Please help anyone !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Please Help Anyone !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answers

Answer 1

The volume of ammonia needed to react completely with 30 Liters of NO at STP is 45 L.

What is the volume of ammonia required in the reaction?

The volume of ammonia needed to react completely with 30 Liters of NO at STP is calculated as follows;

4NH₃  + 6NO    →   5N₂   +  6H₂O

From the reaction given above, we can see that;

4 moles of ammonia ------------> 6 moles of NO

ratio = 4 : 6

The volume of ammonia required is calculated as;

4 -------------- > 6

30 L -----------> ?

? = (30 L x 6 ) / 4

? = 45 L

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Related Questions

How many grams of zinc chloride would be formed if 77.1 grams of zinc reacts?

Zn + HCl -->ZnCl2 + H2

Answers

The amount of zinc chloride that would be formed if 77.1 grams of zinc reacts is approximately 160.77 grams

To determine how many grams of zinc chloride ([tex]ZnCl_2[/tex]) would be formed if 77.1 grams of zinc (Zn) reacts, we'll use stoichiometry.

First, we need the molar masses of the substances involved:

Zn: 65.38 g/mol
[tex]ZnCl_2[/tex] : 136.29 g/mol
Now, we'll convert grams of Zn to moles:
77.1 g Zn × (1 mol Zn / 65.38 g Zn) = 1.179 moles Zn
According to the balanced chemical equation, 1 mole of Zn reacts to form 1 mole of [tex]ZnCl_2[/tex]:
1.179 moles Zn × (1 mol ZnCl₂ / 1 mol Zn) = 1.179 moles ZnCl₂

Finally, convert moles of ZnCl₂ to grams:
1.179 moles ZnCl₂ × (136.29 g ZnCl₂ / 1 mol ZnCl₂) ≈ 160.77 g ZnCl₂
So, approximately 160.77 grams of zinc chloride would be formed if 77.1 grams of zinc reacts.

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Balloons for a New Years Eve party in Fargo, ND, are filled to a volume of 1.90 L at a temperature of 22.0 c and then hung outside. what is the volume of the balloon once they have cooled to the outside temperature of -34.0 c?

Answers

The volume of the balloon once they have cooled to the outside temperature of -34.0 c is 1.53 L.

Charles' law predicts the relationship between the volumes and the temperatures of a sample of an ideal gas at different conditions. For this equation to hold true, the number of molecules and the pressure must remain constant despite changes in the environment.

Determine the volume of the balloon outside, V2. We do this by applying Charles' law, such that we relate the volume, V, and the temperature, T, of a sample of gas as

V₁ /T₁ = V₂/ T₂

at two conditions. We are given the following values for the variables:

• V₁ = 1.90 L

T₁ = 22.0+ 273.15 = 295.15 K

T₂= 34.0+273.15= 239.15 K

We proceed with the solution.

V₁/T₁ = V₂/T₂

V₁ /T₁ × T₂ = V₂

1.90 L/295.15 K x 239.15 K = V₂

1.53 L =V₂

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The complete question is

Balloons for a New Year's Eve party in Fargo, ND, are filled to a volume of 1.90 L at a temperature of 22.0 degrees Celsius and then hung outside where the temperature is -34.0 degrees Celsius. What is the volume of the balloons after they have cooled to the outside temperature? Assume that atmospheric pressure inside and outside the house is the same.

Benzoic acid (C6H5COOH) and aniline (C6H5NH2) are both derivatives of benzene. Benzoic acid is an acid with Ka=6.3×10^(−5) and aniline is a base with Kb=4.3×10^(−10) .What is the value of the equilibrium constant for the following equilibrium? C6H5COOH(aq)+C6H5NH2(aq)⇌C6H5COO−(aq)+C6H5NH3+(aq)
i want an accurate answer

Answers

The reaction C₆H₅COOH(aq) + C₆H₅NH₂(aq) ⇌ C₆H₅COO⁻(aq) + C₆H₅NH₃⁺(aq) has an equilibrium constant of 0.3698.

How to determine equilibrium constant?

The equilibrium constant (Kb) for the reaction can be calculated using the Ka and Kb values of the reactants and the equation:

Kw = Ka x Kb

where Kw = ion product constant of water (1.0 x 10⁻¹⁴ at 25°C).

Calculate the Kb value for aniline:

Kb = Kw/Ka = (1.0 x 10⁻¹⁴)/(4.3 x 10⁻¹⁰) = 2.33 x 10⁻⁵

Use the Kb value for aniline and the Ka value for benzoic acid to calculate the equilibrium constant (K) for the reaction:

K = Kb/Ka = (2.33 x 10⁻⁵)/(6.3 x 10⁻⁵) = 0.3698

Therefore, the equilibrium constant for the reaction C₆H₅COOH(aq) + C₆H₅NH₂(aq) ⇌ C₆H₅COO⁻(aq) + C₆H₅NH₃⁺(aq) is 0.3698.

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Will we ever send humans to another planet? Most believe that if we were to travel to another planet, Mars would be the best option. Which of these would be a potential problem associated with travel to another planet?

Question 1 options:

we already know everything about Mars


no astronauts would ever volunteer for this mission


Mars has such a high gravity that it would crush humans and our spacecraft


the extended time for humans to be in space

Answers

A potential problem associated with travel to another planet is : the extended time for humans to be in space.

What is the potential problem associated with travel to another planet?

It is highly likely that humans will travel to another planet, and Mars is currently considered the most viable option for human exploration. However, there are many potential problems associated with this endeavor, and one of the major issues is the extended time that humans would need to spend in space.

Traveling to Mars would take several months, and once there, astronauts would need to spend significant amount of time on planet before returning to Earth. This means that they would be exposed to high levels of radiation and would need to find ways to survive in harsh and unforgiving environment.

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How many grams of KOH are needed to make 185.5 ml with a concentration of 5 M?
Type your answer...

Answers

To calculate the mass of KOH needed to make a 5 M solution in 185.5 mL, we need to use the formula:

mass = moles × molar mass

where moles is the amount of KOH in moles and molar mass is the mass of one mole of KOH.

We can calculate the moles of KOH as follows:

moles = Molarity × Volume (in liters)

First, we need to convert the volume from milliliters to liters:

185.5 mL = 0.1855 L

Now we can calculate the moles of KOH:

moles = 5 M × 0.1855 L = 0.9275 moles

The molar mass of KOH is 56.11 g/mol. Therefore, the mass of KOH needed is:

mass = 0.9275 moles × 56.11 g/mol = 52.05 g

Therefore, 52.05 grams of KOH are needed to make a 5 M solution in 185.5 mL.

Can someone please help me with chemistry?

Show steps! Thank you

Answers

a. The mass of Cr2O3 is 0.559 g Cr2O3 is the maximum amount of  Cr2O3 produced.

b.The limiting reactant is  Cr(NO3)3  because it produces less moles of Cr2O3 than Na2O.

c. The percent yield is 84%.

How do we calculate?

The balanced equation is shown below:

2 Cr(NO3)3 + 3 Na2O → Cr2O3 + 6 NaNO3

moles of Cr(NO3)3 = 1.75 g / 238.01 g/mol = 0.00735 mol

moles of Na2O = 1.75 g / 61.98 g/mol = 0.0282 mol

moles of Cr2O3 = (0.00735 mol Cr(NO3)3) × (1 mol Cr2O3 / 2 mol Cr(NO3)3) = 0.00368 mol Cr2O3 (theoretical yield)

mass of Cr2O3 = (0.00368 mol Cr2O3) × (151.99 g/mol) = 0.559 g Cr2O3

The percent yield can be calculated by dividing the actual yield by the theoretical yield and multiplying by 100%:

percent yield = (actual yield / theoretical yield) × 100%

percent yield = (0.455 g / 0.559 g) × 100% = 81.4%

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If 50 joules of energy is added to sample of water, the temperature will?

Answers

Explanation:

The temperature change of a substance when it absorbs or loses energy can be calculated using the specific heat capacity of the substance. The specific heat capacity of water is approximately 4.18 J/(g°C), which means that it takes 4.18 joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.

To calculate the temperature change of the water sample when 50 joules of energy is added, we need to use the following equation:

q = m * c * ΔT

where q is the amount of energy absorbed by the water, m is the mass of the water sample, c is the specific heat capacity of water, and ΔT is the resulting temperature change.

Rearranging the equation to solve for ΔT, we get:

ΔT = q / (m * c)

Plugging in the values, we get:

ΔT = 50 J / (m * 4.18 J/(g°C))

We need to know the mass of the water sample to calculate the temperature change. Let's assume a mass of 10 grams:

ΔT = 50 J / (10 g * 4.18 J/(g°C))

ΔT = 1.2°C

Therefore, if 50 joules of energy is added to a 10-gram sample of water, the resulting temperature change will be approximately 1.2 degrees Celsius.

A solution contains 0.0400 M Ca2+ and 0.0990 M Ag+. If solid Na3PO4 is added to this mixture, which of the phosphate species would precipitate out of solution first? Ca3(PO4)2
Ag3PO4
Na3PO4
When the second cation just starts to precipitate, what percentage of the first cation remains in solution?

Answers

When the second cation first starts to precipitate, 80.8% of Ca²⁺ will still be in solution.

What is cation?

A cation is an ion with a positive charge. It is formed when an atom loses one or more of its electrons, resulting in a net positive charge. Cations are attracted to anions (ions with a negative charge) due to their opposite charges. Cations are found in many different substances, including acids, bases, and salts.

Ca₃(PO₄)₂ will be the first species that separates out of solution when solid Na₃PO₄ is introduced to the mixture. This is due to Ca3(PO4)2 having a substantially lower solubility than Ag₃PO₄ and Na₃PO₄.

The proportion of the first cation (Ca²⁺ ) still in solution when the second cation (Ag⁺) is just beginning to precipitate will depend on the starting concentrations of the two cations. In this instance, the starting concentrations of Ca²⁺  and Ag⁺ are 0.0400 M and 0.0990 M, respectively. Therefore, 80.8% of Ca²⁺  will still be in solution when its second cation first begins to precipitate.

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If heat is going INTO the system, that means that energy must have come OUT FROM the ____________

Answers

If heat is going into a system, it means that energy must have come out from the surroundings.

How is energy/heat transferred?

Heat is a form of energy transfer from a hotter object to a cooler one, and the direction of heat flow is always from the hotter object to the cooler one.

Therefore, if heat is entering a system, it must be gaining energy from its surroundings, which are at a lower temperature and therefore have less thermal energy.

Conversely, if heat is leaving a system, it means that energy is being transferred from the system to its surroundings, which are at a higher temperature and therefore have more thermal energy.

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Question 7 of 10
Which variable is unknown until the experiment is performed?
O A. A responding variable
OB. A mathematical variable
OC. A controlled variable
OD. A manipulated variable
SUBI

Answers

The mathematical variable is not a standard term in experimental design and is not typically used to describe variables in scientific experiments.

The variable that is unknown until the experiment is performed is typically the responding variable.

The responding variable is the variable that is observed and measured during the experiment, and its value changes in response to the manipulated variable. In contrast, the manipulated variable is the variable that is intentionally changed by the researcher to observe its effect on the responding variable.

The controlled variable is the variable that is kept constant throughout the experiment to ensure that any changes in the responding variable are due to the manipulated variable and not due to other factors. The mathematical variable is not a standard term in experimental design and is not typically used to describe variables in scientific experiments.

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PLEASE HELP!
Distilled vinegar contains a solution of acetic acid (CH3CO2H) in H2O. Using the formula M1V1=M2V2, solve for the concentration of the solution that results from diluting 0.50 L of 0.839 M vinegar solution to 2.5 L?

Question 4 options:

0.15 M


0.24 M


0.17 M


1.49 M

Answers

0.15M is the answer
Using the formula M1V1=M2V2, where M1 is the initial concentration, V1 is the initial volume, M2 is the final concentration, and V2 is the final volume, we can solve for the final concentration:

M1V1 = M2V2

0.839 M x 0.50 L = M2 x 2.5 L

M2 = (0.839 M x 0.50 L) / 2.5 L

M2 = 0.168 M

Therefore, the concentration of the solution that results from diluting 0.50 L of 0.839 M vinegar solution to 2.5 L is 0.168 M.

Answer: 0.17 M (rounded to two significant figures)

If 12.5 mol
of an ideal gas occupies 50.5 L
at 69.00 ∘C,
what is the pressure of the gas?

Answers

The pressure of a gas that occupies 50.5L at 69.0°C is 6.95 atm.

How to calculate pressure?

The pressure of an ideal gas can be calculated using Avogadro's equation as follows;

PV = nRT

Where;

P = pressureV = volume n = no of molesT = temperatureR = gas law constant

According to this question, 12.5 mol of an ideal gas occupies 50.5 L at 69.00°C. The pressure can be calculated as follows:

P × 50.5 = 12.5 × 0.0821 × 342

50.5P = 350.9775

P = 6.95 atm

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How are models used in chemistry? How does evidence change these models?​

Answers

Answer: As they develop theories, chemists use models to attempt to explain their findings. Chemists assess the model they are using as new evidence becomes available and, if required, continue to refine it by making modifications.

Explanation:

ichiometry in space
A typical space shuttle crew consists of six individuals and
each CCC contains 750 g of LIOH. Assuming that each crew
member expels 42.0 g of CO₂ per hour on average, and that
a mission is scheduled to last 18 days, how many CCCS must
be carried on board the space
shuttle?
- By knowing the recipe (balanced chemical equation), and
some molar masses, I can calculate this answer.

Answers

We need to carry at least 187 CCCs on board the space shuttle to absorb all the CO2 produced by the crew during the 18-day mission.

What is the amount of CO2 absorbed?

To solve this problem, we need to use the following information:

Each crew member expels 42.0 g of CO2 per hour.The mission is scheduled to last 18 days.There are 6 crew members on board.Each CCC contains 750 g of LIOH.

First, we need to calculate the total amount of CO2 that will be expelled during the mission:

Total CO2 = 6 crew members x 42.0 g CO2/hour x 24 hours/day x 18 days = 136,080 g CO2

Next, we need to calculate the amount of LIOH needed to absorb this CO2. The balanced chemical equation for the reaction between CO2 and LIOH is:

CO2 + 2 LIOH → Li2CO3 + H2O

The molar mass of CO2 is 44.01 g/mol, and the molar mass of LIOH is 23.95 g/mol.

This means that 2 moles of LIOH are needed to absorb 1 mole of CO2.

So, to absorb 136,080 g of CO2, we need:

136,080 g CO2 x (1 mol CO2/44.01 g) x (2 mol LIOH/1 mol CO2) x (23.95 g LIOH/1 mol) = 139,648 g LIOH

Since each CCC contains 750 g of LIOH, we need:

139,648 g LIOH / 750 g CCC = 186.2 CCCs

Therefore, we need to carry at least 187 CCCs on board the space shuttle to absorb all the CO2 produced by the crew during the 18-day mission.

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What mass of sulfur must be used to produce 25.7 L of gaseous sulfur dioxide at STP
according to the following equation?
S8 (s) + 8 O2 (g) −→ 8 SO2 (g)
Answer in units of g.

Answers

A mass of 37.0 g of sulfur must be used to produce 25.7 L of gaseous sulfur dioxide at STP.

What is the reactant mass of the sulfur?

The molar ratio of S₈ to SO₂ is 1:8.

At STP, one mole of gas occupies 22.4 L. Therefore, 25.7 L of SO₂ gas will contain;

25.7 L / 22.4 L/mol = 1.15 mol of SO₂.

The number of moles of S₈ needed is calculated as;

= 1.15 mol SO₂ / 8 mol S₈ per 1 mol SO₂

= 0.144 mol S₈.

The mass of S₈ needed is calculated as;

0.144 mol S₈ × 256.6 g/mol = 37.0 g of S₈.

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What mass (grams) of silver oxide would you need to decompose in order to produce 120.6 grams of silver?

Ag2O --> Ag + O2

Answers

The mass of silver oxide needed to decompose in order to produce 120.6 grams of silver is 494.5 grams.

The balanced chemical equation for silver oxide breakdown is:

[tex]Ag_2O[/tex] → [tex]2 Ag[/tex] + [tex]1/2 O_2[/tex]

The equation shows that for every mole of silver oxide that decomposes, two moles of silver are created, and the molar mass of [tex]Ag_2O[/tex] is 231.74 g/mol.

Hence, using stoichiometry, we can calculate the quantity of silver oxide necessary to generate 120.6 grams of silver:

120.6 g Ag × (1 mol Ag / 107.87 g Ag) × (1 mol [tex]Ag_2O[/tex]/ 2 mol Ag) × (231.74 g [tex]Ag_2O[/tex] / 1 mol [tex]Ag_2O[/tex] )

= 494.5 g [tex]Ag_2O[/tex]

As a result, 494.5 grams of silver oxide is needed to decompose in order to produce 120.6 grams of silver.

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Suppose 10.0 g of ice at -10.0C is placed into 300.0 g of water in a 200.0-g copper calorimeter. The final temperature of the water and copper calorimeter is 18.0C.
1) What was the initial common temperature of the water and copper? (Express your answer to three significant figures.)

Answers

The intital common temperature of copper and water is 9.5°C, under the condition that 10.0 g of ice at -10.0C is placed into 300.0 g of water in a 200.0-g copper calorimeter.

Now to evaluate the initial common temperature of the water and copper calorimeter, we have to apply the formula
m1c1(Tk - Ti) + m2c2(Tk - Ti)
= mcopperccopper(Tk - Ti)

Here,
m1 = mass of water,
c1 =specific heat capacity of water,
m2 = mass of copper calorimeter,
c2 = specific heat capacity of copper calorimeter, mcopper = mass of copper block
ccopper =specific heat capacity of copper.

Here, this equation to evaluate Ti
Ti = (m1c1Tk + m2c2Tk - mcopperccopperTk - m1c1Ti - m2c2Ti) / (m1c1 + m2c2 - mcopperccopper)

Staging the given values into this equation
Ti = (-300.0 g)(4.18 J/g°C)(18.0°C) + (200.0 g)(0.385 J/g°C)(18.0°C) + (10.0 g)(0.385 J/g°C)(18.0°C) / [(300.0 g)(4.18 J/g°C) + (200.0 g)(0.385 J/g°C) - (10.0 g)(0.385 J/g°C)]
Ti = 9.5°C

Hence, the initial common temperature of the water and copper calorimeter was 9.5°C.
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bromsted-lowry acids and bases
find out the acids and bases

Answers

Johannes Brsted and Thomas M. Lowry, two chemists, identified the Bromsted-Lowry acids and bases as a particular kind of acid-base reaction in 1923.

Acids are substances that give a base a proton (H+), whereas bases are substances that take a proton from an acid. In a Bromsted-Lowry acid-base reaction, the acid gives the base a proton in order to create the conjugate base and the conjugate acid, two new compounds.

Nitric acid (HNO3), sulfuric acid (H2SO4), and hydrochloric acid (HCl) are a few examples of acids. Sodium hydroxide (NaOH), ammonium hydroxide (NH4OH), and potassium hydroxide (KOH) are a few examples of bases.

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If you mix 30 mL of cold water with 70 mL of hot water in a calorimeter, then calculate that the cold water gained 142 J of heat and the hot water lost 181 J of heat, and the temperature change of the cold water (and calorimeter) was an increase in 1.93°C, then what is the heat capacity of the calorimeter in J/°C (only enter the number, not units, and assume that no heat was lost to the environment around the calorimeter, assume the density of water to be 1.00g/mL and specific heat capacity of water to be 4.184 J/g-°C)?

Answers

First, we need to calculate the heat gained by the cold water and the heat lost by the hot water:

Qcold = mcΔT = (30 g)(4.184 J/g-°C)(1.93°C) = 242.06 J

Qhot = mcΔT = (70 g)(4.184 J/g-°C)(-1.93°C) = -546.53 J

Since energy is conserved, we can assume that the heat gained by the cold water and calorimeter is equal to the heat lost by the hot water:

Qcold + Qcalorimeter = Qhot

Qcalorimeter = Qhot - Qcold

Qcalorimeter = -546.53 J - 242.06 J = -788.59 J

Therefore, the heat capacity of the calorimeter can be calculated as:

Ccalorimeter = Qcalorimeter / ΔT

Ccalorimeter = (-788.59 J) / (1.93°C)

Ccalorimeter ≈ -408.4 J/°C

Note that the negative sign indicates that the calorimeter loses heat when the system gains heat, which is expected since the calorimeter is absorbing some of the heat from the hot water.

Pleas help thanks!!!!!!!!!!!!!!!!!!!!

Answers

The number of molecules of BF₃ present in 2 grams of BF₃ is 1.776×10²² molecules (1st option)

How do i determine the number of molecules of BF₃?

We'll begin by calculating the number of mole of 2 grams of BF₃. Details below:

Mass of BF₃ = 2 grams Molar mass of BF₃ = 67.81 g/molMole of BF₃ =?

Mole = mass / molar mass

Mole of BF₃ = 2 / 67.81

Mole of BF₃ = 0.02949 mole

Finally, we shall determine the number of molecules of BF₃. This is shown below:

Avogadro's hypothesis suggest that:

1 mole of BF₃ = 6.022×10²³ molecules

Therefore,

0.02949 mole of BF₃ = 0.02949 × 6.02×10²³

0.02949 mole of BF₃ = 1.776×10²² molecules

Thus, the number of molecules of BF₃ is 1.776×10²² molecules (1st option)

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A 250.0-mL flask contains 0.2500 g of a volatile oxide of nitrogen. The pressure in the flask is 760.0 mmHg at 17.00°C. How many moles of gas are in the flask?

Answers

Answer:

0.0104 moles of gas in the flask.

Explanation:

To calculate the number of moles of gas in the flask, you can use the ideal gas law equation: PV = nRT. Where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant and T is temperature.

First, you need to convert the pressure from mmHg to atm and the temperature from Celsius to Kelvin. The pressure in atm is 760.0 mmHg / 760 mmHg/atm = 1 atm. The temperature in Kelvin is 17.00°C + 273.15 = 290.15 K.

Next, you need to convert the volume from mL to L. The volume in L is 250.0 mL / 1000 mL/L = 0.2500 L.

Now you can plug all the values into the ideal gas law equation and solve for n: (1 atm)(0.2500 L) = n(0.08206 L·atm/mol·K)(290.15 K). Solving for n gives n = 0.0104 mol.

So there are approximately 0.0104 moles of gas in the flask.

A 0.4856 g sample of solid silver oxide is heated. Find the volume of O2 that can be released at STP.

Answers

The volume of O2 that can be released at STP from the given sample of silver oxide is 23.45 mL.

To solve this problem

Creating the balanced chemical equation for the breakdown of silver oxide is the first step in tackling this issue:

2Ag2O(s) → 4Ag(s) + O2(g)

We can deduce from the equation that 2 moles of AgO will result in 1 mole of O2. Since Ag2O has a molar mass of 231.735 g/mol, 0.4856 g of Ag2O is equivalent to:

0.4856 g Ag2O x (1 mol Ag2O/231.735 g Ag2O) = 0.002095 mol Ag2O

Therefore, the number of moles of O2 that can be produced from 0.4856 g of Ag2O is:

0.002095 mol Ag2O x (1 mol O2/2 mol Ag2O) = 0.0010475 mol O2

1 mole of any gas takes up 22.4 L of space at STP As a result, 0.4856 g of Ag2O can generate the following amount of O2 at STP:

0.0010475 mol O2 x 22.4 L/mol = 0.02345 L or 23.45 mL

Therefore, the volume of O2 that can be released at STP from the given sample of silver oxide is 23.45 mL.

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1. Which is an example of heat being transferred through conduction?

2. 6 C (s) + 3 H2 → C6H12 (l)
ΔH = -903

Therefore, this reaction (loses/gains) heat/energy.

Answers

Answer:

9. B

10. Loses

Explanation:

9. Conduction is The procedure by which thermal energy or electricity is directly transported through a substance without the material moving when there is a variance in temperature between adjacent parts. Only choice B shows this process.

10. In exothermic reactions, energy/heat is lost. Exothermic reactions are characterized by a negative delta H, such as the delta H for the reaction show.

17. An artist took two photographs of the Moon that were several days apart. Images that look like his photographs are shown above. The light part of the Moon appeared to get smaller over time. Why did this happen?

Answers

According to the information, we can infer that the difference between photographs 1 and 2 originate from the translation of the Moon around the earth (option C).

How do we explain the differences between the two images?

To explain the difference between both images we must take into account the movement patterns of the earth and the moon. In the case of the earth, it has 2 main movements, which are rotation on its own axis and translation around the sun.

On the other hand, the moon has a translational movement around the earth, which is what causes the different lunar phases. This motion causes the moon to appear partially shadowed from the earth because the earth blocks the sunlight.

Based on the above, we can infer that the correct answer is option C because this phenomenon is caused by the translation of the moon.

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If the amount of solute present in a solution at a given temperature is less than the maximum amount that can be dissolved at that tempature the solution is said to be

Answers

Answer:

Unsaturated

Explanation:

A solution is unsaturated when it contains less than the maximum amount of solute that is capable of being dissolved.

For the Li2 molecule, rank order the following orbitals from lowest to highest energy: 1s, 2s, σ2s, σ*2s

Answers

The order of the energy levels for the Li2 molecule is:

1s < σ2s < 2s < σ*2s

The 1s orbital is the lowest in energy because it is closest to the nucleus and has the highest electron density. The σ2s orbital is next in energy because it is a bonding orbital that is formed by the overlap of two atomic 2s orbitals. The 2s orbital is higher in energy than the σ2s orbital because it is an atomic orbital that has not participated in bonding. The σ*2s orbital is the highest in energy because it is an antibonding orbital that weakens the bond between the two Li atoms.

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Calculate the energy difference (deltaE, in Joules) of an electron's transition from n = 6 to n = 1 in a hydrogen atom.

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The energy difference of an electron's transition from n = 6 to n = 1 in a hydrogen atom is approximately -2.17 × 10⁻¹⁸ Joules.

To calculate the energy difference (deltaE) of an electron's transition from n = 6 to n = 1 in a hydrogen atom, we can use the following equation:

deltaE = -13.6 * (1/n_final^2 - 1/n_initial^2) eV
where n_initial is the initial energy level (6 in this case), n_final is the final energy level (1 in this case), and -13.6 eV is the ionization energy of hydrogen.

Converting eV to Joules, we get:
1 eV = 1.602 x 10^-19 J
Therefore, deltaE in Joules can be calculated as follows:
deltaE = -13.6 * (1/1^2 - 1/6^2) * 1.602 x 10^-19 J/eV
deltaE = -2.179 x 10^-18 J

Therefore, the energy difference (deltaE) of an electron's transition from n = 6 to n = 1 in a hydrogen atom is -2.179 x 10^-18 J.

To calculate the energy difference (ΔE) for an electron's transition from n = 6 to n = 1 in a hydrogen atom, you can use the following formula:
ΔE = -13.6 eV * (1/nf² - 1/ni²)

Where ΔE is the energy difference in electron volts (eV), nf is the final energy level (1 in this case), and ni is the initial energy level (6 in this case).
ΔE = -13.6 eV * (1/1² - 1/6²)
ΔE ≈ -13.56 eV

Now convert electron volts to Joules:
1 eV = 1.6 × 10⁻¹⁹ J
ΔE ≈ -13.56 eV * 1.6 × 10⁻¹⁹ J/eV
ΔE ≈ -2.17 × 10⁻¹⁸ J

So,  approximately -2.17 × 10⁻¹⁸ Joules is  the energy difference of an electron's transition from n = 6 to n = 1 in a hydrogen atom.

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7 Suppose you weighed a different sample, of 2.500-g, which consisted of a mixture of CuO and potassium chloride and dissolved it in 25.00 mL of 0.437 M H₂SO4 solution. Some acid remains after treatment of the sample. Determine: a) If 35.4-mL of 0.108 M NaOH were required to titrate the excess sulfuric acid, how (6) many moles of CuO were present in the original sample?​

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The initial sample had 0.010925 mol of Copper(II) oxide, or one mole.

What exactly is kinetic-molecular theory?

The kinetic-molecular theory, which describes the states of matter, is based on the presumption that matter is composed of minuscule particles that are constantly in motion. This theory explains the observable properties and behaviours of solids, liquids, and gases. The container's walls and the quickly moving particles' collisions with one another are constant.

Copper(II) oxide + Sulfuric acid → Cupric sulfate + Water

One mole of Copper(II) oxide interacts with one mole of Sulfuric acid, as shown by the equation. The amount of Sulfuric acid that reacted with the Copper(II) oxide in the sample is therefore equal to the amount of Copper(II) oxide in the sample.

We must first determine how many moles of Sulfuric acid interacted with the sample:

moles Sulfuric acid = concentration × volume

moles Sulfuric acid = 0.437 mol/L × 0.025 L

moles Sulfuric acid = 0.010925 mol

Since the acid is in excess, the moles of Sulfuric acid remaining after treatment of the sample is:

moles Sulfuric acid remaining = moles Sulfuric acid added – moles Sulfuric acid reacted

moles Sulfuric acid remaining = 0.437 mol/L × 0.0354 L – 0.010925 mol

moles Sulfuric acid remaining = 0.007571 mol

To determine the number of moles of Copper(II) oxide in the original sample, we can use the following equation:

moles Copper(II) oxide = moles Sulfuric acid reacted = 0.010925 mol

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Please help me with this chemistry investigation I need answers as soon as possible please ​

Answers

B. To plot the data on a bar chart, draw a horizontal axis for metals and a vertical axis for time to complete the reaction. Then, draw bars for each metal that represent the amount of time required to complete the reaction. The height of the bars must match the time values ​​in the table.

C. No, Emilia was not correct in her forecast. According to the data, aluminum reacted 100 seconds faster than magnesium, which reacted in 50 seconds. Thus aluminum reacts more rapidly with hydrochloric acid than magnesium.

From most reactive to least reactive, the metals are as follows:

aluminummagnesiumZincIron

This order is consistent with the reactivity series, which is:

PotassiumSodiumCalciumMagnesiumAluminiumZincIronCopperSilverGold

We are unable to estimate the reactivity of potassium, sodium, calcium, copper, silver, or gold from this experiment because those variables are not present in the data.

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You want to have a barbecue this weekend! But you're worried about global warming. You only want to release a maximum of 0.750 kg of carbon dioxide from your propane grill. Using the below equation to answer the following questions.

CH3(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

ΔHrxn = -2220.1 kJ

a. How many kilojoules will you be able to release?

b. If it requires 1900 kJ to cook one hamburger, how many hamburgers can you cook?

Answers

a. We will be able to release 37,827 kJ. b. You can cook a maximum of 19 hamburgers without exceeding the limit of 0.750 kg of carbon dioxide.

a. We need to use the balanced chemical equation and the enthalpy change of the reaction.

Therefore, moles  [tex]CO_2[/tex] produced are[tex]0.750 kg / 44.01 g/mol = 17.03 mol.[/tex]

The enthalpy change of the reaction is -2220.1 kJ/mol. Thus, the maximum number of kilojoules that can be released is:

[tex]\Delta Hrxn * moles of[/tex] [tex]CO_2[/tex] = [tex]-2220.1 kJ/mol * 17.03 mol = -37,827 kJ[/tex]

We need to reverse the sign of the answer, giving us 37,827 kJ.

b. If it requires 1900 kJ to cook one hamburger, we can divide the maximum number of kilojoules that can be released by the energy required to cook one hamburger:

37,827 kJ / 1900 kJ/hamburger = 19.91 hamburgers

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