A 5.5 gg sample of a substance contains only carbon and oxygen. Carbon makes up 35%% of the mass of the substance. The rest is made of oxygen. You are asked to determine the mass of oxygen in the sample. Which of the following expressions demonstrates a mathematical procedure to solve this problem using the proper order of operations?

a. ((100 - 35)/100) times 5.5 grams =
b. 100 - 35/100 times 5.5 grams =
c. 100 - (35/100) times 5.5 grams=

The numerical value of Kc for the given reaction, with the equilibrium mixture containing 0.0450 M PCl3, 0.116 M Cl2, and 25.8 M PCl5, is approximately 4942.03.

To calculate the numerical value of Kc for the given reaction, we need to set up an expression for the equilibrium constant using the concentrations of the species involved.

The balanced equation for the reaction is:

PCl3 (g) + Cl2 (g) ⇌ PCl5 (g)

The equilibrium constant expression, Kc, is given by:

Kc = [PCI5] / ([PCl3] * [Cl2])

Given the equilibrium concentrations:

[PCl3] = 0.0450 M

[Cl2] = 0.116 M

[PCI5] = 25.8 M

Substituting these values into the equilibrium constant expression, we have:

Kc = (25.8 M) / ((0.0450 M) * (0.116 M))

Calculating Kc:

Kc = 25.8 M / (0.00522 M^2)

Kc = 4942.03

Therefore, the numerical value of Kc for the given reaction, with the equilibrium mixture containing 0.0450 M PCl3, 0.116 M Cl2, and 25.8 M PCl5, is approximately 4942.03.

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The most generally accepted form of conclusive evidence for the theory of Evolution is the fossil record.

Therefore, we can confirm that the most generally accepted form of conclusive evidence for evolution is the presence of fossils that show the genetic correlation between two organisms and the progression of evolution between them.

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Answer:

6x10^-6 M

Explanation:

The relationship between pOH and [OH-] is

pOH = -log[OH-]

Solving this equation for [OH-] is

[OH-] = 10^-pOH

Plugging in the pOH makes

[OH-] = 10^-5.2

         = 6.31x10^-6

Rounding to 1 sig fig since pOH and pH sig figs are only the number of decimals means to round to 1 sig fig or

6 x 10^-6 M

When fossil fuels burn, several forms of energy are produced, including:

Heat energy: The primary form of energy released during fossil fuel combustion is heat. Fossil fuels contain chemical energy stored for millions of years, and when they burn, this energy is released in the form of heat. The heat energy can be harnessed for various purposes, such as heating buildings or generating steam to drive turbines.

Light energy: Burning fossil fuels can also produce light energy in the form of flames or glowing embers. This light energy is a byproduct of combustion.

Mechanical energy: Heat generated by burning fossil fuels can be converted into mechanical energy. This is typically achieved by using heat to produce steam, which drives a turbine connected to a generator. The rotating turbine converts the heat energy into mechanical energy, which is further transformed into electrical energy.

Electrical energy: Through the process described above, burning fossil fuels can ultimately generate electrical energy. The mechanical energy produced by the turbine is converted into electrical energy by the generator. Electrical energy can power various devices, appliances, industries, and infrastructure.

It's critical to note that while burning fossil fuels can produce useful forms of energy, it also results in the release of carbon dioxide and other greenhouse gases. This contributes to climate change and environmental concerns. As a result, there is a global shift towards cleaner and renewable energy sources to mitigate these negative impacts.

Combination Reaction.

Explanation:

Answer:

jayfeather friend me

Explanation:

Table 1. Summary of the Straight-Chain Alkanes

Name Formula Number of Structural Isomers

Octadecane C18H38

Nonadecane C19H40

Eicosane C20H42 366,319 (!)

29 more rows

The formula of the alkanes the molecule contains 28 hydrogen atoms is \(C_1_3H_2_8\), i.e., tridecane.

Alkanes are organic compounds made up entirely of single-bonded carbon and hydrogen atoms with no other functional groups.

Alkanes have the general formula CnH2n+2 and are classified into three types: linear straight-chain alkanes, branched alkanes, and cycloalkanes.

Alkanes are important raw materials in the chemical industry and the main component of gasoline and lubricants.

Natural gas is primarily composed of methane and ethane and is used for heating, cooking, and power generation (gas turbines).

Natural gas and petroleum are separated from the alkanes. Natural gas is primarily composed of methane, with trace amounts of other low molecular weight alkanes.

All other alkanes are derived from petroleum, which is a complex mixture of many compounds.

Here, the formula for 28 hydrogen atom containing alkane is

\(C_1_3H_2_8\)

Thus, this given formula is of tridecane.

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Answer:  Soundwaves travel as longitudinal or transverse waves depending on the temperature of the medium.

Explanation: i think its C

Answer:

Diagram A: Balanced Forces

These forces are balanced, so the motion of the object stays the same. The object in the diagram could be sitting still, or it could be moving. But its motion is not changing. This is because the forces acting on it from opposite sides are exactly the same size.

Explanation:

Answer:

Because a balanced force can keep an object moving at a constant velocity.

Explanation:

Answer:

1. AgNO₃ (aq) + NaCl (aq) ----> NaNO₃ (aq) + AgCl (s)

2. Li₂SO₄ (aq) + BaCl₂ (aq) ----> 2 LiCl (aq) + BaSO₄ (s)

3. 2 NaOH (aq) + MgCl₂ (aq) ----> 2 NaCl (aq) + Mg(OH)₂ (s)

Explanation:

The reaction involving the mixing of two soluble solutions to produce a precipitate is known as a precipitation reaction.

A precipitation reaction is double-replacement reaction (a reaction that exchanges the cations or the anions of two ionic compounds) in which one product is a solid precipitate.

Precipitation reactions at useful in the identification of various ions present in a solution. In order to predict the reactions that will produce a precipitate, solubility rules as given in the solubility table below can be used.

From the tables, the reactions that will produce a precipitate, as well as their balanced molecular equations are as follows:

1. AgNO₃ (aq) + NaCl (aq) ----> NaNO₃ (aq) + AgCl (s)

2. Li₂SO₄ (aq) + BaCl₂ (aq) ----> 2 LiCl (aq) + BaSO₄ (s)

3. 2 NaOH (aq) + MgCl₂ (aq) ----> 2 NaCl (aq) + Mg(OH)₂ (s)

Answer:

The answer is

Explanation:

Density of a substance is given by

From the question

mass of block = 600 g

volume = 82.1 cm³

Substitute the values into the above formula and solve for the Density

That's

We have the final answer as

Hope this helps you

Answer:

Explanation:

B)The compound potassium chloride (KCl) has different properties than potassium (K) and chlorine (Cl).

Answer: all of the water on the Earth

Explanation:

The hydrosphere is made up of all the water that we have on Earth. Included in the hydrosphere are the water that we have underground, those in the Earth's surface, and those that we have in the air.

On Earth, we can find water in the seas, lakes, oceans, streams, rivers etc. Water also comes in form of rain as well.

Answer:

D.

the father parent shares his genetic information with the mother during sexual intercourse( the sperm and egg)

The net ionic equation for the precipitation of calcium phosphate from aqueous solution is  Ca²+ (aq)+ S²- (aq)=>CaS(s)-----> CaS(s).

For a precipitation reaction, the net ionic equation is written by first knowing the precipitate, then use the solubility rules (if there are any).

Then one can write down the precipitate's formula that was made to the right of an arrow. The ions that react to get the precipitate as the reactants to the left of the arrow should have their formulas written down also:

Hence:

Since the molecular fomular is: H2S(aq) + CaCl2(aq) -------> CaS(s) + 2HCl(aq)

Then the full ionic equation will be: 2H^+(aq) + S²-(aq) + Ca²+(aq) + 2Cl^-(aq) -------> CaS(s) + 2H^+(aq) + 2Cl^-(aq)

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For this section:

(d) To determine which acid would be completely dissociated when dissolved in water, consider the strength of the acid and its ability to ionize completely. Strong acids are those that ionize completely in water, while weak acids only partially dissociate.

Among the given options:

(H O)₂SO is sulfuric acid (H₂SO₄), which is a strong acid and completely dissociates in water.

(H O)IO₃ is iodic acid (HIO₃), which is also a strong acid and completely dissociates in water.

(H O)₂SO₂ is not a known acid and cannot be evaluated for dissociation.

HCO₂H is formic acid (HCOOH), which is a weak acid and only partially dissociates in water.

Therefore, the acids that would be completely dissociated when dissolved in water are (H O)₂SO and (H O)IO₃.

(e) To estimate the pH of the given solutions formed by titration, compare the moles of the acid and base used in the reaction.

(i) For the titration of 25.0 mL of 0.10 M aqueous NaOH with 25.0 mL of 0.10 M aqueous HCl, the reaction between a strong acid (HCl) and a strong base (NaOH) forms a neutral salt (NaCl) and water. The resulting solution would have a pH of 7, indicating neutrality.

(ii) For the titration of 25.0 mL of 0.10 M aqueous NaOH with 25.0 mL of 0.10 M aqueous acetic acid, consider the ionization of acetic acid and the formation of its conjugate base. The pKa of acetic acid is given as 4.76.

Since the volumes and concentrations of the acid and base are equal, we have a 1:1 mole ratio between them. This means that half of the acetic acid will be neutralized, and the remaining half will be in the form of the conjugate base (acetate ion, CH₃COO-). The resulting solution will be a buffer solution with a pH close to the pKa of acetic acid, which is 4.76.

(f) To predict the products formed from mixing the given reagents, we need to consider the reactions and the possible chemical reactions that occur.

(i) 3I₂ + PBr₃: This reaction involves the combination of iodine (I₂) with phosphorus tribromide (PBr₃). The balanced equation is:

3I₂ + 2PBr₃ → 6IBr + P₂

(ii) 4LiH + GaCl₃: This reaction involves the combination of lithium hydride (LiH) with gallium trichloride (GaCl₃). The balanced equation is:

4LiH + GaCl₃ → 4LiCl + GaH₃

(iii) NH₃ + BCl₃: This reaction involves the combination of ammonia (NH₃) with boron trichloride (BCl₃). The balanced equation is:

NH₃ + BCl₃ → NH₃BCl₃

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2.32 atoms are there in 1.4 x 10²⁴. The smallest unit of matter with properties like chemical elements is the atom.

The atom represents the smallest portion of material that may be split without producing particles with an electrical charge. Essentially a result, an atom acts as the basic building block of chemistry. An atom is mostly made of space. The remaining material consists of a negatively charged cloud of electrons revolving about a positively charged nucleus composed of protons plus neutrons.

Number of atoms = 1.4 x 10²⁴/ 6.022×10²³

                               = 2.32 atoms

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The normal boiling point of the unknown liquid is 57.4°C.

The normal boiling point occurs when the vapor pressure of the liquid is equal to the atmospheric pressure. At normal boiling point, the temperature of the liquid is called the boiling point.

Using the Clausius-Clapeyron equation:

ln(P₂/P₁) = -(ΔHvap/R) * (1/T₂ - 1/T₁)

where P₁ is the vapor pressure at the given temperature T₁, P₂ is the vapor pressure at the boiling point temperature T₂, ΔHvap is the heat of vaporization, R is the gas constant.

At -170°C, the vapor pressure of the liquid is given as P₁ = 117 torr. At normal boiling point, the vapor pressure of the liquid is P₂ = 760 torr.

Converting all units to SI units, we have:

P₁ = 15.47 Pa

P₂ = 101325 Pa

ΔHvap = 5480 J/mol

R = 8.314 J/(mol*K)

Plugging in the values, we get:

㏑(101325/15.47) = -(5480/8.314) * (1/T₂ - 1/103.15)

Solving for T₂, the boiling point is found to be:

T₂ = 57.4°C

As a result, the unknown liquid's usual boiling point is 57.4°C.

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A student combusted magnesium ribbon in a crucible to produce magnesium oxide according to the reaction: the mass of the oxygen reacted is 0.39 g.

The balanced chemical equations given as :

2Mg   +  O₂   ----->    2MgO

given that :

mass of the oxygen reacted.

Mass of crucible + lid: 23.321 g

Mass of magnesium + crucible + lid: 23.914 g

Mass of magnesium oxide + crucible + lid: 24.821 g

mass of magnesium=(mass of magnesium +crucible+ lid )- (mass of

                                                                                            crucible + lid )

                                 = 23.914 - 23.321

                                = 0.593 g

mass of magnesium = 0.593 g

molar mass of magnesium is = 24 g/mol

number of moles of magnesium = mass / molar mass

                                                    = 0.593 / 24

                                                    = 0.0247 mol

2 moles of Mg react with one mole of  O₂

0.0247 mole react react with = 0.0247 / 2 = 0.0123 mol

mass of oxygen = moles × molar mass

                          = 0.0123 × 32

                          = 0.39 g

Thus, A student combusted magnesium ribbon in a crucible to produce magnesium oxide according to the reaction: the mass of the oxygen reacted is 0.39 g.

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The complete balanced chemical equation  which describes a bacterium's metabolism is CH₃COOH(aq)----->CH₄ + CO₂.

Balanced chemical equation includes the expansion of stoichiometric coefficients to the reactants and items. This is significant in light of the fact that a chemical reaction should submit to the law of protection of mass and the law of steady extents, for example similar number of molecules of every component should exist on the reactant side and the item side of the situation.

Two fast and simple techniques for balanced chemical equation are examined in this article. The main strategy is the conventional adjusting technique and the subsequent one is the mathematical adjusting strategy.

A Balanced chemical equation is an emblematic portrayal of a substance response wherein the reactants and items are indicated by their particular synthetic formulae.

An illustration of a Balanced chemical equation is 2H₂ + O₂ → 2H₂O which depicts the response among hydrogen and oxygen to shape water

The reactant side is the piece of the Balanced chemical equation to the left of the '→' image while the item side is the part to the right of the bolt image.

Number of carbon atoms on both sides of equation is two and number of atoms of hydrogen is 4 on both sides. So, given equation is already balanced.

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(Complete question) is:

Different strains of methanogenic bacteria metabolize different small molecules to produce energy. Complete the balanced chemical equation describing a bacterium's metabolism of the following molecules. 1st attempt Part 1 (1 point) dd See Periodic Table. While typically we do not add coefficients of one. For this question please add coefficients of one. CH₃COOH(aq) → CH₄(g) + CO₂(g)

Answer:

sedimentary rocks

Explanation:

I did this assignment

The amount, in kJ, of heat needed to completely vaporize 50.0g of water at 100°C is 118.8 kJ.

The heat needed to completely vaporize 50.0g of water at 100°C can be calculated using the following formula:

q = m x Hv

where:

First, we need to convert 50.0g to moles by dividing by the molar mass of water which is approximately 18.015 g/mol3:

moles of water = 50.0 g / 18.015 g/mol moles of water = 2.776 mol

Thus:

q = (2.776 mol) x (40.65 kJ/mol) q = 112.8 kJ

In other words, 112.8 kJ of heat is needed to completely vaporize 50.0g of water at 100°C.

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When the grain size is reduced to 0.160 pm, the strength of mild steel is 1851 MPa.There are a number of variables that can affect mild steel's strength, including the grain size, which is controllable.

We may use the above information to determine the constants in the Hall-Petch equation: The strength of mild steel at grain size d1 = 17.43 pm is 1 = 232.9 MPa. Mild steel has a strength of 2 = 874.2 MPa when the grain size is d2 = 0.80 um (or 800 nm). This is the formula for the Hall-Petch equation: = o + Kd(-1/2). We can use the following equation to determine K: K = (σ2 - σ1)(d2^(1/2) - d1^(1/2))^(-1) (-1) When the values are plugged in, we get the following result: K = (874.2 - 232.9)(800(1/2) - 17.43(1/2))(-1) = 276.3 MPa um(1/2) We can use the following equation to determine o: σo = σ1 - Kd1^(-1/2) Plugging in the numbers yields the following result: o = 232.9 - 276.3(17.43(-1/2)) = 62.28 MPa. The Hall-Petch equation's constants are K = 276.3 MPa um(1/2) and o = 62.28 MPa. Hence, the power of the  yield strength of mild steel at 0.160 pm grain size is 1851 MPa.

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The average speed of an object is the ratio of the distance to the time. Hence, the speed of the ball rolling 6 m in 2 seconds is 3 m/s. Therefore, student A is correct.

Speed is a physical quantity measuring the distance covered by an object within unit time. Speed is a scalar quantity. The rate of speed is called velocity. The unit of both quantities in common is m/s.

If an object covers more distance within small time, the object is said to have a greater speed. The speed of the object is the ratio of its change in distance to the change in time.

Given that, distance covered by the ball = 6 m

time taken = 2 seconds

speed = distance/ time

speed = 6 m / 2 s= 3 m/s

Therefor, the speed of the ball is 3 m/s and student A is correct.

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The grams of AgCl are generated when 85 g of BaCl₂ reacts is 10.86 g.

The balanced equation is given as :

BaCl₂    +    2AgNO₃   ----->   2AgCl   +   Ba(NO₃)₂

the mass of the BaCl₂ = 85 g

molar mass of BaCl₂  = 208.23 g/mol

moles of BaCl₂  = mass / molar mass

                          = 85 / 208.23

                          = 0.408 mol

1 mole of BaCl₂  produce 2 moles of AgCl

0.408 moles of BaCl₂  = 2 × 0.408

                                      = 0.816 mol of AgCl

moles of AgCl = 0.816 mol

mass of AgCl = moles × molar mass

                      = 0.816 g × 143.32 g/mol

                      = 10.86 g

Thus,  The grams of AgCl are generated when 85 g of BaCl₂ reacts is 10.86 g.

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A. Cell reaction:\(2 Cd^2+(aq) + 2 Cr(s) - > 2 Cd(s) + 2 Cr^2+(aq) ; E^0cell = -0.51 V\)

B. Cell reaction: \(2 Fe^2+(aq) + Br2(g) - > 2 Fe(s) + 2 Br^-(aq) ; E^0cell = +1.507\)

Let's calculate the standard cell potential, E^0cell, for each pair of half-reactions and write the balanced equations for the overall cell reactions:

A. Half-reactions:

\(Cd^2+(aq) + 2e^- - > Cd(s) (E^0 = -0.403 V)\\Cr^2+(aq) + 2e^- - > Cr(s) (E^0 = -0.913 V)\)

To calculate the standard cell potential, we subtract the reduction potential of the anode (oxidation half-reaction) from the reduction potential of the cathode (reduction half-reaction).

\(E^0cell = E^0cathode - E^0anode\\E^0cell = (-0.913 V) - (-0.403 V) = -0.51 V\)

The balanced equation for the overall cell reaction is obtained by multiplying the half-reactions by coefficients to ensure that the number of electrons transferred is the same:

\(2 Cd^2+(aq) + 2 Cr(s) - > 2 Cd(s) + 2 Cr^2+(aq)\)

B. Half-reactions:

\(Fe^2+(aq) + 2e^- - > Fe(s) (E^0 = -0.447 V)\\Br2(g) + 2e^- - > 2 Br^-(aq) (E^0 = +1.06 V)\\E^0cell = E^0cathode - E^0anode\\E^0cell = (+1.06 V) - (-0.447 V) = +1.507\)V

The balanced equation for the overall cell reaction is:

\(2 Fe^2+(aq) + Br2(g) - > 2 Fe(s) + 2 Br^-(aq)\)

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Answer:  the substance in which another substance

Explanation:

done

Answer:

Is drinks a compound or mixture?

Drinks are usually solutions of various substances dissolved in the water. Since the amount of each of these can vary (sugar, flavoring, color, carbonation), it’s a mixture. The components can be separated by various physical methods such as evaporation. Is smoke a compound or a mixture? Smoke is a mixture.

Answer:

100n

Explanation:

f= 0.05*2000=100 N

The equilibrium constant, Kc, for the response is 0. This approach that at equilibrium, the attention of merchandise is correctly 0 in comparison to the attention of reactants.

The equilibrium constant, Kc, may be calculated the use of the concentrations of the reactants and merchandise at equilibrium, consistent with the regulation of mass action. The balanced chemical equation for the response is:

CO(g) + H2O(g) ⇌ CO2(g) + H2(g)

At equilibrium, the attention of CO2 is given as 0.seventy three M. The preliminary attention of CO is 2.00 M, and the preliminary attention of H2O is likewise 2.00 M. Let x be the alternate in attention of CO and H2O, and y be the alternate in attention of CO2 and H2, then the equilibrium concentrations may be expressed as:

[CO] = 2.00 M - x

[H2O] = 2.00 M - x

[CO2] = 0.seventy three M + y

[H2] = y

The equilibrium expression for the response is:

Kc = ([CO2][H2])/([CO][H2O])

Substituting the equilibrium concentrations into the expression gives:

Kc = ([0.73 + y][y])/([(2.00 - x)][(2.00 - x)])

Assuming that x and y are small in comparison to the preliminary concentrations, we will approximate the equilibrium concentrations as:

[CO] ≈ 2.00 M

[H2O] ≈ 2.00 M

[CO2] ≈ 0.seventy three M

[H2] ≈ 0.00 M

Substituting those values into the equilibrium expression gives:

Kc = (0.730)/(2.002.00) = 0

This suggests that the response does now no longer continue to a enormous quantity withinside the ahead route and is in all likelihood to be an instance of a reversible response with a low equilibrium constant.

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