You have a length of tubing that is closed at one end. You cut the tubing into two pieces of unequal length, giving you a tube open at both ends and a tube closed at one end. You establish that the fundamental frequency for the new open tube is 333 Hz and for the new closed tube is 697 Hz. Determine the fundamental frequency for the original closed tube. (Use 343 m/s as the speed of sound.)

it will be pushed away from the source of the electric field. a negative charge in a cloud induces a positive charge in the ground below the cloud.

explanation:

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures