I need help please:(

Considering the ideal gas law, the volume occupied by 3.25 moles of an ideal gas at a temperature of 18.00°C is 686.71 L.

An ideal gas is the behavior of those gases whose molecules do not interact with each other and move randomly. Under normal conditions and under standard conditions, most gases exhibit ideal gas behavior.

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T), related by a simple formula called the ideal gas law:

P×V = n×R×T

Where:

In this case, you know:

Replacing in the ideal gas law:

1.13 atm×V = 3.25 mol× 0.8205 L.atm/K.mol× 291 K

Solving:

V = (3.25 mol× 0.8205 L.atm/K.mol× 291 K)÷ 1.13 atm

V= 686.71 L

Finally, the volume is 686.71 L.

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Answer: Oxygen gas

Explanation: Gradpoint

Answer:

B. Activated Carbon

Explanation:

Not enough information man

The answer is a                                  bevasue it then becomes a chemical compound

Answer:

a

Explanation:

this would result in a compound and compounds are chemical changes so i think im right....

We can measure the slope of the tangent line to the curve at a certain position on a graph to calculate the pace of a response at that time.

A reaction is a process in which one or more reactants are changed into one or more products, also known as the final products. Atoms are rearranged and chemical bonds are made and broken between the atoms during a chemical reaction, which transforms reactants into products.

The instantaneous rate of the response at that particular time is indicated by the slope of the tangent line.

If we have a graph of the quantity of product produced over time, for instance, we can choose a point on the curve that represents a certain period, draw a tangent line to the curve at that location, and then calculate the slope of the line. The pace of the reaction at that specific moment is shown by the slope of the tangent line.

The average rate of the response over a certain period of time can be calculated by repeating this method at various points on the curve to produce a series of instantaneous rates.

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Answer:

As406(s) + 4MnO_4^- (aq) → 4AsO_4^3- (aq) + 4Mn^2+ (aq)

Explanation:

Answer:

3

Explanation:

co2 has 2 and h2o has 1

product is right side

Temperature of the gas is approximately 570.4 K when there are 1.9 moles of gas at a pressure of 5 ATM and a volume of 5.0 × \(10^{4}\) mL.

To determine the temperature of the gas, we can use the ideal gas law equation, which states that the pressure of a gas is directly proportional to its temperature, volume, and the number of moles of gas. The equation is given by:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

In this case, we are given the pressure (P = 5 ATM), volume (V = 5.0 × 10^4 mL), and number of moles (n = 1.9 moles) of the gas. We can rearrange the ideal gas law equation to solve for temperature:

T = PV / (nR)

Substituting the given values and the value of the ideal gas constant (R = 0.0821 L·atm/(mol·K)), we can calculate the temperature:

T = (5 ATM) × (5.0 × \(10^{4}\) mL) / (1.9 moles × 0.0821 L·atm/(mol·K))

After performing the calculations, we find that the temperature of the gas is approximately 570.4 K.

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Answer:

Osmium

Explanation:

Answer:

a. 181.5 kcal

Explanation:

Step 1: Calculate the enthalpy of the process (ΔH°).

Let's consider the following process.

CO₂(s) → CO₂(g)

We can calculate the enthalpy of the process using the following expression.

ΔH° = ∑ np × ΔH°f(p) - ∑ nr × ΔH°f(r)

ΔH° = 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CO₂(s))

ΔH° = 1 mol × (-393.5 kJ/mol) - 1 mol × (-427.4 kJ/mol) = 33.9 kJ

According to the balanced equation, 33.9 kJ are required to sublime 1 mole of CO₂.

Step 2: Convert 986 g of CO₂ to moles

The molar mass of CO₂ is 44.01 g/mol.

986 g × 1 mol/44.01 g = 22.4 mol

Step 3: Calculate the enthalpy needed to sublime 22.4 moles of CO₂

22.4 mol × 33.9 kJ/1 mol = 759 kJ

We can convert it to Kcal using the conversion factor 1 kcal = 4.184 kJ.

759 kJ × 1 kcal/4.184 kJ = 181.5 kcal

Answer:

First, we need to convert the toxicity value from mg/kg to mg for a 70 kg adult:

Toxic dose for a 70 kg adult = 12,000 mg/kg x 70 kg = 840,000 mg

Next, we need to calculate the number of moles of 7-UP in one can:

Concentration of 7-UP in one can = 0.012 M

Volume of one can = 330 mL = 0.33 L

Number of moles of 7-UP in one can = concentration x volume = 0.012 mol/L x 0.33 L = 0.00396 mol

Finally, we can calculate the number of cans of 7-UP a 70 kg adult would have to drink to reach a toxic dose:

Number of cans of 7-UP = toxic dose / (number of moles in one can x molecular weight of 7-UP)

Molecular weight of 7-UP = 338.14 g/mol

Number of cans of 7-UP = 840,000 mg / (0.00396 mol x 338.14 g/mol) ≈ 619 cans

Therefore, a 70 kg adult would have to drink approximately 619 cans of 7-UP to reach a toxic dose.

Explanation:

Displacement reactions

Answer: you can't really balance it doesn't make anything

Explanation:

Answer:

D. the mass of one mole of a substance

Explanation:

The molar mass of a substance is the mass in grams of one mole of the substance.

Option B, where [OH-] is 1.0 x 10-13 mol dm-³3, is the only one that can be considered basic. Therefore, Option B is the correct answer.

To determine whether a solution is basic or acidic at 25 °C, we can compare the concentration of hydroxide ions ([OH-]) with the concentration of hydronium ions ([\(H_3O\)+]). In a neutral solution, the concentrations of [\(H_3O\)+] and [OH-] are equal, resulting in a pH of 7.

Option A states that the concentration of [\(H_3O\)+] is 1.0 x 10-3 mol dm-3. Since [\(H_3O\)+] represents the concentration of hydronium ions, this solution would be acidic because the concentration of [\(H_3O\)+] is higher than [OH-], indicating an excess of hydronium ions.

Option B states that the concentration of [OH-] is 1.0 x 10-13 mol dm-³3. In this case, [OH-] is higher than [\(H_3O\)+], indicating an excess of hydroxide ions. Therefore, this solution would be considered basic.

Option C states that the solution has a pH of 4.00. A pH of 4.00 is below the neutral pH of 7, indicating an excess of hydronium ions and an acidic solution. Therefore, this option does not represent a basic solution.

Option D states that the concentration of [\(H_3O\)+] is 1.0 x 10-13 mol dm-3. Similar to Option A, this concentration of [\(H_3O\)+] indicates an acidic solution, not a basic one.

Option B

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Answer:

D is the correct answer

Explanation:

In order for a solution to be basic at 25 C, the H+ concentration has to be less than the OH- concentration, and given that H+ times OH- is 10^-14, we deduce that H+ must be less than 10^-7 for the solution to be acidic. Thus, A can be eliminated, and so can C. With B, we calculate an H+ concentration of 0.1M, which also fails to be less than 10^-7
Thus, D is the correct answer and we can verify that as H+ is less than 10^-7.



Note: I do not know why my previous answer was deleted for "being incorrect", and i'm not sure how the incorrect answer was "expert verified", but I am as certain that D is the correct answer as i am sure of 3*(4+5-1) being equal to 24.

Based on the given reactions, the compounds are as follows:

A: The specific product formed from the reaction between prop-1-yne and either 2HBr or H2O2.

B: The product formed when compound A reacts with 2H2O.

C: The product formed when compound B reacts with K2CO3(aq).

D: The product formed from the heat-induced reaction of compound C.

E: The product formed when compound D reacts with HBr.

Based on the given reactions, let's analyze the compounds involved:

Reaction 1: prop-1-yne + 2HBr/H2O2 = A

The reactant prop-1-yne reacts with either 2HBr or H2O2 to form compound A. The specific product formed will depend on the reaction conditions.

Reaction 2: A + 2H2O = B

Compound A reacts with 2H2O (water) to form compound B.

Reaction 3: B + K2CO3(aq) = C

Compound B reacts with K2CO3(aq) (potassium carbonate dissolved in water) to form compound C.

Reaction 4: C + heat = D

Compound C undergoes a heat-induced reaction to form compound D.

Reaction 5: D + HBr = E

Compound D reacts with HBr (hydrobromic acid) to form compound E.

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The boiling point of water is 100.0 °C, the boiling point of the solution will be : 101.49 °C.The correct answer is option (a) 100.00049 °C.

Ideal Solution : An ideal solution is a homogeneous mixture of two or more components that obeys Raoult's law, which states that each component's vapor pressure is proportional to its mole fraction.The boiling point of a solution depends on the solvent's properties and the solute's concentration. It's dependent on the mole fraction of the solvent and solute, as well as the total concentration of the solution. The change in boiling point of a solution is given byΔTb = Kb × m × i, whereKb = ebullioscopic constant, m molarity of the solution, and i = van't Hoff factor.Assuming that the solution's behavior is ideal, we may use the molality of the solution to compute the boiling point elevation of the solution.The molality of the solution is given by the following formula:m = (n₂ / m₂) ÷ (n₁ / m₁), where n is the number of moles, m is the mass, and the subscripts 1 and 2 refer to water and non-volatile solute sucrose, respectively.The molar mass of sucrose (C₁₂H₂₂O₁₁) is342.3 g/mol; therefore, the number of moles of sucrose is115.0 g ÷ 342.3 g/mol = 0.335 mol.m₁ = mass of water = 350.0 g, and m₂ = mass of sucrose = 115.0 g, as given in the problem.Therefore, the molality of the solution is given by:m = (0.335 mol / 0.115 kg) ÷ (1 mol / 1 kg) = 2.91 mol/kg.Substituting these values in the formula for ΔTb, we get:ΔTb = Kb × m = 0.512 °C m⁻¹ × 2.91 mol/kg = 1.49 °C.100.0 °C + 1.49 °C = 101.49 °C.

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The energy of combustion for one mole of butane to be approximately 2888.81 kJ/mol.

To calculate the energy of combustion for one mole of butane (C4H10), we need to use the information provided and apply the principle of calorimetry.

First, we need to convert the mass of the butane sample from grams to moles. The molar mass of butane (C4H10) can be calculated as follows:

C: 12.01 g/mol

H: 1.01 g/mol

Molar mass of C4H10 = (12.01 * 4) + (1.01 * 10) = 58.12 g/mol

Next, we calculate the moles of butane in the sample:

moles of butane = mass of butane sample / molar mass of butane

moles of butane = 0.367 g / 58.12 g/mol ≈ 0.00631 mol

Now, we can calculate the heat released by the combustion of the butane sample using the equation:

q = C * ΔT

where q is the heat released, C is the heat capacity of the calorimeter, and ΔT is the change in temperature.

Given that the heat capacity of the bomb calorimeter is 2.36 kJ/°C and the change in temperature is 7.73 °C, we can substitute these values into the equation:

q = (2.36 kJ/°C) * 7.73 °C = 18.2078 kJ

Since the heat released by the combustion of the butane sample is equal to the heat absorbed by the calorimeter, we can equate this value to the energy of combustion for one mole of butane.

Energy of combustion for one mole of butane = q / moles of butane

Energy of combustion for one mole of butane = 18.2078 kJ / 0.00631 mol ≈ 2888.81 kJ/mol

Therefore, the energy of combustion for one mole of butane is approximately 2888.81 kJ/mol.

In conclusion, by applying the principles of calorimetry and using the given data, we have calculated the energy of combustion for one mole of butane to be approximately 2888.81 kJ/mol.

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Answer:

Female

Explanation:

Answer: A. Female.

Explanation: The word 'heroine' means a female or a woman that is admired and looked upon for her courage, bravery, and noble qualities.

The enthalpy of combustion for 1kg of urea is -1223525.84 J/mol.

Urea is a compound that is used in fertilizers and in some plastics.The enthalpy of combustion for urea is the amount of energy that is released when urea is burned. In order to calculate the enthalpy of combustion for 1kg of urea, we need to use the information that is provided to us in the question. Let us start by writing down the balanced equation for the combustion of urea: CO(NH2)2 + 3/2 O2 → CO2 + 2H2O + N2
The balanced equation shows that 1 mole of urea reacts with 1.5 moles of oxygen gas to produce 1 mole of carbon dioxide, 2 moles of water, and 1 mole of nitrogen gas. The enthalpy change for this reaction is equal to the amount of energy that is released when 1 mole of urea is burned.
The heat of combustion (ΔHc) of urea is -632.6 kJ/mol. This means that 632.6 kJ of energy is released when 1 mole of urea is burned. We know that 12g of urea raised the temperature of water by 30°C. We can use this information to calculate the amount of energy that was released when 12g of urea was burned.
The specific heat capacity of water is 4.18 J/g°C. This means that it takes 4.18 J of energy to raise the temperature of 1 gram of water by 1°C. Therefore, it takes 4.18 x 1000 = 4180 J of energy to raise the temperature of 1 kg of water by 1°C.
We know that 12g of urea raised the temperature of water by 30°C. Therefore, the amount of energy that was released when 12g of urea was burned is:
Energy = mass x specific heat capacity x temperature change
Energy = 0.012 kg x 4180 J/kg°C x 30°C
Energy = 1497.6 J
We can now use this information to calculate the enthalpy of combustion for 1kg of urea:
Enthalpy of combustion = energy released / moles of urea burned
Enthalpy of combustion = 1497.6 J / (0.012 kg / 60.06 g/mol)
Enthalpy of combustion = - 1223525.84 J/mol
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Anwser:

Research scientific principles involved in water availability.

Explanation:

Before building or investment you always want to see and know what your up against and what the best way to help or solve is. Researching why there's a problem in water availability or/and the principals to water availability will help you see what the problem is with what's happening to and around the process and it's environment.

The change which shift the equilibrium to produce the maximum possible amount of SO₃ is decreasing the temperature.

According to this principle if any external stress is applied in the equilibrium state then equilibrium will shift towards that side where the external stress will decreases.

In the given chemical equation energy is releases in the form of heat, then in this condition:

Hence decreasing the temperature will produce maximum SO₃.

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The manner in which the transfer of energy would be different if the bars were in an open system is as follows: Energy transfer would occur between the copper bars and the surroundings (option B).

The law of conservation of energy principle stating that energy may not be created or destroyed.

An isolated system exchanges neither energy nor matter with the surroundings. According to this question, two copper bars at different temperatures transfer energy until both are at the same temperature.

However, in an open system, some of the energy would be transferred to the surroundings.

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Answer: 20.9 billion years.

Explanation: To estimate the age of the rock, we can use the fact that the half-life of uranium-238 is 4.47 billion years. This means that half of the uranium-238 in the rock would have decayed into lead-206 after 4.47 billion years, and half of what remains would decay in the next 4.47 billion years, and so on.

If 20% of the uranium-238 in the rock has decayed into lead-206, then 80% of the original uranium-238 is still present in the rock. This means that the rock has gone through one half-life of uranium-238 decay, and we can estimate its age by using the graph on page 253.

According to the graph, the ratio of lead-206 to uranium-238 after one half-life is approximately 0.027. This means that the rock contains 0.027 times as much uranium-238 as it did originally, and the remaining 80% of the uranium-238 corresponds to 0.8 times the original amount.

Therefore, we can estimate the original amount of uranium-238 in the rock as follows:

Original amount of uranium-238 = (0.8) / 0.027 = 29.63 times the current amount

Next, we can use the fact that each half-life corresponds to a reduction in the amount of uranium-238 by a factor of 2, to estimate the number of half-lives that have passed since the rock formed:

Number of half-lives = log2(29.63) = 4.88

Finally, we can estimate the age of the rock as follows:

Age of the rock = (4.88) x (4.47 billion years per half-life) = 20.9 billion years

Therefore, we can estimate that the age of the rock is approximately 20.9 billion years.

The correct option that represents the constants of the reaction vessel after the N₂ and Cl₂ react as completely as possible is A.

The balanced chemical equation for the reaction is:

N₂(g) + 3Cl₂(g) → 2NCl₃(g)

This means that 1 mole of nitrogen gas and 3 moles of chlorine gas react to produce 2 moles of nitrogen trichloride gas.

The contents of the two vessels are:

Vessel 1: 1 mole of N₂

Vessel 2: 3 moles of Cl₂

When these two vessels are combined, the chlorine gas will react completely with the nitrogen gas to produce nitrogen trichloride gas. The remaining gas in the vessel will be nitrogen trichloride gas.

Which is:

[N₂] = 0 mol

[Cl₂] = 0 mol

[NCl₃] = 2 mol

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Complete question:

11 Question 3 A =N O=CI N2 +3 Cl2 2 NC13 The contents of two vessels are combined, as shown. The balanced chemical equation for the reaction that occurs is also shown. Which of the following represents the contents of the reaction vessel after the N2 and Cl2 react as completely as possible?

Answer:

The answer to your question is given below.

Explanation:

The following data were obtained from the question:

Speed limit = 60 km/h

To know if the car was speeding, let us calculate the speed of the car.

This can be obtained as follow:

Distance (d) = 5 m

Convert 5 m to km

Recall:

1000 m = 1 km

5 m = 5/1000

5 m = 0.005 km

Time (t) = 0.2 sec.

Convert 0.2 sec to hour.

Recall:

3600 secs = 1 h

Therefore,

0.2 sec = 0.2/3600

0.2 sec = 5.56×10¯⁵ h

Thus,

Distance (d) = 0.005 km

Time (t) = 5.56×10¯⁵ h

Speed of car =?

Speed = Distance /time

Speed = 0.005/5.56×10¯⁵

Speed of the car = 90 km/h

Comparing the speed of the car (i.e 90 km/h) to the speed limit of the zone (i.e 60 km/h), we can see that the car is on a high speed since its speed (i.e 90 km/h) is higher than the speed limit of the zone (i.e 60 km/h).

Speed of the car = 90 km/h

Speed limit = 60 km/h

Difference = 90 – 60

Difference = 30 km/h

Therefore, the car is speeding by 30 km/h more compared to the speed limit.