How many liters would you need to make a 1 m solution if you have 6 mil of sodium hydroxide
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Detonation of nitroglycerin:
<span>Mass nitroglycerin = 2.00 mL x 1.592 g/mL = 3.184 g
</span>
Moles = mass / molar mass = 3.184<span> g/ 227.0872 g/mol = 0.01402
</span>
the ratio between nitroglycerin and Carbon dioxide is 4 : 12
So, moles CO2 = 0.01402 x 12 / 4 =0.0420
the ratio between nitroglycerin and N2 is 4 : 6
moles N2 = 0.01402 x 6 / 4 =0.0841
<span>the ratio between nitroglycerin and O2 is 4 : 1 </span>
moles O2 = 0.01402 x 1 / 4 = 0.0035
<span>the ratio between nitroglycerin and water is 4 : 1 </span>
<span>in the same way moles water = 0.005258 </span>
total moles = 0.0420 + 0.0841 + 0.0035 + 0.005258 = 0.130758
0.130758<span> x 55 = 5.78 L </span>
Mass N2 = 0.0841 mol x 28.0134 g/mol = 2.3548 g
<span>Mass nitroglycerin = 2.00 mL x 1.592 g/mL = 3.184 g
</span>
Moles = mass / molar mass = 3.184<span> g/ 227.0872 g/mol = 0.01402
</span>
the ratio between nitroglycerin and Carbon dioxide is 4 : 12
So, moles CO2 = 0.01402 x 12 / 4 =0.0420
the ratio between nitroglycerin and N2 is 4 : 6
moles N2 = 0.01402 x 6 / 4 =0.0841
<span>the ratio between nitroglycerin and O2 is 4 : 1 </span>
moles O2 = 0.01402 x 1 / 4 = 0.0035
<span>the ratio between nitroglycerin and water is 4 : 1 </span>
<span>in the same way moles water = 0.005258 </span>
total moles = 0.0420 + 0.0841 + 0.0035 + 0.005258 = 0.130758
0.130758<span> x 55 = 5.78 L </span>
Mass N2 = 0.0841 mol x 28.0134 g/mol = 2.3548 g