Sheila lives in the seashore community of Beachview. At 6:00 a. m. one morning, she noticed that the local ocean water was at high tide. At approximately noon that same day, she noticed the water was at low tide. Which of the following describes what caused the change in tides between 6:00 a. m. and noon that day? A. The Earth completed one full rotation (360°), and the gravitational attraction between the Moon and the Beachview area increased. B. The Earth revolved approximately 90° around the Sun, and the gravitational attraction between the Moon and Sun increased. C. The Earth rotated approximately 90°, and the gravitational attraction between the Moon and the Beachview area decreased. D. The Earth completed one full revolution around the Sun (360°), and the gravitational attraction between the Moon and Sun decreased.

The earth rotated approximately by 90 degrees and the gravitational attraction between the moon and the Beachview area decreased. This caused a change in tides between 6:00 a.m and noon that day.

Option C.

The high tides occur in oceans that are on the side of the earth which is directly facing the moon. Since moon is closer to earth, the gravitational pull it exerts on the earth's surface is much stronger than that of the sun.

As one side of the earth turns further away from the moon, the gravitational pull on that side decreases. Hence, there is seen a difference between the size of the tides on different time of the day.

The gravitational pull by sun and moon over the oceans of earth cause tides. These are bulges of water on the coastal areas which are created as the moon and sun pull earth towards them. Tides are of two types:

  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