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At a certain temperature, 0.4011 mol of N, and 1.541 mol of H, are placed in a 2.00 L container. N2(g) + 3H2(g) = 2 NH3(g) At

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Answer #1

Na(g) + 3H2(g) - 2NH3(g) (U= 22) initially moles present o you1 1.541 at equilibrium 1.541-3x o 4011-x and moles of Na presen2. ke [ equilibrium coustoint] [NH3] [Na] [42] 3 2 [ 0-1201) 00:39105] [136657 3 -2 Х S (0.12018² (0.34105) C136085)? 2 0.016

answered by: ANURANJAN SARSAM
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Answer #2

Na(g) + 3H2(g) - 2NH3(g) (U= 22) initially moles present o you1 1.541 at equilibrium 1.541-3x o 4011-x and moles of Na presen2. ke [ equilibrium coustoint] [NH3] [Na] [42] 3 2 [ 0-1201) 00:39105] [136657 3 -2 Х S (0.12018² (0.34105) C136085)? 2 0.016

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