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LiH可作飞船的燃料,已知下列反应:①2Li(s)+H2(g)===2LiH(s)ΔH=-182kJ·mol-1;②2H2(g)+O2(g)===2H2O(l)ΔH=-572kJ·mol-1;

题目详情

LiH 可作飞船的燃料,已知下列反应:

2Li(s) H2(g)===2LiH(s)   ΔH =- 182 kJ·mol 1

2H2(g) O2(g)===2H2O(l)   Δ H =- 572 kJ·mol 1

4Li(s) O2(g)===2Li2O(s)   Δ H =- 1 196 kJ·mol 1

LiH O2 中燃烧的热化学方程式为 (    )

A 2LiH(s) O2(g)===Li2O(s) H2O(l)   Δ H =- 702 kJ·mol 1

B 2LiH(s) O2(g)===Li2O(s) H2O(l)   Δ H =- 1 950 kJ·mol 1

C 2LiH(s) O2(g)===Li2O(s) H2O(l)   Δ H =- 1 586 kJ·mol 1

D 2LiH(s) O2(g)===Li2O(s) H2O(g)   Δ H =- 988 kJ·mol 1

LiH 可作飞船的燃料,已知下列反应:

2Li(s) H2(g)===2LiH(s)   ΔH =- 182 kJ·mol 1

2H2(g) O2(g)===2H2O(l)   Δ H =- 572 kJ·mol 1

4Li(s) O2(g)===2Li2O(s)   Δ H =- 1 196 kJ·mol 1

LiH O2 中燃烧的热化学方程式为 (    )

A 2LiH(s) O2(g)===Li2O(s) H2O(l)   Δ H =- 702 kJ·mol 1

B 2LiH(s) O2(g)===Li2O(s) H2O(l)   Δ H =- 1 950 kJ·mol 1

C 2LiH(s) O2(g)===Li2O(s) H2O(l)   Δ H =- 1 586 kJ·mol 1

D 2LiH(s) O2(g)===Li2O(s) H2O(g)   Δ H =- 988 kJ·mol 1

LiH 可作飞船的燃料,已知下列反应:

LiH 可作飞船的燃料,已知下列反应: 可作飞船的燃料,已知下列反应:

2Li(s) H2(g)===2LiH(s)   ΔH =- 182 kJ·mol 1

2Li(s) H2(g)===2LiH(s)   ΔH =- 182 kJ·mol 1 2Li(s) H2(g)===2LiH(s)   ΔH =- 182 kJ·mol 1 H2(g)===2LiH(s)   ΔH =- 182 kJ·mol 1 H2(g)===2LiH(s)   ΔH =- 182 kJ·mol 1   ΔH =- 182 kJ·mol 1 ΔH =- 182 kJ·mol 1 =- 182 kJ·mol 1 182 kJ·mol 1 1 1

2H2(g) O2(g)===2H2O(l)   Δ H =- 572 kJ·mol 1

2H2(g) O2(g)===2H2O(l)   Δ H =- 572 kJ·mol 1 2H2(g) O2(g)===2H2O(l)   Δ H =- 572 kJ·mol 1 O2(g)===2H2O(l)   Δ H =- 572 kJ·mol 1 O2(g)===2H2O(l)   Δ H =- 572 kJ·mol 1   Δ H =- 572 kJ·mol 1 Δ H =- 572 kJ·mol 1 H =- 572 kJ·mol 1 =- 572 kJ·mol 1 572 kJ·mol 1 1 1

4Li(s) O2(g)===2Li2O(s)   Δ H =- 1 196 kJ·mol 1

4Li(s) O2(g)===2Li2O(s)   Δ H =- 1 196 kJ·mol 1 4Li(s) O2(g)===2Li2O(s)   Δ H =- 1 196 kJ·mol 1 O2(g)===2Li2O(s)   Δ H =- 1 196 kJ·mol 1 O2(g)===2Li2O(s)   Δ H =- 1 196 kJ·mol 1   Δ H =- 1 196 kJ·mol 1 Δ H =- 1 196 kJ·mol 1 H =- 1 196 kJ·mol 1 =- 1 196 kJ·mol 1 1 196 kJ·mol 1 1 1

LiH O2 中燃烧的热化学方程式为 (    )

LiH O2 中燃烧的热化学方程式为 (    ) LiH O2 中燃烧的热化学方程式为 (    ) O2 中燃烧的热化学方程式为 (    ) O2 中燃烧的热化学方程式为 (    ) 中燃烧的热化学方程式为 (    ) (    )    ) )

A 2LiH(s) O2(g)===Li2O(s) H2O(l)   Δ H =- 702 kJ·mol 1

A 2LiH(s) O2(g)===Li2O(s) H2O(l)   Δ H =- 702 kJ·mol 1 2LiH(s) O2(g)===Li2O(s) H2O(l)   Δ H =- 702 kJ·mol 1 2LiH(s) O2(g)===Li2O(s) H2O(l)   Δ H =- 702 kJ·mol 1 O2(g)===Li2O(s) H2O(l)   Δ H =- 702 kJ·mol 1 O2(g)===Li2O(s) H2O(l)   Δ H =- 702 kJ·mol 1 H2O(l)   Δ H =- 702 kJ·mol 1 H2O(l)   Δ H =- 702 kJ·mol 1   Δ H =- 702 kJ·mol 1 Δ H =- 702 kJ·mol 1 H =- 702 kJ·mol 1 =- 702 kJ·mol 1 702 kJ·mol 1 1 1

B 2LiH(s) O2(g)===Li2O(s) H2O(l)   Δ H =- 1 950 kJ·mol 1

B 2LiH(s) O2(g)===Li2O(s) H2O(l)   Δ H =- 1 950 kJ·mol 1 2LiH(s) O2(g)===Li2O(s) H2O(l)   Δ H =- 1 950 kJ·mol 1 2LiH(s) O2(g)===Li2O(s) H2O(l)   Δ H =- 1 950 kJ·mol 1 O2(g)===Li2O(s) H2O(l)   Δ H =- 1 950 kJ·mol 1 O2(g)===Li2O(s) H2O(l)   Δ H =- 1 950 kJ·mol 1 H2O(l)   Δ H =- 1 950 kJ·mol 1 H2O(l)   Δ H =- 1 950 kJ·mol 1   Δ H =- 1 950 kJ·mol 1 Δ H =- 1 950 kJ·mol 1 H =- 1 950 kJ·mol 1 =- 1 950 kJ·mol 1 1 950 kJ·mol 1 1 1

C 2LiH(s) O2(g)===Li2O(s) H2O(l)   Δ H =- 1 586 kJ·mol 1

C 2LiH(s) O2(g)===Li2O(s) H2O(l)   Δ H =- 1 586 kJ·mol 1 2LiH(s) O2(g)===Li2O(s) H2O(l)   Δ H =- 1 586 kJ·mol 1 2LiH(s) O2(g)===Li2O(s) H2O(l)   Δ H =- 1 586 kJ·mol 1 O2(g)===Li2O(s) H2O(l)   Δ H =- 1 586 kJ·mol 1 O2(g)===Li2O(s) H2O(l)   Δ H =- 1 586 kJ·mol 1 H2O(l)   Δ H =- 1 586 kJ·mol 1 H2O(l)   Δ H =- 1 586 kJ·mol 1   Δ H =- 1 586 kJ·mol 1 Δ H =- 1 586 kJ·mol 1 H =- 1 586 kJ·mol 1 =- 1 586 kJ·mol 1 1 586 kJ·mol 1 1 1

D 2LiH(s) O2(g)===Li2O(s) H2O(g)   Δ H =- 988 kJ·mol 1

D 2LiH(s) O2(g)===Li2O(s) H2O(g)   Δ H =- 988 kJ·mol 1 2LiH(s) O2(g)===Li2O(s) H2O(g)   Δ H =- 988 kJ·mol 1 2LiH(s) O2(g)===Li2O(s) H2O(g)   Δ H =- 988 kJ·mol 1 O2(g)===Li2O(s) H2O(g)   Δ H =- 988 kJ·mol 1 O2(g)===Li2O(s) H2O(g)   Δ H =- 988 kJ·mol 1 H2O(g)   Δ H =- 988 kJ·mol 1 H2O(g)   Δ H =- 988 kJ·mol 1   Δ H =- 988 kJ·mol 1 Δ H =- 988 kJ·mol 1 H =- 988 kJ·mol 1 =- 988 kJ·mol 1 988 kJ·mol 1 1 1

▼优质解答
答案和解析

解析 根据盖斯定律 ( ③-① ×2 +② )/2 的热化学方程式 2LiH(s) + O2(g)===Li2O(s) + H2O(l)   Δ H =- 702 kJ·mol - 1 。

答案  A

解析 根据盖斯定律 ( ③-① ×2 +② )/2 的热化学方程式 2LiH(s) + O2(g)===Li2O(s) + H2O(l)   Δ H =- 702 kJ·mol - 1 。

解析 根据盖斯定律 ( ③-① ×2 +② )/2 的热化学方程式 2LiH(s) + O2(g)===Li2O(s) + H2O(l)   Δ H =- 702 kJ·mol - 1 。

答案  A

答案  A