Экспорт данных из запроса Access в Excel-Шаблон с определенной строки

sub ToExcelTabl(nameTabl)
   dim app as object, wrk as object
   dim rst as dao.recordset
   set app=createobject("excel.application")
   set wrk=app.workbooks.open(nameTabl)
   set rst=currentdb.openrecordset("select * from ИмяЗапроса")
   app.range("a10").copyfromrecordset rst
   app.visible=true
end sub

Public Sub ToExcelTabl()
   Dim app As Object
   Dim wrk As Object
   Dim rst As dao.Recordset
   Set app = CreateObject("excel.application")
   Set wrk = app.Workbooks.Open(CurrentProject.Path & "\B_S_JD.xltm")
   Set rst = CurrentDb.OpenRecordset("select * from 0180_x01_JD_SITTENSEN")
   app.Range("a10").CopyFromRecordset rst
   app.Visible = True
End Sub

ToExcelTabl Me.ИмяПоляПути

wrk.saves Me.ПолеФормыПапкиХраненияОтчет\ИмяОтчета.xlsx

wrk.saves (CurrentProject.Path & "\B_S_JD.xlsx")

Public Sub ToExcelTabl(nameQuery)
   Dim app As Object
   Dim wrk As Object
   Dim rst As dao.Recordset
   Set app = CreateObject("excel.application")
   Set wrk = app.Workbooks.Open(CurrentProject.Path & "\B_S_JD.xltm")
   Set rst = CurrentDb.OpenRecordset("select * from [" & nameQuery & "]")
   app.Range("a10").CopyFromRecordset rst
   app.Visible = True
   wrk.SaveAs Filename:=Me.ИмяПапкиХраненияОтчета & "\ИмяОтчета.xlsx", _
       FileFormat:=xlOpenXMLWorkbook
End Sub

Public Sub ToExcelTabl()
   Dim app As Object
   Dim wrk As Object
   Dim rst As dao.Recordset
   Set app = CreateObject("excel.application")
   Set wrk = app.Workbooks.Open(CurrentProject.Path & "\B_S_JD.xltm")
   Set rst = CurrentDb.OpenRecordset("select * from 0180_x01_JD_SITTENSEN")
   app.Range("a10").CopyFromRecordset rst
   app.Visible = False
   wrk.SaveAs CurrentProject.Path & "\B_S_JD.xlsx"
            wrk.Close True
            Set wrk = Nothing
            app.Quit
            Set app = Nothing
End Sub