Answer:
39.7
Explanation:
Therefore, the final temperature of the water is 39.7°C.
. A ring with a mass of 25.5 g appears to be pure silver. Rather than test for density, you can confirm the ring's composition by determining its specific heat. Suppose the ring is heated to a temperature of 84.0°C and then immersed in a container of water until the ring's temperature is 25.0°C. If the ring gives up 667.5 J of energy to the water, what is its specific heat? Is the ring made of silver (C = 0.234 J/g °C), nickel (C = 0.444 J/g. °C), or palladium (C = 0.244 J/g °C) help me
The specific heat capacity of the ring, given that the ring gives up 667.5 J of energy to the water is 0.444 J/gºC. The ring is made of nickel.
How do i determine the specific heat capacity of the ring?The specific heat capacity of the ring can be obtain as illustrated below:
Heat absorbed by water (Q) = 667.5 JHeat released by ring (Q) = -667.5 JMass of ring (M) = 25.5 gInitial temperature of ring (T₁) = 84.0 °CFinal temperature (T₂) = 25.0 °CChange in temperature (ΔT) = 25.0 - 84.0 = -59 °CSpecific heat capacity of ring (C) = ?Q = MCΔT
-667.5 = 25.5 × C × -59
-667.5 = -1504.5 × C
Divide both sides by -1504.5
C = -667.5 / -1504.5
C = 0.444 J/gºC
Thus, the specific heat capacity of the ring is 0.444 J/gºC. Hence, the ring is made of nickel
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A student conducted three trails to determine the concentration of an unknown concentration of HCI. In the first trail the calculated concentration was 0.104 M, the second was 0.113 M and the third trail was 0.108 M. What is the percent difference between the first two trials and based on the lab procedures procedures guidelines what would the average molarity be?
The percent difference between the first two trials is 8.3% and final outcome would be an average molarity of 0.108 M.
How to calculate percent difference and average molarity?To calculate the percent difference between the first two trials, use the formula:
% Difference = |(Value 1 - Value 2) / ((Value 1 + Value 2) / 2)| x 100%
% Difference = |(0.104 M - 0.113 M) / ((0.104 M + 0.113 M) / 2)| x 100%
% Difference = |-0.009 M / 0.1085 M| x 100%
% Difference = 8.3%
The percent difference between the first two trials is 8.3%.
To find the average molarity, add the three calculated concentrations together and divide by the number of trials:
Average Molarity = (0.104 M + 0.113 M + 0.108 M) / 3
Average Molarity = 0.108 M
Based on the lab procedures guidelines, the average molarity would be the most accurate representation of the unknown concentration of HCI. Therefore, the average molarity of 0.108 M would be the final result.
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1. Which metal is the most reactive? How do you know this?
2. Rank the metals in order of increasing reactivity.
3. Give the chemical equations for each single replacement reaction that took place.
4. Was Fe^3+ reduced? Of so what metal(s) acted as reducing agents?
Calculate the volume of hydrogen produced at s.t.p. When 25g of zinc are added to excess dilute hydrochloride acid at 31°c and 778mm Hg pressure. (H=1, Zn=65, Cl=35.5, molar volume of a gas at s.t.p = 22.4 dm3
To solve this problem, we need to use the balanced chemical equation for the reaction between zinc (Zn) and hydrochloric acid (HCl):
[tex]Zn + 2HCl - > ZnCl_2 + H_2[/tex]
According to the stoichiometry of this equation, one mole of Zn reacts with two moles of HCl to produce one mole of H2. Therefore, we need to determine the number of moles of Zn in 25 g, and then use the mole ratio to find the number of moles of H2 produced.
Finally, we can convert the number of moles of H2 to volume at STP using the molar volume of a gas.
First, we need to calculate the number of moles of Zn in 25 g:
The molar mass of Zn is 65.38 g/mol
The number of moles of Zn in 25 g is:
25 g / 65.38 g/mol = 0.383 mol Zn
Next, we use the mole ratio from the balanced equation to find the number of moles of H2 produced:
According to the balanced equation, one mole of Zn reacts with one-half mole of H2, so we produce 0.5 x 0.383 = 0.192 mol H2.
Finally, we can use the molar volume of a gas at STP to convert the number of moles of H2 to volume:
The molar volume of a gas at STP is 22.4 dm3/mol
Therefore, the volume of H2 produced is:
V = (0.192 mol) x (22.4 dm3/mol) = 4.30 dm3 or 4,300 ml
Therefore, the volume of hydrogen gas produced at STP is 4.30 dm3 or 4,300 ml when 25 g of zinc is added to excess dilute hydrochloric acid at 31°C and 778 mm Hg pressure.
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An unknown alkene is ozonolyzed and worked up under oxidizing conditions. The H NMR spectrum of the only product obtained is shown. Identify the alkene.
To identify the unknown alkene based on its H NMR spectrum, a qualified organic chemist would need to analyze the chemical shifts, integration values, and splitting patterns of the peaks in the spectrum, and compare them with known reference data and other spectroscopic techniques (such as C NMR, IR, and mass spectrometry) to make an accurate determination.
The alkene is likely to be a symmetrical alkene with two equivalent methyl groups attached to the double bond. This can be seen from the singlet at 1.7 ppm, which is characteristic of a methyl group, appearing twice in the spectrum. The ozonolysis of the alkene would lead to the formation of two carbonyl compounds, which are then oxidized to carboxylic acids under the given oxidizing conditions. Therefore, the alkene in question is likely to be cis-2-butene.
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The table below shows the vapor pressure of water at various temperatures.
Temp(degC) Vapor Pressure (mmHg)
17
14.5
18
15.5
19
16.5
20
17.5
21
18.7
22
19.8
During an experiment 675 mL of helium gas is collected over water at 22 degC. The air pressure in the lab is 0.926 atm. What is the partial pressure of the dry helium collected?
The partial pressure of the Helium gas is 0.9 atm.
What is the partial pressure of gas collected over water?The pressure that a gas exerts on its own when it is collected over water, independent of the pressure that the water vapor in the collecting vessel also produces, is known as its partial pressure.
When gas is collected over water, some of the water vapor will dissolve in it and change the overall pressure in the collecting vessel. Water vapor has its own partial pressure, which is affected by the relative humidity and temperature of the air around it. This is why it behaves in this way.
We have that;
Vapor pressure of the gas = 19.8 mmHg or 0.026 atm
Partial pressure of the Helium gas = 0.926 atm - 0.026 atm
= 0.9 atm
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A sample of an ideal gas has a volume of 2.31 L
at 279 K
and 1.01 atm.
Calculate the pressure when the volume is 1.09 L
and the temperature is 308 K.
The pressure of the gas when the volume is 1.09 L and the temperature is 308 K is 2.36 atm.
What is the final pressure of the gas?The final pressure of the gas is calculated by applying ideal gas law as follows;
(P₁V₁)/T₁ = (P₂V₂)/T₂
where
P₁, V₁, and T₁ are the initial pressure, volume, and temperature, P₂, V₂, and T₂ are the final pressure, volume, and temperature,P₂ = (P₁V₁ x T₂)/(V₂ x T₁)
P₂ = (1.01 atm x 2.31 L x 308 K) / (1.09 L x 279 K)
P₂ = 2.36 atm
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A student is tasked with writing the net ionic equation for the following
reaction:
4
Al(s) + 3 AgNO3(aq) → Al(NO3)3(aq) + 3 Ag(s)
What is the net ionic equation?
The net ionic equation of the reaction is as follows:
4 Al3+(aq) + 12 NO3-(aq) + 3 Ag(s) = 4 Al(s) + 12 Ag+(aq) + 12 NO3-(aq)
Ions which remain in their ground state and do not take part in the reaction are called spectator ions. The net ionic equation cancels out these ions, which are present on both the reactant and product sides of the equation.
Spectator ions, which can be found on both the reactant and product sides, but are not included in the finished reaction from the net ionic equation. The [tex]NO^3^-[/tex] ions are spectator ions in this example, thus taking them out of the equation. The net ionic equation makes up the rest of the species.
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The value of H for the following reaction is +128.1kJ: CH3OH(I) —> CO(g)+2H2(g) . Calculate the value of H (in kJ) when 5.10g of H2(g) is formed.
A) 653 B)326 C)-162.0 D)128 E)162
The value of H (in kJ) when 5.10 g of H2(g) is formed is 326 kJ (option B).
The given reaction is: CH3OH(I) —> CO(g)+2H2(g)
From the given value of H, we know that when one mole of CH3OH reacts, 128.1 kJ of heat energy is absorbed.
The molar mass of H2 is 2 g/mol. So, 5.10 g of H2 is equivalent to 5.10/2 = 2.55 moles of H2.
From the balanced equation, we can see that two moles of H2 are produced for each mole of CH3OH that reacts.
So, 2.55 moles of H2 are produced by 1.275 moles of CH3OH reacting (2.55/2).
Therefore, the amount of heat energy absorbed when 1.275 moles of CH3OH reacts can be calculated as:
Q = n x ΔH = 1.275 mol x 128.1 kJ/mol = 163.28 kJ
Since this amount of heat energy is absorbed when 1.275 moles of CH3OH reacts, to find the amount of heat energy absorbed when 2.55 moles of H2 is formed, we can simply double the value of Q:
Q = 2 x 163.28 kJ = 326.56 kJ
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Mark needs to determine how much the temperature of a solution changes during a chemical reaction. Which tool does he need?
Mark would need a thermometer to determine the temperature change of a solution during a chemical reaction. A thermometer is a tool used to measure temperature and can be used to monitor and record changes in temperature during a chemical reaction. So the answer is thermometer .
There are different types of thermometers, such as liquid-in-glass thermometers, bimetallic strip thermometers, digital thermometers, and infrared thermometers, among others. The choice of thermometer depends on the specific requirements of the experiment or process being carried out.
By measuring the initial and final temperatures of the solution before and after the chemical reaction, Mark can determine the temperature change, which is an important parameter in many chemical reactions as it provides information about the heat energy involved in the reaction, and helps in understanding the thermodynamics and kinetics of the process. Therefore the answer is thermometer .
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A gas‑filled weather balloon has a volume of 56.0 L
at ground level, where the pressure is 761 mmHg
and the temperature is 23.1 ∘C.
After being released, the balloon rises to an altitude where the temperature is −6.97 ∘C
and the pressure is 0.0772 atm.
What is the weather balloon's volume at the higher altitude?
We can use the combined gas law to determine the volume of the balloon at a higher altitude. The combined gas law relates the pressure, volume, and temperature of a gas:
(P1 x V1) / T1 = (P2 x V2) / T2
where P1, V1, and T1 are the pressure, volume, and temperature of the gas at the initial state, and P2, V2, and T2 are the pressure, volume, and temperature of the gas at the final state.
We are given the initial pressure (P1 = 761 mmHg), volume (V1 = 56.0 L), and temperature (T1 = 23.1 °C = 296.25 K) of the gas, and the final pressure (P2 = 0.0772 atm), and temperature (T2 = -6.97 °C = 266.18 K) of the gas. We can solve for V2, the final volume of the gas:
(P1 x V1) / T1 = (P2 x V2) / T2
V2 = (P1 x V1 x T2) / (P2 x T1)
V2 = (761 mmHg x 56.0 L x 266.18 K) / (0.0772 atm x 296.25 K)
V2 = 2,040 L (rounded to three significant figures)
Therefore, the volume of the weather balloon at the higher altitude is approximately 2,040 L.
Three of the primary components of air are
carbon dioxide, nitrogen, and oxygen. In a
sample containing a mixture of only these
gases at exactly one-atmosphere pressure, the partial pressures of carbon dioxide and nitrogen are given as PCO2 = 0.285 torr and
PN2 = 580.502 torr. What is the partial pressure of oxygen?
Answer in units of torr.
The partial pressure of the oxygen is 0.236 atm.
What is partial pressure?The pressure that one gas component in a mixture of gases exerts is known as partial pressure. It is the pressure that the gas would experience if it took up the same amount of space in the mixture at the same temperature on its own.
We know that;
P[tex]CO_{2}[/tex] = 0.285 torr or 0.000375 atm
P[tex]N_{2}[/tex] = 580.502 torr or 0.764 atm
P[tex]O_{2}[/tex] = ?
Total pressure = 1 atm
Then we have that;
PT =P[tex]CO_{2}[/tex] +P[tex]N_{2}[/tex]+ P[tex]O_{2}[/tex]
P[tex]O_{2}[/tex] = PT - (P[tex]CO_{2}[/tex] + P[tex]N_{2}[/tex])
P[tex]O_{2}[/tex] = 1 - (0.000375 + 0.764)
P[tex]O_{2}[/tex]= 0.236 atm
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what is the ph of a .100M naclo solution
The pH of a 0.100M NaClO solution is 1.
How to calculate pH?pH, meaning power of hydrogen, is a measure of how acidic/basic a solution is. The range goes from 0 - 14, with 7 being neutral. pHs of less than 7 indicate acidity, whereas a pH of greater than 7 indicates a base.
pH is really a measure of the relative amount of free hydrogen and hydroxyl ions in the water. It can be estimated using the following formula;
pH = - log {H+}
Where;
H+ = hydrogen ion concentrationpH = - log {0.100}
pH = 1
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HELP ASAP!! 50 POINT AND A BRAINLIEST FOR THE CORRECT ANSWER
FeO (s) + Fe (s) + O2(g) →
Fe2O3 (s)
Given the following table of thermodynamic data at 298 K:
Substance ΔHf° (kJ/mol) S° (J/K⋅mol)
FeO (s) -271.9 60.75
Fe (s) 0 27.15
O2 (g) 0 205.0
Fe2O3 (s) -822.16 89.96
The value K for the reaction at 25 °C is ________.
Consider the reaction:
FeO (s) + Fe (s) + O2(g) Fe2O3 (s)
Given the following table of thermodynamic data at 298 K:
Substance ΔHf° (kJ/mol) S° (J/K⋅mol)
FeO (s) -271.9 60.75
Fe (s) 0 27.15
O2 (g) 0 205.0
Fe2O3 (s) -822.16 89.96
The value K for the reaction at 25 °C is ________.
8.1 *10^19
5.9 *10^4
3.8 ⋅*10^-14
370
7.1 *10^85
Answer:
3.8 ⋅*10^-14
Explanation:
The standard free energy change (ΔG°) for the reaction at 298 K can be calculated using the following equation:
ΔG° = ΣnΔGf°(products) - ΣnΔGf°(reactants)
where ΔGf° is the standard molar free energy of formation of the species and n is the stoichiometric coefficient.
ΔG° = [1×ΔGf°(Fe2O3)] - [1×ΔGf°(FeO) + 1×ΔGf°(Fe) + 1×ΔGf°(O2)]
ΔG° = [1×(-822.16 kJ/mol)] - [1×(-271.9 kJ/mol) + 1×(0 kJ/mol) + 1×(0 kJ/mol)]
ΔG° = -550.26 kJ/mol
The standard enthalpy change (ΔH°) and standard entropy change (ΔS°) can be used to calculate the standard free energy change (ΔG°) at any temperature using the following equation:
ΔG° = ΔH° - TΔS°
where T is the temperature in Kelvin.
ΔG° = ΔH° - TΔS° = (-550.26 kJ/mol) - (298 K)(-0.08996 kJ/K/mol) = -524.05 kJ/mol
Now, we can calculate the equilibrium constant (K) for the reaction at 298 K using the following equation:
ΔG° = -RTlnK
where R is the gas constant (8.314 J/K/mol) and T is the temperature in Kelvin.
-524.05 kJ/mol = -(8.314 J/K/mol)(298 K)lnK
lnK = -200.16
K = e^(-200.16) = 3.89×10^(-87)
Therefore, the value of K for the reaction at 25 °C is 3.89×10^(-87). Answer: 3.8 ⋅*10^-14.
Three gases (8.00 g of methane, CH4, 18.0 g of ethane, C2H6 , and an unknown amount of propane, C3H8 ) were added to the same 10.0- L container. At 23.0 ∘C, the total pressure in the container is 3.70 atm. Calculate the partial pressure of each gas in the container.
The partial pressure of each gas are:
Partial pressure of CH₄ is 1.22 atmPartial pressure of C₂H₆ is 1.46 atmPartial pressure of C₃H₈ is 1.02 atmHow do i determine the partial pressure of each gas?First, we shall determine the mole of 8.00 g of methane, CH₄ and 18.0 g of ethane, C₂H₆. Details below:
For methane, CH₄
Mass of CH₄ = 8 g Molar mass of CH₄ = 16 g/mol Mole of CH₄ =?Mole = mass / molar mass
Mole of CH₄ = 8 / 16
Mole of CH₄ = 0.5 mole
For ethane, C₂H₆
Mass of C₂H₆ = 18 g Molar mass of C₂H₆ = 30 g/mol Mole of C₂H₆ =?Mole = mass / molar mass
Mole of C₂H₆ = 18 / 30
Mole of C₂H₆ = 0.6 mole
Next, we shall determine the total mole. Details below:
Volume (V) = 750 mL = 10 LTemperature (T) = 23 °C = 23 + 273 = 296 KPressure (P) = 3.70Gas constant (R) = 0.0821 atm.L/mol KTotal of mole (n) =?PV = nRT
3.70 × 10 = n × 0.0821 × 293
Divide both sides by (0.0821 × 293)
n = (3.70 × 10) / (0.0821 × 293)
n = 1.52 mole
Finally, we shall determine the partial pressure of each gas. Details below:
For methane, CH₄
Mole of CH₄ = 0.5 moleTotal mole = 1.52 moleTotal pressure = 3.70 atmPartial pressure of CH₄ =?Partial pressure = (Mole / total mole) × total pressure
Partial pressure of CH₄ = (0.5 / 1.52) × 3.70
Partial pressure of CH₄ = 1.22 atm
For ethane, C₂H₆
Mole of C₂H₆ = 0.6 moleTotal mole = 1.52 moleTotal pressure = 3.70 atmPartial pressure of C₂H₆ =?Partial pressure = (Mole / total mole) × total pressure
Partial pressure of C₂H₆ = (0.6 / 1.52) × 3.70
Partial pressure of C₂H₆ = 1.46 atm
For propane, C₃H₈
Partial pressure of CH₄ = 1.22 atmPartial pressure of C₂H₆ = 1.46 atmTotal pressure = 3.70 atmPartial pressure of C₃H₈ =?Partial pressure of C₃H₈ = Total pressure - (Partial pressure of CH₄ + Partial pressure of C₂H₆)
Partial pressure of C₃H₈ = 3.7 - (1.22 + 1.46)
Partial pressure of C₃H₈ = 1.02 atm
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help with questions 1-5 pls??
In comparison to towns located inland, cities close to water features like lakes or oceans typically experience cooler summer temperatures.
Why is a city not so hot in summer when the city is close to water?Since water has a higher specific heat capacity than land, this is the case. The quantity of energy needed to raise a substance's temperature by a specific amount is known as its specific heat capacity. Compared to land, raising the temperature of water requires more energy because water has a higher specific heat capacity.
The summer sun warms both land and water, but due to land's lower specific heat capacity, land warms up more quickly than water. As a result, communities farther from water bodies tend to be hotter than cities closer to water.
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please help me pair pka values with displayed molecules
If we label the compounds ABCD from left to right;
A - 12.10
B - 15.90
C - 12.66
D - 12.35
What is the pKa?A molecule or compound's acidity is quantified by the pKa, which is the negative logarithm (base 10) of the acid dissociation constant (Ka). The lower the pKa value, the stronger the acid; it reflects a compound's propensity to give a proton (H+) in a solution.
The compound that has the highest number of attachment of the most electronegative elements would have the greatest pKa.
The justification of the answer above is that, seeing that the compound labelled B has three highly electronegative atoms hence it would have the most or the highest pKa of about 15.90 among the other compounds. The other compounds A, C and D have fewer electronegative atoms attached and thus a lower pKa as shown
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help!!!!!!!!!!!!!!!!!!!!!!!!
Answer:
two
Explanation:
the answer is two
option b
A piece of iron at 408 grams is heated in a flame and is then plunged into a beaker containing 1.00 kg of water. The original temperature of the water was 20.0°C, but it was 32.8°C after the iron bar is dropped in. What was the original temperature of the hot iron bar?
Note: The specific heat of iron is 0.45 J/g °C.
Do not round your answer in the middle of the problem. Round at the very end.
Round your answer to the correct number of sig figs. Your units should be degrees Celsius.
the original temperature of the h ot iron bar was 327.9°C.
We can use the specific heat of iron to do this:
Q1 = m1 * C1 * (Ti - 32.8°C)
Q1 = 408 g * 0.45 J/g °C * (Ti - 32.8°C)
Q1 = 183.6 J/g °C * (Ti - 32.8°C)
where m1 is the mass of the iron bar, C1 is the specific heat of iron, and Ti is the initial temperature of the iron bar.
Next, let's calculate the heat gained by the cold water when it is heated from 20.0°C to 32.8°C:
Q2 = m2 * C2 * (32.8°C - 20.0°C)
Q2 = 1000 g * 4.184 J/g °C * (32.8°C - 20.0°C)
Q2 = 52272 J
where m2 is the mass of the water, C2 is the specific heat of water.
Since the energy lost by the iron bar is gained by the water, we can set Q1 equal to Q2:
Q1 = Q2
183.6 J/g °C * (Ti - 32.8°C) = 52272 J
Now, let's solve for Ti:
183.6 J/g °C * Ti - 60236.8 J = 0
183.6 J/g °C * Ti = 60236.8 J
Ti = 327.9°C
Therefore, the original temperature of the h ot iron bar was 327.9°C.
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How many grams of air are in a 2.35 L balloon when its density is 1.4 g/L?
Answer:
3.29 grams
Explanation:
This is found by multiply 2.35 L by 1.4 g/L that is because the liters will cancel each other out leaving just grams. [tex]\frac{g}{L} * \frac{L}{1}[/tex]
3. In a lab, students mixed HCI acid with a Mg strip. The Mg started to bubble and dissolved within a few seconds. The rate at which the reaction occurs is determined by the A. number of effective collisions B. large AH C. the stabilization of the reactants D. mass of the products after the reaction
Answer:It might exposed
Explanation: or a spayed H2O might change because different water change over time
your answer to the following question on the information below and you knowledge of chemistry.
A 100. -gram sample of liquid water is heated from 30.0°C to 80.0°C. Enough KCIO:(s) is dissolved in the sample of water at 80.0°C to form a saturated solution.
Based on Table H, determine the vapor pressure of the water sample at its final temperature.
Explanation:
Table H lists vapor pressure data for pure water at various temperatures. We can use this data to estimate the vapor pressure of the water in the given system at its final temperature of 80.0°C.
First, we need to calculate the heat absorbed by the water sample during the heating process. We can use the specific heat capacity of water to do this:
q = m * c * ΔT
where q is the heat absorbed, m is the mass of water (100 g), c is the specific heat capacity of water, and ΔT is the temperature change (80°C - 30°C = 50°C).
Plugging in the values, we get:
q = 100 g * 4.18 J/(g*C) * 50 C
q = 20900 J
This tells us that 20,900 joules of energy were absorbed by the water sample during heating.
Next, we need to consider the saturated solution of KCIO3 in the water sample. At 80.0°C, the water is already close to boiling, so it is likely that the vapor pressure of the water in the system is close to the vapor pressure of pure water at this temperature. From Table H, we can see that the vapor pressure of pure water is approximately 356 mmHg at 80.0°C.
Therefore, the vapor pressure of the water in the given system at its final temperature of 80.0°C is approximately 356 mmHg.
A gas sample originally occupies 436 mL at 24 C. When the volume is expanded to 612 mL and the temperature is increased to 97 C, the pressure becomes 526 mm Hg. What was the original pressure?
Initially, there was a 266.8 mm Hg pressure.
solve this problem, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas sample. The formula is:
(P1 × V1) ÷ (T1) = (P2 × V2) ÷ (T2)
where P1 and P2 are the initial and final pressures, V1 and V2 are the initial and final volumes, and T1 and T2 are the initial and final temperatures.
We are given that:
- V1 = 436 mL
- V2 = 612 mL
- T1 = 24 C + 273.15 = 297.15 K (convert from Celsius to Kelvin)
- T2 = 97 C + 273.15 = 370.15 K
- P2 = 526 mm Hg
We want to find P1, the original pressure.
Plugging in the values, we get:
(P1 × 436 mL) ÷ (297.15 K) = (526 mm Hg × 612 mL) ÷ (370.15 K)
Solving for P1, we get:
P1 = (526 mm Hg × 612 mL × 297.15 K) ÷ (436 mL × 370.15 K) = 266.8 mm Hg
Therefore, the original pressure was 266.8 mm Hg.
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Gaseous butane (CH3(CH2)2CH3) will react with gaseous oxygen (02) to produce carbon dioxide (CO2) and gaseous water (H2O). Suppose 34.g of butane s mixed with 200. g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.
The maximum mass of water that can be produced by the reaction is 43.3 g, rounded to three significant figures.
Determining the maximum mass of water producedThe balanced chemical equation for the reaction between butane and oxygen is:
C4H10 + 13/2 O2 → 4 CO2 + 5 H2O
From the equation, we can see that 1 mole of butane reacts with 13/2 moles of oxygen to produce 5 moles of water.
moles of butane = 34. g / 58.12 g/mol = 0.585 mol
moles of oxygen = 200. g / 32.00 g/mol = 6.25 mol
Determining the limiting reactant.
butane : oxygen = 0.585 mol : 6.25 mol
= 0.0936 : 1.00
stoichiometric ratio = 1 : 13/2
= 0.7692 : 1.00
Since the actual ratio is lower than the stoichiometric ratio for oxygen, it is the limiting reactant.
The maximum amount of water that can be produced is determined by the amount of limiting reactant (oxygen).
moles of water = 5/13 * 6.25 mol
= 2.403 mol
Finally, we can convert the moles of water to grams:
mass of water = 2.403 mol * 18.015 g/mol
= 43.3 g
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One way to cool down your cup of coffee is to plunge an ice-cold piece of aluminum into it. Suppose you store an 18 g piece of aluminum in the refrigerator at 4.4°C, and drop it into your coffee. The coffee temperature drops from 90.0°C to 55.0°C. How much kJ of heat energy did the aluminum block absorb?
Note: The specific heat of aluminum is 0.89 J/g °C.
Do not round your answer in the middle of the problem, round at the very end.
Round your answer to the proper number of sig figs. Don't forget your units.
The aluminum block absorbed 0.875 kJ of heat energy when it was dropped into the coffee.
let's calculate the heat lost by the coffee when it is cooled from its initial temperature of 90.0°C to its final temperature of 55.0°C:
Q1 = m1 * C1 * (90.0°C - 55.0°C)
Q1 = 850 g * 4.184 J/g °C * (90.0°C - 55.0°C)
Q1 = 125660 J
where m1 is the mass of the coffee, C1 is the specific heat of water.
Next, let's calculate the heat gained by the aluminum block when it is heated from 4.4°C to the final temperature of the mixture, which is 55.0°C:
Q2 = m2 * C2 * (55.0°C - 4.4°C)
Q2 = 18 g * 0.89 J/g °C * (55.0°C - 4.4°C)
Q2 = 875.16 J
where m2 is the mass of the aluminum block, and C2 is the specific heat of aluminum.
Since the energy lost by the coffee is gained by the aluminum block, we can set Q1 equal to Q2:
Q1 = Q2
125660 J = 875.16 J + m2 * C2 * (55.0°C - 4.4°C)
Solving for m2, we get:
m2 = (125660 J - 875.16 J) / (0.89 J/g °C * (55.0°C - 4.4°C))
m2 = 152.2 g
Therefore, the mass of the aluminum block that was dropped into the coffee is 152.2 g. To calculate the heat energy absorbed by the aluminum block, we can use the heat gained by the aluminum block that we calculated earlier:
Q2 = 875.16 J
Converting this to kJ, we get:
Q2 = 0.875 kJ
Therefore, the aluminum block absorbed 0.875 kJ of heat energy when it was dropped into the coffee.
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Five types begging the question
Five types of begging the question include: Circular reasoning, Loaded question, False analogy, Suppressed evidence and Appeal to authority.
Begging the question is a logical fallacy that occurs when someone assumes the truth of a premise in their argument, without providing evidence or proof. There are several types of begging the question:
1. Circular reasoning: This occurs when someone uses their conclusion as one of their premises, essentially assuming what they are trying to prove.
Example: "God exists because the Bible says so, and the Bible is the word of God."
2. Loaded question: This occurs when someone asks a question that assumes a particular answer or perspective.
Example: "Have you stopped beating your spouse yet?" This question assumes that the person being asked was previously beating their spouse.
3. False analogy: This occurs when someone uses an analogy that is not relevant or applicable to the argument at hand.
Example: "Banning guns is like banning cars because both can be used to kill people." This analogy is false because cars have a primary function of transportation, whereas guns have a primary function of killing.
4. Suppressed evidence: This occurs when someone ignores or dismisses evidence that contradicts their argument.
Example: "I don't believe in climate change because it's cold outside today." This argument suppresses evidence that shows long-term trends of warming temperatures.
5. Appeal to authority: This occurs when someone uses an authority figure or expert as evidence, without providing any other support for their argument.
Example: "Dr. Smith says that this diet is the best for losing weight, so it must be true." This argument appeals to Dr. Smith's authority without providing any evidence or research to support the claim.
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How much aluminum can be produced from 9.00 ton of Al2O3?
To calculate the amount of aluminum produced from 9.00 tons of Al2O3, we need to use stoichiometry. First, we'll convert the mass of Al2O3 to moles, and then use the balanced chemical equation to find the moles of aluminum. Finally, we'll convert the moles of aluminum back to mass.
1. Convert mass of Al2O3 to moles:
9.00 tons = 9,000 kg
Molar mass of Al2O3 = (2 * 26.98) + (3 * 16.00) = 101.96 g/mol
9,000 kg * (1000 g/kg) = 9,000,000 g
moles of Al2O3 = 9,000,000 g / 101.96 g/mol = 88,258 moles
2. Use balanced chemical equation to find moles of aluminum:
The balanced chemical equation is:
2 Al2O3 → 4 Al + 3 O2
Using stoichiometry, we find the ratio of Al2O3 to Al is 2:4 or 1:2.
moles of Al = 88,258 moles Al2O3 * (2 moles Al / 1 mole Al2O3) = 176,516 moles
3. Convert moles of aluminum back to mass:
Molar mass of Al = 26.98 g/mol
Mass of Al = 176,516 moles * 26.98 g/mol = 4,762,984 g
Mass of Al in tons = 4,762,984 g / (1000 g/kg) / (1000 kg/ton) = 4.76 tons
So, 4.76 tons of aluminum can be produced from 9.00 tons of Al2O3.
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How many grams of oxygen would be produced by electrolysis of 83.7 grams of water?
H2O --> O2 + H2
The balanced chemical equation for the electrolysis of water is:
2H2O → 2H2 + O2
This equation shows that for every two moles of water that are electrolyzed, one mole of oxygen gas is produced. To solve this problem, we need to first convert the given mass of water (83.7 grams) to moles of water.
The molar mass of water (H2O) is:
2(1.008 g/mol H) + 15.999 g/mol O = 18.015 g/mol
So, 83.7 grams of water is equal to:
83.7 g / 18.015 g/mol = 4.646 mol H2O
Next, we need to determine how many moles of oxygen gas will be produced when 4.646 moles of water are electrolyzed. Since the mole ratio of water to oxygen is 2:1, we can use the following proportion:
2 mol H2O : 1 mol O2 = 4.646 mol H2O : x mol O2
Solving for x, we get:
x mol O2 = (1 mol O2 / 2 mol H2O) * 4.646 mol H2O = 2.323 mol O2
Finally, we can convert the moles of oxygen gas produced to grams using the molar mass of oxygen:
2.323 mol O2 * 32.00 g/mol O2 = 74.3 g O2
Therefore, 83.7 grams of water will produce 74.3 grams of oxygen gas by electrolysis.
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2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2(g)
Moles of NA = Given Mass (g) ÷ Molecular Mass (g/mol)
= 27.5 ÷ 22.9897
= 1.196 mol
Moles of H2 Produced = Mol of NA × 1 mol H2 ÷ 2 Mol NA
= 1.196 × 1 ÷ 2
= 0.60 mol
Number of Molecules = Moles × Avogadro's Number
= 0.60 × 6.023 × 10²³ mol - 1
= 3.61 × 10²³
The Number of Molecules of Hydrogen Gas Produced When Added To Water Is 3.61 × 10²³
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The compounds labeled benzophenone-3 (C14H12O3) and benzophenone-5 (C14H11NaO6S) are found in certain sunscreens. Would you expect a sunscreen made with benzophenone-3 or benzophenone-5 be more waterproof? Explain your choice.
A sunscreen made with Benzophenone-5 ([tex]C_1_4H_1_1NaO_6S[/tex]) would be expected to be more waterproof than benzophenone-3 ([tex]C_1_4H_1_2O_3[/tex]).
This is due to the presence of a sodium salt group (Na) and a sulfonic acid group ([tex]SO_3H[/tex] ) in benzophenone-5, which makes it more polar than benzophenone-3. Polar molecules interact more strongly with water molecules and are less likely to dissolve in nonpolar solvents such as oils.
Because sunscreen is designed to be water-resistant, the more polar benzophenone-5 should have stronger interactions with water and give more water resistance than benzophenone-3.
Moreover, the sulfonic acid group in benzophenone-5 may allow it to make stronger hydrogen bonds with water, increasing its water resistance even further.
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