At high temperatures one mole of hydrogen gas reacts with one mole of bromine gas to form hydrogen bromide. At a given temperature the equilibrium constant is 57.6. If at the same temperature, a mixture of 4.67 × 10^-3M bromine gas, 2.14 × 10^−3 hydrogen gas, and 2.40 × 10^−2M hydrogen bromide gas is made, then:

a. the system is at equilibrium.
b. the system is far from equilibrium and will shift to form more hydrogen gas.
c. the system is far from equilibrium and will shift to form more hydrogen bromide gas.
d. nothing can be deduced since we do not know whether the reaction is endothermic or exothermic.
e. nothing can be deduced since we do not know whether the equilibrium constant is Kc or Kp.

Answer:

the answer is ionic

Such a bond forms when the valence (outermost) electrons of one atom are transferred permanently to another atom. ... The atom that loses the electrons becomes a positively charged ion (cation), while the one that gains them becomes a negatively charged ion (anion). A brief treatment of ionic bonds follows

A pH of 4 is ten times more acidic than a pH of 5.

Yes, pH can have decimal values. In fact, pH values can range from 0 to 14 and can have any value between them including decimals. pH is a measure of the acidity or alkalinity of a solution, and it is determined by the concentration of hydrogen ions (H+) in the solution.A solution with a pH of 7 is neutral, which means that it has an equal concentration of hydrogen ions (H+) and hydroxide ions (OH-). An acidic solution has a pH below 7 and a high concentration of H+ ions. On the other hand, an alkaline solution has a pH above 7 and a low concentration of H+ ions.A pH that is less than 7.0 indicates acidity. pH less than 7.0 is acidic while pH greater than 7.0 is alkaline. Each number on the pH scale represents a ten-fold change in the acidity/alkalinity of the solution. For example, a pH of 5 is 10 times more acidic than a pH of 6, and 100 times more acidic than a pH of 7.A pH of 0 indicates a very strong acidic solution while a pH of 14 indicates a very strong alkaline solution. It's worth noting that pH is a logarithmic scale, meaning that a change of one pH unit corresponds to a ten-fold change in hydrogen ion concentration.

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Answer:

Explanation:

From the given information:

The two reactions occurring in the reactor are:

\(\mathtt{CH_3COOH + 2H_2 \to C_2H_5OH + H_2O}\)

\(\mathtt{2C_2H_5OH \to (C_2H_5)_2O+H_2O}\)

Let molar flowrate of \(\mathtt{CH_3COOH}\) entering reactor = \(\mathtt{x_n}\)

From the first reaction:

\(\mathtt{CH_3COOH + 2H_2 \to C_2H_5OH + H_2O}\)

Since there is a 71.6% conversion of \(\mathtt{CH_3COOH}\);

Then, amount of \(\mathtt{CH_3COOH}\) remaining unreacted and exiting the reactor is:

= x - 0.716x

= 0.284x

Recall that; the component flow rate exiting the reactor = 3.81 mols

∴

0.284x = 3.81

x = 3.81/0.284

x = 13.42 moles

Thus, the molar flowrate of \(\mathtt{CH_3COOH}\) entering the reactor is 13.42 moles

Answer

the answer is p block

Experiment: Does it Dissolve?

Materials Needed:

Step 1: Form a question before starting the experiment.

Step 2: Make a hypothesis for each substance.

Step 3: Scoop a teaspoon of each substance in the jars, only adding one substance per jar. Stir it up!

Step 4: Observe whether or not each substance dissolves and record the findings.

Ammonia (\(NH_3\)) and ammonium chloride (\(NH_4Cl\)) mixture could be a useful buffer in a solution. Option C

A buffer is a solution that can resist changes in pH when small amounts of acid or base are added. It consists of a weak acid and its conjugate base or a weak base and its conjugate acid. The buffer system works by the principle of Le Chatelier's principle, where the equilibrium is shifted to counteract the changes caused by the addition of an acid or a base.

In option A, acetic acid (\(CH_3CO_2H\)) is a weak acid, but hydrochloric acid (HCl) is a strong acid. This combination does not form a buffer because HCl is completely dissociated in water and cannot provide a significant concentration of its conjugate base.

Option B consists of sodium hydroxide (NaOH), which is a strong base, and elemental sodium (Na), which is a metal. This combination does not form a buffer as there is no weak acid-base pair involved.

Option D contains acetic acid (\(CH_3CO_2H\)), a weak acid, and ammonia (\(NH_3\)), a weak base. Although they are weak acid and base, they do not form a buffer system together as they are both weak acids or bases and lack the required conjugate acid-base pair.

Option C, ammonia (\(NH_3\)), is a weak base, and ammonium chloride (\(NH_4Cl\)) is its conjugate acid. This combination can form a buffer system. When ammonia reacts with water, it forms ammonium ions (NH4+) and hydroxide ions (OH-).

The ammonium ions act as the weak acid, while the ammonia acts as the weak base. The addition of a small amount of acid will be counteracted by the ammonium ions, and the addition of a small amount of base will be counteracted by the ammonia, thus maintaining the pH of the solution relatively stable.

Therefore, option C, consisting of ammonia (\(NH_3\)) and ammonium chloride (\(NH_4Cl\)), is the suitable mixture that could be a useful buffer in a solution.

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Answer:

no of moles=mass in gm÷molar mass so let x be the mass in gm

4.5=x÷35.5×2

x=4.5×35.5×2 grams

x=319.5 gm

Explanation:

formula

Answer:

and can cause changes in motion. Inertia.

Explanation:

Answer:  change its motion.

Explanation: hope this helps :)

Answer:

1. Least

2. Most

Answer:

1st one is less 2nd is greater

Explanation:

Answer:

true i am pretty sure

Explanation:

the ans. is false (:) ok

Answer:

A.  fossil fuel combustion

Explanation:

Explanation:

The microwave receives, from a transformer (energy source), a fixed voltage and generates, through the vibration of electrons in its resonant cavities, electromagnetic waves inside the device.

Answer: D. Electrical to electromagnetic

Answer:

Take 1ml of the stock solution (covert(100ug/ml) and add 9mls of water to produce a100ug/10ml solution.

Explanation:

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Avogadro's number lets you calculate the number of anything in a mole:

0.250 mol Br2 X 6.02 X 10^23 molecules/mole = 1.50 X 10^23 molecules

One mole of Br₂ contains 6.02 × 10²³  bromine molecules  or formula units. Hence, 0.250  moles of Br₂  contains 1.5 × 10²³ Br₂  molecules.

One mole of a substance is its amount containing 6.02 × 10²³  atoms or molecules. This number is called Avogadro number. One mole of every elements contains Avogadro number of atoms.

Similarly one mole of every compound contains 6.02 × 10²³  molecules. The mass of one mole of compound is called its molar mass.

Br is 35th element in periodic table. One mole of bromine molecule  contains 6.02 × 10²³ Br₂ molecules . Hence number of molecules  of 0.250 moles is:

0.250  × 6.02 × 10²³  = 1.5 × 10²³ .

Hence, there will be 1.5 × 10²³  Br₂  atoms in 0.250  moles.

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The intermolecular force between \(NH_{3}\)  (ammonia) and HF (hydrogen fluoride) molecules is hydrogen bonding.

\(NH_{3}\) and HF molecules hydrogen bind. \(NH_{3}\) and HF are polar molecules having a significant electronegativity difference between hydrogen and nitrogen and fluorine, respectively. Hydrogen bonding strengthens the dipole-dipole interaction between the molecules.

Hydrogen bonding occurs when a hydrogen atom is covalently bound to a strongly electronegative atom like N, O, or F and attracts an electronegative atom in a neighbouring molecule. NH3's hydrogen atom bonds with HF's fluorine.

No ions are present, although ion-dipole forces exist between ions and polar molecules. NH3 and HF are polar, hence dispersion (London) forces do not apply. NH3 and HF, neutral chemicals, do not connect through ionic bonding. Dipole-dipole forces between NH3 and HF are not the greatest intermolecular force.

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Metal reactivity is influenced by their propensity to lose electrons. The valence electrons are simply eliminated as the atomic size grows, resulting in the formation of positive ions.

A valence electron is an electron in an atom's outer shell that can take part in the production of a chemical bond if the outer shell is not closed. This concept is used in both chemistry and physics. One valence electron from each atom in the link is contributed to form a shared pair in a single covalent bond. The electrons in an atom's outermost electron shell are known as valence electrons.

Metal reactivity is related to their capacity to oxidise, produce basic hydroxides, and combine with non-metals to produce ionic compounds. Nuclear charge, atomic radius, shielding effect, and sublevel arrangement are the four main variables that govern metal reactivity (of electrons).

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As you can see the picture, in the three given compounds i.e. ethane, ethene and ethyne, ethyne have shortest bond. Bond length of ethyne is very short when compared to the ethane and ethene. Shorter the bond, bond strength will be more. Hence, our answer is ethyne.

Answer:

ethyne has shortest bond

No, Ag₂SO₄ will not precipitate when 100ml of .050M AgNO3 is mixed with 10ml of 5x10-2 m Na₂SO₄ solution because the precipitate is made only in an aqueous solution.

Precipitation is the solid extract that is collect in a place. Precipitate is the concentration of the substance in a solution in a specific place.

Thus, No, Ag₂SO₄ will not precipitate when 100ml of .050M AgNO3 is mixed with 10ml of 5x10-2 m Na₂SO₄ solution because the precipitate is made only in an aqueous solution.

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The volume of hydrogen gas required to produce 25.0 grams of water at STP is 31.2 liters.

The balanced chemical equation for the reaction between hydrogen gas and oxygen gas to produce water is:

2H2(g) + O2(g) → 2H2O(l)

From the equation, we can see that 2 moles of hydrogen gas are required to produce 2 moles of water. The molar mass of water is 18 g/mol, so 2 moles of water has a mass of 36 g.

Therefore, to produce 25.0 grams of water, we need to use:

(25.0 g / 36.0 g/mol) * 2 mol H2/mol H2O = 1.39 mol of H2

At STP (standard temperature and pressure), 1 mole of gas occupies 22.4 L of volume.

So, 1.39 moles of H2 will occupy:

1.39 mol x 22.4 L/mol = 31.2 L

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Answer:

the volume reduces to 6 L

Explanation:

Boyle's law states that the volume of a gas is inversely proportional to the pressure, as long as temperature and other physical conditions are kept constant. This means that, as pressure increases, volum decreases and vice versa.

the formula is: (in blue pen in attached image)

P1 = 3 atm

P2 = 4 atm

V1 = 8 L

V2 = x

making V2 subject of the formula (in red pen in attached image),

V2 = 3*8/4

= 6L

have a great day!!!

Answer

a. Combustion Reaction

Explanation

C3H8 + 5O2 --> 3CO2 + 4H2O

The reaction above involves an organic compound combusting in the presence of oxygen to produce carbon dioxide and water. This type of reaction is known as a combustion reaction.

Molar mass of 0₂= 16 and H₂= 2 g.

Molar mass is defined as the mass per mole. In other words, molar mass is the sum of the masses of all the atoms that make up a mole of a material. It is expressed in values of grams per mole.

The molar mass of the elements or molecules is shown. In the case of single elements or individual atoms, the molar mass is just the element's mass given in atomic mass units. In other words, the atomic mass and molar mass of an atom are same. Because mol mass is comparable to atomic mass for individual atoms, it may be used to identify a particle.

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Explanation:

lo que tú as puesto es falso

Answer:

Explanation:

From the information given:

\(CaF_2 \to Ca^{2+} + 2F^-\)

\(Ksp = 3.2 \times 10^{-11}\)

no of moles of \(Ca^{2+}\) = 0.01 L × 0.0010 mol/L

no of moles of \(Ca^{2+}\) = \(1 \times 10^{-5} \ mol\)

no of moles of \(F^-\) = 0.01 L × 0.00010 mol/L

no of moles of \(F^-\) = \(1 \times 10^{-6}\ mol\)

Total volume = 0.02 L

\([Ca^{2+}}] = \dfrac{1\times10^{-5} \ mol}{0.02 \ L} \\ \\ \\ \[[Ca^{2+}}] = 0.0005 \ mol/L\)

\([F^{-}] = \dfrac{(1\times 10^{-6} \ mol)}{0.02 \ L}\)

\([F^{-}] = 5 \times 10^{-5} \ mol/L\)

\(Q = [Ca^{2+}][F^-]^2 \\ \\ Q = 0.0005 \times (5\times 10^{-5})^2 \\ \\ Q = 1.25 \times 10^{-12}\)

Since Q<ksp, then there will no be any precipitation of CaF2

The student needs to add approximately 83.3 mL of 12 M HCl solution to the existing 1 L of 2 M HCl solution to increase the concentration to 3 M. It is important to handle concentrated acids with caution and follow proper safety procedures.

To increase the concentration of a 1 L solution of 2 M HCl to 3 M, the student needs to calculate the volume of concentrated HCl needed and add it to the existing solution. Here's how the calculation can be done:

Given:

Initial concentration of HCl solution = 2 M

Final concentration desired = 3 M

Initial volume of HCl solution = 1 L

Step 1: Calculate the moles of HCl in the initial solution.

Moles of HCl = Initial concentration × Initial volume = 2 M × 1 L = 2 moles

Step 2: Calculate the moles of HCl needed for the desired concentration.

Moles of HCl needed = Final concentration × Final volume = 3 M × 1 L = 3 moles

Step 3: Calculate the moles of HCl to be added.

Moles of HCl to be added = Moles needed - Moles present = 3 moles - 2 moles = 1 mole

Step 4: Convert the moles of HCl to the required volume of concentrated HCl.

To calculate the volume, we need to know the concentration of the concentrated HCl solution. Assuming it is 12 M, we can use the following formula:

Volume of concentrated HCl = Moles of HCl to be added / Concentration of concentrated HCl

Volume of concentrated HCl = 1 mole / 12 M = 0.0833 L or 83.3 mL

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Answer:

Heat it slowly, gently, and carefully.

Explanation:

Centrifugation, a technique that separates particles based on their density, is used to separate a mixture of solid components of blood.

A important biological fluid called blood circulates throughout the body, carrying nutrients and oxygen to the tissues and eliminating waste. It is made up of platelets, plasma, and red and white blood cells.

The centrifuge, a device that spins the blood sample at high speeds to separate the solid from the liquid components of the sample, is used to collect a blood sample.

White blood cells and platelets form a thin layer above red blood cells, while red blood cells, which are denser than other blood components, settle at the bottom of the tube. At the top of the tube is the plasma, which is the liquid portion of blood.

The layers of the sample can be delicately scraped off using a pipette after centrifugation, allowing the liquid and solid components of blood to be separated. Certain blood components, such as red blood cells or platelets, are isolated with this approach in lab settings for further examination or therapy.

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