Answer:
1. 25 moles water.
2. 41.2 grams of sodium hydroxide.
3. 0.25 grams of sugar.
4. 340.6 grams of ammonia.
5. 4.5x10²³ molecules of sulfur dioxide.
Explanation:
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In this case, since the mole-mass-particles relationships are studied by considering the Avogadro's number for the formula units and the molar mass for the mass of one mole of substance, we proceed as shown below:
1. Here, we use the Avogadro's number to obtain the moles in the given molecules of water:
[tex]1.5x10^{25}molecules*\frac{1mol}{6.022x10^{23}molecules} =25 molH_2O[/tex]
2. Here, since the molar mass of NaOH is 40.00 g/mol, we obtain:
[tex]1.2mol*\frac{40.00g}{1mol} =41.2g[/tex]
3. Here, since the molar mass of C6H12O6 is 180.15 g/mol:
[tex]45g*\frac{1mol}{180.15g}=0.25g[/tex]
4. Here, since the molar mass of ammonia is 17.03 g/mol:
[tex]20mol*\frac{17.03g}{1mol}=340.6g[/tex]
5. Here, since the molar mass of SO2 is 64.06 g/mol:
[tex]48g*\frac{1mol}{64.06g} *\frac{6.022x10^{23}molecules}{1mol} =4.5x10^{23}molecules[/tex]
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