The stopcock connecting a 1.30 L bulb containing xenon gas at a pressure of 6.15 atm, and a 4.25 L bulb containing oxygen gas at a pressure of 3.53 atm, is opened and the gases are allowed to mix. Assuming that the temperature remains constant, the final pressure in the system is 8.33 atm. True or false?

Answer:

False

Explanation:

Genes are found inside chromosomes and chromosomes are in cells. So technically genes are in cells

Answer:

Base

Explanation:

acid + base = salt + water

The compound reacts with an acid to form a salt and water is base.

A base is a substances that is slippery to touch,corrosive and sour taste which react with acid to give salt and water. It turns red lithmus paper to blue. It is the degree of hydroxide ion in a solution.

The compound reacts with an acid to form a salt and water is base.

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Answer:

17.08 moles of manganese dioxide

Explanation:

From the balanced reaction equation;

Zn + 2 MnO2 + 1 H2O → Zn(OH)2 + Mn2O3 (notice that I did not put 1 as a stoichiometric coefficient. It is expected that any specie written without a coefficient should have a coefficient of 1)

It is clear from the reaction equation that 2 moles of manganese dioxide produced 1 mole of zinc II hydroxide

Hence x moles of manganese dioxide will produce 8.54 moles of zinc II hydroxide

x= 2× 8.54/1

x= 17.08 moles of manganese dioxide

Answer:

N₂ = 0.7515atm

O₂ = 0.1715atm

NO = 0.0770atm

Explanation:

For the reaction:

N₂(g) + O₂(g) ⇄ 2NO(g)

Where Kp is defined as:

[tex]Kp = \frac{P_{NO}^2}{P_{N_2}P_{O_2}}}[/tex]

Pressures in equilibrium are:

N₂ = 0.790atm - X

O₂ = 0.210atm - X

NO = 2X

Replacing in Kp:

0.0460 = [2X]² / [0.790atm - X] [0.210atm - X]

0.0460 = 4X² / 0.1659 - X + X²

0.0460X² - 0.0460X + 7.6314x10⁻³ = 4X²

-3.954X² - 0.0460X + 7.6314x10⁻³ = 0

Solving for X:

X = - 0.050 → False answer. There is no negative concentrations.

X = 0.0385 atm → Right answer.

Replacing for pressures in equilibrium:

N₂ = 0.790atm - X = 0.7515atm

O₂ = 0.210atm - X = 0.1715atm

NO = 2X = 0.0770atm

Answer:

partial pressure N2 =  0.7515 atm

partial pressure O2 =  0.1715 atm

partial pressure NO =  0.077 atm

Explanation:

Step 1: Data given

Temperature = 2200 °C

Pressure of N2 = 0.790 atm

Pressure of O2 = 0.210 atm

Kp = 0.0460

Step 2: The balanced equation

N2(g) + O2(g) ⇆ 2NO(g)

Step 3: The pressure at equilibrium

pN2 = 0.790 - X atm

pO2 = 0.210 - X atm

pNO = 2X

Step 4: Define Kp and the partial pressures

Kp = (pNO)² / (pO2 * pN2)

0.0460 = 4X² / (0.210 - X)(0.790 - X)

X = 0.0385

pN2 = 0.790 - 0.0385 =  0.7515 atm

pO2 = 0.210 - 0.0385 = 0.1715 atm

pNO = 2*0.0385 = 0.077 atm

Answer:

12 moles Pb(NO₄)₂ needed.

Explanation:

                3Pb(NO₃)₂ +2AlCl₃     => 3PbCl₂ + 2Al(NO₃)₃

Given =>   ? moles      8 moles

from reaction stoichiometry, 2 moles AlCl₃ requires 3 moles Pb(NO₄)₂ then 8 moles AlCl₃ requires 3/2(8) moles of the Pb(NO₄)₂ => 12 moles Pb(NO₄)₂ needed.

Answer:hope we can be friends

can i please get brainliest

Although phlorizin inhibition of Na+-glucose cotransport occurs within a few seconds, 3H-phlorizin binding to the sodium-coupled glucose transport protein(s) requires several minutes to reach equilibrium (the fast-acting slow-binding paradigm). Using kinetic models of arbitrary dimension that can be reduced to a two-state diagram according to Cha’s formalism, we show that three basic mechanisms of inhibitor binding can be identified whereby the inhibitor binding step either (A) represents, (B) precedes, or (C) follows the rate-limiting step in a binding reaction. We demonstrate that each of mechanisms A–C is associated with a set of unique kinetic properties, and that the time scale over which one may expect to observe mechanism C is conditioned by the turnover number of the catalytic cycle. In contrast, mechanisms A and B may be relevant to either fast-acting or slow-binding inhibitors.

Explanation:

Answer : The final volume of gas will be, 26.3 mL

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas = 0.974 atm

[tex]P_2[/tex] = final pressure of gas = 0.993 atm

[tex]V_1[/tex] = initial volume of gas = 27.5 mL

[tex]V_2[/tex] = final volume of gas = ?

[tex]T_1[/tex] = initial temperature of gas = [tex]22.0^oC=273+22.0=295K[/tex]

[tex]T_2[/tex] = final temperature of gas = [tex]15.0^oC=273+15.0=288K[/tex]

Now put all the given values in the above equation, we get:

[tex]\frac{0.974 atm\times 27.5 mL}{295K}=\frac{0.993 atm\times V_2}{288K}[/tex]

[tex]V_2=26.3mL[/tex]

Therefore, the final volume of gas will be, 26.3 mL

Answer:

volume of gas will be, 26.3 mL

Explanation:

Answer:

D All of these

Explanation:

hope this helped

Answer: 0

Explanation:

The oxidation number for nitrogen in NH3 (ammonia) is -3, determined by setting up the equation x + 3(+1) = 0, where x represents the oxidation number of nitrogen.

The oxidation number for NH3 (ammonia) can be found by considering the usual oxidation states of nitrogen and hydrogen. Hydrogen typically has an oxidation number of +1, except when it forms hydride compounds with metals. Since ammonia consists of one nitrogen atom and three hydrogen atoms, and since each hydrogen has an oxidation number of +1, the total oxidation number contributed by the hydrogen atoms is +3 (3 x +1).

To find the oxidation number of nitrogen in NH3, let the oxidation number be represented as x. The sum of the oxidation numbers in a neutral compound is zero. Therefore, if we have x as the oxidation number of nitrogen and +3 from the hydrogen atoms, we can set up the equation x + 3(+1) = 0 to solve for x. Simplifying, we get x = -3. Thus, the oxidation number of nitrogen in NH3 is -3.

Following this approach, we can understand how oxidation numbers reflect the degree of electron transfer between atoms in a chemical compound or during a chemical reaction, such as NH3 reacting with O2 to form N2 and H2O. The reaction 4 NH3 + 3 O2 → 2 N2 + 6 H2O illustrates this electron transfer process and the involvement of oxidation states in balancing chemical equations.

Answer:

D

Explanation:

well, just need to remember

Answer:

Catalyst decreases activation energy

Explanation:

Consider the attached diagram and note the annotation => top of transition diagram for catalyzed reaction is lower than uncatalyzed reaction.

Answer:

the rectives are just not doing anything so just do sodium

Explanation:

Answer:

Limiting reactant is H2.

Excess reactant is O2.

Maximum 5.6 mol of H2O can be produced.

Explanation:

                               2H2 + O2 ----> 2 H2O

from reaction         2 mol   1 mol    

given                    5.6mol   4.7mol

calculated            5.6mol   2.8 mol

We can see that  for 5.6 mol H2 only 2.8 mol O2 needed, but we have 4.7 mol O2 given, so we have excess of O2.

Then limiting reactant is H2.

Excess reactant is O2.

                               2H2 + O2 ----> 2 H2O

from reaction         2 mol                2 mol

given                      5.6 mol             x mol = 5.6 mol

Final answer:

The maximum number of moles of H₂O that can be produced from 5.6 mol H₂ and 4.7 mol O₂ is 5.6 mol H₂O, with hydrogen (H₂) as the limiting reactant and oxygen (O₂) as the excess reactant.

Explanation:

To answer the question of the maximum number of moles of H₂O that can be produced from the reaction of 5.6 mol H₂ and 4.7 mol O₂, we must first look at the balanced chemical equation for the reaction which is 2H₂ + O₂ ightarrow 2H₂O. From this equation, we can see that every 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.

Now, let's see if we have enough of each reactant:

Hydrogen: 5.6 moles H₂ is available

Oxygen: 4.7 moles O₂ is available

According to the stoichiometry of the equation, oxygen will run out first since 4.7 moles of oxygen can react completely with (4.7 times 2) = 9.4 moles of hydrogen. But only 5.6 moles of hydrogen are available, which is less than 9.4 moles, so actually, hydrogen will limit the reaction.

Thus, the limiting reactant is hydrogen (H₂) and the excess reactant is oxygen (O₂). The maximum number o2f moles of water that can be produced is therefore equal to the moles of hydrogen available, which is 5.6 moles of H₂, resulting in 5.6 moles of H₂O being produced.

Final answer:

Nuclear fusion is a reaction in which two nuclei combine to form a larger nucleus, releasing energy. It is the process that powers the sun and stars.

Explanation:

Nuclear fusion is a reaction in which two nuclei are combined to form a larger nucleus, releasing energy. It is the process that powers the sun and other stars. In the sun, hydrogen nuclei combine to form helium, releasing a significant amount of energy.

Option B is correct. The statement that describes all solids is that they have a definite shape and volume

States of matter are one of the ways in which matter exists. Matter can exist as a solid, liquid, gas, and plasma.

The states of matter have different characteristics. Some of the properties of solids are:

The liquid contains loosely packed atoms eliminating the first option. Based on the explanations above, we can conclude that solids have a definite shape and volume.

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The compound MnCL2 called

Manganese cloride

Answer:

Explanation:

Cloruro de manganeso

Answer:

55J

Explanation:

∆E = 400-345 = 55J

Answer:

Energy dissipated as sound is 55 J

Explanation:

Here we have the principle of conservation of energy which states that energy can neither be created nor destroyed but can be transformed from one form to another

Where there is an initial 400 J of thermal energy in the heating element of the electric kettle we have;

Total available energy = 400 J

The energy (heat) transferred to the water is given as 345 J

The heat dissipated as sound of the kettle during heating is then found as follows;

Total available energy = Heat transferred to water + Energy dissipated as sound

400 = 345 + Energy dissipated as sound

∴ Energy dissipated as sound = 400 - 345 = 55 J.

Final answer:

From 5.0 grams of silver nitrate reacting with barium chloride, 4.21 grams of silver chloride are produced. The exact amount of barium chloride is not calculated because it is in excess.

Explanation:

To determine how many grams of silver chloride are produced from the reaction of 5.0g of silver nitrate (AgNO3) with an excess of barium chloride (BaCl2), we use the stoichiometry of the balanced chemical equation. The balanced equation is:

AgNO3 (aq) + BaCl2 (aq) → AgCl (s) + Ba(NO3)2 (aq)

To calculate the mol product, we follow the equation:

(mol product) = (mol reactant) × (stoichiometric mole ratio)

For AgNO3 with a molar mass of 169.88 g/mol, we calculate the moles of reactant:

moles of AgNO3 = mass of AgNO3 / molar mass of AgNO3

= 5.0 g / 169.88 g/mol

= 0.0294 moles of AgNO3

For silver chloride (AgCl) with a molar mass of 143.32 g/mol, we calculate the mass of AgCl produced:

mass of AgCl = moles of AgCl x molar mass of AgCl

= 0.0294 moles × 143.32 g/mol

= 4.2094 grams of AgCl (rounded to 4.21 grams)

Regarding the amount of barium chloride needed, since it is in excess and not limiting the reaction, the exact amount needed is not calculated. However, if one wanted to calculate it, we would use the stoichiometry of the reaction based on the amount of silver nitrate, ensuring that barium chloride is in excess.

Answer:

9.03*10^23 atoms  of C

Explanation:

1 mol of any substance contains 6.02*10^23 particle of this substance.

1 mol C ---    6.02*10^23 atoms of C

1.50 mol C ---  x atoms of C

x = 1.50*6.02*10^23 = 9.03*10^23 atoms of C

The critical concentration is key in determining whether actin monomers will form filaments or disassemble. Above this concentration, polymerization of actin occurs, while below it, monomers dissociate. A steady state is reached at the critical concentration, essential for cell motility and structure.

The critical concentration is the threshold at which actin filaments will either form or dissociate. In the presence of actin monomers above this concentration, the actin will polymerize into filaments; below it, the filaments will disassemble into monomers. At the critical concentration, there is a dynamic state where the rates of polymerization and depolymerization are equal, leading to a steady state in the reaction containing F-actin (filamentous actin).

Actin dynamics are essential for various cellular functions, including muscle contraction and cell motility. The polymerization and depolymerization processes are regulated by ATP-binding and hydrolysis, with critical concentration playing a crucial role in achieving a steady state of F-actin within cells. This balance affects the cell's ability to exert forces on itself and its environment, a key aspect of cellular motility and structure.

Answer: Mrs. Salge can make 6 ice cream cones

Explanation:

Mrs. Salge's recipe:

1 cone

1 scoop blue ice cream

2 scoops red ice cream

1 cherry

Now we will find what the limiting ingredient is:

We know she has:

10 cones.    →    10 ice cream cones

12 scoops blue ice cream.    →    12 ice cream cones

12 scoops red ice cream.    →    6 ice cream cones

10 cherries.    →    10 ice cream cones

The red ice cream is the limiting factor. Mrs. Salge can make 6 ice cream cones.

Mrs. Salge can make a total of 6 ice cream cones with her available ingredients, as the red ice cream scoops are the limiting factor.

To determine how many ice cream cones Mrs. Salge can make given her resources, we must find out which ingredient limits the number of cones she can make. Her recipe requires one cone, one scoop of blue ice cream, two scoops of red ice cream, and one cherry for each ice cream cone. Therefore, we will check each ingredient to see which will run out first if she keeps making the ice cream cones as per the recipe.

Since each cone requires 2 scoops of red ice cream, the number of cones she can make will be limited by the red ice cream. With 12 scoops of red ice cream, she can make 6 cones because each cone requires 2 scoops. This is the limiting factor because even though she has 10 cones and 10 cherries, after making 6 cones, she will run out of red ice cream.

Therefore, Mrs. Salge can make a total of 6 ice cream cones with the ingredients available to her before one of the ingredients runs out and prevents her from making more.

The equilibrium constant for the decomposition of water is approximately 1.01 × 10^-13.

The equilibrium constant (Kc) for the decomposition of water can be calculated using the equation: Kc = [H2]2[O2]/[H2O]2.

Given that ΔG°f(H2O(l)) = -237.2 kJ/mol, we can use the equation ΔG° = -RTlnK to find the equilibrium constant. R is the ideal gas constant (8.314 J/(mol·K)) and T is the temperature in Kelvin (25 °C + 273.15 = 298.15 K). Plugging in the values, we can solve for K.

ΔG° = -RTlnK

-237.2 kJ/mol = -(8.314 J/(mol·K) × 298.15 K) × lnK

lnK = -237.2 kJ/mol ÷ (8.314 J/(mol·K) × 298.15 K)

lnK ≈ -29.155

K ≈ e-29.155

K ≈ 1.01 × 10-13

Therefore, the equilibrium constant for the decomposition of water at 25 °C is approximately 1.01 × 10-13.

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Answer:

11.4 moles of H₂SO₄ are needed to completely react the 7.6 moles of Al

Explanation:

The equation indicates that 2 moles of aluminum react to 3 moles of sulfuric acid in order to produce 1 mol of aluminum sulfate and 3 moles of hydrogen gas.

The reaction is:  2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

This question can be solved with an easy rule of three. Ratio in the reaciton is 2:3, so we propose:

2 moles of Al react with 3 moles of sulfuric acid

Then, 7.6 moles of Al will react with 11.4 moles of H₂SO₄

I believe that leading zeros, and tail zeros without a decimal point are not significant

Final answer:

When the volume of a chamber containing CH₄ increased from 7.00L to 8.20L, 0.260 moles of CH₄ were added, assuming constant temperature and pressure.

Explanation:

To determine how many moles of CH₄ (methane) were added when the volume of the chamber increased from 7.00L to 8.20L, we can use the molar volume concept under the assumption that temperature and pressure remain constant, therefore following Avogadro's Law. According to Avogadro's Law, equal volumes of gases at the same temperature and pressure contain an equal number of moles. You can calculate the number of moles in the new volume by setting up a proportion based on the known conditions (3.00 L contains 0.650 moles) and then solving for the number of moles in the new volume of 8.20 L.

First, determine the number of moles in the 7.00 L chamber:

moles in 3.00 L / 3.00 L = moles in 7.00 L / 7.00 L

0.650 moles / 3.00 L = x moles / 7.00 L

x = (0.650 moles / 3.00 L) × 7.00 L

x = 1.517 moles in 7.00 L

Now, calculate the number of moles in the increased volume of 8.20 L:

moles in 3.00 L / 3.00 L = moles in 8.20 L / 8.20 L

0.650 moles / 3.00 L = y moles / 8.20 L

y = (0.650 moles / 3.00 L) × 8.20 L

y = 1.777 moles in 8.20 L

The number of moles added is the difference between the moles in 8.20 L and the moles in 7.00 L:

moles added = 1.777 moles - 1.517 moles

moles added = 0.260 moles

Therefore, when the volume of the CH₄ chamber increased from 7.00L to 8.20L, 0.260 moles of CH₄ were added.

Answer:

2

Explanation:

well since the question has given you the concentration of the solution 0.2mol/L and the wanted amount (moles) of naoh(0.4mol) you are able put this into the formula n=cV; where n is the moles of naoh the solution, c is the concentraton of the solution and V is the volume of the solution in litres.

therefore the for the solution to have 0.4 moles of naoh you put it into the formula, giving you:

0.4 = 0.2V

V = 2

V = 2 litres

Final answer:

To obtain 0.4 moles of NaOH from a 0.2 M NaOH solution, you would need to measure out 2 liters of the solution.

Explanation:

To calculate the volume of a 0.2 M NaOH solution needed to have 0.4 moles of NaOH, we can use the molarity equation, which is:

Molarity (M) = Moles of solute / Volume of solution in liters (L)

We can rearrange this equation to solve for the volume:

Volume of solution (L) = Moles of solute / Molarity (M)

Substituting the given values:

Volume of solution (L) = 0.4 moles NaOH / 0.2 M NaOH

Volume of solution (L) = 2 liters

Therefore, you would need 2 liters of a 0.2 M NaOH solution to have 0.4 moles of NaOH.

Volumetric flask are calibrated to contain.

A volumetric flask is calibrated to contain a specific volume of solution when filled to its calibration mark, while pipettes are designed to deliver precise volumes of liquids.

A volumetric flask is calibrated 'to contain' (T. C.) a specific volume of solution rather than to dispense it. This means when the flask is filled to its calibration mark, it is accurate to a specific volume such as 10.00 mL ± 0.02 mL for a 10-mL volumetric flask or 250.0 mL ± 0.12 mL for a 250-mL volumetric flask. Unlike volumetric flasks, pipettes like volumetric and graduated pipettes are designed to deliver a known volume of liquid, either a single volume in the case of volumetric pipettes or variable volumes for graduated pipettes. To achieve accurate measurements, it is essential that both pipets and volumetric flasks are clean because any residue can affect the volume of liquids either delivered or contained.

To determine the Ksp of Ag2CrO4 at 25 °C, you would first need to determine the standard reduction potentials and use the free energy equation ∆G° = -nFE°. Next, use the relationship between the equilibrium constant and ∆G° given by ∆G° = -RT ln Ksp to find the Ksp. However, without the necessary information, a precise value cannot be given.

To find the solubility product constant (Ksp) of Ag2CrO4 at 25 °C given the provided standard reduction potentials, we would need to look at the relationship between the equilibrium constant, K, and ∆G° (standard free energy). The equation ∆G° = -RT ln K is used in these cases, where R is the gas constant, T is the temperature in Kelvin, and K is the equilibrium constant that we're interested in. Note that in order to use this equation, we first need to calculate ∆G° using the given reduction potentials and the equation ∆G° = -nFE°, where n is the number of electrons transferred, F is Faraday's constant, and E° is the standard reduction potential.

For example, if the ∆G° for the reaction came out to be around 167.9 kJ as per the details provided, and we can use ∆G° = -RT ln Ksp to find the Ksp. However, without the complete standard reduction potentials, a precise Ksp value for Ag2CrO4 at 25 °C cannot be provided here.

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Answer : The molarity of a solution is, 0.0337 M

Explanation : Given,

Mass of [tex]MgI_2[/tex] = 468 mg  = 0.468 g   (1 mg = 0.001 g)

Volume of solution = 50.0 mL

Molar mass of [tex]MgI_2[/tex] = 278 g/mole

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

[tex]\text{Molarity}=\frac{\text{Mass of }MgI_2\times 1000}{\text{Molar mass of }MgI_2\times \text{Volume of solution (in mL)}}[/tex]

Now put all the given values in this formula, we get:

[tex]\text{Molarity}=\frac{0.468g\times 1000}{278g/mole\times 50.0mL}=0.0337mole/L=0.0337M[/tex]

Therefore, the molarity of a solution is, 0.0337 M

The molarity of the solution formed is 0.0337 M

From the question,

We are to determine the molarity of the solution formed.

First, we will determine the number of moles MgI₂ dissolved

Mass of MgI₂ dissolved = 468 mg = 0.468 g

From the formula

[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass}[/tex]

Molar mass of MgI₂ = 278.1139 g/mol

∴ Number of moles of MgI₂ present = [tex]\frac{0.468}{278.1139}[/tex]

Number of moles of MgI₂ present = 0.001682764 mole

Now,

For the molarity of the solution formed,

Using the formula

[tex]Molarity = \frac{Number\ of\ moles}{Volume}[/tex]

Volume of the solution = 50.0 mL = 0.050 L

∴ Molarity of the solution = [tex]\frac{0.001682764}{0.050}[/tex]

Molarity of the solution = 0.033655

Molarity of the solution ≅ 0.0337 M

Hence, the molarity of the solution formed is 0.0337 M

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Answer:

P₂ = 1.0 atm

Explanation:

Boyles Law problem => P ∝ 1/V at constant temperature (T).

Empirical equation

P ∝ 1/V => P = k(1/V) => k = P·V => for comparing two different case conditions, k₁ = k₂ => P₁V₁ = P₂V₂

Given

P₁ = 1.6 atm

V₁ = 312 ml

P₂ = ?

V₂ = 500 ml

P₁V₁ = P₂V₂ => P₂ = P₁V₁/V₂ =1.6 atm x 312 ml / 500ml = 1.0 atm