1 N204(l) + 2 N2H4(l)

1 N204(l) + 2 N2H4(l) --> 3 N2(g) + 4 H2O(g)If you start with: 5.0 x 10^4 g hydrazine (N2H4) "in the tank": 1) How many moles of
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2. asked by Lauren K.
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Balanced Chemical Equation:1 N204(l) + 2 N2H4(l) --> 3 N2(g) + 4 H20(g) If you start with: 5.0 x 10^4g hydrazine (N2H4) "in the
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2. asked by Roger
3. 746 views
Balanced Chemical Equation:1 N204(l) + 2 N2H4(l) --> 3 N2(g) + 4 H20(g) If you start with: 5.0 x 10^4g hydrazine (N2H4) "in the
1. 1 answer
2. asked by Roger
3. 811 views
1 N204(l) + 2 N2H4(l) --> 3 N2(g) + 4 H2O(g)If you start with: 5.0 x 10^4 hydrazine (N2H4) "in the tank": 1) How many moles of
1. 4 answers
2. asked by Lauren K.
3. 627 views
N204 (l) Delta Hf (kJ/mol) = -20E cell S (J/mol.K) = 209 N204(g) Delta Hf (kJ/mol) = 10 E cell S (j/mol.k) = 304.1 Given the
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2. asked by Sara
3. 477 views
5. The industrial production of hydroiodic acid takes place by treatment of iodine with hydrazine (N2H4):2I2 + N2H4 → 4HI + N2
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I am really confused on this problem pleas help me.N2H4(l)+O2(g)(arrow)N2(g)+2H2O(g) a. How many liters of N2 (at STP) form when
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2. asked by Alekya
3. 4,116 views
Consider the equilibrium system: N204 (g) = 2 NO2 (g) for which the Kp = 0.1134 at 25 C and deltaH rx is 58.03 kJ/mol. Assume
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2. asked by Vee
3. 2,298 views
Please help!N2H4 is decomposed in a closed container according to this reaction: N2H4 → N2 + 2H2 . If the temperature is
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2. asked by Anonymous
3. 580 views
N2H4 is decomposed in a closed container according to this reaction: N2H4 → N2 + 2H2 . If the temperature is constant and the
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2. asked by tomas
3. 873 views