1) <span> 2.7 g water + 6,6 g carbon dioxide </span>→<span> 9.3 g carbonic acid.
According to </span><span>principle of mass conservation mass of reactants and products are the same after chemical reactio. 2,7 g + 6,6 g = 9,3 g.
2) </span><span>32.0 g sodium hydroxide + 16.0 g hydrofluoric acid --> 14,4 g water + 33.6 g sodium fluoride.
m(water) = 32 g + 16 g - 33,6 g.
3) </span><span>0.60 g calcium carbonate + 0.48 g sodium hydroxide --> 0,63 g sodium carbonate + 0.45 g calcium.
m(sodium carbonate) = 0,6 g + 0,48 g - 0,45 g.
4) </span><span>0.53 g sodium hydroxide + 0.37 g carbon dioxide --> 0,9 g sodium hydrogen carbonate.
m(sodium hydrogen carbonate) = 0,53 g + 0,37 g = 0,9 g.</span>
Answer:
222.30 L
Explanation:
We'll begin by calculating the number of mole in 100 g of ammonia (NH₃). This can be obtained as follow:
Mass of NH₃ = 100 g
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mole of NH₃ =?
Mole = mass /molar mass
Mole of NH₃ = 100 / 17
Mole of NH₃ = 5.88 moles
Next, we shall determine the number of mole of Hydrogen needed to produce 5.88 moles of NH₃. This can be obtained as follow:
N₂ + 3H₂ —> 2NH₃
From the balanced equation above,
3 moles of H₂ reacted to produce 2 moles NH₃.
Therefore, Xmol of H₂ is required to p 5.88 moles of NH₃ i.e
Xmol of H₂ = (3 × 5.88)/2
Xmol of H₂ = 8.82 moles
Finally, we shall determine the volume (in litre) of Hydrogen needed to produce 100 g (i.e 5.88 moles) of NH₃. This can be obtained as follow:
Pressure (P) = 95 KPa
Temperature (T) = 15 °C = 15 + 273 = 288 K
Number of mole of H₂ (n) = 8.82 moles
Gas constant (R) = 8.314 KPa.L/Kmol
Volume (V) =?
PV = nRT
95 × V = 8.82 × 8.314 × 288
95 × V = 21118.89024
Divide both side by 95
V = 21118.89024 / 95
V = 222.30 L
Thus the volume of Hydrogen needed for the reaction is 222.30 L
[H+] = 0.030 + 0.085 = 0.115M
pH = -log(0.115) <<< use a calculator
pOH = 14 - pH <<< just subtract
pH(HCl) = 1.5 => pOH = 12.5
pH(HBr) = 1.1 => pOH = 12.9
We gots to write the equation
2Mg + O2 -------> 2MgO
moles of Mg = 97.2 / (24.3 x 2)
= 2
therefore there are 2 moles of MgO
Mr of MgO = 40.3
mass MgO = 40.3 x 2 = 80.6
D Direction of motion I hope this helps