how can you find the Frequency if you only have the wavelength? can you solve this example what is the Frequency of the wavelength 4km? can you please show your steps thanks.

Answer:

19.59 g MgCl2 to the nearest hundredth.

Explanation:

Mg (s) + 2HCI (aq) → MgCl₂ (aq) + H₂ (g)

1 mole of Mg produces 1 mole of MgCl2

Using the molar masses of the elements:

24.305 g  Mg produces (24.305 + 2 * 35.45) g MgCl2

So 5g --->  5 *  (24.305 + 2 * 35.45) / 24.305

=  19.59 g MgCl2 to the nearest hundredth.

Answer:

its B

Explanation:

a p e x

Answer:

the answer is b. allows electricity to flow easily

Answer:

Get 3 glass vessels that are sealed except for the opening. If they are test tubes, get 2 rubber stoppers to fit two of the three. If they are clean, dry jelly jars with screw-on lids, then have two lids available for two of the three and use some stretchable plastic wrap to put over the mouth of the jars before affixing the caps (later). Get a piece of painted sheetmetal (iron) , or similar iron like nails that are not coated with zinc galvanization, nor plated with brass or nickel. Whatever they are, make sure they fit within the glass containers you choose, and are all the same size., and from the same sheet or batch of nails. Using some rough, 80-grit sandpaper, abrade the surface until the steel is bared of any paint or coating. throw one piece into the open trst tube ( or jar). In the second container put moisture-absorbing desiccant, place the stretchable wrap over the jar’s mouth then screw on the cap, or put metal, desiccant, then plug into the test tube. This will be the test for moisture significance. In the third test tube, place the metal in, insert a flaming wood splint or kitchen match until the fire goes out then withdraw the splint (match) and tightly plug it shut. If using the jar, place a small birthday-cake candle in it by lighting the candle, dripping the wax, and sticking the base of the candle to the hot wax at the inside jar’s bottom, until it sets upright. Then light it, place the metal in the jar in a place where the soot won’t contaminate the metal (off to one side) then with the candle burning, place a stretchable seal over the top and screw the lid on. The candle will consume a lot of the oxygen so this will be the test for lack of oxygen. The open container is the ‘control’ It is exposed to the daily oxygen and moisture content naturally occurring in your area. Keep the jars separated by a foot or so from each other so no one can say that it was magnetic effects or electric charge that was responsible for the rusting. Start your observation by logging each day, the progress of rusting. Try and quantify it by color, and percentage of area covered by rust, then perhaps apparent depth of rust penetration. You might be able to get desiccant at a pharmacy, or in a hardware store where people use it to remove excess humidity from closets

Explanation:

Choose the products that complete the reaction. The chemical equation may not be balanced.

NaHCO3 + HC2H3O2

Answer: NaC2H3O2 + CO2 + H2O

Answer:

1.64 moles O₂

Explanation:

Part A:

Remember 1 mole of particles = 6.02 x 10²³ particles

So, the question becomes, how many  '6.02 x 10²³'s are there in 9.88 x 10²³ molecules of O₂?

This implies a division of given number of particles by 6.02 x 10²³ particles/mole.

∴moles O₂ = 9.88 x 10²³ molecules O₂ / 6.02 x 10²³ molecules O₂ · mole⁻¹ = 1.64 mole O₂

_______________

Part B needs an equation (usually a combustion of a hydrocarbon).

Collecting DNA sample sin plastic container will rupture the originality of DNA , and will decrease the authenticity of the sample.

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Since we are starting from the number of atoms of Sulfur, we need to know two sets of formulas:

        ⇒ mass = (atoms ÷ Avogadro's Number) ×  molar mass

  mass = [(2.01 × 10²⁴ atoms) ÷ (6.022 × 10²³ atoms/mole)] × (32.06 g/mol)  

            = 107.01 g

The molar mass of sulfur S is 32.06 g/mol, then the mass in grams of a sample of S containing 2.01x10^24 atoms  is 107.01 g

The molar mass can be defined as the weight of one sample mole, by Multiplying  the subscript means  the number of atoms times that element’s atomic mass and add the masses of all the elements in the molecule to obtain the molecular mass.

Molar mass is  expressed in either gram ( g) or kilograms (kg).

Mass  = moles × molar mass

moles = atoms ÷ Avogadro's Number

       ⇒ mass = (atoms ÷ Avogadro's Number) ×  molar mass

 mass = [(2.01 × 10²⁴ atoms) ÷ (6.022 × 10²³ atoms/mole)] × (32.06 g/mol)  

          = 107.01 g

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Answer:

Robot 1 : 150 J

Robot 2: 40 J

Explanation:

30 N x 5m = 150 J

20 N x 2m = 40 J

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The coefficients of Fe^3+ and Fe in the balanced equation are 2 and 2, respectively.

The balanced chemical equation for the given reaction is:

3 Mg + 2 Fe^3+ ---> 3 Mg^2+ + 2 Fe

This means that for every 3 moles of magnesium (Mg) that react, 2 moles of iron(III) ions (Fe^3+) are required. In the balanced equation, two Fe^3+ ions react with three Mg atoms to produce two Fe atoms and three Mg^2+ ions.

Therefore, the coefficients of Fe^3+ and Fe in the balanced equation are determined by the requirement to maintain the conservation of mass and charge on both sides of the equation.

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Answer:

A) Chemical Change

Explanation:

Chemical changes occur when bonds are broken and formed between molecules or atoms. This means that one substance with a particular set of properties (such as melting point, color, taste, etc.) is turned into a different substance with different properties.

The number of moles of oxygen gas needed to form the iron (II) oxide is 0.517 moles.

From the balanced chemical equation:

iron(s) + oxygen(g) → iron(II) oxide(s)

The mole ratio between iron and oxygen is 1:1, which means that for every 1 mole of iron that reacts, 1 mole of oxygen is needed.

Therefore, to form 0.517 moles of iron(II) oxide, 0.517 moles of oxygen gas are also needed.

In conclusion, the number of moles of Oxygen gas needed to form 0.517 moles of Iron Oxide is the same 0.517 moles.

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Approximately 12.45 grams of solid silver chromate will precipitate when 150 mL of 0.500 M silver nitrate solution is added to excess potassium chromate.

To determine the number of moles of AgNO3 present in 150 mL of a 0.500 M AgNO3 solution, we can use the formula:

moles = concentration × volume

Given:

Concentration of AgNO3 solution = 0.500 M

Volume of AgNO3 solution = 150 mL

First, we need to convert the volume from milliliters (mL) to liters (L) since the concentration is given in moles per liter (M).

1 L = 1000 mL

Therefore, the volume of the AgNO3 solution in liters is:

150 mL × (1 L / 1000 mL) = 0.150 L

Now we can calculate the moles of AgNO3 using the formula:

moles = concentration × volume

moles = 0.500 M × 0.150 L

moles = 0.075 mol

So, there are 0.075 moles of AgNO3 present in 150 mL of the 0.500 M AgNO3 solution.

Now, let's proceed to determine how many grams of solid silver chromate (Ag2CrO4) will precipitate when the AgNO3 solution reacts with excess potassium chromate (K2CrO4).

From the balanced chemical equation:

2AgNO3(aq) + K2CrO4(aq) → Ag2CrO4(s) + 2KNO3(aq)

We can see that the molar ratio between AgNO3 and Ag2CrO4 is 2:1. Therefore, for every 2 moles of AgNO3, we will form 1 mole of Ag2CrO4.

Since we have 0.075 moles of AgNO3, we can calculate the moles of Ag2CrO4 formed:

moles of Ag2CrO4 = 0.075 mol / 2 = 0.0375 mol

To determine the mass of Ag2CrO4, we need to multiply the moles by its molar mass. The molar mass of Ag2CrO4 is calculated by summing the atomic masses of each element in the compound:

Ag2CrO4 = 2(Ag) + 1(Cr) + 4(O) = 2(107.87 g/mol) + 1(52.00 g/mol) + 4(16.00 g/mol) = 331.87 g/mol

mass of Ag2CrO4 = moles of Ag2CrO4 × molar mass of Ag2CrO4

mass of Ag2CrO4 = 0.0375 mol × 331.87 g/mol = 12.45 g

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Answer:

sodium

Explanation:

sodium has 11 electrons.

Answer:

they both revolve or rotate around something

Explanation:

The earth revolves around the sun and the moon rotates around the earth.

The number of atoms of hydrogen in 12.26 pounds of sugar (C₆H₁₂O₆) is 2.23 × 10²⁶.

Given information,

Mass of sugar = 12.26 pounds

12.26 lb ÷ 2.20 lb/kg = 5.57 kg = 5570 grams

The molar mass of sugar = 12 × 6 + 1 × 12 + 16 × 6

Total molar mass = 72.06 + 12.12 + 96.00 = 180.18 g/mol

Number of moles of sugar = Mass / Molar mass

Number of moles = 5570 g / 180.18 g/mol = 30.89 mol

Number of moles of hydrogen = 30.89 mol × 12 = 370.68 mol

Number of atoms of hydrogen = The number of moles × Avogadro's number

Number of atoms of hydrogen = 370.68 × 6.022 × 10²³ ≈ 2.23 × 10²⁶ atoms

Therefore, there are approximately 2.23 × 10²⁶ atoms of hydrogen in 12.26 pounds of sugar (C₆H₁₂O₆).

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Answer:

12 H, 18 O, 6 C

Explanation:

\(6H_{2} O + 6CO_{2} -- > C_{6}H_{12} + 6O_{2}\\\)

___________________

first, we'll fo the h on the first equation: \(6H_{2}\)

multiply the 6 and 2 to get twelve.

in the second equation, we see that there is \(H_{12}\)

so, in both there are 12 H.

__________________

Now we'll do O.

in the first equation we have \(6O + 6O_{2}\)

so we have 6 + (6*2) or 6 + 12, which gives us 18

in the second equation we have \(O_{6} + 6O_{2}\)

it's basically the same thing, just written differently, so we have 18.

__________________

Finally, C.

In the first equation, we have \(6C\)

in the second \(C_{6}\), again, same thing expressed differently.

both have 6 C.

Answer:

i think is probably A

Explanation:

because at constant temperature, the product of the pressure and volume of a given mass of and ideal gass in a closed system is always constant

Answer:

I believe the answer is A or D

Explanation:

I hope this helps you out!

Answer:

Where are the answer choices

?

Answer: Your answer is Chemical weathering.

Hope this helps!

Answer:

chemical

Explanation:

I hope it help.........

Answer:

No.

Explanation:

In order to find the density of an object, you have to divide the weight of the object (mass) by the amount of space the object takes up (volume) by each other and then you get the density of the object. So, "no density is NOT a ratio of a substances volume to its mass, but a ratio of the substances mass to volume."

\(p=\frac{m}{v}\)
p = density
m = mass
v = volume

Hope this helps.

Expressing this answer in scientific notation, the amount of N2 produced according to the engineer's design would be approximately 1.467 x 10^1 mol.

To determine the amount of N2 produced in the reaction between dinitrogen tetroxide (N2O4) and excess hydrazine (N2H4), we need to consider the stoichiometry of the reaction.

The balanced equation for the reaction is:

N2H4 + N2O4 → N2 + 2H2O

According to the stoichiometry of the reaction, for every one mole of N2H4, one mole of N2 is produced. The molar mass of N2H4 is approximately 32.05 g/mol.

Given that the rocket is designed to hold 1.35 kg (1350 g) of N2O4, we can calculate the moles of N2H4 required:

Moles of N2H4 = Mass of N2O4 / Molar mass of N2O4

Moles of N2H4 = 1350 g / 92.01 g/mol ≈ 14.67 mol

Since the stoichiometry is 1:1, the amount of N2 produced will be equal to the moles of N2H4:

Moles of N2 produced = Moles of N2H4 ≈ 14.67 mol

Expressing this answer in scientific notation, the amount of N2 produced according to the engineer's design would be approximately 1.467 x 10^1 mol.

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Answer:

One triple bond and four non bonding electrons

Explanation:

In considering the lewis structure of carbon monoxide, we must remember that the molecule contains a total of ten valence electrons. Four are the valence electrons that are present on the valence shell of carbon while six are the valence electrons on oxygen. Some of these valence electrons participate in bonding in the CO molecule.

Out of the six valence electrons on oxygen, two valence electrons participate in bonding with carbon while the other four electrons remain localized on the oxygen atom as two lone pairs of electrons.

Hence there are four nonbonding electrons in the lewis structure of CO as well as one triple bond.

Answer: Energy is removed from the particles in each change, but more energy is removed in melting than in boiling because the attractions are weaker. Energy is removed from the particles in each change, but more energy is removed in boiling than in melting because the attractions are weaker. Energy is added to the particles in each change, but boiling requires more energy than melting because the attractions must be completely overcome. Energy is added to the particles in each change, but melting requires more energy than boiling because the attractions must be completely overcome.  

Answer:

9.78 liters of stock and 15.22 liters of water.

Explanation:

We can use the formula:

M1V1 = M2V2

where M1 is the initial concentration, V1 is the initial volume, M2 is the final concentration, and V2 is the final volume.

We know that the final volume is 25.0 liters and the final concentration is 4.50 M. We also know the initial concentration is 11.5 M.

Let's solve for the initial volume of stock solution (V1):

M1V1 = M2V2

11.5 V1 = 4.50 (25.0)

V1 = (4.50 x 25.0) / 11.5

V1 = 9.78 liters

So the pharmacist needs to mix 9.78 liters of the 11.5 M stock solution with water to make a 25.0 liter 4.50 M solution.

The remaining volume must be water, so:

Volume of water = Total volume - Volume of stock solution

Volume of water = 25.0 - 9.78

Volume of water = 15.22 liters

Therefore, the answer is:

9.78 liters of stock and 15.22 liters of water.

Answer:

D) 9.78 liters of stock and 15.22 liters of water.

Explanation:

I took the test

Answer:

Coloroplasts I hope it's right ❤️

Answer:

i. If the area is doubled keeping the force constant, then pressure will be halved.

ii. If force is doubled keeping area constant, then pressure will also be doubled.

I hope that my answer helped you!!

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The fax header typically includes details such as the date and time the fax was sent, the fax number of the sender and recipient, and sometimes additional information like the sender’s name, company, and contact information. It may also include a unique identification number or code specific to that fax transmission.The purpose of the fax header is to provide essential information about the fax, allowing recipients to identify the sender, verify the authenticity and integrity of the fax, and track the fax for reference or archival purposes. The header can be used for comparison and authentication purposes, as it provides a record of the fax transmission details that can be cross-referenced or compared with other records.

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