Calculate the wavelength in meters of electromagnetic radiation that has a frequency of 640.0 kHz. (c = 3.00 X 108 m/s)

Answer:

-114kJ mol-1

Explanation:

Definition of the enthalpy change of neutralization: the energy released with the formation of 1 mole of water when neutralization takes place between an acid and a base.

Since both reactions yield the same number of moles of water, the answer should be -114kJ mol-1.

The statement for the enthalpy change of reaction 2 is "-114 kJ"

The change in enthalpy (ΔH) is a quantity of heat of a system. The enthalpy change is the amount of heat that enters or exits a system during a reaction.

One equivalent of hydrogen ions is neutralised with one equivalent of hydroxide ions in the reaction 1,

NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

The change in enthalpy is given as -114 kJ.

Two equivalents of hydrogen ions are neutralised with two equivalents of hydroxide ions in the reaction 2,

Ba(OH)2 + H2SO4 (aq) → BaSO4 (s) + 2H2O (l)

But, that primary ionic reaction is same for both the reaction in which hydrogen ion combines with hydroxide ion to generate a water molecule. So, the enthalpy change of reaction 1 would be exactly same as for reaction 2.

The neutralization enthalpy comes out to be -114 kJ.

Hence the correct answer is -114 kJ.

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9.31 g should be the same mass as 9,310 mg.

Given that,

Based on the above information, the calculation is as follows:

Since, 1 gram (g) = 103 milligram (mg)

So,  

i.e \(9.31 g = 9.31 \times 103 mg\)

= 9310 mg  

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There is no reaction between zinc II sulfate and magnesium sulfate.

We know that a reaction is said to occur when two species are combined and there are new products that appear in the system. This implies that if there are no new substances that appear in the system we can not say that a chemical reaction has taken place.

In this case, we can see that we can not be able to observe any change when wee mix a solution of magnesium sulfate and zinc II sulfate since the both of them have the same anion.

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The general formula for many YBCO compounds, also known as Y123, is YBa2Cu3O7x. However, materials with other Y: Ba: There are Cu ratios like YBa2Cu4Oy (Y124) and Y2Ba4Cu7Oy (Y247).

A compound is a substance composed of two or more distinct chemical elements arranged in a predetermined ratio in chemistry. Chemical bonds that are difficult to break form when the elements combine and react with one another. Atomic electron sharing or exchanging results in the formation of these bonds.

compound Include in the list Share When a problem is compounded, something is added to make it worse, like pouring water on a grease fire. The combination is a compound word; A compound is a mixture or combination of at least two things.

A substance in science that is made by chemically joining two or more distinct elements. Table salt (NaCl), which is made from the elements sodium and chloride, and water (H2O), which is made from the elements hydrogen and oxygen, are examples of compounds.

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If 4.0 L of a 4.6M SrCl2 solution is diluted to 45L, the molarity of the diluted solution is 0.41M.

The molarity of a diluted solution can be calculated using the following formula:

M1V1 = M2V2

Where;

According to this question, 4.0 L of a 4.6M SrCl2 solution is diluted to 45L, the molarity of the diluted solution is calculated as follows:

4.6 × 4 = 45 × M2

18.4 = 45M2

M2 = 18.4/45

M2 = 0.41M

Therefore, if 4.0 L of a 4.6M SrCl2 solution is diluted to 45L, the molarity of the diluted solution is 0.41M.

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Answer:

the rate at which reactants form products is the same as the rate at which products form reactants

Explanation:

There is still  a reaction happening just that the second one happens the opposite happens and keeps it at net 0

In dynamic equilibrium conditions, the vapor pressure is unique, and the rate of vaporization and condensation are at equal rates. Thus, option b is accurate.

A dynamic equilibrium is a reaction state where the rate of the forward and the backward reaction are equal. In the above case, the vapourization and the condensation will occur simultaneously at the same rate.

The vapor pressure of the liquid to gas and vice versa has a distinctive value as the temperatures are different making the pressure change directly.

Therefore, option b. II and III are the correct options.

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Our sample has 1.626 g, and 1.060 g is As.

Amount of Oxygen = 1.626 - 1.060 = 0.566 g O

Part 1) We must calculate the number of moles of each element:

Please, get the atomic masses of As and O from the periodic table.

For As) 1.060 g As x (1 mole As/74.92 g) = 0.01415 moles As

For O) 0.566 g O x (1 mole O/16.00 g) = 0.03537 moles O

Part 2) We need to calculate the subindexes we put in the empirical formula:

Let's divide the number of moles by the smallest one of them

For As) 0.01415 moles As/0.01415 moles As = 1

For O) 0.03537 moles O/0.01415 moles As = 2.5

We convert these 2 values into whole numbers, multiplying by 2:

1 x 2 = 2

2.5 x 2 = 5

The empirical formula: As2O5

Explanation:

a. 1 nm = 10Å

683nm=683×10Å

=6830 Å

b. 1 mi/h=0.44704

520 mi/h=520×0.44704

=232.4608

c. 1 g/cm3=62.427961 lb/ft3

0.714 g/cm3=0.714×62.427961 lb/ft3

=44.573564

d. 1°C =9/5 +32 =33.8°F

-164°C = -164×9/5+32 = -263.2°F

Marks as brainest one.

Answer:  **

H-N-H

   |

  H

Explanation:

Look at a periodic table to determine how many electrons you need to account for. Hydrogen (H) only has 1 electron, while Nitrogen (N) has 5. We have three Hydrogen atoms and one Nitrogen atom, so the total number of electrons will be 3 * 1 + 5 = 8 e-.

Now, place the center atom, which will be Nitrogen and place the three Hydrogens on three sides of it as above in the answer. You should use single bonds for this. Each single bond is a pair of electrons, so since we have three single bonds so far, we have accounted for 2 * 3 = 6 electrons. However, we need 2 more electrons for the total of 8. We put these electrons in as a lone pair above Nitrogen.

We check to see if everything follows the octet rule: Nitrogen has three single bonds, so that's 6 e-, as well as one lone pair, so that's another 2 e- for a total of 8 e-. Check. Now look at Hydrogen: H is the only element whose full orbital is 2 e-. Each H has a single bond with Nitrogen, so each does have 2 e-.

Thus, we know this is the correct diagram, and we are done.

Explanation:

A bond line structure of a compound has H N H in linear plane and a hydrogen is branching upward, and the compound is H N (H) H. The nitrogen has two dots at its bottom represents a lone pair of electrons. So ,the correct answer is option C.

The correct Lewis structure for ammonia (\(NH_3\)) is option C. It shows a bond line structure with three hydrogen atoms (H) bonded to a central nitrogen atom (N) in a linear plane.

One hydrogen atom branches upward from the plane. Additionally, the nitrogen atom in this structure has two dots at its bottom, indicating a lone pair of electrons. This arrangement follows the octet rule, as nitrogen has formed three covalent bonds with hydrogen, completing its valence shell. The lone pair on nitrogen gives ammonia its characteristic properties.

Thus, option C accurately represents the Lewis structure of ammonia, showing the bonding and lone pair arrangement of its atoms.

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Answer:

metallic bonds

Explanation:

atoms in a metallic solid loose their outer electrons and form a regular lattice of positive metallic ions.

The chemical bondings are present between the atoms due to the attractive forces. The bond shown in the diagram represents the metallic bonds. Thus, option A is correct.

A metallic bond is a chemical bonding present due to the electrostatic attractive forces present between the delocalized electrons and the ions of the metals.

The metal produces cations that bond with the electrons delocalized around them. This type of bonding accounts for the malleability and conductivity of the metallic species.

The delocalized electrons are shared by the positively charged metal ions. The cations are largely spread in space. It is seen in the elements of aluminum, magnesium, copper, sodium, zinc, calcium, etc.

Therefore, option A. the metallic bond is seen in the diagram.

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Answer:

127529 KJ

Explanation:

Since 1gallon = 3.78 L= 3780 ml

The density of C8H18= 0.703 g/mL

Density = mass/volume

Mass= Density × volume

Mass= 0.703 g/mL × 3780 ml

Mass= 2657.34 g

Molar mass of C8H18= 114 g/mol

Number of moles= mass/molar mass

Number of moles= 2657.34/114

Number of moles= 23.31 moles

Since 1 mole evolved=5471 kJ

23.31 moles = 5471 kJ × 23.31 = 127529 KJ

Answer:

The percent composition of fluorine is 65.67%

Explanation:

Percent Composition is a measure of the amount of mass an element occupies in a compound. It is measured in percentage of mass.

That is, the percentage composition is the percentage by mass of each of the elements present in a compound.

The calculation of the percentage composition of an element is made by:

\(percent composition element A=\frac{total mass of element A}{mass of compound} *100\)

In this case, the percent composition of fluorine is:

\(percent composition of fluorine=\frac{39.6 g}{60.3 g} *100\)

percent composition of fluorine= 65.67%

The percent composition of fluorine is 65.67%

Answer:

The percent composition of fluorine is 65.67%

Explanation:

Percent Composition is a measure of the amount of mass an element occupies in a compound. It is measured in percentage of mass.

That is, the percentage composition is the percentage by mass of each of the elements present in a compound.

The calculation of the percentage composition of an element is made by:

In this case, the percent composition of fluorine is:

percent composition of fluorine= 65.67%

The percent composition of fluorine is 65.67%

Answer:

The characteristics of unicellular organisms are as follows:

The unicellular organisms usually reproduce by asexual means.

They can be eukaryotes or prokaryotes.

They are found in almost all habitats, from hot springs to frozen tundra.

They possess whip-like structures for movement.

The nutrients enter or leave the cell by the process of diffusion.

Explanation:

Answer:

involved in recycling

Makes oxygen we breathe

Explanation:

Answer:

A. Argon

Explanation:

it is a gas that makes up to 1% of the atmosphere

From the titration curve that have been shown in the image, the equivalence point is 50 mL

At the equivalence point on a titration curve, the amount of titrant added is chemically equivalent to the amount of analyte in the sample being evaluated. As a result of the reaction between the titrant and analyte at this point, the entire analyte has been neutralized by the titrant.

You can locate the equivalence point by plotting the pH or any relevant aspect of the sample under examination as a function of the volume of titrant used.

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H2SO4 is referred to as a strong acid and is denoted as option A.

This refers to any substance which tastes sour when in water and changes the color of blue litmus paper to red. It is usually very corrosive and are used in industries for different functions.

H2SO4 is referred to as a strong acid because it dissociates completely in its aqueous solution or water.

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The enthalpy change for the reaction \(C2H4(g) + H2(g) → C2H6(g)\) is -137.15 kJ/mol.

Given:C2H4(g) + H2(g) → C2H6(g)The enthalpy of formation of C2H6(g) is -84.68 kJ mol-1The enthalpy of formation of C2H4(g) is 52.47 kJ mol-1The enthalpy of formation of H2(g) is 0 kJ mol-1Hence, using Hess's Law, the enthalpy change for the reaction \(C2H4(g) + H2(g) → C2H6(g)\) can be calculated by considering the formation of reactants and products from their respective elements. It can be given as:

\($$C_2H_4 + H_2 → C_2H_6$$$$\Delta H = H_f(C_2H_6) - [H_f(C_2H_4) + H_f(H_2)]$$$$\Delta H = -84.68 - [52.47 + 0]$$$$\Delta H = -84.68 - 52.47$$$$\Delta H = -137.15 kJ/mol$$.\)

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150 grams of NaOH is approximately equal to 2.256 x 10^24 particles of NaOH.

To convert grams of NaOH to particles of NaOH, we need to use the concept of molar mass and Avogadro's number. The molar mass of NaOH is calculated by adding the atomic masses of sodium (Na), oxygen (O), and hydrogen (H) together. It can be determined as follows:

Na: 22.99 g/mol

O: 16.00 g/mol

H: 1.01 g/mol

Molar mass of NaOH = (22.99 g/mol) + (16.00 g/mol) + (1.01 g/mol) = 40.00 g/mol

Now, we can use the molar mass to convert grams of NaOH to moles. Since 1 mole contains Avogadro's number (approximately 6.022 x 10^23) particles, we can determine the number of particles as follows:

150 g NaOH * (1 mol NaOH / 40.00 g NaOH) * (6.022 x 10^23 particles / 1 mol NaOH) ≈ 2.256 x 10^24 particles

It's important to note that this calculation assumes the substance is pure NaOH and that the molar mass and Avogadro's number are accurate.

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The energy of combustion for one mole of butane to be approximately 2888.81 kJ/mol.

To calculate the energy of combustion for one mole of butane (C4H10), we need to use the information provided and apply the principle of calorimetry.

First, we need to convert the mass of the butane sample from grams to moles. The molar mass of butane (C4H10) can be calculated as follows:

C: 12.01 g/mol

H: 1.01 g/mol

Molar mass of C4H10 = (12.01 * 4) + (1.01 * 10) = 58.12 g/mol

Next, we calculate the moles of butane in the sample:

moles of butane = mass of butane sample / molar mass of butane

moles of butane = 0.367 g / 58.12 g/mol ≈ 0.00631 mol

Now, we can calculate the heat released by the combustion of the butane sample using the equation:

q = C * ΔT

where q is the heat released, C is the heat capacity of the calorimeter, and ΔT is the change in temperature.

Given that the heat capacity of the bomb calorimeter is 2.36 kJ/°C and the change in temperature is 7.73 °C, we can substitute these values into the equation:

q = (2.36 kJ/°C) * 7.73 °C = 18.2078 kJ

Since the heat released by the combustion of the butane sample is equal to the heat absorbed by the calorimeter, we can equate this value to the energy of combustion for one mole of butane.

Energy of combustion for one mole of butane = q / moles of butane

Energy of combustion for one mole of butane = 18.2078 kJ / 0.00631 mol ≈ 2888.81 kJ/mol

Therefore, the energy of combustion for one mole of butane is approximately 2888.81 kJ/mol.

In conclusion, by applying the principles of calorimetry and using the given data, we have calculated the energy of combustion for one mole of butane to be approximately 2888.81 kJ/mol.

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The difference between the numbers,

(2.260×104)−(8.098×103) = 598.539.

Subtraction is a mathematic operation. Which is used to remove terms or objects in an expression.

Given:

Two numbers,

(2.260×104) and (8.098×103).

First, we simplify the numbers,

(2.260×104) = 235.04

And  (8.098×103) = 833.579.

The difference between the numbers,

(2.260×104)−(8.098×103) = 833.579 - 235.04.

(2.260×104)−(8.098×103) = 598.539

Therefore, then value is 598.539.

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Answer:

D)

Explanation:

A tendon is a fibrous connective tissue that attaches muscle to bone. Tendons may also attach muscles to structures such as the eyeball

Answer:

c. all matter is composed of small particles,which combine to make larger objects

The new volume of the balloon after heating it to a temperature of 51 °C is approximately 2.39 L.

Charles's law states that "the volume occupied by a definite quantity of gas is directly proportional to its absolute temperature.

It is expressed as;

\(\frac{V_1}{T_1} =\frac{V_2}{T_2}\)

Given that:

Substituting the given values and solve for V₂:

\(V_1T_2 = V_2T_1\\\\V_2 = \frac{V_1T_2}{T_1} \\\\V_2 = \frac{2.2\ *\ 324.15}{298.15 }\\ \\V_2 = 2.39 \ L\)

Therefore, the final volume is 2.39 litres.

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To prepare  1.000 / of pH 9.20 buffer made from NH3 and NH4Cl we need add 1.990 L of 3.00 M NH3 and 250 g of NH4Cl solid to water to make a final volume of 1.000 L.

To prepare a 1.000 L of pH 9.20 buffer made from NH3 and NH4Cl, we need to determine the amount of NH3 and NH4Cl required.

First, we can calculate the pKb of NH3 using the Kb value given:

Kb = [NH4+][OH-]/[NH3]

pKb = -log(Kb) = -log(1.75 x 10^-5) = 4.756

The Henderson-Hasselbalch equation for a buffer made of NH3 and NH4Cl is:

pH = pKb + log([NH4+]/[NH3])

We can rearrange this equation to solve for the ratio of [NH4+] to [NH3]:

[NH4+]/[NH3] = 10^(pH - pKb)

Substituting the given values:

[NH4+]/[NH3] = 10^(9.20 - 4.756) = 395.9

This means that the ratio of [NH4+] to [NH3] in the buffer solution should be 395.9.

Next, we can use this ratio to calculate the amounts of NH3 and NH4Cl required to make the buffer solution.

Let x be the amount of NH3 needed in moles.

Then, the amount of NH4Cl needed in moles is also x, since we need the same number of moles of NH4+ and NH3.

From the given information, we have:

2.000 L of 3.00 M NH3, which contains 6.00 moles of NH3.

250 g of NH4Cl, which is equivalent to 2.465 moles of NH4Cl (using the molar mass of NH4Cl, which is 53.49 g/mol).

Using the ratio of [NH4+] to [NH3] calculated above, we can write:

[NH4+]/[NH3] = x/(6.00 - x)

Solving for x:

x = 6.00/[1 + 1/395.9] = 5.970 moles

Therefore, we need 5.970 moles of NH3 and 5.970 moles of NH4Cl to make the buffer solution.

To calculate the volumes required, we need to convert the moles of NH3 to volume using its concentration of 3.00 M:

Volume of NH3 = moles of NH3 / concentration of NH3

= 5.970 moles / 3.00 M = 1.990 L

Since the addition of solid does not result in a volume change, we can add the 250 g of NH4Cl directly to the 1.000 L of solution.

Therefore, to prepare 1.000 L of pH 9.20 buffer made from NH3 and NH4Cl, we need to add 1.990 L of 3.00 M NH3 and 250 g of NH4Cl solid to water to make a final volume of 1.000 L.

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Explanation:

Stars aren't visible during the sunlit hours of daytime because the light-scattering properties of our atmosphere spread sunlight across the sky. Seeing the dim light of a distant star in the blanket of photons from our Sun becomes as difficult as spotting a single snowflake in a blizzard. (put in your own words this is from online)

Answer:

40.88 kPa.

Explanation:

Given that,

Volume, V₁= 4.2 L

Pressure, P₁ = 110 kPa

We need to find the pressure of the gas when the volume is changed to 11.3L at constant temperature.

According to Boyle's law, at constant temperature, volume is inversely proportional to the pressure. Mathematically,

\(P_1V_1=P_2V_2\\\\P_2=\dfrac{P_1V_1}{V_2}\\\\P_2=\dfrac{4.2\ L\times 110\ kPa}{11.3\ L}\\\\P_2=40.88\ kPa\)

So, the new pressure is 40.88 kPa.