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3 years ago

What are the two different types of ocean currents, and how are they formed?

Chemistry

1 answer:

suter [353]3 years ago

3 0

Answer:

Explanation:

There are two main types of ocean currents: currents driven mainly by wind and currents mainly driven by density differences. Density depends on temperature and salinity of the water. Cold and salty water is dense and will sink. Warm and less salty water will float.

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A man heats a balloon in the oven. If the balloon initially has a volume of 0.4 liters and a temperature of 20 degrees celsius,

Airida [17]

Answer:

0.714 liter.

Explanation:

Given:

The balloon initially has a volume of 0.4 liters and a temperature of 20 degrees Celsius.

It is heated to a temperature of 250 degrees Celsius.

Question asked:

What will be the volume of the balloon after he heats it to a temperature of 250 degrees Celsius ?

Solution:

By using:

$PV=nRT$

Assuming pressure as constant,

V∝ T

Now, let K is the constant.

V = KT

Let initial volume of balloon , $V_{1}$ = 0.4 liter

1000 liter = 1 meter cube

1 liter = $\frac{1}{1000} m^{3} = 10^{-3} m^{3$

0.4 liter = $0.4\times10^{-3}=4\times10^{-4} m^{3}$

And initial temperature of balloon, $T_{1}$ = 20°C = (273 + 20)K

= 293 K

Let the final volume of balloon is $V_{2}$

And a given, final temperature of balloon, $T_{2}$ is 250°C = (273 + 250)K

= 523 K

Now, $V_{1}$ = $KT_{1}$

$4\times10^{-4}=K\times293\ (equation\ 1 )$

$V_{2}$ = $KT_{2}$

$=K\times523\ (equation 2)$

Dividing equation 1 and 2,

$\frac{4\times10^{-4}}{V_{2} } =\frac{K\times293}{K\times523}$

K cancelled by K.

By cross multiplication:

$293V_{2} =4\times10^{-4} \times523\\V_{2} =\frac{ 4\times10^{-4} \times523\\}{293} \\ = \frac{2092\times10^{-4}}{293} \\ =7.14\times10^{-4}m^{3}$

Now convert it into liter with the help of calculation done above.

$7.14\times10^{-4} \times1000\\7.14\times10^{-4} \times10^{3} \\0.714\ liter$

Therefore, the volume of the balloon be after he heats it to a temperature of 250 degrees Celsius is 0.714 liter.

5 0

3 years ago

Which of the following is a product of aerobic respiration?

sasho [114]

Answer:

The product of aerobic respiration is Carbon dioxide.

Explanation:

The process of breaking down glucose to produce energy and waste products is called respiration. Livings beings need respiration process to generate energy so that they can survive.
The types of respiration are : Anaerobic and aerobic respiration.
Aerobic respiration takes place in presence of oxygen and produces large amount of energy.
The final product of aerobic respiration are carbon dioxide, water and 38 ATP of energy.

3 0

3 years ago

Read 2 more answers

Si Units Worksheet. Use si units to complete

Keith_Richards [23]

$\\ \bull\tt\dashrightarrow 2000m=2km$

$\\ \bull\tt\dashrightarrow 9000mL=9L$

$\\ \bull\tt\dashrightarrow 6L=6000mL$

$\\ \bull\tt\dashrightarrow 6cm=60mm$

$\\ \bull\tt\dashrightarrow 1000m=1km$

$\\ \bull\tt\dashrightarrow 50mm=5cm$

$\\ \bull\tt\dashrightarrow 4L=4000mL$

$\\ \bull\tt\dashrightarrow 10000g=10kg$

$\\ \bull\tt\dashrightarrow 10000m=10km$

$\\ \bull\tt\dashrightarrow 4000g=4kg$

$\\ \bull\tt\dashrightarrow 9km=9000m$

$\\ \bull\tt\dashrightarrow 3kg=3000g$

$\\ \bull\tt\dashrightarrow 90mm=9cm$

$\\ \bull\tt\dashrightarrow 4km=4000m$

$\\ \bull\tt\dashrightarrow 2000g=2kg$

$\\ \bull\tt\dashrightarrow 400cm=4m$

$\\ \bull\tt\dashrightarrow 3km=3000m$

$\\ \bull\tt\dashrightarrow 2cm=20mm$

$\\ \bull\tt\dashrightarrow 9000g=9kg$

8 0

3 years ago

What is the molarity of a solution that contains 0.082 mol KI in a 2.03 L solution?

astraxan [27]

Answer:

molarity of the KI solution = 0.04 mol/L

Explanation:

Molarity (M) is the concentration of a solution expressed as the number of moles of solute per liter of solution.

The law that we can applied to calculate the M is:

M = n / V

n - number of moles

V- volume of the solution (liters)

Then insert in the equation the values from the question;

M = 0.082 mol / 2.03 L = 0.04 mol/L

4 0

3 years ago

Choose the correct statement regarding the behavior of water.Group of answer choicesThe heat capacity of liquid water is greater

vodka [1.7K]

Answer:

The water phase with the smallest temperature increase when adding 10 kcal of heat is solid ice.

Explanation:

The rest of the statements are incorrect. The density of ice is lower than the density of water. The heat capacity of solid ice is greater almost twice the heat capacity of the liquid water. The heat capacity of vapors is less than heat capacity of liquid.

3 0

3 years ago