O) = V(N₂O) ÷ Vm.
n(N₂O) = 8.4 L ÷ 22.4 L/mol.
n(N₂O) = 0.375 mol.<span>
Vm - molare volume on STP.</span>
