This is the balanced eq
N2 + 3H2 -> 2NH3
first you need to find mole of N2 by using
mol = mass ÷ molar mass.
mol N2= 20g ÷ (14.01×2)g/mol
=0.7138mol
then look at the coefficient between H2 and NH3.
it is N2:NH3
1:2
0.7138:0.7138×2
0.7138:1.4276 moles
moles of NH3 = 1.4276 moles
The balanced equation for the neutralisation reaction is as follows
2H₃PO₄ + 3Mg(OH)₂ --> Mg₃(PO₄)₂ + 6H₂O
stoichiometry of H₃PO₄ to H₂O is 2:6
number of H₃PO₄ moles reacted - 0.24 mol
if 2 mol of H₃PO₄ form 6 mol of H₂O
then 0.24 mol of H₃PO₄ forms - 6/2 x 0.24 = 0.72 mol of H₂O
therefore 0.72 mol of H₂O are formed
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High explosive: Ignite almost instantly, like dynamite and TNT. Two different types are primary and secondary.
<em>Primary: easily ignited, very sensitive to heat and friction. often used to ignite other explosives. </em>
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The answer for this is 26.6°c
0.000132 g of hydrated sodium borate (Na₂B₄O₇ · 10 H₂O)
Explanation:
First we need to find the number of moles of sodium borate (Na₂B₄O₇) in the solution:
molar concentration = number of moles / volume (L)
number of moles = molar concentration × volume (L)
number of moles of Na₂B₄O₇ = 0.1 × 0.5 = 0.05 moles
We know now that we need 0.05 moles of hydrated sodium borate (Na₂B₄O₇ · 10 H₂O) to make the solution.
Now to find the mass of hydrated sodium borate we use the following formula:
number of moles = mass / molar weight
mass = number of moles × molar weight
mass of hydrated sodium borate = 0.05 / 381 = 0.000132 g
Learn more about:
molar concentration
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