Helpppp

describe an incident that shows how you were doing "selection" during the process of perception. Then, give your own example about a stereotype that you do.

Answer:

We would expect stars near the edge of the galaxy to rotate slower than stars near the center. This is because the rotational speed of objects in a rotating system depends on their distance from the center of rotation. Stars near the center of the galaxy are closer to the gravitational center of the galaxy's mass, so they feel a stronger gravitational pull and rotate faster. Stars near the edge are farther from the center, so they feel a weaker gravitational pull and rotate slower. This results in a gradient of rotational speeds, with the fastest speeds near the galactic center and slower speeds near the edges.

Answer:

To find the molarity of the new solution, we need to use the equation:

M1V1 = M2V2

Where:

M1 = initial molarity = 3.00 M

V1 = initial volume = 15.0 mL

M2 = final molarity (what we're solving for)

V2 = final volume = 30.0 mL

Rearranging the equation to solve for M2:

M2 = (M1V1)/V2

M2 = (3.00 M * 15.0 mL)/30.0 mL

M2 = 1.50 M

Therefore, the molarity of the new solution is 1.50 M.

Answer:

1.00 M

Explanation:

Molarity is defined as the number of moles of solute per liter of solution. When a solution is diluted, the number of moles of solute remains constant, but the volume of the solution increases. Therefore, the molarity of the solution decreases.

In this case, the initial number of moles of solute in the 15.0 mL of 3.00 M hydrochloric acid solution is (15.0 mL) * (3.00 mol/L) * (1 L/1000 mL) = 0.045 mol.

After dilution, the volume of the solution increases to 30.0 mL + 15.0 mL = 45.0 mL. The molarity of the diluted solution is (0.045 mol) / (45.0 mL) * (1000 mL/L) = 1.00 M.

So, the molarity of 30.0 mL of hydrochloric acid solution after 15.0 mL of a 3.00 M solution has been diluted is 1.00 M.

To solve this problem, we need to use the concept of weak acid-base equilibrium and the Henderson-Hasselbalch equation. The reaction between NH3 and HNO3 can be written as:

NH3 + HNO3 → NH4+ + NO3-

Before any HNO3 is added, the NH3 solution is a weak base with the following equilibrium equation:

NH3 + H2O ⇌ NH4+ + OH-

where the Kb of NH3 is 1.8 × 10-5. At the start, we have 0.10 M NH3, and the concentration of OH- can be calculated using the Kb expression:

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

[OH-] = Kb[NH3]/[NH4+] = 1.8 × 10^-5 × 0.10 / x, where x is the concentration of NH4+.

At the equivalence point, the moles of HNO3 added equals the moles of NH3 initially present, and the solution contains only NH4+ and NO3-. Therefore, the concentration of NH4+ is:

x = 0.10 - 0.05 = 0.05 M

At this point, all the OH- ions have been consumed by the HNO3, so the pH of the solution depends on the concentration of NH4+. The Henderson-Hasselbalch equation for a weak acid-base system is:

pH = pKa + log([base]/[acid])

where pKa is the negative logarithm of the acid dissociation constant (Ka) and [base]/[acid] is the ratio of the concentrations of the weak base and its conjugate acid.

For NH3, the conjugate acid is NH4+ and the pKa can be calculated as:

pKa = -log(Ka) = -log(1.8 × 10^-5) = 4.74

Plugging in the values, we get:

pH = 4.74 + log(0.05/0.05) = 4.74

Therefore, the pH of the solution after the addition of 50.0 mL of HNO3 is 4.74.

The molarity of potassium hydroxide is 1 mol/dm³.

Molarity is a measure of concentration, expressing the number of moles of a solute per litre of solution. It is denoted by the symbol M and is an important concept in chemistry, especially when dealing with solutions. Molarity is related to the molar mass of the solute and the density of the solution. It is a useful tool for measuring the amount of a particular solute in a given solution.

To calculate the molarity of potassium hydroxide, we must first calculate the moles of HCl and KOH.

First, we calculate the moles of HCl. We use the formula moles = concentration x volume.

HCl: 0.15 moldm³ x (31.40/1000) = 0.00471 moles

Next, we calculate the moles of KOH.

KOH: (20/1000) = 0.02 moles

Now we can calculate the molarity of KOH. We use the formula molarity = moles/volume.

KOH: 0.02/0.02 = 1 mol/dm³

Therefore, the molarity of potassium hydroxide is 1 mol/dm³.

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The half-life of radon-222 is 4 days, which means that after each 4-day period, the amount of radon-222 in a sample is halved.

After 8 days, only 25 g of the 100 g sample of radon-222 would be remaining. After 8 days, there have been two half-life periods. Therefore, we can find the amount of radon-222 remaining after 8 days using the following formula:

Amount remaining = (Initial amount) x (1/2)^(number of half-life periods)

Initial amount = 100 g

Number of half-life periods = 8 days / 4 days per half-life = 2 half-life periods

Substituting these values into the formula gives:

Amount remaining = 100 g x (1/2)² = 100 g x (1/4) = 25 g

Therefore, after 8 days, only 25 g of the 100 g sample of radon-222 would be remaining.

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The balanced chemical equation for the reaction between antimony trioxide (Sb2O3) and carbon monoxide (CO) is:

Sb2O3 + 3CO → 2Sb + 3CO2

From the equation, we can see that 1 mole of Sb2O3 reacts with 3 moles of CO to produce 3 moles of CO2. The molar mass of Sb2O3 is 291.52 g/mol, so 0.465 g of Sb2O3 is equal to 0.465 g / 291.52 g/mol = 0.001592 mol.

Since 1 mole of Sb2O3 reacts with 3 moles of CO, we need 3 * 0.001592 = 0.004776 mol of CO to react completely with 0.465 g of Sb2O3.

The molar mass of CO is 28.01 g/mol, so 0.004776 mol of CO is equal to 0.004776 mol * 28.01 g/mol = 0.1338 g of CO.

Therefore, 0.465 g of Sb2O3 will produce 0.1338 g of CO. To convert this to liters of CO at a given temperature and pressure, we would need to know the volume of 0.1338 g of CO under those conditions using the ideal gas law.

hope this helped (:

The chemical formula for the product.

(a)Orbital diagram for Li:

1s² 2s¹

Orbital diagram for S:

1s² 2s² 2p⁶ 3s² 3p⁴

Lewis structure for Li:

Li: [Li]+

Lewis structure for S:

:S:::S:

Combination of Li and S:

Li₂S

(b)

Orbital diagram for Ca:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s²

Orbital diagram for Cl:

1s² 2s² 2p⁶ 3s² 3p⁵

Lewis structure for Ca:

Ca: [Ca]²⁺

Lewis structure for Cl:

:Cl:

Combination of Ca and Cl:

CaCl₂

(c)

Orbital diagram for K:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹

Orbital diagram for Cl:

1s² 2s² 2p⁶ 3s² 3p⁵

Lewis structure for K:

K: [K]+

Lewis structure for Cl:

:Cl:

Combination of K and Cl:

KCl

(d) Orbital diagram for Na:

1s² 2s² 2p⁶ 3s¹

Orbital diagram for N:

1s² 2s² 2p³

Lewis structure for Na:

Na: [Na]+

Lewis structure for N:

:N:::N:

Combination of Na and N:

Na₃N

An orbital diagram is a visual depiction of the electrons located in an atom's or molecule's orbitals. Each electron is represented by an arrow, while each orbital is illustrated by a line.

The two electrons in each orbital's two lines are drawn in pairs to represent their opposing spins. Lewis structures, on the other hand, are schematics that display the interactions between the atoms in a molecule as well as any potential lone pairs of electrons.

Each atom's valence electrons are shown as dots, and the connections between atoms are shown as lines. The kind of bond that can be created between two elements depends on the number of valence electrons.

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The data table supports the idea that valence electrons affect the chemical characteristics of elements and may be used to forecast chemical reactions by showing how the amount of valence electrons follows different patterns on the periodic table.

The number of valence electrons in groups 1-2 and 13–18 rises by one from one element to the next throughout each row, or period, of the periodic table.

The ability of an atom to make covalent bonds with other atoms is known as its "valency." Valence electrons, on the other hand, are the quantity of electrons required in a compound's entire outer shell in order for bonds to form.

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It is important to note that lead poisoning is a serious condition that requires prompt medical attention. If you or someone you know has ingested lead nitrate, seek medical attention immediately.

What is Lead Nitrate?

Lead nitrate is an inorganic compound with the chemical formula Pb(NO3)2. It is a colorless, odorless, and crystalline solid that is highly soluble in water. Lead nitrate is commonly used in various industrial processes, including the manufacture of lead-based explosives, pigments, and pyrotechnics.

Swallowing soluble lead nitrate, Pb(NO3)2, can lead to lead poisoning, which can cause various health problems, including abdominal pain, vomiting, diarrhea, seizures, and in severe cases, coma or death. If someone has swallowed this compound, it is important to seek medical attention immediately.

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The bond energy is -183 kJ/ mol for the given reaction and bond energies for the reaction. The reaction is exothermic.

Atoms bond together to create compounds because doing so allows them to achieve lower energies than they would have as individual atoms. A quantity of energy equivalent to the difference between the energies of the bonded and separated atoms is released, typically as heat. That is, linked atoms have less energy than individual atoms. Energy is always released when atoms combine to form a compound, and the compound has a reduced overall energy.

When a chemical reaction happens, molecular bonds are broken and new bonds are formed, resulting in the formation of new molecules.

Change is energy(ΔE) = ΣΔBE (Products) - ΣΔBE (Reactant)

ΔE = (432+ 239) - 2×427

ΔE = -183 kJ/ mol

Therefore it is exothermic reaction.

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The amount of lithium nitrate needed to make 230 grams of lithium sulfate depends on the amount of lead (IV) sulfate provided and is equal to half of the moles of lithium sulfate produced, which is 2.091/2 = 1.046 mol. The mass of lithium nitrate required can be calculated using its molar mass.

To calculate the amount of lithium nitrate required to make 230 grams of lithium sulfate, we can use the following steps:

Calculate the molar mass of lithium sulfate:

Li2SO4: 2(6.94 g/mol) + 1(32.06 g/mol) + 4(16.00 g/mol) = 109.94 g/mol

Determine the number of moles of lithium sulfate:

n = m/M = 230 g / 109.94 g/mol = 2.091 mol

Since 2 moles of lithium sulfate are produced for every 1 mole of lead (IV) sulfate, we need 2.091/2 = 1.046 mol of lead (IV) sulfate to react with the lithium sulfate.

Calculate the mass of lead (IV) sulfate required:

m = nM = 1.046 mol x Pb(SO4)2 molar mass (assuming it's provided)

From the balanced equation, we know that for every 2 moles of lithium sulfate, we need 1 mole of lithium nitrate.

The amount of lithium nitrate needed to make 230 grams of lithium sulfate depends on the amount of lead (IV) sulfate provided and is equal to half of the moles of lithium sulfate produced, which is 2.091/2 = 1.046 mol. The mass of lithium nitrate required can be calculated using its molar mass.

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The net ionic equation for the reaction is:

2Fe3+(aq) + 3S2-(aq) → Fe2S3(s)

What is Ionic Equation?

An ionic equation is a chemical equation that shows the chemical species as their respective ions in a solution. In other words, an ionic equation only includes the ions that participate in the chemical reaction, and excludes any spectator ions that do not participate in the reaction.

The net ionic equation shows only the species that participate in the reaction and omits the spectator ions (ions that appear on both sides of the equation without undergoing a change).

First, let's write the balanced chemical equation for the reaction:

Fe2(SO4)3(aq) + 3K2S(aq) → Fe2S3(s) + 3K2SO4(aq)

To write the net ionic equation, we need to identify the ions that are present in aqueous solution and that participate in the reaction. These are:

Fe2+(aq), SO42-(aq), K+(aq), and S2-(aq)

The balanced ionic equation is:

2Fe3+(aq) + 3SO42-(aq) + 6K+(aq) + 3S2-(aq) → Fe2S3(s) + 3K2SO4(aq)

To obtain the net ionic equation, we eliminate the spectator ions, which are the potassium and sulfate ions that appear on both sides of the equation:

2Fe3+(aq) + 3S2-(aq) → Fe2S3(s)

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The products of the reaction are iron (II) carbonate and copper (II) nitrate.

[tex]Fe(NO_3)_2 + Cu_2CO_3 - > FeCO_3 + 2Cu(NO_3)_2[/tex]

An inorganic substance with the molecular formula Fe(NO₃)₂ is iron (II) nitrate. It is a crystalline green substance that can be dissolved in water. Iron or iron oxide can be used to combine with nitric acid to create iron (II) nitrate, which is a salt of both iron and nitric acid.

It is frequently used as an iron supply in chemical processes as well as a starting point for the creation of other iron compounds. Additionally, the creation of pigments, dyes, and catalysts can be accomplished using iron (II) nitrate.

However, it needs to be handled carefully because it is a potent oxidizer and, if not treated correctly, can irritate the skin and eyes.

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Explanation:

Formic acid (HCOOH) reacts with sodium hydroxide (NaOH) to form sodium formate (HCOONa) and water. The balanced chemical equation is:

HCOOH + NaOH → HCOONa + H2O

The reaction is a strong acid-strong base titration. We can use the following equation to calculate the concentration of formate ion (HCOO^-) in the resulting solution:

[HCOO^-] = [OH^-] - [HCOOH]

where [OH^-] is the concentration of hydroxide ion and [HCOOH] is the concentration of formic acid before the reaction.

Before the reaction, the solution contains 0.25 mol/L of formic acid in 55.0 mL, or 0.25 mol/L × 0.055 L = 0.01375 mol of formic acid. The solution also contains 0.12 mol/L of sodium hydroxide in 75.0 mL, or 0.12 mol/L × 0.075 L = 0.009 mol of sodium hydroxide.

Since the reaction between formic acid and sodium hydroxide is a 1:1 reaction, all the 0.009 mol of sodium hydroxide will react with 0.009 mol of formic acid, leaving 0.00475 mol of formic acid unreacted.

[HCOO^-] = [OH^-] - [HCOOH]

[OH^-] = [NaOH] = 0.12 mol/L × 0.075 L / 0.13 L = 0.0692 mol/L

[HCOO^-] = 0.0692 mol/L - 0.00475 mol/L = 0.0645 mol/L

Now we can calculate the pH of the resulting solution using the Ka expression for formic acid:

Ka = [HCOO^-][H3O^+]/[HCOOH]

[H3O^+] = Ka × [HCOOH] / [HCOO^-]

[H3O^+] = 1.77 × 10^-4 × 0.00475 mol/L / 0.0645 mol/L

[H3O^+] = 1.29 × 10^-5 mol/L

pH = -log[H3O^+]

pH = -log(1.29 × 10^-5)

pH = 4.89

Therefore, the resulting pH is 4.89.

Explanation:

The stovetop example would be an open system, because heat and water vapor can be lost to the air. A closed system, on the other hand, can exchange only energy with its surroundings, not matter.

The dissociation of 500 molecules of carbonic acid would produce 200 + 15 = 215 ions.

What is Dissociation?

Dissociation is a chemical process in which a compound breaks down into two or more simpler components, usually ions, when it is exposed to a suitable solvent or energy source such as heat or light. In other words, it is the separation of a molecule or compound into smaller particles such as atoms, ions, or radicals.

The dissociation of carbonic acid (H2CO3) in the first degree produces two ions (H+ and HCO3-) per molecule, while the dissociation in the second degree produces three ions (H+, CO32-, and HCO3-) per molecule.

If 500 molecules of carbonic acid dissociate in the first degree by 20%, then 20% of the molecules (0.2 x 500 = 100) will dissociate, producing 100 x 2 = 200 ions (H+ and HCO3-).

If 500 molecules of carbonic acid dissociate in the second degree by 1%, then 1% of the molecules (0.01 x 500 = 5) will dissociate, producing 5 x 3 = 15 ions (H+, CO32-, and HCO3-).

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Answer:

Products will be Ga2Br2 and CaS3 (Double Displacement Reaction)

To solve this problem,  the ideal gas law equation can be used. The volume of the balloon will come to 79.9 liters.

The ideal gas law is a fundamental equation in physics and chemistry that describes the behavior of ideal gases under various conditions. The law is expressed mathematically as:

PV = nRT

where P is the pressure of the gas, V is the volume it occupies, n is the number of moles of gas, T is the temperature of the gas in Kelvin, and R is the universal gas constant.

First, we need to convert the temperature from Celsius to Kelvin:

T = 20°C + 273.15 = 293.15 K

Next, we need to find the value of R, which is 0.0821 L·atm/mol·K for ideal gases.

We also know that the pressure is standard pressure, which is 1 atm.

Plugging in all the values, we get:

V = (nRT) / P

V = (3.2 mol * 0.0821 L·atm/mol·K * 293.15 K) / 1 atm

V = 79.9 L

Therefore, the volume of the balloon is 79.9 liters.

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Answer:

Divide the mass of the water lost by the mass of hydrate and multiply by 100. The theoretical (actual) percent hydration (percent water) can be calculated from the formula of the hydrate by dividing the mass of water in one mole of the hydrate by the molar mass of the hydrate and multiplying by 100.

The number 38.7 x 10⁷ is already in scientific notation

The coefficient is 38.7 and the exponent is 7.

In the given number, 38.7 x  10⁷,  the coefficient is 38.7, which is a decimal number between 1 and 10.

The exponent of 10 is 7, which tells us to move the decimal point seven places to the right to get the actual value of the number.

So, 38.7 x 10⁷ can be expanded as follows:

38.7 x  10⁷= 38.7 000 000

we moved the decimal point seven places to the right and filled the empty spaces with zeros. This gives us the actual value of the number in standard form.

Scientific notation, also referred to as standard form or exponential notation, is a format for succinctly expressing very large or very tiny numbers. It is predicated on the notion that a number can be represented as the result of a coefficient and a multiple of 10.

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the pH of the aqueous solution is 8.51 with a hydrogen ion concentration of [H+]=3.1×[tex]10^-9[/tex]M

The pH of the aqueous solution can be calculated using the formula:

pH = -log[H+]

where [H+] is the hydrogen ion concentration of the solution.

Substituting the given value, we get:

pH = -log(3.1×[tex]10^-9[/tex])

pH = 8.51

An aqueous solution is one in which water serves as the solvent. It is utilised in a variety of applications, including analytical chemistry, biochemistry, and industrial chemistry. It is the most prevalent kind of solution used in chemical reactions. Water serves as both the solvent and the solute in an aqueous solution, where the solute is often a solid, liquid, or gas. Due to its high polarity and capacity to make hydrogen bonds with other molecules, water is an excellent solvent that can dissolve a variety of materials, including polar molecules and ionic compounds. Acid-base reactions, redox reactions, and precipitation reactions are just a few of the numerous chemical processes that take place in aqueous solutions. A variety of variables can have an impact on an aqueous solution's characteristics.

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The ΔH for the given reaction is +204.8 kJ. This indicates that the reaction is endothermic, which absorbs heat from the surroundings.

To determine the AH for the given reaction, we can use the following steps:

Write the balanced chemical equation for the reaction.

2CO(g) + 2NO(g) ⇒ 2CO₂(g) + N₂(g)

Use the given equations to write the overall reaction as a combination of the given reactions. We can do this by reversing equation (a) and multiplying equation (b) by 2 so that the reactants and products match the overall reaction.

2CO₂(g) ⇒ 2CO(g) + O₂(g) ΔH = +566.0 kJ (reversed)

2N₂(g) + 2O₂(g) ⇒ 4NO(g) ΔH = -2(180.6 kJ) = -361.2 kJ (multiplied by 2)

Overall reaction:

2CO(g) + 2NO(g) ⇒ 2CO₂(g) + N₂(g) ΔH = ?

Step 3: Add the ΔH values for the individual reactions to obtain the ΔH for the overall reaction.

ΔH = (+566.0 kJ) + (-361.2 kJ) = +204.8 kJ

Therefore,  the ΔH for the given reaction is +204.8 kJ. This indicates that the reaction is endothermic, meaning that it absorbs heat from the surroundings., meaning that it absorbs heat from the surroundings.

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The number of moles of CO contained in the 20.0 L tank at 93 °C and 4.52 atm is 3.01 moles

Ideal gas equation is given as follow:

PV = nRT

Where

With the above formula, we can obtain the number of mole of CO in the tank. This is shown below:

PV = nRT

4.52 × 20 = n × 0.0821 × 366

Divide both sides by (0.0821 × 366)

n = (4.52 × 20) / (0.0821 × 366)

n = 3.01 moles

Thus, we can conclude that the number of mole of the gas is 3.01 moles. The correct answer is the 3rd option

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Your speed could change in several ways as you jog along a park's path. Option 2.

One possible way is that you could speed up if you increase your pace or start running instead of jogging. This could happen if you feel more energized, motivated, or if you need to catch up with someone.

Conversely, your speed could slow down if you get tired, experience muscle fatigue, or encounter an uphill section of the path that requires more effort. Other factors such as weather conditions, terrain, or distractions can also affect your speed.

Therefore, the correct answer would be "all of these, except none of these," as there are various factors that can cause changes in your jogging speed along a park's path.

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The basic laboratory techniques that are used for separating components of the mixture are:-

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The length of side BC is approximately 1.82 units.

To find the length of side BC, we can use trigonometry. Since we know two angles of the triangle, we can use the fact that the sum of the angles in a triangle is 180 degrees to find the measure of angle ABC:

Angle ABC = 180 - 90 - 70 = 20 degrees

We can now use the trigonometric function tangent to find the length of BC:

tan(20) = BC/5

Solving for BC, we get:

BC = 5*tan(20) ≈ 1.82 units

The length of side BC is approximately 1.82 units.

The connections between the sides and angles of triangles are the subject of the mathematical discipline of trigonometry. It is helpful in several disciplines, including navigation, physics, engineering, and building.

Sine, cosine, and tangent are the three fundamental trigonometric functions that link the ratios of the lengths of the sides of a right triangle to the angles opposite those sides.

The ratio of the length of the side opposing the angle to the length of the hypotenuse is known as the sine of an angle. (the longest side of the right triangle).

The proportion of the neighboring side's length to the hypotenuse's length is known as the cosine of an angle.

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It's easy

Explanation:

We can use the equilibrium constant expression to calculate the equilibrium concentration of O2:

Kc = [SO3]^2 / ([SO2]^2 [O2])

At equilibrium, the value of Kc is constant, so we can use the equilibrium concentrations of SO2 and SO3 to solve for [O2]:

Kc = [SO3]^2 / ([SO2]^2 [O2])

Kc = (11.0 M)^2 / ((4.20 M)^2 [O2])

Simplifying:

[O2] = (11.0 M)^2 / (Kc (4.20 M)^2)

The value of Kc for this reaction is 4.67 × 10^1, as determined by experiment.

[O2] = (11.0 M)^2 / (4.67 × 10^1 (4.20 M)^2)

[O2] = 0.153 M

Therefore, the equilibrium concentration of O2 in this mixture is 0.153 M.

Answer:

C₅H₁₀.

Explanation:

To determine the molecular formula of the compound, we need to know the molar mass of the compound. Since the empirical formula is CH₂, the empirical molar mass can be calculated as:

Empirical molar mass = 12.01 g/mol (atomic mass of C) + 2(1.01 g/mol) (atomic mass of H)

Empirical molar mass = 14.03 g/mol

The molecular mass of the compound is given as 70.15 g/mol. To find the molecular formula, we need to know how many empirical units are present in the molecule. This can be calculated by dividing the molecular mass by the empirical molar mass:

Number of empirical units = Molecular mass / Empirical molar mass

Number of empirical units = 70.15 g/mol / 14.03 g/mol

Number of empirical units = 5

This means that there are 5 empirical units (CH₂) present in the molecular formula of the compound. Therefore, the molecular formula is:

Molecular formula = 5(CH₂) = C₅H₁₀

Thus, the molecular formula of the compound is C₅H₁₀.

We would need to measure 7.5 mL of the 2.0M solution of KNO₃ and then add enough solvent to get the total volume up to 100.0 mL.

Volume is a unit used to describe how much three-dimensional space an object or substance occupies. By multiplying the length, breadth, and height of an object or substance, as well as additional mathematical formulas tailored to the shape of the object or substance, one can determine the volume of the thing or substance.

We can use the following formula to create a solution:

M1V1 = M2V2

where M1 denotes the starting concentration, V1 the starting volume, M2 the ending concentration, and V2 the ending volume.

In this instance, we want to make a 100.0 mL 0.15M KNO₃ solution using a 2.0M KNO₃ solution.

When these values are added to the formula, we obtain:

(2.0 M) V1 = (0.15 M) (100.0 mL) 

When we solve for V1, we get:

V1 = (0.15 M) (100.0 mL) / (2.0 M) (2.0 M)

V1 = 7.5 mL

So, to prepare 100.0 mL of a 0.15M solution of KNO₃, we would need to measure 7.5 mL of the 2.0M solution of KNO₃ and then add enough solvent to get the total volume up to 100.0 mL.

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