<em>Let </em><em>the </em><em>mass </em><em>be </em><em>X </em><em>g</em>
<em>percentage </em><em>=</em><em> </em><em>X/</em><em> </em><em>6.</em><em>5</em><em>0</em><em> </em><em>*</em><em> </em><em>100 </em><em>=</em><em>2.</em><em>2</em><em>%</em>
<em>X=</em><em> </em><em>0.</em><em>1</em><em>4</em><em>3</em><em> </em><em>g</em>
<em>The </em><em>mass </em><em>is </em><em>0.</em><em>1</em><em>4</em><em>3</em><em> </em><em>g</em>
3Na2O(at) + 2Al(NO3)3(aq) —> 6NaNO3(aq) + Al2O3(s)
This is a double replacement reaction and NaNO3 is aqueous because Na is an alkali metal, plus nitrate is in the solution. Both of these are soluble. Al2O3 is not soluble because it does not contain any element that is soluble and is hence the precipitate.
Hope this helped!
Answer:
A. It's a 'clean' energy. Your cooch need a cleanup
Answer:
ΔH = 57.04 Kj/mole H₂O
Explanation:
60ml(0.300M Ba(OH)₂(aq) + 60ml(0.600M HCl(aq)
=> 0.06(0.3)mole Ba(OH)₂(aq) + 0.60(0.6)mole HCl(aq)
=> 0.018mole Ba(OH)₂(aq) + 0.036mole HCl(aq)
=> 100% conversion of reactants => 0.018mole BaCl₂(aq) + 0.036mole H₂O(l) + Heat
ΔH = mcΔT/moles H₂O <==> Heat Transfer / mole H₂O
=(120g)(4.0184j/g°C)(27.74°C - 23.65°C)/(0.036mole H₂O)
ΔH = 57,042 j/mole H₂O = 57.04 Kj/mole H₂O
1.question, 2.observe, 3.hypothesize, 4.experiment 5.conclusion, 6. record.
