5F2 + 2NH3 --> N2F4 + 6HF
<span>60.1g NH3 / 17g/mole = 3.54moles NH3 </span>
<span>3.54moles NH3 x (5 F2 / 2NH3) x 38g/mole = 335.85g required </span>
<span>5.25g HF / 20g/mole = 0.262moles HF </span>
<span>0.262moles HF x (2NH3 / 6HF) x 17g/mole = 1.49g required </span>
<span>209g / 38g/mole = 5.5moles F2 </span>
<span>5.5moles F2 (1 N2F4 / 5F2) x 66g/mole = 72.6g produced </span>
<span>Li3N + 3H2O --> NH3 + 3LiOH </span>
<span>(37.7g / 34.7g/mole) x (3H2O / 1 Li3N) x 18g/mole = 58.67g required </span>
<span>1.08moles Li3N (1NH3 / 1Li3N) x 6.022x10^23molecules/mole = 6.54x10^23 molecules </span>
<span>10.3L at STP: 10.3L / 22.4L/mole = 0.46moles NH3 produced </span>
<span>0.46moles NH3 x (1Li3N / 1NH3) x 34.7g/mole = 15.96g</span>
Answer:
Explanation:
<em>Endothermi</em>c processes absorb energy. The final state contains more energy than the initial state.
Since ice absorbs heat energy <em>in the process of completely melting</em> this is an <em>endothermic</em> process.
The process involves two stages: 1) heating the ice up to the melting point, which is 0ºC, and 2) melting the ice.
1. Heating the ice from -15ºC to 0ºC
a) Formula: Q = m×C×ΔT
- C = 2.108 kJ/kg.ºC (specific heat of ice)
b) Calculations:
- Q = m×C×ΔT = 1.6 kg × 2.108 kJ/kg.ºC × 15ºC = 50.592 kJ
2. Melting the ice at 0ºC
a) Formula: L = m × ΔHf
- ΔHf = 334 kJ/kg (latent heat of fussion)
b) Calculations
- L = m × ΔHf = 1.6 kg × 334 kJ/kg = 534.40 kJ
<u />
<u>2. Total heat</u>
<u />
- 50.592 kJ + 534.40 kJ = 584.992 kJ ≈ 590 kJ (rounded to 2 significant figures)
Explanation:
<em>5</em><em>0</em><em>0</em><em> </em><em>degre</em><em>e</em><em> </em><em>Celsius</em><em> </em><em> </em><em>be</em><em>cause</em><em> </em><em>we</em><em> </em><em>can</em><em> </em><em>see</em><em> </em><em>i</em><em>t</em><em> in</em><em> the</em><em> </em><em>tube</em>
Answer:
look up chemical board it will help you
Explanation:
Answer:
The answer is b
Explanation:
have common properties and are good conductors of heat and electricity. They reflect light and are malleable, and ductile.
