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3 years ago

If a wave has a wave height of 5 meters, in what depth of water would it begin to break?

Chemistry

1 answer:

Marysya12 [62]3 years ago

4 0

Answer:

6.5 meters

Explanation:

Background Knowledge:

The most familiar example of breaking wave is the breaking of water surface waves on a coastline.

Explanation:

The process of Wave breaking in most of the cases occurs when the amplitude reaches the point that the crest of the wave actually overturns

Factors by which waves are produced:

Different type of waves are produced and varied according to different type of factors. These factors are

Wind direction

Type of swell

Sea floor features

Slope of sea bed

Direct answer of the question:

In general rule, Waves start to break on reaching a water depth of 1.3 times the wave height.

In this question wave height is 5 meters so wave breaks on

5 × 1.3 = 6.5 meters

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The vapor pressure of liquid iodomethane, CH3I, is 40.0 mm Hg at 249 K.A sample of CH3I is placed in a closed, evacuated contain

lora16 [44]

Explanation:

It is given that vapor pressure of liquid iodomethane is 40.0 mm Hg. So, if we calculate the vapor pressure according to the given values and if its value will be greater than the the given vapor pressure of iodomethane then it means that some of the vapors has converted into liquid state.

As the given values are as follows.

$P_{1}$ = 72.0 mm Hg, $T_{1}$ = 404 K

$P_{2}$ = ? , $T_{2}$ = 249 K

As volume is constant so, according to Gay-Lussac's law pressure is directly proportional to temperature.

$P \propto T$ (at constant volume)

or, $\frac{P}{T}$ = k

Therefore, the formula to calculate the value of $P_{2}$ is as follows.

$\frac{P_{1}}{T_{1}}$ = $\frac{P_{2}}{T_{2}}$

$\frac{72.0 mm Hg}{404 K}$ = $\frac{P_{2}}{249 K}$

$P_{2}$ = 44.37 mm Hg

As calculated vapor pressure is more than the given vapor pressure. Hence, the liquid will convert into gas.

As a result, no condensation will occur and only vapors of iodomethane will be present.

5 0

2 years ago

100 PIONTSSSSS HELP ASAP

valina [46]

Left Panel

Short answer A

Solution

Since you have been given choices, my sloppy numbers will do, but it anyone is going to see this, YOU SHOULD CLEAN THEM UP WITH THE NUMBERS THAT COME FROM YOUR PERIODIC TABLE.

Equation

Sodium Phosphate + Calcium Chloride ===> Sodium Chloride + Calcium Phosphate.

Na3PO4 + CaCl2 ===> NaCl + Ca3(PO4)2

Step One

Balance the Equation

2Na2PO4 + 3CaCl2 ==> 6NaCl + Ca3(PO4)2

Step Two

Find the molar mass of CaCl2

Ca = 40

2Cl = 71

Molar Mass = 40 + 71 = 111 grams/mole

Step Three

Find the number of moles of CaCl2

Given mass = 379.4

Molar Mass = 111

moles = given Mass / molar Mass

moles of CaCl2 = 379.4/111 = 3.418 moles

Step Four

Find the number of moles of Ca3(PO4)2 needed.

This requires that you use the balance numbers from the balanced equation.

For every 3 moles of CaCl2 you have, you get 1 mole of Ca3(PO4)2

n_moles of Ca3(PO4)2 = 3.418 / 3 = 1.13933 moles

Step Five

Find the molar mass of Ca3(PO4)2

From the periodic table,

3Ca = 3 * 40 = 120

2 P = 2 * 31 = 62

8 O = 8 * 16 =128

Molar Mass = 120 + 62 + 128= 310 grams per mole.

Step Six

1 mole of Ca3(PO4)2 has a molar mass of 310 gram

1.13933 moles of Ca3(PO4)2 = x

x = 1.13933 moles * 310 grams /mole

x = 353.2 grams. As you can see, even with my rounding I'm only out 0.3 of a gram. DON'T FORGET TO PUT THIS TO THE PROPER SIG DIGS IF SOMEONE ELSE IS GOING TO SEE IT.

Middle Panel

Short Answer C

Equation

2HCl + Mg ===> H2 + MgCl2

The object of the first part of the game is to find the number of moles of H2.

Step One

Find the moles of HCl

1 mole HCl = 35.5 + 1 = 36.5

n = given mass divided by molar mass

n = 49 grams / 36.5 = 1.34 moles.

The balanced equation tells you that for ever mole of H2 produced, you need 2 moles of HCl. That's what the balance numbers are for.

So the number of moles of H2 is 1.34 / 2 = 0.671 moles of H2.

Now we come to Part II. We have to use an new friend of yours that I have seen only once before from you.

Find V using PV = nRT

R is going to be in kPa so the value of R = 8.314

V = ???

n = 0.671 moles

T = 25 + 273 = 298oK

P = 101.3 kPa

101.3 * V= 0.671*8.314 * 298

V = 0.671 * 8.314 * 298 / 101.3

V = 16.4

The answer is C and again, I have rounded almost everything except R, although it can go out to 8 places.

Right Panel

I can't see the panel. I don't know what the problem is. Never mind I got it. I'm going to be a little skimpy on this one since I've done two like it and they are long.

LiOH + HBr ===> LiBr + H2O and the equation is balanced.

You have to figure out the moles of LiOH and HBr. Use the LOWEST number of moles

n_LiOH = given mass / molar mass = 117/(7 + 16 + 1) = 117 / 24 = 4.875 moles

n_HBr = given mass / molar mass = 141/(1 + 80) = 141 / 81 = 1.741 moles

HBr is the lower number. That's all the LiBr you are going to get is 1.741. There is no adjustment to be made from the balance equation.

n = given mass / molar mass multiply both sides by the molar mass

n * Molar mass (LiBr) = n * (7 + 80) = 1.741 * 87 = 151 grams of

The answer is C

6 0

2 years ago

Which body changes occur during adolescence?

cupoosta [38]

The answer is C,growth spurts,puberty,& sexual maturity

6 0

2 years ago

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Calculate the moles of calcium chloride (CaCl2) needed to react in order to produce 85.00 grams of calcium carbonate (CaCO3). us

BartSMP [9]

Answer:

0.85 mole

Explanation:

Step 1:

The balanced equation for the reaction of CaCl2 to produce CaCO3. This is illustrated below:

When CaCl2 react with Na2CO3, CaCO3 is produced according to the balanced equation:

CaCl2 + Na2CO3 -> CaCO3 + 2NaCl

Step 2:

Conversion of 85g of CaCO3 to mole. This is illustrated below:

Molar Mass of CaCO3 = 40 + 12 + (16x3) = 40 + 12 + 48 = 100g/mol

Mass of CaCO3 = 85g

Moles of CaCO3 =?

Number of mole = Mass /Molar Mass

Mole of CaCO3 = 85/100

Mole of caco= 0.85 mole

Step 3:

Determination of the number of mole of CaCl2 needed to produce 85g (i.e 0. 85 mole) of CaCO3.

This is illustrated below :

From the balanced equation above,

1 mole of CaCl2 reacted to produced 1 mole of CaCO3.

Therefore, 0.85 mole of CaCl2 will also react to produce 0.85 mole of CaCO3.

From the calculations made above, 0.85 mole of CaCl2 is needed to produce 85g of CaCO3

5 0

2 years ago

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What is a liquid mixture whose parts are evenly blended?.

Ilia_Sergeevich [38]

Homogeneous mixture

5 0

1 year ago

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