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《天谕手游》孤高的浮云乐谱代码分享

桃子1年前 (2023-12-17)阅读数 10#综合百科
文章标签氢原子氧原子

天谕手游当中,玩家可以把自己喜欢的音乐写进乐谱,很多玩家想知道孤高的浮云乐谱代码是怎样的。接下来就让我给大家带来《天谕手游》孤高的浮云乐谱代码分享,感兴趣的玩家一起来看看吧。

我推荐: 《天谕手游》乐谱代码汇总

《天谕手游》孤高的浮云乐谱代码分享

t127

A轨

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B轨

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乐谱代码使用方法

1.首先找到一首歌曲的代码,然后复制下来;

2.然后打开天谕手游-游戏中界面-演奏的功能键(在相机右边,只有乐师职业才会有);

3.点击这个功能键展开,就会有谱曲的功能,点击进入,选择新建曲谱,然后把保存代码复制进去;

4.曲谱分为音轨A、音轨B、音轨C,玩家可以都复制粘贴,然后点击右侧的保存键即可导入成功,之后在自动演奏中就可以播放了。

(1)由图可知,水的分解化学键断裂先生成H原子与O原子,氢原子结合生成氢气,氧原子结合生成氧气,由水的分子式可知氢原子物质的量是氧原子2倍,故A为氢原子、B为氧原子;

氧原子和氢原子生成水时不需要断裂化学键,所以其活化能=0,

故答案为:氢原子、氧原子;=;

(2)含有6.02×1022个氢原子的氢气的物质的量=

6.02×1022
2×6.02×1023/mol
=0.05mol,

标况下氢气的体积=0.05mol×22.4L/mol=1.12L=1120mL;

负极上失电子发生氧化反应,电极反应式为LaNi5H6+6OH--6e-═LaN5+6H2O,

故答案为:1120;LaNi5H6+6OH--6e-═LaN5+6H2O;

(3)①根据图象知,二氧化碳的反应速率=

(1.00?0.25)mol/L
10min
=0.075mol/(L?min),

故答案为:0.075mol/(L?min);

②根据反应方程式知,平衡时c(CH3OH)=c(H2O)=0.75mol/L,c(CO2)=0.25mol/L,根据甲醇和氢气的关系式知,平衡时氢气浓度=3mol/L-3×0.75mol/L=0.75mol/L,

平衡常数K=

c(CH3OH)?c(H2O)
c(CO2)?c3(H2)
=
0.75×0.75
0.25×(0.75)3
=
16
3

故答案为:

16
3

《天谕手游》孤高的浮云乐谱代码分享

③该反应是一个反应前后气体体积减小的、放热的可逆反应,降低温度、增大压强、增大氢气的量或减少生成物的量都增大二氧化碳的转化率,

故选:CE.

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