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3 years ago

PLZ HELP Which factors play a role in creating ocean waves?

Chemistry

2 answers:

Irina18 [472]3 years ago

5 0

Answer:

<em>Hello, The factors that play a role in creating ocean waves are the wind speed, direction, duration, and fetch. Ocean waves are usually produced by the wind that moves its energy to the water. The size of the wave can be determined by the wind speed, direction, duration, and fetch (area in which the wind blows). This changeability resulted in waves of different sizes and shapes. So, The answer is B. Hope That Helps!</em>

Yuki888 [10]3 years ago

4 0

Answer:

all of these

Explanation:

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A 45-g aluminium spoon(specific heat 0.80 / J/gdegree Celsius) at 24 degree celsius is placed in 180 ml(180 grams) of coffee at

Vlad [161]

Explanation:

a) The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, $heat_{absorbed}=heat_{released}$

To calculate the amount of heat released or absorbed, we use the equation:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

Also,

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ..........(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminium = 45 g

$m_2$ = mass of coffee = 180 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of aluminium = $24^oC$

$T_2$ = temperature of coffee = $85^oC$

$c_1$ = specific heat of aluminium = $0.80J/g^oC$

$c_2$ = specific heat of coffee= $4.186 J/g^oC$

Putting all the values in equation 1, we get:

$45 g\times 0.80J/g^oC\times (T_{final}-24^oC)=-[180 g\times 4.186J/g^oC\times (T_{final}-83^oC)]$

$T_{final}=80.30^oC$

80.30 °C is the final temperature.

b) Energy flows from higher temperature to lower temperature.Whenever two bodies with different energies and temperature come in contact. And the resulting temperature of both bodies will less then the body with high temperature and will be more then the body with lower temperature.

So, is our final temperature of both aluminium and coffee that is 80°C less than initial temperature of coffee and more than the initial temperature of the aluminum.

8 0

3 years ago

You used a calorimeter in the heat transfer lab. Explain how the calorimeter works, and how to calculate the heat given off or a

Butoxors [25]

A calorimeter contains reactants and a substance to absorb the heat absorbed. The initial temperature (before the reaction) of the heat absorbent is measured and then the final temperature (after the reaction) is also measured. The absorbent's specific heat capacity and mass are also known. Given all of this data, the equation:
Q = mcΔT
To find the heat released.

3 0

3 years ago

Calculate the ratio of the mass ratio of SS to OO in SOSO to the mass ratio of SS to OO in SO2SO2. Consider a sample of SOSO in

Veronika [31]

Answer:

The ratio of the mass ratio of S to O; in SO, to the mass ratio of S to O; in SO₂, is 2:1

Explanation:

According to the consideration, let us first find the ratio of S and O in both the compounds

For SO:

$\frac{m_{S} }{m_{O} }= \frac{32}{16}\\\\ \frac{m_{S} }{m_{O} }= 2$

Let us express it as

$SO_{\frac{m_{S} }{m_{O} }} = 2$

For SO₂,

Due to two oxygen atoms in the molecule, the mass of oxygen will be taken two times

$\frac{m_{S} }{m_{O} }= \frac{32}{(2)(16)}\\\\ \frac{m_{S} }{m_{O} }= 1$

Let us express it as

$SO_2_{\frac{m_{S} }{m_{O} }}= 1$

Now, for the ratio of both the above-calculated ratios,

$\frac{SO_{\frac{m_{S} }{m_{O} }}}{SO_2_{\frac{m_{S} }{m_{O} }}}=\frac{2}{1}$

The required ratio is 2:1

3 0

3 years ago

how do the properties of compounds compare to the properties of the elements from which they are composed

Crank

<span>The elements that form the compound don't compare because it doesn't matter about the elements when they are combined. The elements that formed it together is completely different than the outcome compound. </span>

6 0

3 years ago

what is the limiting reactant when 1.50g of lithium and 1.50 g of nitrogen combine to form lithium nitride, a component of advan

PSYCHO15rus [73]

<h3>Answer:</h3>

Limiting reactant is Lithium

<h3>Explanation:</h3>

<u>We are given;</u>

Mass of Lithium as 1.50 g
Mass of nitrogen is 1.50 g

We are required to determine the rate limiting reagent.

First, we write the balanced equation for the reaction

6Li(s) + N₂(g) → 2Li₃N

From the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.

Second, we determine moles of Lithium and nitrogen given.

Moles = Mass ÷ Molar mass

Moles of Lithium

Molar mass of Li = 6.941 g/mol

Moles of Li = 1.50 g ÷ 6.941 g/mol

= 0.216 moles

Moles of nitrogen gas

Molar mass of Nitrogen gas is 28.0 g/mol

Moles of nitrogen gas = 1.50 g ÷ 28.0 g/mol

= 0.054 moles

According to the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.
Therefore, 0.216 moles of lithium will require 0.036 moles (0.216 moles ÷6) of nitrogen gas.
On the other hand, 0.054 moles of nitrogen, would require 0.324 moles of Lithium.

Thus, Lithium is the limiting reagent while nitrogen is in excess.

7 0

3 years ago