freezing-energy lost (exothermic)
sublimation-energy gain (endothermic)
evaporation- energy gain(endothermic)
Melting- energy gain(endothermic)
deposition- energy lost(exothermic)
condensation-energy lost(exothermic)
Its C, Outermost Electrons.
Answer:
1. 0.125 mole
2. 42.5 g
3. 0.61 mole
Explanation:
1. Determination of the number of mole of NaOH.
Mass of NaOH = 5 g
Molar mass of NaOH = 23 + 16 + 1
= 40 g/mol
Mole of NaOH =?
Mole = mass /molar mass
Mole of NaOH = 5/40
Mole NaOH = 0.125 mole
2. Determination of the mass of NH₃.
Mole of NH₃ = 2.5 moles
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mass of NH₃ =?
Mass = mole × molar mass
Mass of NH₃ = 2.5 × 17
Mass of NH₃ = 42.5 g
3. Determination of the number of mole of Ca(NO₃)₂.
Mass of Ca(NO₃)₂ = 100 g
Molar mass of Ca(NO₃)₂ = 40 + 2[14 + (3×16)]
= 40 + 2[14 + 48]
= 40 + 2[62]
= 40 + 124
= 164 g/mol
Mole of Ca(NO₃)₂ =?
Mole = mass /molar mass
Mole of Ca(NO₃)₂ = 100 / 164
Mole of Ca(NO₃)₂ = 0.61 mole
Answer:
Ca3(PO4)2 + 3H2SO4 —> 3CaSO4 + 2H3PO4
The coefficients are 1, 3, 3, 2
Explanation:
Ca3(PO4)2 + H2SO4 —> CaSO4 + H3PO4
From the above equation,
There are 3 atoms of Ca on the left and 1 atom of Ca on the right. To balance Ca, put 3 in front of CaSO4 as shown below
Ca3(PO4)2 + H2SO4 —> 3CaSO4 + H3PO4
Now, we have 3 atoms of SO4 on the right and 1atom on the left. To balance SO4, put 3 in front of H2SO4 as shown below:
Ca3(PO4)2 + 3H2SO4 —> 3CaSO4 + H3PO4
Looking closely, there are 6 atoms of H on the left and 3 on the right. Therefore, it is balanced by by putting 2 in front of H3PO4 as shown below:
Ca3(PO4)2 + 3H2SO4 —> 3CaSO4 + 2H3PO4
The coefficients are 1, 3, 3, 2