如何保证转化出口CH4<0.8%?

如何保证转化出口CH4<0.8%?


相关考题:

下列过程定义语句中,参数不是对象的定义语句是( )。A.Sub Pro4(x As Form)B.Sub Pro4(y As Control)C.Sub Pro4(Form1 As Form,Labell As Control)D.Sub Pro4(x As Currency)

( 14 )执行以下程序后输出的是Private Sub Command1_Click()Ch$= ” AABCDEFGH ”Print Mid(Righ(ch$,6),Len(left(ch$,4)),2)End SubA)CDEFGHB)ABCDC)FGD)AB

( 30 )下面程序的输出结果是Private Sub Command1_Click()ch$= ” ABCDEF ”proc chPrint chEnd SubPrivate Sub proc(ch As String)S= ””For k=Len(ch) To 1 Step-1s=sMid(ch,k,1)Next kch=sEnd SubA ) ABCDEFB ) FEDCBAC ) AD ) F

把窗体的KeyPreview属性设置为Tree,然后编写如下事件过程 Private Sub Form_KeyPress(KeyAscii As Integer) Dim ch As String ch=Chr(KeyAscii) KeyAscii=Asc(UCase(ch)) Print Chr(KeyAscii+2) End Sub 程序运行后,按键盘上的“A”键,则在窗体上显示的内容是______。A. AB.BC.CD.D

在通用声明中定义a,在窗体中添加一个命令按钮Command1,编写如下程序代码:DimaAsIntegerSubtest()a=a+1:b=b+1:c=c+1PrintSub:;a;,;b;,;cEndSubPrivateSubCommand1_Click()a=2:b=3:c=4CalltestCalltestEndSub程序运行后,单击命令按钮,窗体中将显示( )A.Sub:3,4,5Sub:4,5,6B.Sub:2,3,4Sub:2,3,4C.Sub:3,1,1Sub:4,1,1D.Sub:1,1,1Sub:1,1,1

(12)有下列Sub过程: Sub Sub(x As Single,y As Single) t=x x=t/y y=t Mody End Sub 在窗体上的命令按钮Commandl中,编写下列事件过程,执行该事件过程调用Sun过程,结果是( )。 Private Sub Commandl_Click() Dim a As Single Dim b As Single a=5 b=4 Sun a,b Print a;b End Sub A.1.25 1 B.5 4 C.4 5 D.1 1.25

运行以下程序后,输出结果为_____________。 Private Sub Command1_Click() a=1:b=2:c=3 Call test(a,b+3,(c)) Print "main:";a;b;c End Sub Private Function test(p,m,n) p=p+1:m=m+1:n=n+1 Print "sub:";p;m;n End Function:A. sub:2 6 4 main:1 2 3B. sub:2 6 4 main:2 2 3C. sub:2 6 4 main:2 6 4D. sub:2 6 4 main:1 6 4

下列程序的功能是把文件C:\a1.txt复制成C:\a2.txt,请填空。Private Sub Form. C1ick()Dim ch As StringOpen "C:\a1.txt" For【 】Open "C:\a2.txt" For【 】Do While Not【 】ch=Input(1,10)Print 20,ch;LoopClose 10,20End Sub

在窗体中添加一个命令按钮,编写如下程序: Private Sub Subl(p,m,n) p=p+1:m=m+1:n=n+1 Print"subl:";p;m;n End Sub Private Sub Command1_Click() al=1:b=2:c1=3 Call Subl(a,b1+3,c1) Print"Main:";a1;b1;c1 End Sub 程序运行后,输出结果为A.Sub:2 6 4 Main:2 6 4B.Sub:2 6 4 Main:2 6 4C.Sub:2 6 4 Main:1 2 3D.Sub:2 6 4 Main:2 2 3

若有以下变量和函数说明:includecharCh='*';void sub(int x,int y,char ch,double* 若有以下变量和函数说明: #include<iostream.h> charCh='*'; void sub(int x,int y,char ch,double*Z) { switch(ch) { case'+':*Z=x+y;break; case'-':*Z=x-y;break: case'*':*Z=x*y;break; case'/':*z=x/y;break: } } 以下合法的函数调用语句是( )。A.sub(10,20,Ch,y);B.sub(1.2+3,2*2,'+',Z);C.sub(sub(1,2,'+',y),sub(3,4'+',x),'-',y);D.sub(a,b,x,ch);

窗体上画一个名称为Text1的文本框,然后编写如下事件过程: Private Sub Form. Load() Show TextSetFocus End Sub Private Sub Text1_ KeyPress (KeyAscii As Integer) Dim ch As String ch = Chr (KeyAscii) KeyAscii = Asc(UCase(ch)) Pint Chr(KeyAscii + 4) End Sub程序运行后,在文本框中输入abcde,则在窗体上显示的内容是______。A.abcdeB.efghiC.EFGHID.ABCDE

执行以下程序后输出的是。 Private Sub Commandl_Click ch$ = "ABCDEFGH": Print Mid(Right(ch$, 6), Len(Left(ch$, 4)), 2) End Sub A.CDEFGH B.ABCD C.FG D.AB

写出程序运行的结果Public class BasePublic virtual string Hello() {return “Base”;}Public class Sub:BasePublic override string Hello() {return “Sub”;}1. Base b = new Base(); b.Hello;2. Sub s = new Sub(); s.Hello;3. Base b = new Sub (); b.Hello;4. Sub s = new Base(); s.Hello;

下面程序的输出结果是。 Private Sub Commandl_Click ch$=“ABCDEF” proc ch:Print ch End Sub Private Sub proc(ch As Stnng) s=“” For k=Len(ch) TO 1 Step -1 s=sMid(ch,k,1) Next k ch=s End Sub A.ABCDEF B.FEDCBA C.A D.F

在窗体中添加一个命令按钮,编写如下程序: Private Sub Test(p,m,n) p=p+1:m=m+1:n=n+1 Print "Sub: ";p;m;n End Sub Private Sub Command1.Click() a1=1:b=2:c1=3 Call Test((a,b1+3,(c1)) Print "Main:";a1;b1;c1 End Sub 程序运行后,输出结果为A.Sub: 2 6 4 Main: 2 6 4B.Sub: 2 6 4 Main: 2 6 4C.Sub: 2 6 4 Main: 1 2 3D.Sub: 2 6 4 Main: 2 2 3

在窗体中添加一个命令按钮,编写如下程序:Private Sub Sub1(p,m,n)p=p+1:m=m+1:n=n+1Print "sub1:";p;m;nEnd SubPrivate Sub Command1_Click()a1=1:b=2:c1=3Call Sub1(a,b1+3,c1)Print"Main:";a1;b1;c1End Sub程序运行后,输出结果为A.Sub: 2 6 4 Main: 2 6 4B.Sub: 2 6 4 Main: 2 6 4C.Sub: 2 6 4 Main: 1 2 3D.Sub: 2 6 4 Main: 2 2 3

在窗体中添加一个命令按钮,编写如下程序: Private Sub Sub1(p,m,n) p=p+1:m=m+1:n=n+1 Print "sub1:";p;m;n End Sub Private Sub Command1_Click() a1=1:b=2:c1=3 Call Sub1(a,b1+3,c1) Print"Main:";a1;b1;c1 End Sub 程序运行后,输出结果为A.Sub: 2 6 4 Main: 2 6 4B.Sub: 2 6 4 Main: 2 6 4C.Sub: 2 6 4 Main: 1 2 3D.Sub: 2 6 4 Main: 2 2 3

class super {  public int getLength()  {return 4;}  }  public class Sub extends Super {  public long getLength() {return 5;}  public static void main (String[]args)  {  super sooper = new Super ();  Sub sub = new Sub();  System.out.printIn(  sooper.getLength()+ “,” + sub.getLength()   };  }  What is the output?()  A、 4, 4B、 4, 5C、 5, 4D、 5, 5E、 The code will not compile.

class Super { public int getLenght() { return 4; } } public class Sub extends Super { public long getLenght() { return 5; } public static void main(String[] args) {Super sooper = new Super(); Sub sub = new Sub(); System.out.println( sooper.getLenght() + “,” + sub.getLenght() ); } } What is the output? ()A、 4,4B、 4,5C、 5,4D、 5,5E、 Compilation fails.

转化气CH4含量为什么控制<0.8%?

单选题干熄焦循环气体的主要成分为N2,其他成分还有H2、CH4、CO等气体,其中属于不可燃组分是( )。ApNsub2/sub /pBpHsub2/sub/pCpCHsub4/sub/pDCO

单选题煤矿井下的有害气体主要是由()、CO2、H2S、NO2、H2、NH3气体组成。ACOBpCHsub4/sub、SOsub2/sub/pCpSOsub2/sub、CO/pDpCO、CHsub4/sub、SOsub2/sub/p

单选题酶促反应速度(v)达到最大反应速度(Vmax)的80%时,底物浓度[S]为Ap1Ksubm/sub/pBp2Ksubm/sub/pCp3Ksubm/sub/pDp4Ksubm/sub/pEp5Ksubm/sub/p

单选题class super {  public int getLength()  {return 4;}  }  public class Sub extends Super {  public long getLength() {return 5;}  public static void main (String[]args)  {  super sooper = new Super ();  Sub sub = new Sub();  System.out.printIn(  sooper.getLength()+ “,” + sub.getLength()   };  }  What is the output?()A 4, 4B 4, 5C 5, 4D 5, 5E The code will not compile.

单选题设有函数定义:void sub(int k,char ch){…}则以下对函数sub的调用语句中,正确的是(  )。Asub(1,97);Bsub(2,'97');Ch=sub(3,'a');Dsub(4,a);

单选题class Super { public int getLenght() { return 4; } } public class Sub extends Super { public long getLenght() { return 5; } public static void main(String[] args) {Super sooper = new Super(); Sub sub = new Sub(); System.out.println( sooper.getLenght() + “,” + sub.getLenght() ); } } What is the output? ()A 4,4B 4,5C 5,4D 5,5E Compilation fails.

单选题A pIsub2/sub=Isub3/sub/pB pIsub2/sub=4Isub3/sub/pC pIsub2/sub=2Isub3/sub/pD pIsub3/sub=4Isub2/sub/p