I did some of the boxes just need to figure out what the rest are. (LeChatelier’s Principle)

The new pressure of the gas is 203 psi.

Boyle's law is a gas law that states that the pressure of a gas is inversely proportional to its volume at a constant temperature. The law can be expressed mathematically as:

P * V = k

where P is the pressure of the gas, V is its volume, and k is a constant. This law states that as the volume of a gas increases, its pressure decreases, and as its volume decreases, its pressure increases.

We know that;

P1V1 = P2V2

P1 = initial pressure

P2 = final pressure

V1 = initial temperature

V2 = final temperature

P1 = 29

V1 = V

P2 = ?

V2 = P/7

Then;

P2 = P1V1/V2

P2 = 29 * V/V/7

P2 = 203 psi

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Taking into account the reaction stoichiometry, 380.12 grams of O₂ are produced as the result of the decomposition of 843 g of NaClO₃.

In first place, the balanced reaction iS:

2 NaClO₃ → 2 NaCl + 3 O₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

The molar mass of the compounds is:

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

The following rule of three can be applied: if by reaction stoichiometry 212.9 grams of NaClO₃ form 96 grams of O₂, 843 grams of NaClO₃ form how much mass of O₂?

mass of O₂= (843 grams of NaClO₃× 96 grams of O₂) ÷212.9 grams of NaClO₃

mass of O₂= 380.12 grams

Finally, 380.12 grams of O₂ are formed.

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If the reaction yield is 95.7%, then 2.37 grams of lead (II) oxide will be produced by the decomposition of 2.50 grams of lead (II) carbonate.

Answer:

2.54g

Explanation:

Lead carbonate (PbCO3) can be decomposed to produce lead oxide (PbO) and carbon dioxide (CO2). The equation for this reaction is as follows:

PbCO3(s) -> PbO(s) + CO2(g)

Let's assume that the reaction yield is 95.7%, meaning that 95.7% of the theoretical amount of lead oxide produced in the reaction is actually obtained. To find the actual amount of lead oxide produced, we first need to find the theoretical amount of lead oxide produced by the reaction.

The theoretical amount of lead oxide can be calculated using the stoichiometry of the reaction, where the number of moles of reactants is balanced with the number of moles of products. If we assume that 2.50g of lead carbonate is decomposed, the number of moles of lead carbonate can be calculated as follows:

n = m/M

where n is the number of moles, m is the mass of the substance, and M is the molar mass of the substance. For lead carbonate, the molar mass is:

M = 207.19 g/mol

So, the number of moles of lead carbonate is:

n = 2.50 g / 207.19 g/mol = 0.01202 mol

Since the reaction is balanced, the number of moles of lead oxide produced should be equal to the number of moles of lead carbonate. The mass of lead oxide produced can be calculated using the number of moles and the molar mass of lead oxide:

m = n x M

where M = 223.20 g/mol is the molar mass of lead oxide. So, the mass of lead oxide produced is:

m = 0.01202 mol x 223.20 g/mol = 2.68 g

Since the reaction yield is 95.7%, the actual amount of lead oxide produced is:

actual_mass = 0.957 x 2.68 g = 2.54 g

So, approximately 2.54g of lead oxide will be produced by the decomposition of 2.50g of lead carbonate.

The missing nucleus in the equation is 234/90 Th

A nuclear reaction is a procedure wherein an atom's nucleus is altered by being divided or fused with the nucleus of another atom. We know that the masses and the charges that we have in the nuclear reaction must be balanced according to what have been shown in the image that is attached here.

We can then see that the nuclear reaction that is shown in the image is an alpha decay and there is the loss of the helium atom from the system as has been typified in the equation.

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The cell would burst as water moved into the cell along a concentration gradient from high to low.

If a salt-filled cell is placed in fresh water, osmosis will occur. Osmosis is the diffusion of water molecules from an area of high concentration to an area of low concentration through a semi-permeable membrane. In this case, the semi-permeable membrane is the cell membrane that surrounds the salt-filled cell.

Since the concentration of salt inside the cell is higher than the concentration of salt in the fresh water, water molecules will move from the fresh water into the cell in an attempt to balance the concentration of salt on both sides of the membrane. This results in an increase in the volume of the cell, which can cause the cell to swell or even burst.

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the molarity of the final solution is 0.045 M If 60. 0 ml of a 1. 5 m hcl solution is put into a flask and diluted with water to make 2000 ml of solution.

To solve this problem, we can use the formula:

M1V1 = M2V2

Where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.

We are given that the initial volume is 60.0 ml and the initial molarity is 1.5 M. We are also given that the final volume is 2000 ml (which is the sum of the initial volume and the volume of water added).

Using the formula above, we can solve for the final molarity:

M1V1 = M2V2

(1.5 M) (60.0 ml) = M2 (2000 ml)

M2 = (1.5 M) (60.0 ml) / (2000 ml)

M2 = 0.045 M

Therefore, the molarity of the final solution is 0.045 M.

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The balanced form of the chemical equation given above is as follows: C₂H₁₆ + 60₂ -> 6CO₂ (g) + 6H₂O(g).

A chemical equation is a symbolic representation of a chemical reaction where reactants are represented on the left, and products on the right.

A chemical equation is said to be balanced when the number of atoms of each element on both sides of the equation is the same.

According to this question, the equation for photosynthesis is given. The balanced form is as follows:

C₂H₁₆ + 60₂ -> 6CO₂ (g) + 6H₂O(g)

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Some strategies which are used to improve the powerless language include the reduction of the use of the following below:

This is referred to as the type of language which suggests to listeners that you are uncertain, insufficient, and not entirely confident with oneself due ti various reasons.

Examples of the reasons are the use of hedges , intensifiers etc which have negative effects on how the speaker is perceived.  It is usually characterized by Hesitation like “Well” or “Um” etc and it should be avoided to improve it thereby making it he correct choice.

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Activation energy is used to describe the energy needed for a chemical reaction to begin.

Define chemical reactions .

Breaking chemical bonds between reactant molecules (particles) and creating new bonds between atoms in product particles are two aspects of chemical reactions (molecules). While there are the same number of atoms both before and after the chemical transition, there will be a change in the number of molecules.

The least amount of energy necessary to start a process, such a chemical reaction, is known as the activation energy. Exothermic reactions require activation energy to begin, just like all other chemical reactions. Reactants require activation energy in order to move collectively, overcome repulsive forces, and begin bond breakdown.

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Answer:

divide it by

its molar mass

Explanation:

a. We may use the equilibrium constants for the two reactions provided to get the equilibrium constant for the reaction HbO2(aq) + CO(aq) -> HbCO(aq) + O2(aq). O2(aq) + Hb(aq).

= 1.8 Hb(aq) + CO(aq) = HbO2(aq) Kc Kc HbCO(aq) = 306 Considering the forward and reverse.  reaction ratios for the whole response, we get: ([HbCO][O2])/ K eq ([HbO2][CO]) Substituting in the equilibrium constants for the two specified reactions yields: K eq = (306 * [O2]) / 1.8 K eq = (306 * [O2]) / 1.8 The concentration of O2 in the circulation at body temperature is around 0.13 mM, or 0.13/1000 moles per liter. Substituting this value yields: (306 * 0.13/1000) / 1.8 = 0.022 K eq At body temperature, the equilibrium constant for the reaction HbO2(aq) + CO(aq) -> HbCO(aq) + O2(aq) is around 0.022. b. If 0.10% of the air mixture includes carbon monoxide and 20.0% oxygen, which indicates that there is one molecule of carbon monoxide and 200 molecules of oxygen in every 1000 molecules of air. Considering that the oxygen and carbon monoxide concentrations in blood are the same as those in air, this means that for every 200 molecules of HbO2 that join with oxygen, 1 molecule of HbO2 will bond with carbon monoxide to generate HbCO. The circulatory HbCO/HbO2 ratio can be computed as follows: [CO]/([O2]/200) = [CO] / ([O2]/200) Substituting the provided values yields: (1/1000) / (20/100 * 200) = [HbCO]/[HbO2 1/20000 [HbCO]/[HbO2] As a result, the circulatory HbCO/HbO2 ratio is 1:20000, or 0.005%. Carbon monoxide is extremely harmful because it bonds to Since it binds to hemoglobin considerably more firmly than oxygen, it can displace oxygen from hemoglobin and prevent it from reaching essential organs. This can result in tissue damage, organ failure, and even death. Low quantities of carbon monoxide can be hazardous, causing headaches, dizziness, nausea, and other symptoms.

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statements that describe Diels-Alder reaction. are A. The reaction is initiated by heat. D. The reaction involves the formation of three new C-C σ bonds. E. The reaction is concerted.

The Diels-Alder reaction is initiated by heat or light and involves the formation of new C-C sigma (σ) bonds. It is a concerted reaction, meaning that the breaking of the π bonds and the formation of the σ bonds occur simultaneously. The reaction is exothermic, not endothermic, as it releases energy upon the formation of the new bonds. Therefore, statements B and C are incorrect.

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In the given reaction, carbon disulfide acts as an excess reagent and oxygen acts as a limiting reagent.

a) Oxygen is the limiting reactant.

b) 0.667 moles of carbon disulfide will still be present.

c) The formation of 0.333 moles of carbon dioxide and 0.667 moles of sulfur dioxide.

provided here,

To make 1 mole of carbon dioxide and 2 moles of sulfur dioxide from 1 mole of carbon disulfide, 3 moles of oxygen are required.

1.00 ol of carbon disulfide is the number of moles.

One mole of oxygen is equal to 1.00 mol.

The limiting reactant is oxygen. It will be totally eaten. (1.00 mol).

(B) An excess reactant is carbon disulfide. It will respond with

Oxygen reacts with 0.33 moles of carbon disulfide.

The extra reagent still present:

0.667 moles There will be a residue of carbon disulfide.

In order to make one mole of carbon disulfide and one mole each of carbon dioxide and sulfur dioxide, you need three moles of oxygen.

The carbon dioxide equivalent for 1 mole of oxygen is 1 mole/3, or 0.333 moles.

Sulfur dioxide is equal to 0.667 moles per 1.00 mol of oxygen (1.00 mol/(3/2)) mol.

Since carbon disulfide is an excess reagent in the above reaction, oxygen is a limiting reagent.

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Sometimes a neutral oxygen atoms gains electrons. The oxygen then becomes an anion with a negative charge.

All the atoms undergoes chemical reactions to attain the octet electronic configuration. It means that it tries to attain the configuration of noble gases, which has the most stable atoms.

Oxygen has 6 electrons in its outer shell. It accepts two more electrons to attain the configuration of neon. As it attains electrons the atoms become negatively charged ions, called anions. It will now become O²⁻.

So the oxygen atom attains electrons to become an anion, with negative charge.

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Answer:

0.54 M NaCl.

Explanation:

The final concentration of the mixture can be found using the formula for the dilution of a solution:

C1 * V1 + C2 * V2 = C_final * (V1 + V2)

where C1 and C2 are the initial concentrations of the two solutions, V1 and V2 are their volumes, and C_final is the final concentration of the mixture. Plugging in the given values, we get:

0.680 M * 279 mL + 0.450 M * 462 mL = C_final * (279 mL + 462 mL)

0.680 M * 279 mL + 0.450 M * 462 mL = C_final * 741 mL

0.680 M * 279 mL + 0.450 M * 462 mL = C_final * 0.741 L

Converting volumes to liters:

0.680 M * 0.279 L + 0.450 M * 0.462 L = C_final * 0.741 L

0.680 M * 0.279 L + 0.450 M * 0.462 L = C_final * 0.741 L

0.190982 + 0.209719 = C_final * 0.741 L

0.400701 = C_final * 0.741 L

Dividing both sides by 0.741 L:

0.400701 / 0.741 L = C_final

So, the final concentration of the mixture is 0.54 M NaCl.

79942.6 calorie is the number of calories of energy absorbed by the water. The energy of an isolated system holds that energy could be transformed.

In physics, energy is indeed the quantitative quality that is imparted to a body or a physical system and is visible in the form of light and heat.

The energy of an isolated system holds that energy could be transformed in form but cannot be generated or destroyed. The joule is the International System of Units (SI) unit of measurement for energy (J).

Q = 2 kg × 4181 J / kg °C × 40°C

   = 334,480 J

   =79942.6 calorie

Therefore, 79942.6 calorie is the number of calories of energy absorbed by the water.

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Answer:

A = 252 * 0.04 * 0.0028 = 0.275584

Therefore, the absorbance of the solution in a 4.00 mm pathlength cell is approximately 0.28.

Explanation:

Answer:

0.28

Explanation:

The formula to calculate the absorbance of a solution is given by:

A = ε * l * c

Where:

A = absorbance

ε = molar absorptivity (L/mol cm)

l = pathlength (cm)

c = concentration (mol/L)

The molar absorptivity of the solute at 500 nm is 252 M^-1cm^-1, and the concentration of the solute in the solution is 0.00140 moles in 500.0 mL, so c = 0.00140 / 0.500 = 0.0028 M.

A = ε * l * c = 252 M^-1 cm^-1 * 4.00 mm * 0.00280 M = 2.16

So, the absorbance in a 4.00 mm pathlength cell would be 2.16.

Answer:

Chromium(III) phosphate trihydrate

Explanation:

The correct IUPAC name for CrPO4 · 3H2O is Chromium(III) phosphate trihydrate.

The mass (in grams) of Mn present in 1.07×10²⁴ atoms is 97.735 grams

We'll begin by obtaining the number of mole of Mn. This can be obtained as follow:

From Avogadro's hypothesis,

6.02×10²³ atoms = 1 mole of Mn

Thus,

1.07×10²⁴ atoms = (1.07×10²⁴ atoms × 1 mole) / 6.02×10²³ atoms

1.07×10²⁴ atoms = 1.777 moles of Mn

Now, we shall determine the mass (in grams) of Mn. Details below:

Mole = mass / molar mass

Cross multiply

Mass = Mole × molar mass

Mass of Mn = 1.777 × 55

Mass of Mn = 97.735 grams

Thus, from the above calculation, it is evident that the mass of Mn is 97.735 grams

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Answer:

(C3H8) produces 660 g of CO2 and 360 g of H2O

Explanation:

The balanced chemical equation for the combustion of propane (C3H8) is:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

This equation tells us that for every molecule of propane (C3H8) that reacts with 5 molecules of oxygen (O2), 3 molecules of carbon dioxide (CO2) and 4 molecules of water (H2O) are produced.

So, if we have 220. g of propane (C3H8), we can find the amount of CO2 and H2O produced by using the mole ratio from the balanced equation:

1 mole C3H8 reacts with 5 moles of O2 to produce 3 moles of CO2 and 4 moles of H2O

We can find the number of moles of C3H8 by dividing the mass by the molar mass of C3H8 (44 g/mol):

220 g / 44 g/mol = 5 moles C3H8

So, the number of moles of CO2 and H2O produced can be found by multiplying the number of moles of C3H8 by the mole ratio:

3 moles CO2 = 3 moles CO2/1 mole C3H8 * 5 moles C3H8 = 15 moles CO2

4 moles H2O = 4 moles H2O/1 mole C3H8 * 5 moles C3H8 = 20 moles H2O

Finally, we can convert the number of moles of CO2 and H2O to grams by multiplying by their molar masses (44 g/mol for CO2 and 18 g/mol for H2O):

15 moles CO2 * 44 g/mol = 660 g CO2

20 moles H2O * 18 g/mol = 360 g H2O

So, the combustion of 220 g of propane (C3H8) produces 660 g of CO2 and 360 g of H2O.

Answer: Single Replacement

Explanation:

This reaction is single replacement because the Zinc pushes out the Copper from the CuCl2 to form ZnCl. This happens because Zinc is above Copper in the activity series.

If you coated the surface of a penny with a thin layer of a nonpolar liquid (oil) and then dropped alcohol and water onto the penny what would occur is that the alcohol will mix with the water and spread out to form a thin layer on top of the oil.

Any molecule without any electrical charges or partial charges is said to be a nonpolar substance. Atoms with similar levels of electronegativity make up nonpolar compounds.

Alcohol is a polar substance and water is also a polar substance.

Hence, the water and alcohol will not mix with the oil because oil is nonpolar and will not dissolve in a polar solvent like water or alcohol.

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Answer:

All molecules have same speed.

Explanation:

Molecules of an ideal gas move randomly with different speeds.

Answer:

142 can be expressed in scientific notation as 1.42 x 10².

Explanation:

Answer:

157.48 g

Explanation:

To calculate the mass of a piece of iron with a volume of 20.00 cm³, we can use the formula:

mass = density x volume

where density is the density of pure iron (7.874 g/cm³) and volume is the volume of the iron piece (20.00 cm³).

mass = 7.874 g/cm³ x 20.00 cm³ = 157.48 g

So, the mass of the iron piece is 157.48 g.

The student needs to weigh out 18.055 g of chromium(II) nitrate.

What is mass?

Mass is a measure of the amount of matter in an object, typically measured in units of kilograms (kg), grams (g), or milligrams (mg). It is a scalar quantity, meaning it has magnitude but no direction.

(a) To prepare a 0.235 M solution of chromium(II) nitrate in 500 mL of water, we need to calculate the amount of chromium(II) nitrate that needs to be weighed out.

The formula weight of chromium(II) nitrate is Cr(NO3)2. Using the periodic table, we can find the atomic weights of the elements in this compound:

Cr: 52

N: 14

O: 16

So, the formula weight of chromium(II) nitrate is:

52 + 2(14 + 3(16)) = 154 g/mol

To prepare a 0.235 M solution, we need 0.235 moles of chromium(II) nitrate in 1 liter (1000 mL) of solution. In 500 mL of solution, we need:

0.235 moles/L x 0.5 L = 0.1175 moles

The mass of chromium(II) nitrate that contains 0.1175 moles can be calculated using the formula:

mass = moles x formula weight

mass = 0.1175 mol x 154 g/mol = 18.055 g

Therefore, the student needs to weigh out 18.055 g of chromium(II) nitrate.

(b) Option C, "Add a small amount of water to the volumetric flask and swirl until the salt sample has dissolved," would not be an expected step in the procedure used by the student. When preparing a solution in a volumetric flask, it is important to first add a portion of the solvent (in this case, water) to the flask, and then add the solute (chromium(II) nitrate) and mix it thoroughly to dissolve it. This is to ensure that the final volume of the solution is accurate, since the volumetric flask is calibrated to contain a specific volume of liquid only when filled to the mark after adding the solute.

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The empirical formula of a compound can be determined from the results of a combustion analysis. The empirical formula represents the simplest whole-number ratio of atoms in the molecule.

Here's how you can determine the empirical formula of the hydrocarbon based on the given information:

Determine the number of moles of CO2 produced:

moles = mass / molar mass

moles = [tex]22.0 g / (44.0 g/mol) = 0.5 mol[/tex]

Determine the number of moles of carbon in the sample:

moles of carbon = moles of CO2 produced / ratio of C:CO2

moles of carbon = [tex]0.5 mol / 1 : 1 = 0.5 mol[/tex]

Determine the number of moles of hydrogen in the sample:

moles of hydrogen = moles of CO2 produced / ratio of H:CO2

moles of hydrogen = [tex]0.5 mol / 1 : 2 = 0.25 mol[/tex]

Express the number of moles of each element as a whole number ratio:

moles of carbon / smallest number of moles = [tex]0.5 mol / 0.5 mol = 1[/tex]

moles of hydrogen / smallest number of moles =[tex]0.25 mol / 0.25 mol = 1[/tex]

So, the empirical formula of the hydrocarbon is CH, indicating that for every one carbon atom, there is one hydrogen atom in the molecule.

To find the molecular formula, you can use the empirical formula and the molar mass of the compound. The molecular formula is a multiple of the empirical formula and can be determined by dividing the molar mass of the compound by the empirical formula weight:

Empirical formula weight =[tex]12.0 g/mol + 1.0 g/mol = 13.0 g/mol[/tex]

Molecular formula multiple = molar mass / empirical formula weight

Molecular formula multiple = [tex]70 g/mol / 13.0 g/mol = 5.3846[/tex]

The molecular formula multiple is not a whole number, so we round up to the nearest whole number, which is 6. The molecular formula of the hydrocarbon is therefore 6 times the empirical formula, or C6H6.

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In precipitation reaction, two reactants in aqueous solution form an insoluble product, or precipitate, that separates from the solution.

What do you mean by precipitation reaction?

Precipitation reactions are chemical reactions in which two or more aqueous solutions react to produce an insoluble product, known as a precipitate. Precipitation reactions can occur between two ionic compounds, or between an ionic compound and a molecular compound.

The reaction products would be:

LiCl + AgOH → LiOH + AgCl

NaI + AgNO3 → AgI + NaNO3

Ca(OH)2 + K2CO3 → KOH + CaCO3

The products of these reactions are the insoluble compound formed, often referred to as the precipitate, and the two aqueous solutions that remain after the reaction has taken place. Examples of common precipitation reactions include those involving two salts, such as silver nitrate and sodium chloride, which form silver chloride, and the reaction between lead nitrate and potassium iodide, which form lead iodide.

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In the years after the civil war   texas farmers produced cotton the most  among all crops,

In the years following the Civil War, cotton was the most important cash crop in Texas and the primary source of income for many farmers. Texas had favorable climatic conditions and fertile soils that were well-suited for cotton cultivation. Additionally, the demand for cotton was high both domestically and internationally, which made it a lucrative crop for farmers. As a result, cotton production in Texas increased significantly in the years after the Civil War, with the state becoming one of the largest cotton producers in the country. However, this dependence on a single crop also left farmers vulnerable to market fluctuations and crop failures, which led to economic hardship and debt for many.

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Answer:

168.64 grams of O2

Explanation:

To burn 46.8 grams of propane (C3H8), we need to find the amount of oxygen that reacts with it according to the balanced chemical equation:

C3H8 + 5O2 -> 3CO2 + 4H2O

From the equation, 1 mole of propane (C3H8) reacts with 5 moles of oxygen to form 3 moles of carbon dioxide and 4 moles of water. The number of moles of propane can be calculated as follows:

46.8 g C3H8 / (44.1 g/mol) = 1.054 mol C3H8

Therefore, 1.054 mol of propane (C3H8) requires 5 x 1.054 = 5.27 moles of oxygen to burn completely. To find the mass of the oxygen in grams, we multiply the number of moles by the molar mass of oxygen (O2):

5.27 moles O2 * 32.0 g/mol = 168.64 g O2.

So, 168.64 grams of oxygen are required to burn 46.8 grams of propane (C3H8).