Materials
 Computer with Internet access  14 Days: A Timeline article
Assignment Instructions
For this project, you are expected to submit two (2) items:
1. Completed Tracking Hurricane Katrina chart
2. Brief essay to summarize your findings about hurricane forecasting and advance warning
systems for hurricanes
Step 1: Prepare for the project.
a) Read through the guide before you begin so you know the expectations for this project.
b) If anything is not clear to you, be sure to ask your teacher.
Step 2: Read the background information about hurricanes.
a) Be sure you can answer these questions:
1.) How do hurricanes form?
2.) How are hurricanes classified?
3.) What kind of damage can hurricanes cause?
Step 3: Read the 14 Days: A Timeline article.
a) As you read, complete the Tracking Hurricane Katrina chart, which can be found at the end of this
document. Not all the dates will have every box filled.
b) Be sure that you note the date from the timeline along with the other information shown in the
chart.
Step 4: Dig deeper thr

Answer:

840 J

Explanation:

c ≈ 4200 J / (kg * °C)

m = 20 g = 0,02 kg

[tex]t_{1}[/tex] = 30 °C

[tex]t_{2}[/tex] = 40 °C

The formula is: Q = c * m * ([tex]t_{2} - t_{1}[/tex])

Calculating:

Q = 4200 * 0,02 * (40 - 30) = 840 (J)

Answer:

Moles of carbon atoms  = 3.33  ×  [tex]10^{-6}[/tex] mol

No. of atoms of C in Diamond  = 2.007 ×   [tex]10^{28}[/tex] atom

Atoms of graphite = 6.27 × [tex]10^{23}[/tex] Atoms

Explanation:

given data

Cost of 0.2g of diamond = $5000

Cost of 25 g of graphite = $ 2

solution

we know cost of 0.2g of diamond is $ 5000 so that for 1$

if buy 1$ = [tex]\frac{0.20}{5000}[/tex]

1$ = 4.0 × [tex]10^{-5}[/tex] g Carbon

and Moles of carbon atoms  is express as

Moles of carbon atoms = Given mass of Carbon ÷ atomic mass of C      .........1

Moles of carbon atoms  = 4.0  ×  [tex]10^{-5}[/tex]    g/ 2.0g

Moles of carbon atoms  = 3.33  ×  [tex]10^{-6}[/tex] mol

and

No. of atoms of C in Diamond = No. of moles × Avogadro NO    ..............2

No. of atoms of C in Diamond  = 3.33 ×   [tex]10^{-6}[/tex] mol × 6.022 ×   [tex]10^{28}[/tex]

No. of atoms of C in Diamond  = 2.007 ×   [tex]10^{28}[/tex] atom

Graphite

and wew have given Cost of 25 g of graphite is $2 so for but 1$ we get

for buy $1 = 25÷2  = 12.5 g Of graphite

Moles of graphite = 12.5÷12 = 1.04 mol

Atoms of graphite = 1.04 × 6.022 × 1023

Atoms of graphite = 6.27 × [tex]10^{23}[/tex] Atoms

The empirical and  molecular formula : C₂₁H₂₂N₂O₂

Given

The percent composition of an unknown substance is 75.42% Carbon, 6.63 % Hydrogen, 8.38 % Nitrogen, and 9.57 % Oxygen

Required

The empirical and  molecular formula

Solution

C : 75.42 : 12 = 6.285

H : 6.63 : 1 = 6.63

N : 8.38 : 14 = 0.599

O : 9.57 : 16 = 0.598

Divide by 0.598

C : H : N : O = 10.5 : 11 : 1 : 1 = 21 : 22 : 2 : 2

The empirical formula : C₂₁H₂₂N₂O₂

(C₂₁H₂₂N₂O₂)n = 334

(334)n=334

n = 1

Answer:

1 product is made. The left overs are 1 piece of bread and 1 slice of cheese

Explanation:

Answer:

There are 1.8021 ⋅ 1024 molecules of CH4 in 48 grams of CH4. To answer this question, you must understand how to convert grams of a molecule into the number of molecules. To do this, you have to utilize the concepts of moles and molar mass. A mole is just a unit of measurement. Avogadro's number is equal to 6.022 ⋅1023 molecules/mole. i think please dont complain to me if its wrong im sorry

Explanation:

Answer:

Follows are the solution to the given points:

Explanation:

In part 1:

As described and in the query, they become precipitated whenever the solutions are refrozen to [tex]25^{\circ} \ C[/tex].

Afterward, certain precipitate becomes replaced as well as the remaining water is evaporated, it implies that certain precipitate remained throughout the solution to just the container when the entire balance is evaporated.

The unrecoverable salt precipitates whenever the solvent is cooled at [tex]25^{\circ} \ C[/tex]and the remaining salt dissolves. It dissolved salt remains whenever the water is evaporated because as dissolved salt value is given that results can be achieved.

In part 2:

They have precipitation weight = [tex]0.063\ g[/tex]. They have a [tex]2 \ L[/tex] the solution, they may disregard the volume increases due to its precipitation. The intensity therefore is [tex]\frac{0.063}{2} = 0.0315 \ \frac{g}{L}[/tex]

Answer:

The answer is "Strontium iodide and [tex]SrI_2[/tex]".

Explanation:

In the flames test, two ionic metals give red fire. It was calcium (Ca, which gives a red brick flame) and strontium (Sr, which gives a persistent red flame). Both Ca and Sr respond to insoluble carbonate with ammonium carbonate. Ca and Sr produce insoluble ammonium phosphate phosphates. Even then, the water is reactive with calcium sulfate (CaSO4) whereas the strontium sulfate (SrSO4) is illiquid. Because the metal ion formed 3 precipitates, they can't get Ca, and we have Sr as its metal.

The hexane of its halide gives a violet color which would be typical of iodide (I-).

Strontium iodide is the chemical name of the unknown ionic compound.

Strontium iodide's chemical formula is [tex]SrI_2[/tex].

Answer:

Kc=[[CO][H2]3[CH4][H2O]

3.90=(0.30)(0.10)3[CH4]×0.02

[CH4]=0.023.90×0.30×(0.10)3=5.85×10−2 M

Thus, the concentration of methane in the mixture is 5.85×10−2 M.

Answer:

Love for Music

Explanation:

This is one example of many non-physical traits. In the womb a mother can listen to her favorite music and the growing baby could grow to like it in the womb!

This is just one of the many other traits that could be passed to their offspring.

Hope this Helps!

Answer:

Explanation:

In a gaseous reaction mixture partial pressure is proportion to mole of the gas concerned .

Pressure of the reactant gas from gas equation

PV = nRT

P = nRT / V

= .166 x .082 x ( 273+179) / 10

= .615 atm

C₃H₇OH   =     (CH₃)₂CO     +    H₂

before reaction moles in terms of pressure

.615                       0                      0

After reaction

.615 - x                     x                     x

.444 =  x² / ( .615 - x )

.273 - .444 x = x²

x² + .444 x - .273 = 0

x = .361 atm

So partial pressure of acetone is .361 atm at equilibrium.

Answer:

12408 feet per minute

Explanation:

Given: Speed is 141 mi/h

To find: speed in units of feet per minute

Solution:

Use the following units to convert the given speed into feet per minute.

1 mile = 5280 foot

1 h = 60 minutes

Therefore,

141 mi/h [tex]=\frac{141(5280)}{60} =12408[/tex] feet per minute.

Answer: i think its A diatomic compound..

Explanation: hope i helped! sorry if im wrong!

Answer:

SnCl4 * 5H2O

Explanation:

Answer:

A) { + } Ge-Se { -}

B) { + } Ge-Br { - }

C) { - } Br-Se { + }

Explanation:

The (-)ive sign shall be placed for the atom with higher electronegativity, while the other atom will be electropositive.  

a) Electronegativity of Ge = 2.01

Electronegativity of Se = 2.55

{ + } Ge-Se { -}

b) Electronegativity of Ge = 2.01

Electronegativity of Br = 2.95

{ + } Ge-Br { - }

c) Electronegativity of Br = 2.95

Electronegativity of Se = 2.55

{ - } Br-Se { + }

Answer:

m = 1.0164 g

% = 65.03%

Explanation:

First of all, we need to write the chemical reaction that is taking place here. We have the camphor being reduced to isoborneol:

C₁₀H₁₆O + NaBH₄ -----------> C₁₀H₁₈O + NaBH₂

We have a 1:1 mole ratio between Camphor and isoborneol, so, the moles of the camphor will be the same moles produced of isoborneol.

To get the theorical yield we need to calculate the theorical moles produced of isoborneol, then, the mass and compare it to the given mass. In that way we will get the %yield.

The Molar mass of camphor and isoborneol are:

MM C₁₀H₁₆O = (16*1) + (12*10) + 16 = 152 g/mol

MM C₁₀H₁₈O = (18*1) + (12*10) + 16 = 154 g/mol

The moles of camphor will be:

molesC₁₀H₁₆O = 1 / 152 = 0.0066 moles

The mass produced then of isoborneol should be:

mC₁₀H₁₈O = 0.0066 * 154

Now, the %yield would be:

% = (0.661 / 1.0164) * 100

Hope this helps

Answer:

0.591 g of magnesium phosphate is the theoretical yield.

Magnesium nitrate is the limiting reactant.

Explanation:

Hello!

In this case, since the balanced reaction turns out:

[tex]3Mg(NO_3)_2+2Na_3PO_4\rightarrow Mg_3(PO_4)_2+6NaNO_3[/tex]

Next, we compute the grams of magnesium phosphate yielded by each reactant, considering the present mole ratios and molar masses:

[tex]m_{Mg_3(PO_4)_2}^{by\ Mg(NO_3)_2}=1.00gMg(NO_3)_2*\frac{1molMg(NO_3)_2}{148.31gMg(NO_3)_2}*\frac{1molMg_3(PO_4)_2}{3molMg(NO_3)_2} *\frac{262.86gMg_3(PO_4)_2}{1molMg_3(PO_4)_2} \\\\m_{Mg_3(PO_4)_2}^{by\ Mg(NO_3)_2}= 0.591gMg_3(PO_4)_2\\\\m_{Mg_3(PO_4)_2}^{by\ Na_3PO_4}=1.00gNa_3PO_4*\frac{1molNa_3PO_4}{163.94gNa_3PO_4}*\frac{1molMg_3(PO_4)_2}{2molNa_3PO_4} *\frac{262.86gMg_3(PO_4)_2}{1molMg_3(PO_4)_2} \\\\m_{Mg_3(PO_4)_2}^{by\ Na_3PO_4} = 0.802gMg_3(PO_4)_2[/tex]

Thus, we infer that the correct theoretical yielded mass is 0.591 g as magnesium nitrate is the limiting reactant for which it produces the fewest grams of product.

However, is not possible to compute the percent yield since no actual yield is given, and must be provided or indicated by the problem or an experiment and it not here, nevertheless, you may compute the percent yield by dividing the actual yield by the theoretical and then multiplying by 100:

[tex]Y=\frac{actual}{0.591g}*100\%[/tex]

Best regards!

[tex]\tt -\dfrac{1}{2}\dfrac{d[N_2O]}{dt}=\dfrac{1}{2}\dfrac{d[N_2]}{dt}=\dfrac{1}{1}\dfrac{d[O_2]}{dt}[/tex]

Reaction

2N2O(g) — 2N2(g) + O2(g)

Required

relative rate

Solution

The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.

so the relative rates for the reaction above are :

[tex]\tt -\dfrac{1}{2}\dfrac{d[N_2O]}{dt}=\dfrac{1}{2}\dfrac{d[N_2]}{dt}=\dfrac{1}{1}\dfrac{d[O_2]}{dt}[/tex]

[tex]{ \boxed{ \bold{ \blue{Question}}}}[/tex]

Is H2SO4 an element, compound or molecule?

[tex]{ \boxed{ \bold{ \blue{Answer}}}}[/tex]

H2SO4 (sulphuric acid) is a compound. It is made up of 2 hydrogen atoms, 1 sulphur atom, and 4 oxygen atoms.

_______________________________

[tex] \underline \bold \orange{hope \: it \: helps}[/tex]

A chemical compound is any substance that consists of two or more different elements combined together.

I think that H2SO4 fits this definition of a compound .

HOPE IT HELPS

PLEASE MARK ME BRAINLIEST ☺️

The molar concentration, often called molarity, describes how much of a substance (a solute) is present per unit of solvent. By definition, the molar concentration (M) is equal to the number of moles (n) of solute divided by the number of liters (the volume, or V) of the solution.

Here, your solute is potassium nitrate, or KNO3. You're given the mass of KNO3 (505 g), but you need to convert this quantity to moles before you can find the molarity. To go from mass to moles, simply divide the mass of the substance by its molar mass (given to you as 101.1 g/mol).

[tex]505 \text{ g KNO}_\text{3} \div 101.1 \text{ g KNO}_3/\text{mol KNO}_{3} = 4.995 \text{ mol KNO}_3[/tex]

Now that we have the moles of solute, we divide by the liters of solution. We're given the volume of solution in milliliters, so to convert to liters, simply divide by 1000 (1 L = 1000 mL, so 1 mL = 1/1000 mL). Our volume of solution is thus 0.250 L.

Finally, we can calculate the molar concentration of the KNO3 solution:

[tex]4.995 \text{ mol KNO}_{3 }\div 0.250 \text{ L} = 19.98 \text{ }\frac{\text{moles}}{\text{liters}}\text{ of KNO}_3[/tex]

But, we're told to round our answer to three sig figs. Thus, our rounded and final answer would be 20.0 moles/liters of KNO3.

Answer:

[tex]m_{leftover}=37g[/tex]

Explanation:

Hello!

In this case, since the melting process involves the heat added to the system, the heat of fusion and the mass of water:

[tex]Q=m\Delta H_{fus}[/tex]

Thus, as it has been reported that the heat of fusion of ice is 333.6 J/g, we can compute the melted mass of ice as shown below:

[tex]m=\frac{Q}{\Delta H_{fus}} =\frac{46000J}{333.6J/g}\\\\m=138g[/tex]

Thus, the ice leftover results:

[tex]m_{leftover}=175g-138g\\\\m_{leftover}=37g[/tex]

Regards!

Answer:

Yes it is because it effects the food chain in many ways like it takes more time to get rid of them and they eat parasites that other living things need. We can get rid of zebra mussels by removing them little by little or putting animals in the water that eats them.

Explanation:

Answer:

By encouraging boaters to carefully clen, drain and dry their boats before launching them in different bodies of water.

Explanation:

Answer:

0.73g/cm^3

Explanation:

d=m/v

d=11/15

d=0.73

Answer:

1.01 grams/ cm^3

Explanation:

because that's just how it is

91.2 g Mn

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Atomic Structure

Stoichiometry

Step 1: Define

[Given] 1.00 × 10²⁴ atoms Mn

Step 2: Identify Conversions

Avogadro's Numer

[PT] Molar Mass of Mn - 54.94 g/mol

Step 3: Convert

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

91.2321 g Mn ≈ 91.2 g Mn

Answer:

BUDDY YO

Explanation:

The boiling point of water generally increases as the amount of impurities (which a solute like KCl technically can be thought of) dissolved increases. This relation can be quantified using the equation,

[tex]\Delta T_b = i \times K_b \times m[/tex]

where [tex]\Delta{T}_{b}[/tex] is the change in the water's boiling point (normally taken to be 100 °C), [tex]i[/tex] is the Van 't Hoff factor (the number of particles a single formula unit of the solute dissociates into in water), [tex]K_b[/tex] is the boiling point elevation constant, and [tex]m[/tex] is the molality (moles of solute/kilogram(s) of solvent) of the solution.

We are forming a solution by dissolving KCl in water. KCl is an electrolyte that, in water, will dissociate into K⁺ and Cl⁻ ions. So, for every formula unit, KCl, we obtain two particles. Thus, the Van 't Hoff factor, or [tex]i[/tex], will be 2.

The molality of the solution can be calculated by dividing the number of moles of KCl by the mass of water in kilograms. Since we have 1.00 kg of water, we would be dividing 0.75 mol KCl by 1, giving us a molality (m) of 0.75 m.

We aren't provided the boiling point elevation constant for water. Several authoritative sources give the value 0.512 °C/m, so we will adopt that as our [tex]K_b[/tex].

Note: m = mol/kg as used in this problem.

Plugging everything in,

[tex]\Delta T_b = i \times K_b \times m \\\Delta T_b = 2 \times 0.512 \text{ } \frac{^oC}{mol/kg} \times 0.75 \text{ } \frac{mol}{kg} \\\Delta T_b = 0.768 \text{ } \mathrm{ ^oC}[/tex]

As you can see, our change in boiling point is positive (the boiling point is elevated), and it is also quite modest. Taking 100 °C to be the boiling point of pure water, the boiling point of our solution would be 100 ⁰C + 0.768 ⁰C, or 100.768 ⁰C.

If we are considering significant figures, then we must give our answer to two significant figures (since 0.75 has two sig figs). We can regard the boiling point of water (100 ⁰C) as a defined value. Since our final answer is a sum, the boiling point of our solution to two significant figures would be 100.77 ⁰C.

Given:

We know,

The molality of the solution will be:

= [tex]\frac{Mole}{Mass}[/tex]

= [tex]\frac{0.75}{1}[/tex]

= [tex]0.75 \ m[/tex]

Now,

→ [tex]T_{solution}-T_{water} = Kb\times m\times i[/tex]

By putting the values, we get

                              [tex]= 0.51\times 0.75\times 2[/tex]

                              [tex]= 0.765[/tex]

hence,

Solution's boiling point will be:

→ [tex]T_{solution} = 100+0.765[/tex]

                [tex]= 100.765^{\circ} C[/tex]

Thus the above approach is right.

Learn more about boiling point here:

https://brainly.com/question/23549697

Answer:

Formulas

3.2 Determining Empirical and Molecular Formulas

Learning Objectives

By the end of this section, you will be able to:

Compute the percent composition of a compound

Determine the empirical formula of a compound

Determine the molecular formula of a compound

The previous section discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, one may determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, these same principles will be applied to derive the chemical formulas of unknown substances from experimental mass measurements.

Percent Composition

The elemental makeup of a compound defines its chemical identity, and chemical formulas are the most succinct way of representing this elemental makeup. When a compound’s formula is unknown, measuring the mass of each of its constituent elements is often the first step in the process of determining the formula experimentally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:

%H=mass Hmass compound×100%

%C=mass Cmass compound×100%

If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:

%H=2.5g H10.0g compound×100%=25%

%C=7.5g C10.0g compound×100%=75%

EXAMPLE 3.9

Calculation of Percent Composition

Analysis of a 12.04-g sample of a liquid compound composed of carbon, hydrogen, and nitrogen showed it to contain 7.34 g C, 1.85 g H, and 2.85 g N. What is the percent composition of this compound?

Solution

To calculate percent composition, divide the experimentally derived mass of each element by the overall mass of the compound, and then convert to a percentage:

%C=7.34g C12.04g compound×100%=61.0%%H=1.85g H12.04g compound×100%=15.4%%N=2.85g N12.04g compound×100%=23.7%

The analysis results indicate that the compound is 61.0% C, 15.4% H, and 23.7% N by mass.

Check Your Learning

A 24.81-g sample of a gaseous compound containing only carbon, oxygen, and chlorine is determined to contain 3.01 g C, 4.00 g O, and 17.81 g Cl. What is this compound’s percent composition?

ANSWER:

12.1% C, 16.1% O, 71.8% Cl

Determining Percent Composition from Molecular or Empirical Formulas

Percent composition is also useful for evaluating the relative abundance of a given element in different compounds of known formulas. As one example, consider the common nitrogen-containing fertilizers ammonia (NH3), ammonium nitrate (NH4NO3), and urea (CH4N2O). The element nitrogen is the active ingredient for agricultural purposes, so the mass percentage of nitrogen in the compound is a practical and economic concern for consumers choosing among these fertilizers. For these sorts of applications, the percent composition of a compound is easily derived from its formula mass and the atomic masses of its constituent elements. A molecule of NH3 contains one N atom weighing 14.01 amu and three H atoms weighing a total of (3 × 1.008 amu) = 3.024 amu. The formula mass of ammonia is therefore (14.01 amu + 3.024 amu) = 17.03 amu, and its percent composition is:

%N=14.01amu N17.03amuNH3×100%=82.27%%H=3.024amu H17.03amuNH3×100%=17.76%

This same approach may be taken considering a pair of molecules, a dozen molecules, or a mole of molecules, etc. The latter amount is most convenient and would simply involve the use of molar masses instead of atomic and formula masses, as demonstrated Example 3.10. As long as the molecular or empirical formula of the compound in question is known, the percent composition may be derived from the atomic or molar masses of the

Answer:

Given:

New pressure: approximately [tex]3.4\times 10\; \rm atm[/tex] ([tex]34\; \rm atm[/tex].) (Assuming that the gas is an ideal gas.)

Explanation:

Both the volume and the temperature of this gas has changed. Consider the two changes in two separate steps:

By Boyle's Law, the pressure of an ideal gas is inversely proportional to the volume of this gas (assuming constant temperature and that no gas particles escaped or was added.)

For this gas, [tex]V_0 = 45\; \rm m^{3}[/tex] while [tex]V_1 = 5.0\; \rm m^{3}[/tex].

Let [tex]P_0[/tex] denote the pressure of this gas before the volume change ([tex]P_0 = 3.65\; \rm atm[/tex].) Let [tex]P_1[/tex] denote the pressure of this gas after the volume change (but before changing the temperature.) Apply Boyle's Law to find the ratio between [tex]P_1\![/tex] and [tex]P_0\![/tex]:

[tex]\displaystyle \frac{P_1}{P_0} = \frac{V_0}{V_1} = \frac{45\; \rm m^{3}}{5.0\; \rm m^{3}} = 9.0[/tex].

In other words, because the final volume is [tex](1/9)[/tex] of the initial volume, the final pressure is [tex]9[/tex] times the initial pressure. Therefore:

[tex]\displaystyle P_1 = 9.0\times P_0 = 32.85\; \rm atm[/tex].

On the other hand, by Amonton's Law, the pressure of an ideal gas is directly proportional to the temperature (in degrees Kelvins) of this gas (assuming constant volume and that no gas particle escaped or was added.)

Convert the unit of the temperature of this gas to degrees Kelvins:

[tex]T_1 = (25 + 273.15)\; \rm K = 298.15\; \rm K[/tex].

[tex]T_2 = (35 + 273.15)\; \rm K = 308.15\; \rm K[/tex].

Let [tex]P_1[/tex] denote the pressure of this gas before this temperature change ([tex]P_1 = 32.85\; \rm atm[/tex].) Let [tex]P_2[/tex] denote the pressure of this gas after the temperature change. The volume of this gas is kept constant at [tex]V_2 = V_1 = 5.0\; \rm m^{3}[/tex].

Apply Amonton's Law to find the ratio between [tex]P_2[/tex] and [tex]P_1[/tex]:

[tex]\displaystyle \frac{P_2}{P_1} = \frac{T_2}{T_1} = \frac{308.16\; \rm K}{298.15\; \rm K}[/tex].

Calculate [tex]P_2[/tex], the final pressure of this gas:

[tex]\begin{aligned} P_2 &= \frac{308.15\; \rm K}{298.15\; \rm K} \times P_1 \\ &= \frac{308.15\; \rm K}{298.15\; \rm K} \times 32.85\; \rm atm \approx 3.4 \times 10\; \rm atm\end{aligned}[/tex].

In other words, the pressure of this gas after the volume and the temperature changes would be approximately [tex]3.4\times 10\; \rm atm[/tex].