Suppose that in an equilibrium mixture of HCl, Cl2, and H2, the concentration of H2 is 1.0 x 10-11 mol-L-1and that of Cl2 is 2.0 x 10-10 mol-L-1. What is the equilibrium molar concentration of HCl at 500 K, given Kc = 4.0 x 1018 for H2(g) +Cl2(g) ⇆ 2HCl(g).

Answer:

43.93642

Explanation:

Phosphorus P 30.973762 1 86.0061

Hydrogen H 1.00794 5 13.9939

I don't know physics, but I hope this helps :)

(I'm a seventh grader so don't judge if this is wrong please)

Answer:

atom

Explanation:

The sodium atom has a single valence electron that it can easily lose. (If the sodium atom loses its valence electron, it achieves the stable electron configuration of neon.) The chlorine atom has seven valence electrons and can easily gain one electron.

Answer:

78n

Explanation:

The output force exerted on the nail in the claw is equal to 78 N which has a mechanical advantage of 5.2.

The mechanical advantage can be demonstrated as the ratio of the output force to the Input force. The mechanical advantage of any machine can be expressed in the form of the ratio of the forces utilized to do the work.

The ratio of the resistance to the effort is said to be the actual mechanical advantage which will be less. The efficiency of a machine can be evaluated by equating the ratio of the output to its input.

Given, the input force = 15 N

The mechanical advantage of the hammer = 5.2

Mechanical advantage = Output force/ Input force

5.2 = Output/15

Output force = 15 ×5.2 = 78 N

Therefore, the force is exerted on the nail in the claw is equal to 78 N.

Learn more about the Mechanical advantage,  here:

brainly.com/question/16617083

#SPJ2

Explanation:

Carbon dioxide and water

I hope it helps

Answer:

The usual products of combustion reactions are carbon dioxide and water.

Explanation:

Combustion reaction is when a substance reacts with oxygen gas, resulting in a release of energy in the form of light and heat. Combustion reactions must have oxygen (O2) as one of the reactants.

The reducing agent in the reaction 2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) is lithium (Li).

The general reaction is:

2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s)   (1)

We can write the above reaction in two reactions, one for oxidation and the other for reduction:

Li⁰(s) → Li⁺(aq) + e⁻   (2)

Fe²⁺(aq) + 2e⁻ → Fe⁰(s)    (3)

We can see that Li⁰ is oxidizing to Li⁺ (by losing one electron) in the lithium acetate (reaction 2) and that Fe²⁺ in iron(II) acetate is reducing to Fe⁰ (by gaining two electrons) (reaction 3).  

We must remember that the reducing agent is the one that will be oxidized by reducing another element and that the oxidizing agent is the one that will be reduced by oxidizing another species.

In reaction (1), the reducing agent is Li (it is oxidizing to Li⁺), and the oxidizing agent is Fe(CH₃COO)₂ (it is reducing to Fe⁰).  

Therefore, the reducing agent in reaction (1) is lithium (Li).  

Learn more here:

I hope it helps you!

Answer:

if you drop a water balloon onto the ground, its kinetic energy is converted mostly to thermal energy. If the balloon weighs 1 kilogram and you drop it from about 2 meters, it will heat up by less.

Explanation:

As you say, kinetic energy of large objects can be converted into this thermal energy. For example, if you drop a water balloon onto the ground, its kinetic energy is converted mostly to thermal energy. If the balloon weighs 1 kilogram and you drop it from about 2 meters, it will heat up by less than.

Answer:

Gypsum

Explanation:

Calcium sulphate, is a naturally occurring calcium salt. It is commonly known in its dihydrate form, a white or colourless powder called gypsum.

Answer:

Ptotal=P1+P2+… +Pn. + P nExplanation:

its c

Answer: yes it is correct

Explanation: the higher it is the cooler.

The solubility of nitrogen gas in water is 1.90 mL/dL at 1.00 atm and 13.3 mL/dL at 7.00 atm.

We want to relate the solubility of a gas with its partial pressure.

We can do so using Henry's law.

Henry's law states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid.

C = k × P

where,

The solubility of nitrogen gas is 1.90 mL/dL of blood at 1.00 atm.

Since the solvent is basically water, we can understand that the concentration of nitrogen gas is 1.90 mL/dL at 1.00 atm.

We can use this information to calculate Henry's Law constant.

k = C/P = (1.90 mL/dL)/1.00 atm = 1.90 mL/dL.atm

We want to calculate the solubility of nitrogen gas at a pressure of 7.00 atm.

We will use Henry's law.

C = k × P = (1.90 mL/dL.atm) × 7.00 atm = 13.3 mL/dL

The solubility of nitrogen gas in water is 1.90 mL/dL at 1.00 atm and 13.3 mL/dL at 7.00 atm.

Learn more about solubility here: https://brainly.com/question/11963573

Answer:

chlorine

Explanation:

Answer:chlorine reacts with hydrogen peroxide

Explanation:

Explanation:

H3PO4 is a weak acid so it partially dissociates in water

Ka1

H3PO4 (aq) + H2O(l) <----> H2PO4-(aq) + H3O+ (aq)

Ka2

H2PO4- (aq) + H2O(l) <----> HPO4 2- (aq) + H3O+ (aq)

Ka3

HPO4 2- (aq) + H2O(l) < ---> PO4 3- (aq) + H3O+ (aq)

Answer: The correct answer is black because the product of its side and mass is lower.

Explanation: The density of a substance is defined as the amount of matter that can be stored in a given volume.

Hence, the black cube will be denser because the product of its side and mass is lower.

Answer: it was wrong on my test

Explanation:

literally dont believe them

Answer:

[tex]\blue{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}[/tex]

[tex]\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}[/tex]

This problem, is describing two scenarios, the first one, a reaction between potassium carbonate and nitric acid, and also this very same acid with sodium carbonate. In addition, it is asked to set up two equations whereby you can get the grams of nitric acid needed to react with the unknown grams of both carbonates.

In this case, we can start off by designating the unknown mass of potassium carbonate as X and that of sodium carbonate as Y, so that we will be able to provide a reliable answer. Next, we write the corresponding chemical equations that take place:

K2CO3 + 2HNO3 --> 2KNO3 + H2O + CO2

Na2CO3 + 2HNO3 --> 2NaNO3 + H2O + CO2

After that, we can set up the conversion by considering the following calcultion track:

g Carbonate --> mol Carbonate --> mol HNO3 --> g HNO3

The conversion from grams to moles involve the carbonates' molar mass and the conversion from moles of nitric acid to grams, its molar mass well. In addition, we need the 1:2 mole ratio of the carbonates to nitric acid that it is evidenced in the reaction.

Therefore, the resulting equations that can be set up are shown as follows:

[tex]X g K_2CO_3 *\frac{1mol K_2CO_3 }{138.2 gK_2CO_3 } \frac{2molHNO_3}{1molK_2CO_3 } *\frac{63.1gHNO_3}{1molHNO_3} \\\\Y g Na_2CO_3 *\frac{1mol K_2CO_3 }{105.99 gK_2CO_3 } \frac{2molHNO_3}{1molNa_2CO_3 } *\frac{63.1gHNO_3}{1molHNO_3}[/tex]

Learn more:

Answer:

[tex]11 \times 6.022 \times {10}^{23} \\ = 66.242\times {10}^{23} \: of \: \\ water \: molecules[/tex]

This problem is describing the equilibrium whereby hydrofluoric acid decomposes to hydrogen and fluorine gases at 298 K whose equilibrium constant is 8.70x10⁻³, the equilibrium concentrations of all the reactants are both 1.33x10⁻³ M and asks for the initial concentration of hydrofluoric acid which turns out to be 2.86x10⁻³ M.

Then, we can write the following equilibrium expression for hydrofluoric acid once the change, [tex]x[/tex], has taken place:

[tex][HF]=[HF]_0-2x[/tex]

Now, since both products are 1.33x10⁻³ M we infer the reaction extent is also 1.33x10⁻³ M, and thus, we can calculate the equilibrium concentration of HF via the law of mass action (equilibrium expression):

[tex]8.70x10^{-3}=\frac{(1.33x10^{-3} M)^2}{[HF]} }[/tex]

[tex][HF]=\frac{(1.33x10^{-3} M)^2}{8.70x10^{-3}} }=2.03x10^{-4}M[/tex]

Finally, the initial concentration of HF is calculated as follows:

[tex][HF]_0=[HF]+2x=2.033x10^{-4}+2*(1.33x10^{-3})=2.86x10^{-3}M[/tex]

Learn more:

Answer:

you hear

Explanation:

beautiful hear can be wonderful day in your life so to be rotations is not a big deal

0.04 moles of iodide is required to precipitate all the lead II ions from  25.0 mL of a 1.6 M Pb(NO3)2 solution.

The reaction equation is;

2KI(aq) + Pb(NO3)2(aq) -------> PbI2(s) + 2KNO3(aq)

The net ionic equation is;

Pb^2+(aq) + 2I^-(aq) ------> PbI2(s)

Number of moles of Pb(NO3)2  = 25/1000 L ×  1.6 M = 0.04 moles

Number of moles of Pb^2+ = 0.04 moles /2 = 0.02 moles

Since 2 moles of iodide reacts with 1 mole of Pb^2+

x moles of iodide reacts with 0.02 moles of Pb^2+

x =  2 moles × 0.02 moles/ 1 mole

x = 0.04 moles of iodide

Hence,  0.04 moles of iodide is required to precipitate all the lead II ions from  25.0 mL of a 1.6 M Pb(NO3)2 solution.

Learn more: https://brainly.com/question/9743981

Answer:

The pulling atom is oxidized while the pulled atom is reduced. Grade 9 Chemistry

Explanation:

Answer:

Explanation:

The material that is not a mixture; it has the same properties all the way through is called a substance. Thus the material that is not a mixture; it has the same properties all the way through is called a substance.

ALL THE BEST :)

This  problem provides the molar mass and radius of a metal that has an FCC unit cell and the density is required.

Firstly, we begin with the formula that relates the aforementioned variables and also includes the Avogadro's number and the volume of the unit cell:

[tex]\rho=\frac{Z*M}{V*N_A}[/tex]

Whereas Z stands for the number of atoms in the unit cell, M the molar mass, V the volume and NA the Avogadro's number. Next, since FFC is able to hold 4 atoms and M and NA are given. Next, we calculate the volume of the atom in the unit given the radius in meters:

[tex]V=a^3=(2*1.92x10^-10m*\sqrt{2} )^3=1.60x10^{-28}m^3/atom[/tex]

And finally the required density in g/cm³:

[tex]\rho=\frac{4*241.5g/mol}{1.60x10^{-28}m^3/atom*6.022x10^{23}atom/mol} \\\\\rho=10025739g/m^3=10.03 g/cm^3[/tex]

Learn more:

When the oil is added to the heated copper, the energy in the system is

conserved.

Reasons:

The question parameters are;

Temperature of the oil in the cup = 25.00°C

Final temperature of the oil and copper, T₂ = 27.33 °C

Specific heat of copper, c₂ = 0.387 J/(g·°C)

Specific heat capacity of oil, c₁ = 1.74 J/(g·°C)

Required:

The mass of oil in the cup.

Solution:

The mass of the copper, m₂ = 17.920 g

Temperature of copper after heating, T₂ = 65.17°C

Temperature of the copper after being placed in the cup of oil, T₂ = 27.33°C

Heat lost by copper = Heat gained by the oil

Therefore, we get;

17.920 × 0.387 × (65.17 - 27.33) = m₁ × 1.74 × (27.33 - 25)

262.4219136 = 4.0542·m₁

m₁ ≈ 64.73

Learn more here:

https://brainly.com/question/21406849

Possible part of the question obtained from a similar question online, are;

The mass of the copper, m₂ = 17.920 g

Temperature of copper after heating = 65.17°C

This question is providing the exothermic heat of neutralization of acetic acid in units of kilojoules per mollimole (-49,8 kj/mmol) and asks for the same value but in calories per millimole which results -11,902.5 cal/mmol.

In this case, according to the given problem, it turns out necessary to solve a two-factor conversion in order to convert the kilojoules to joules and finally to calories as shown below:

[tex]-49.8\frac{kJ}{mmol}*\frac{1000J}{1kJ}*\frac{1cal}{4.184J}[/tex]

Thus, we cancel out the kJ and J, to obtain the following result, rounded to one decimal place:

[tex]-11,902.5\frac{cal}{mmol}[/tex]

Learn more:

Answer:

Tristan wraps some gifts and then brings them to the post office where they are delivered to people in different parts of the country. Which organelle is Tristan most like?

Answer: Tristan is most like the Golgi Body

Explanation: