at 800 mm Hg, a gas has a volume of 760 L, what is its volume at standard pressure?

Answers

Answer 1

At 800 mm Hg, a gas has a volume of 760 L,  its volume at standard pressure is 800 L. This is according to Boyle's Law.

What is Boyle's Law?

A gas law known as Boyle's law asserts that a gas's pressure is inversely proportional to its volume when it is held at a fixed temperature and of a given mass. To put it another way, as long as the temperature and volume of the gas remain constant, the pressure and volume of the gas are inversely proportional to one another. The Anglo-Irish chemist Robert Boyle proposed Boyle's law in the year 1662.

According to Boyle's law, a gas's pressure will change if its volume changes while remaining the same quantity and temperature. To put it another way, the ratio of a gas's beginning pressure to its initial volume is equal to the ratio of the gas' final pressure to its final volume (at constant temperature and number of moles).

Using the formula:

P₁V₁ = P₂V₂

Substituting the values and solving for V₂:

V₂= 800L

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Related Questions

element which are placed is group VIIIA don't take part in chemical reaction why​

Answers

Answer:

Elements placed in Group VIIIA of the periodic table are also known as the noble gases, and they are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn).

These elements do not take part in chemical reactions because they have a full outermost electron shell, also known as the valence shell. This means that their valence shell contains the maximum number of electrons that it can hold, and therefore, these elements do not have any electron available for bonding with other atoms.

In other words, their outermost electron shell is already stable, which means they have very low reactivity with other elements. Because of their unreactive nature, noble gases are often referred to as inert gases.

two isotopes of carbon, carbon-12 and carbon-13, have masses of 19.93 x 10^-27 kg and 21.59 x 10^-27 kg, respectively. these two isotopes are singly ionized ( e) and each is given a speed of 6.667 x 10^5 m/s. the ions then enter the bending region of a mass spectrometer where the magnetic field is 0.85 t. determine the spatial separation between the two isotopes after they have traveled through a half-circle

Answers

The spatial separation between Carbon-12 and Carbon-13 each with a given speed 6.667 x 10^5 m/s is 28.2 x 10^-7 m

The spatial separation between two isotopes of carbon, carbon-12 and carbon-13, can be determined using Mass Spectrometry. Mass Spectrometry uses a combination of magnetic and electric fields to separate ions of different masses based on their charge-to-mass ratio.

The two isotopes of carbon, each singly ionized (e), have a speed of 6.667 x 10^5 m/s and enter the bending region of a mass spectrometer, where the magnetic field is 0.85 t. After they have traveled through a half-circle, the spatial separation between the two isotopes will be calculated using the following formula:

Separation (S) = Speed (V) x Magnetic Field (B) x Time (T)

In this case, Time (T) = (π/2)/V = (π/2)/(6.667 x 10^5 m/s) = 4.5 x 10^-7 s

Therefore, the spatial separation between the two isotopes after they have traveled through a half-circle will be:

Separation (S) = 6.667 x 10^5 m/s x 0.85 t x 4.5 x 10^-7 s = 28.2 x 10^-7 m

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Look at the image below and answer the following questions:

What kind of circuit has been built?

What will happen if one of the light bulbs is burnt out and no longer lights up?

Answers

Answer: series circuit

Explanation:

when one bulb burns out the other bulb will go off because the circuit breaks

A chocolate chip cookie recipe calls for 0. 050 moles of baking soda (sodium bicarbonate, NaHCO3).

How many grams should the chef mass out?

Answers

4.2 gram should the chef mass out of a chocolate chip cookie recipe calls for 0. 050 moles of baking soda. This calculated by using mole concept.

Mass is equal to the number of moles multiplied by the molar mass of substance. We know that the number of moles is equal to the actual given mass of the substance over its molar mass.

      m = n x M.

The term molar mass of a chemical compound is defined as the ratio between the mass and the amount of substance of any sample of said compound. It is a bulk, not molecular, property of a substance.

According to the mole concept, one mole of a compound contains Avogadro's number that is 6.022 x 1023 of molecules or formula units of ionic compound. The molar mass of a compound defined as the mass of 1 mole of that substance. Generally we can say the number of grams per mole of a compound.

putting the value in the expression we get,

mass =84 × 0.050

        =4.2 g.

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Which set of coefficients, when used in the order listed, will balance the following skeleton equation for the combustion of benzene, C6H6(l)?

___C6H6(l) + ___O2(g) --> ___CO2(g) + ___H2O(g)

Answers

Answer:

2, 15, 12, 6.

Explanation:

Tick the correct answer. power is needed to act. ( )
plants get power from the sun.( ) firewood is not a source of power.( )

Answers

According to this statement, Power is required to act. (correct) Plants get strength from the sun. (correct) Firewood isn't a source of power. (incorrect).

What is firewood and its uses?

Firewood is characterized by the Food and Agricultural Organization of the United Nations, or FAO, as "wood in the rough (from trunks and branches of trees) to be used as fuel for uses such as heating, cooking, or power production."

What is the benefit of firewood?

Burning firewood lowers the need over electric heating in your house. If the electricity goes out, you can still heat your house.Wintertime is a crucial time to remember this. Storms can knock out the power for days, putting people's homes, plumbing, and health at danger.

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I WILL GIVE 35 POINTS TO THOSE WHO ANSWER THIS QUESTION RIGHT NOOOO SCAMS PLEASE

Answers

Answer:2, 7, 4, 6

Explanation:

2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

What is the pH at the equivalence point in the titration of a 15. 4 mL sample of a 0. 344 M aqueous hypochlorous acid solution with a 0. 412 M aqueous sodium hydroxide solution

Answers

Therefore, the pH at the equivalence point in the titration of a 15.4 mL sample of a 0.344 M aqueous hypochlorous acid solution with a 0.412 M aqueous sodium hydroxide solution is 7.542.

The pH at the equivalence point in the titration of a 15.4 mL sample of a 0.344 M aqueous hypochlorous acid solution with a 0.412 M aqueous sodium hydroxide solution can be calculated using the following steps:

1. First, determine the number of moles of hypochlorous acid (HOCl) in the 15.4 mL sample:
[tex]0.344 M * 0.0154 L = 0.00530[/tex] moles of HOCl

2. Next, determine the number of moles of sodium hydroxide (NaOH) needed to reach the equivalence point:
0.00530 moles of HOCl * (1 mole of NaOH / 1 mole of HOCl) = 0.00530 moles of NaOH

3. Calculate the volume of the 0.412 M NaOH solution needed to reach the equivalence point:
0.00530 moles of NaOH / 0.412 M = 0.0129 L or 12.9 mL

4. At the equivalence point, the moles of HOCl and NaOH are equal, and the resulting solution contains only water and the salt sodium hypochlorite (NaOCl). The pH of the solution will depend on the dissociation of the salt:
NaOCl → Na+ + OCl-

5. The hypochlorite ion (OCl-) can undergo hydrolysis to form hypochlorous acid and hydroxide ions:
OCl- + H2O → HOCl + OH-

6. The presence of the hydroxide ions will increase the pH of the solution. To calculate the pH, we can use the Kb value for the hypochlorite ion, which is [tex]2.9 x 10^{-7}[/tex]:
[tex]Kb = [HOCl][OH-] / [OCl-][/tex]

7. Assuming that the initial concentration of OCl- is equal to the concentration of NaOH at the equivalence point (0.412 M), we can rearrange the Kb equation and solve for the concentration of OH-:
[tex][OH-] = Kb * [OCl-] / [HOCl]\\\\ [OH-] = (2.9 * 10^{-7} ) * (0.412) / (0.344)\\\\ [OH-] = 3.48 * 10^{-7} 10^-7 M[/tex]

8. Finally, we can use the concentration of OH- to calculate the pH of the solution:
[tex]pOH = -log[OH-] = -log(3.48*10^{-7} ) = 6.458[/tex]
[tex]pH = 14 - pOH = 14 - 6.458 = 7.542[/tex]

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citric acid is a triprotic weak acid. if 11.9 ml of 0.150 m naoh is required to reach the first equivalence point of a solution of citric acid how many ml of naoh are required to completely neutralize this solution?

Answers

35.7 ml of 0.150 M NaOH will be needed to completely neutralize the citric acid solution.

Citric acid is a triprotic weak acid, and 11.9 ml of 0.150 M NaOH is required to reach the first equivalence point of a solution of citric acid.

Citric acid, being a weak acid, is not completely ionized in solution. As a result, it will require more than 1 equivalent of base to neutralize it completely. So, it is necessary to calculate the number of equivalents present in the solution to answer the question. Let's first write the balanced chemical equation between citric acid (H3C6H5O7) and NaOH as follows:

H3C6H5O7 + NaOH → Na3C6H5O7 + H2O

For each mole of citric acid, three moles of NaOH are required to neutralize it completely. Since the citric acid solution is 0.150 M, there will be 0.150 moles of citric acid per liter of solution (Molarity = moles/liter).

Thus, the number of moles of citric acid present in 11.9 ml of solution can be calculated as follows: 0.150 moles of citric acid / 1000 ml of solution = X moles of citric acid / 11.9 ml of solution X = 0.001785 moles of citric acid Therefore, 3 times the number of moles of citric acid will be the number of moles of NaOH required to neutralize the citric acid completely. Therefore, the number of moles of NaOH required to neutralize the citric acid completely is as follows: 3 X 0.001785 moles of citric acid = 0.005355 moles of NaOH.

The volume of 0.150 M NaOH needed to contain 0.005355 moles of NaOH can be calculated as follows:

Molarity = moles/volume

Volume = moles / Molarity = 0.005355 moles / 0.150 M = 0.0357 L or 35.7 ml.

Therefore, 35.7 ml of 0.150 M NaOH will be needed to completely neutralize the citric acid solution.

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The half-life for Carbon-14 is 5614 years. An ancient piece of cloth is found to contain ¼ of its original Carbon-14. How old is the cloth? Describe or show in detail how you solved this.

Answers

Answer:

To determine the age of the ancient cloth, we can use the concept of radioactive decay and the half-life of Carbon-14.

Carbon-14 is a radioactive isotope of carbon, which decays over time into nitrogen-14 through beta decay. The half-life of Carbon-14 is 5614 years, which means that after 5614 years, half of the original amount of Carbon-14 in a sample will have decayed.

In this case, the cloth contains only ¼ of its original Carbon-14. This means that three half-lives have passed since the cloth was first created, as each half-life reduces the amount of Carbon-14 by half.

To determine the age of the cloth, we can use the following formula:

N = N0(1/2)^t/T

where N is the current amount of Carbon-14 in the cloth, N0 is the original amount of Carbon-14 in the cloth, t is the time that has passed, and T is the half-life of Carbon-14.

We know that N = ¼ N0, and T = 5614 years. Plugging these values into the formula, we get:

¼ N0 = N0(1/2)^(3/T)

Solving for t, we get:

t = (3/T) * log(2)

Substituting in T = 5614 years, we get:

t = (3/5614) * log(2) ≈ 1,684 years

Therefore, the cloth is approximately 1,684 years old.

In summary, we can use the concept of radioactive decay and the half-life of Carbon-14 to determine the age of the ancient cloth. By knowing the current amount of Carbon-14 in the cloth, we can calculate the time that has passed since it was first created using a simple formula. In this case, the cloth is approximately 1,684 years old.

A mass of 100 g of NAN03 is dissolved in 100 g of water at 80°C is the solution is saturated or unsaturated?

Answers

NaNO3 has a solubility of 80 grams per 100 grams of water at 10oC. (It seems to be somewhat less than 80 grams, but after rounded off, it's practically 80 grams.)

What kind of solution is formed when 80 g of NaNO3 is dissolved in 100 g of water at 10 degrees Celsius?

• Assume we just required 60 mL of NaNO3 solution at 10oC. Further Information: Because the solubility of solids rises with temperature, solutes whose curves go higher with increasing temperature are often solids.

Thus the solution in issue is saturated, or even somewhat supersaturated — but you're probably correct.

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nick has been the head footbal coach at a prminent university for many years. he has always led his team to success. another university believes that it must be nicks leadership that has led his team to victory so many times. they offer nick a position leading their own football team, but it ends up being the worst season the team has ever had. university officals realize that nicks leadership mustve worked for his former team and didnt translate to theirs. this situation could be used as an example of what kind of leadership theory?
a. great man theory
b. behavioral theory
c. contingency theory
c. transactional theory

Answers

Answer:

a

Explanation:

a gas contains 70.0 wt% methane, 15.0% ethane, 5.0% ethylene, and the balance water. a) calculate the molar composition of this gas on both a wet and a dry basis and the ratio (mol h2o/mol dry gas). b) if 100 kg/h of this fuel is to be burned with 30% excess air, what is the required air feed rate (kmol/h)? how would the answer change if the combustion were only 75% complete? c) if the fuel b

Answers

A- the ratio of mol water to mol dry gas is 10/100, B- The required air feed rate is 166.67 kmol/h, C- composition of exhaust gas steam is 100% of water.

a) On a wet basis, the molar composition of the gas is 70.0 mol % methane, 15.0 mol % ethane, 5.0 mol % ethylene, and 10.0 mol % water. On a dry basis, the molar composition is 78.26 mol % methane, 17.74 mol % ethane, 4.26 mol % ethylene, and 0 mol % water. The ratio of mol water to mol dry gas is 10.0/100.


b) If 100 kg/h of the fuel is to be burned with 30% excess air, the required air feed rate is 133.33 kmol/h. If the combustion is only 75% complete, the air feed rate would be 166.67 kmol/h.


c) If the fuel burns to 100% completion, the composition of the exhaust gas stream will be 0 mol % methane, 0 mol % ethane, 0 mol % ethylene, and 100 mol % water.

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(16) Balance the shown equation. If the mass of C6H₁0 is 25 g and the mass of O₂ is 25
g, what is the limiting reagent?
C6H10+
0₂_CO₂+
H₂O
20
Folg).
h
mal
C

Answers

H2 is the limiting reactant in this reaction. Water is generated when 5.00 g of H2 and 10.0 g of O2 react. Because there is 10.0 g of surplus reagent, 10.0 g of H2O is generated.

How to find maximum amount of product using limiting reactant?

Calculate the amount of moles of product that can be generated from the limiting reactant using mole ratios. To get the equivalent mass of the product, multiply the number of moles by its molar mass. Divide the molar masses of each component to get the mass in moles. Create a balanced equation and calculate the molar ratio.

Examine the formula and compare the moles. In a chemical process, the limiting reagent is the reactant that will be totally consumed. The reaction cannot continue after that reactant is depleted. As a result, it prevents the reaction from proceeding. The surplus reagent is the reactant that may continue to react if the other was not consumed.

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the enthalpy of formation of caesium chloride is Cs(s)------->Cs(g) ΔH⁰= –44.28 kj mol-¹
the enthalpy of sublimation of caesium is Cs(s) ------>Cs(g) ΔH⁰=+77.6 kj mol-¹​

Answers

The enthalpy of the formation of cesium chloride is -68.4 kJ/mol.

What is the enthalpy of formation of cesium chloride?

The enthalpy of formation of caesium chloride (CsCl) can be calculated using the following equation:

ΔHf⁰(CsCl) = ΔHf⁰(Cs) + ΔHf⁰(Cl) - ΔHf⁰(CsCl)

We are given the enthalpy of sublimation of caesium, which tells us the energy required to convert solid caesium into gaseous caesium:

ΔHsub⁰(Cs) = +77.6 kJ/mol

Since the enthalpy of formation of an element in its standard state is zero, we can assume that:

ΔHf⁰(Cs) = 0

The enthalpy of formation of chlorine gas (Cl₂) is -95.7 kJ/mol. The enthalpy change for the reaction that forms CsCl can be written as:

Cs(s) + 1/2 Cl₂(g) → CsCl(s)

The enthalpy change for this reaction is equal to the enthalpy of formation of CsCl:

ΔHf⁰(CsCl) = ΔHf⁰(Cs) + 1/2 ΔHf⁰(Cl₂) - ΔHrxn

where ΔHrxn is the enthalpy change for the reaction. We can use Hess's Law to calculate ΔHrxn by considering the following two steps:

Cs(s) → Cs(g) ΔHsub⁰(Cs) = +77.6 kJ/mol (endothermic)

1/2 Cl₂(g) → Cl(g) ΔHdiss⁰(Cl) = +121 kJ/mol (endothermic)

Adding these two reactions gives:

Cs(s) + 1/2 Cl₂(g) → Cs(g) + Cl(g) ΔHrxn = ΔHsub⁰(Cs) + 1/2 ΔHdiss⁰(Cl)

Substituting the values we have for ΔHsub⁰(Cs) and ΔHdiss⁰(Cl) gives:

ΔHrxn = +77.6 kJ/mol + 1/2(+121 kJ/mol) = +138.3 kJ/mol

Substituting all the values into the equation for ΔHf⁰(CsCl) gives:

ΔHf⁰(CsCl) = 0 + 1/2(-95.7 kJ/mol) - (-138.3 kJ/mol)

ΔHf⁰(CsCl) = -68.4 kJ/mol

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Help with this chem work ASAP PLEASE

Answers

Answer: you have to do x+axis than you will get ur answer

Explanation:During a chemical reaction, all of the atoms that make up the reactants rearrange to form the products.

Hello can someone please help me with this please
7. The blood on the right side in Model 1 only contains 50% oxygen, but it has 95% total gases.
a. What gas other than oxygen do you think might be dissolved in the blood on the right side of the heart?
b. What process produced this gas?
c. What happens to this gas before the blood enters the left side of the heart?
8. Looking at the arrows on Model 1, how would you describe the flow pattern of the blood inside the circulatory system?
9. What features might the entrances and exits to the heart need in order to maintain this flow pattern?

Answers

7. a. Carbon dioxide (CO2) is the gas that might be dissolved in the blood on the right side of the heart.

b. CO2 is produced as a waste product of cellular respiration, which is the process by which cells produce energy.

c. Before the blood enters the left side of the heart, CO2 is transported from the tissues to the lungs, where it is exchanged for oxygen.

8. The flow pattern of the blood inside the circulatory system is unidirectional,  with the blood flowing from the right atrium to the right ventricle, then to the lungs, and from the lungs to the left atrium and left ventricle, and finally to the body.

What features are needed by the entrances and exits to the heart in order to maintain a unidirectional flow pattern??

The entrances and exits to the heart need valves to maintain the unidirectional flow pattern of the blood.

The atrioventricular valves (tricuspid valve on the right side and mitral valve on the left side) prevent the backflow of blood from the ventricles to the atria, while the semilunar valves (pulmonary valve on the right side and aortic valve on the left side) prevent the backflow of blood from the arteries to the ventricles.

The heart also needs a pacemaker to coordinate the contraction of the atria and the ventricles to ensure the efficient pumping of blood.

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What is Keq for the reaction 2SO₂(g) + O₂(g) O₂(g) = 2SO3(g)? O A. Kea OB. Keq о C. Kea- OD. Keq= [SO₂ 1² [SO₂ 1² [0₂] [SO,] [SO, [0,] [So, 1² [So, 1² [0₂] 2[SO] 2[S0₂ ][0₂] ​

Answers

The equilibrium constant (Keq) for the reaction 2SO2(g) + O2(g) 2SO3(g) is correctly expressed as Keq = [SO3]2 / ([SO2]2 [O2]).

In a reply, what does Keq mean?

The ratio of the mathematical product of the concentrations of the products of a reaction to the mathematical product of the concentrations of the reactants of the reaction is known as the equilibrium constant (Keq). The balanced chemical equation causes each concentration to grow to the power of its corresponding coefficient.

What is the Keq value?

Keq's "K" stands for "Constant," thus. At a specific temperature, the "eq" indicates that the reaction is at equilibrium. No matter what we would try to do with the concentrations, the value of this ratio remains at 0.0183.

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What is the name of the formula AuCN?​

Answers

Answer:

The name of the formula AuCN is gold cyanide.

find the ph of the equivalence points when 28.9 ml of 0.0850 m h2so3 is titrated with 0.0372 m naoh.

Answers

At the equivalence point of the titration, the pH is 12.88. When 28.9 ml of 0.0850 m H₂SO₄ is titrated with 0.0372 m NaOH.

Given:

Volume of H₂SO₃ solution (V₁) = 28.9 mL

= 0.0289 L

Concentration of H₂SO₃ (C₁) = 0.0850 M

Volume of NaOH solution (V₂) = unknown

Concentration of NaOH (C₂) = 0.0372 M

H₂SO₄ is a diprotic acid, so it can donate two protons. The balanced equation for the reaction between H₂SO₃ and NaOH is as follows:

H₂SO₃ + 2NaOH → Na₂SO₃ + 2H₂O

From the equation, 1 mole of H₂SO₃ reacts with 2 moles of NaOH.

Use the stoichiometry of the reaction:

C₁V₁ = C₂V₂

0.0850 M × 0.0289 L = 0.0372 M × V2

V2 = (0.0850 M × 0.0289 L) / 0.0372 M

V2 = 0.0658 L

= 65.8 mL

Therefore, the volume of NaOH required to reach the equivalence point is 65.8 mL.

Since NaOH is a strong base, it fully dissociates in water to produce OH- ions. At the equivalence point, the moles of OH- ions will be equal to twice the moles of H₂SO₃ initially present.

Moles of H₂SO₃ = C1 × V1

Moles of H₂SO₃ = 0.0850 M × 0.0289 L

= 0.00246 moles

Moles of OH- ions = 2 × Moles of H₂SO₃

Moles of OH- ions = 2 × 0.00246 moles

= 0.00492 moles

Concentration of OH- ions = Moles of OH- ions / V₂

Concentration of OH- ions = 0.00492 moles / 0.0658 L

Concentration of OH- ions = 0.0749 M

pOH = -log10[OH-]

pOH = -log10(0.0749)

pOH = 1.12

pH + pOH = 14:

pH + 1.12 = 14

pH = 14 - 1.12

pH = 12.88

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Calculate the molarity of a solution that contains 0. 07 mol of K2S in 1. 98 L of solution

Answers

The molarity of a solution that contains 0.07 mol of K2S in 1.98 L of solution is 0.035 M.

The amount of moles of solute per litre of solution is known as molarity (M).

To calculate the molarity of the given solution, we first need to determine the number of moles of K2S in the solution:

moles of K2S = 0.07 mol

The solution's volume in litres must then be calculated:

volume of solution = 1.98 L

Now we can use the formula for molarity:

Molarity = moles of solute / volume of solution

Substituting the given values:

Molarity = 0.07 mol / 1.98 L

Molarity = 0.035 M

Therefore, the molarity of the solution is 0.035 M.

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Hydrogen gas was cooled from 15 C to -15 C. Its new volume is 75. 0mL. What was its original volume

Answers

When Hydrogen gas was cooled from 15 C to -15 C by use of the formula we get the value of the original volume as 671.875 mL.

Charles' Law describes the link that exists between the volume and temperature of a specific amount of ideal gas kept at constant pressure.

This rule says that for a given amount of gas at constant pressure, as the temperature rises, the volume of the gas rises, and as the temperature falls, the volume of the gas falls because temperature is directly proportional to the energy of movement of the gas molecules. That is, the volume of the gas is precisely proportional to its temperature.

The kinetic energy of a gas's molecules is exactly proportional to its absolute temperature, according to the kinetic theory of gases. The kinetic energy of the component molecules reduces as the temperature of the gas falls.

We have,

v1 t1 = v2 t2

v1 = x

v2 = 750 mL

t1 = 15 + 273 = 288 K

t2 = -15 + 273 = 258 K

putting the values in the equation,

x(288) = (750)(258)

x = 193500 / 288

x = 671.875 mL

So the original volume was 671.875 mL.

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As the temperature of a reaction is decreased, the rate of the reaction decreases because thegroup of answer choicesreactant molecules collide less frequentlyreactant molecules collide more frequently and with greater energy per collisionactivation energy is higherreactant molecules collide more frequently with less energy per collision

Answers

Answer:

Reactant molecules collide less frequently.

Explanation:

The decrease in temperature decreases the kinetic energy of the reactant molecules leading to decreased chances of collisions of the reactants for the formation of the products.

If a solution has a concentration of 12 M when the volume is 150 mL. What is the new concentration if you increase the volume to 600 mL?

Answers

A 150 mL solution with a concentration of 12 M whose volume is increased to 600 mL will have a new concentration of 3 M.

Stoichiometric problem

Assuming the amount of solute remains constant, we can use the formula:

C1 x V1 = C2 x V2

where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.

Plugging in the values, we get:

12 M x 150 mL = C2 x 600 mL

Solving for C2, we get:

C2 = (12 M x 150 mL) / 600 mL

C2 = 3 M

Therefore, the new concentration is 3 M.

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20. at the same temperature, will the ph of a 0.001 m ca(oh)2 aqueous solution be the same, larger, or smaller than that of a 0.001 m naoh(aq)? briefly explain your reasoning

Answers

At the same temperature, the pH of a 0.001 M Ca(OH)2 aqueous solution will be higher than that of a 0.001 M NaOH(aq).

The pH value of a solution is determined by the concentration of hydrogen ions in the solution. Higher hydrogen ion concentration in a solution leads to lower pH values, while lower hydrogen ion concentration results in higher pH values. The chemical formulas of the two solutions are: Ca(OH)2(aq) → Ca2+(aq) + 2OH-(aq) NaOH(aq) → Na+(aq) + OH-(aq). The hydroxide ion (OH-) concentration of a Ca(OH)2 solution is twice that of a NaOH solution of equal concentration. This is due to the fact that 1 mole of Ca(OH)2 produces two moles of OH- ions, whereas 1 mole of NaOH produces only 1 mole of OH- ions. As a result, the pH of Ca(OH)2 is higher than that of NaOH.

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If 5. 00L of nitrogen reacts completely with hydrogen at a pressure of 3. 00atm and a temperature of 298k, how much ammonia, in gram, is produced?

Answers

If 5.00L of nitrogen completely reacts with hydrogen at a pressure of 3.00atm and a temperature of 298k,approximately 20.67 grams of ammonia will be produced.

The balanced chemical equation for the reaction between nitrogen and hydrogen to form ammonia is:

N2 + 3H2 → 2NH3

We can use the ideal gas law to calculate the number of moles of nitrogen:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.0821 L atm mol K), and T is the temperature in Kelvin.

Rearranging the equation, we get:

n = PV/RT

Substituting the given values, we get:

n = (3.00 atm) (5.00 L) / (0.0821 L atm mol K) (298 K)

n = 0.607 mol

n(NH3) = 2 × n(N2) = 2 × 0.607 mol = 1.214 mol

Mass(NH3) = n(NH3) × M(NH3) = 1.214 mol × 17.03 g/mol ≈ 20.67 g

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65g of NaCl are placed in a beaker and enough water is added to fill the beaker to 1 liter. What is the molar out of this solution? Round answer to the nearest hundredths

Answers

The molarity of the solution is 1.11 M (rounded to 2 decimal places). To find the molarity (M) of the solution, we need to know the number of moles of NaCl and the volume of the solution in liters.

How do molality and molarity differ?

Although a solution's molarity is determined by the moles of solute divided by the volume of the solution in litres, a solution's molality is determined by the moles of solute divided by the mass of the solvent in kilogrammes.

The number of moles of NaCl can be calculated using its molar mass and the given mass of NaCl:

Number of moles = mass / molar mass

The molar mass of NaCl is 58.44 g/mol (22.99 g/mol for Na + 35.45 g/mol for Cl).

Number of moles of NaCl = 65 g / 58.44 g/mol = 1.1138 mol (rounded to 4 decimal places)

The volume of the solution is given as 1 liter.

Now we can use the formula for molarity:

Molarity (M) = number of moles / volume (in liters)

M = 1.1138 mol / 1 L = 1.1138 M (rounded to 2 decimal places)

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Barium sulfide decomposes into its elements when heat and electricity areapplied.Which reaction shows the balanced equation for the decomposition?A. 8BaS→ 8Ba + SgB. Ba₂S → 2Ba + SC. BaS2 -> Ba + S₂D. 2Ba₂S -> 4Ba + 2S

Answers

The correct answer is option B. Ba₂S → 2Ba + S. This is the balanced equation for the decomposition reaction of barium sulfide into its elements.

In a decomposition reaction, a compound breaks down into simpler substances, often its elements. The balanced equation for a decomposition reaction shows the same number of atoms of each element on both sides of the equation.
In option A, there are 8 atoms of barium and 8 atoms of sulfur on the left side of the equation, but only 8 atoms of barium and 1 atom of sulfur on the right side. In option C, there are 1 atom of barium and 2 atoms of sulfur on the left side, but only 1 atom of barium and 2 atoms of sulfur on the right side. In option D, there are 4 atoms of barium and 2 atoms of sulfur on the left side, but 4 atoms of barium and 2 atoms of sulfur on the right side. Only option B has the same number of atoms of each element on both sides of the equation, making it the correct answer.

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the masses of species involved are given in atomic mass units below each species, and 1 amu can create 932 mev of energy. what is the energy liberated due to transfomration of mass into energy during this reaction

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Given that the masses of species involved are given in atomic mass units below each species, and 1 amu can create 932 meV of energy. Therefore, the energy liberated due to the transformation of mass into energy during this reaction is 1285.44 MeV.

The energy liberated due to the transformation of mass into energy during this reaction can be found by using the following formula: E = (Δm)c² Where E = Energy liberated during the transformation of mass into energy.

Δm = Change in mass during the transformation.

c = Speed of light in a vacuum (3.0 x 10⁸ m/s)

Given that, the masses of species involved are given in atomic mass units below each species, and 1 amu can create 932 meV of energy.

Therefore, first find the mass difference between the reactant and product as shown below: N + O → P + Q

Species Mass N 14 amu

O 16 amu

P 16 amu

Q 14 amu

Total 60 amu

Mass of reactants = 14 + 16 = 30 amu

Mass of products = 16 + 14 = 30 amu

Δm = (Mass of reactants - Mass of products)Δm = 30 - 30Δm = 0amu

Now, calculate the energy liberated as follows: E = (Δm)c²E = 0 x (3 x 10⁸)²E = 0 x 9 x 10¹⁶E = 0 MeV

Therefore, the energy liberated during the transformation of mass into energy is 0 MeV, which indicates that the reaction is not exothermic or endothermic.

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22. 0 mL of stock solution is used

to produce a. 458 M solution after

dilution with 50. 0 mL of water. What

is the molarity of the stock solution?

Answers

The molarity of the stock solution is 1.05 M.

We can use the equation for dilution to solve this problem:

M₁V₁ = M₂V₂

where M₁ is the initial molarity (of the stock solution), V₁ is the initial volume (in mL) of the stock solution, M₂ is the final molarity (after dilution), and V₂ is final volume (in mL) of diluted solution.

In this case, we know that V₁ = 22.0 mL, V₂ = 50.0 mL, and M₂ = 0.458 M. We want to solve for M₁.

Plugging in the values, we get:

M₁(22.0 mL) = (0.458 M)(50.0 mL)

Simplifying and solving for M₁, we get:

M₁ = (0.458 M)(50.0 mL) / (22.0 mL)

M₁ = 1.05 M

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