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Furkat [3]
3 years ago
12

Every day, about 1 4 0 , 0 0 0 , 0 0 0 140,000,000140, comma, 000, comma, 000 plastic water bottles are bought in the United Sta

tes. Each bottle is about 0 . 2 5 0.250, point, 25 meters long, and the equator of the Earth is about 4 0 , 0 0 0 , 0 0 0 40,000,00040, comma, 000, comma, 000 meters around. How many times can the Earth's equator be circled by a line of plastic bottles used in the United States every day?
Chemistry
1 answer:
jeka943 years ago
8 0

Answer:

0.875 time.

Explanation:

The length of plastic water bottles are bought in the United States every day = (the no. of plastic water bottles are bought every day)(length of each bottle) = (140.0 million bottle)(0.25 m) = 35.0 million meter.

The length of the equator of the earth = 40.0 million meter.

∴ The times can the Earth's equator be circled by a line of plastic bottles used in the United States every day = (The length of plastic water bottles are bought in the United States every day)/(The length of the equator of the earth) = (35.0 million meter)/(40.0 million meter) = 0.875 times.

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Because ocean currents circulate water worldwide, they have a significant impact on the movement of energy and moisture between the oceans.

Explanation:

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fomenos

Answer:

Forces acting on an object may be balanced or unbalanced. When the forces  acting on an object have <u>equal</u> strength and act in opposite directions, they  are <u>balanced</u>. These forces cancel out one another, and the <u>motion</u> of the  object does not <u>change</u>.  When the forces acting on an object are <u>unbalanced</u>, they do not cancel out  one another. An unbalanced force acting on an object results in the object’s  motion <u>changing</u>. The object may change its <u>speed</u> (speed up or slow down), or it may change its <u>direction</u>.  <u>Friction</u> is a force that resists the motion or the tendency toward motion  between two objects in contact with each other.  <u>Gravity</u> is a force that pulls objects toward one another. For example, Earth  pulls all objects toward it.

Explanation:

3 0
3 years ago
A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4
Anna71 [15]

Answer:

31.652g of Na3PO4

Explanation:

We'll begin by calculating the molarity of Na3PO4 solution. This can be achieved as shown below:

Na3PO4 will dessicate in solution as follow:

Na3PO4(aq) —> 3Na+(aq) + PO4³¯(aq)

From the balanced equation above,

1 mole of Na3PO4 produce 3 moles of sodium ion, Na+.

Therefore, xM Na3PO4 will produce 1.10M sodium ion, Na+ i.e

xM Na3PO4 = (1.10 x 1)/3

xM Na3PO4 = 0.367M

Therefore, the molarity of Na3PO4 is 0.367M.

Next, we shall determine the number of mole of Na3PO4 in the solution. This is illustrated below:

Molarity of Na3PO4 = 0.367M

Volume = 525mL = 525/1000 = 0.525L

Mole of Na3PO4 =..?

Molarity = mole /Volume

0.367 = mole /0.525

Cross multiply

Mole of Na3PO4 = 0.367 x 0.525

Mole of Na3PO4 = 0.193 mole.

Finally, we shall convert 0.193 mole of Na3PO4 to grams. This is illustrated below:

Molar mass of Na3PO4 = (23x3) + 31 + (16x4) = 164g/mol

Mole of Na3PO4 = 0.193 mole

Mass of Na3PO4 =.?

Mass = mole x molar mass

Mass of Na3PO4 = 0.193 x 164

Mass of Na3PO4 = 31.652g

Therefore, 31.652g of Na3PO4 is needed to prepare the solution.

6 0
3 years ago
A tank contains 90 kg of salt and 2000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain
sashaice [31]

Answer:

a) 90 kg

b) 68.4 kg

c) 0 kg/L

Explanation:

Mass balance:

-w=\frac{dm}{dt}

w is the mass flow

m is the mass of salt

-v*C=\frac{dm}{dt}

v is the volume flow

C is the concentration

C=\frac{m}{V+(6-3)*L/min*t}

-v*\frac{m}{V+(6-3)*L/min*t}=\frac{dm}{dt}

-3*L/min*\frac{m}{2000L+(3)*L/min*t}=\frac{dm}{dt}

-3*L/min*\frac{dt}{2000L+(3)*L/min*t}=\frac{dm}{m}

-3*L/min*\int_{0}^{t}\frac{dt}{2000L+(3)*L/min*t}=\int_{90kg}^{m}\frac{dm}{m}

-[ln(2000L+3*L/min*t)-ln(2000L)]=ln(m)-ln(90kg)

-ln[(2000L+3*L/min*t)/2000L]=ln(m/90kg)

m=90kg*[2000L/(2000L+3*L/min*t)]

a) Initially: t=0

m=90kg*[2000L/(2000L+3*L/min*0)]=90kg

b) t=210 min (3.5 hr)

m=90kg*[2000L/(2000L+3*L/min*210min)]=68.4kg

c) If time trends to infinity the  division trends to 0 and, therefore, m trends to 0. So, the concentration at infinit time is 0 kg/L.

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