1.03 g of butane will yield 1.59 g of water when it reacts with too much oxygen.
What is mass?The amount of matter in an item is measured by the fundamental physical quantity known as mass. It is a scalar quantity that is measured in grams (g) or kilograms (kg) (g). No matter where it is or what force is pushing against it, an object's mass always remains constant.
How do you determine it?For butane and oxygen to burn, the chemical equation is balanced as follows:
2C4H10+13 O2→ 8 CO2+ 10 H2O
According to the equation, 1 mole of butane (C4H10) and 13/2 moles of oxygen (O2) combine to form 5 moles of water (H2O).
To begin with, we must count the butane moles that are present:
Mass of butane divided by its molar mass yields moles of butane.
1.03 g/58.12 g/mol is the formula for butane.
A mole of butane weighs 0.0177 mol.
Then, we can calculate the quantity of water created using the butane-to-water mole ratio:
Moles of water = [tex]\frac{5 mol H2O}{(1 mol C4H10 ×0.0177 mol C4H10)}[/tex] = 0.0885 mol.
Eventually, we can determine how much water was generated:
moles of water equal 0.0885 mol when 5 mol H2O is divided by 1 mol C4H10 and multiplied by 0.0177 mol C4H10.
Lastly, we can determine the mass of created water:
Water's mass is equal to its moles times its molar mass= 0.0885 mol × 18.02 g/mol = 1.59 g.
As a result, 1.03 g of butane will yield 1.59 g of water when it reacts with to moles of water equal 0.0885 mol when 5 mol H2O is divided by 1 mol C4H10 and multiplied by 0.0177 mol C4H10.
Lastly, we can determine the mass of created water:
Water's mass is equal to its moles times its molar mass = 0.0885 mol × 18.02 g/mol = 1.59 g.
As a result, 1.03 g of butane will yield 1.59 g of water when it reacts with too much oxygen.
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Locating the epicenter of an earthquake lab
Eruption triangulation is a method for locating an earthquake's epicenter.
What is epicenter?The area right above the spot in the Earth's crust where an earthquake originates or begins is known as the epicenter. It marks the spot on the Earth's surface where the earthquake's seismic waves first touchdown.
The distance between the earthquake's epicenter and a number of seismograph stations is calculated using seismograms, which are records of the ground motion brought on by an earthquake.
The following are the steps to find an earthquake's epicenter:
A minimum of three separate seismograph stations should have data collected. The epicenter's distance from each station is determined by keeping track of the time the earthquake waves arrived at each station.
Map out the positions of each seismograph station.
Draw circles with an equal radius around each seismograph station using the distance information. Each circle's radius is equal to how far away the station is from the epicenter.
At the place where the circles converge, there is an epicenter.
It is crucial to keep in mind that determining the epicenter of an earthquake is not an exact science, and the precision of the position will vary depending on a number of variables, such as the caliber of the seismograms and the distance between the earthquake and the seismograph stations.
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5.0g of hydrochloric acid is mixed with 24.0g of magnesium hydroxide.
what mass of water is produced?
From the balanced chemical equation we can see tha the mass of water produced is 2.47 grams.
What mass of water is produced?The balanced chemical equation for the reaction between hydrochloric acid (HCl) and magnesium hydroxide (Mg(OH)2) is:
2 HCl + Mg(OH)2 → MgCl2 + 2 H2O
From the equation, we can see that 2 moles of hydrochloric acid react with 1 mole of magnesium hydroxide to produce 2 moles of water. To calculate the mass of water produced, we need to determine the limiting reactant, which is the reactant that is completely consumed first and determines the amount of product formed.
First, let's calculate the number of moles of each reactant:
Mass of HCl = 5.0 g
Molar mass of HCl = 36.46 g/mol
Number of moles of HCl = 5.0 g / 36.46 g/mol = 0.137 mol
Mass of Mg(OH)2 = 24.0 g
Molar mass of Mg(OH)2 = 58.33 g/mol
Number of moles of Mg(OH)2 = 24.0 g / 58.33 g/mol = 0.411 mol
According to the balanced chemical equation, 2 moles of HCl react with 1 mole of Mg(OH)2 to produce 2 moles of water. Therefore, the stoichiometry of the reaction requires 2 moles of HCl for every 1 mole of Mg(OH)2. Since we have only 0.137 mol of HCl and 0.411 mol of Mg(OH)2, the HCl is the limiting reactant because it is completely consumed first.
The molar ratio of HCl to water is 2:2, which simplifies to 1:1. Therefore, the number of moles of water produced is also 0.137 mol.
Now, let's calculate the mass of water produced using the molar mass of water:
Molar mass of water (H2O) = 18.02 g/mol
Mass of water produced = 0.137 mol x 18.02 g/mol = 2.47 g
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Calculate the cell potential (Ecell) at 25oC (298 K) for the following reaction if the Cu2+ ion concentration is 0.064 M and the Fe2+ ion concentration is 0.645 M.
Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s)
Half-reaction Standard Reduction Potential (V)
Fe2+(aq) + 2e−→ Fe(s) −0.440
Cu2+(aq) + 2e−→ Cu(s) +0.337
R = 8.31 V/mol·K
F = 96500 C/mol
The cell potential (Ecell) at 25°C (298 K) for the given reaction is 1.065 V which means that the reaction is spontaneous because the calculated Ecell is positive
The cell potential (Ecell) for the given reaction can be calculated using the Nernst equation:
Ecell = E°cell - (RT ÷ nF) × ln(Q)
where E°cell = standard cell potential, R = gas constant, T = temperature in Kelvin, n = number of electrons transferred in the reaction, F = Faraday constant, and Q is the reaction quotient.
Since two electrons are transported in the half-reactions, n in this instance equals 2. With the help of the species concentrations, it is possible to determine the reaction quotient Q:
Q = [Fe2+] ÷ [Cu2+]
Q = 0.645 ÷ 0.064
Q = 10.078
Now, we can calculate the Ecell:
Ecell = E°cell - (RT ÷ nF) × ln(Q)
Ecell = (0.337 - (-0.440)) - (8.31 × 298 ÷ (2 × 96500)) × ln(10.078)
Ecell = 1.065 V
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26. The normal boiling point of argon is 21.9K and its latent heat of vaporization is 1.57kJ/mol. Calculate it's boiling point at 1.2 atm.
Using the Clausius-Clapeyron equation, we can see that the new boling point is 34.6 K.
How to find the new boiling point?To calculate the boiling point of argon at 1.2 atm, we can use the Clausius-Clapeyron equation, which relates the boiling point of a substance at one pressure to its boiling point at another pressure, along with the latent heat of vaporization.
The Clausius-Clapeyron equation is given as:
ln(P1/P2) = ΔHvap/R * (1/T1 - 1/T2)
where:
P1 and P2 are the initial and final pressures, respectively,ΔHvap is the latent heat of vaporization,R is the ideal gas constant (8.314 J/(mol*K)),T1 and T2 are the initial and final temperatures in Kelvin, respectively.Given:
P1 = 1 atm (normal pressure)P2 = 1.2 atm (given pressure)T1 = 21.9 K (normal boiling point of argon)ΔHvap = 1.57 kJ/mol = 1.57 * 10^3 J/mol (latent heat of vaporization)We can rearrange the Clausius-Clapeyron equation to solve for T2 (the boiling point at 1.2 atm):
ln(P1/P2) = ΔHvap/R * (1/T1 - 1/T2)
Rearranging further:
1/T2 = (ln(P1/P2) * R) / ΔHvap + 1/T1
Plugging in the given values:
1/T2 = (ln(1 atm / 1.2 atm) * (8.314 J/(mol*K))) / (1.57 * 10^3 J/mol) + 1/21.9 K
Simplifying:
1/T2 = -0.0624 + 0.0456 + 0.0457
1/T2 = 0.0289
T2 = 1 / 0.0289
T2 = 34.6 K
That is the new one.
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Provide feedback to me if this assisted in giving you a better understanding of the history of chemistry, and what could be done differently (three paragraph maximum).
Lavoisier has been considered by many scholars to be the father of chemistry. Chemists continued to discover new compounds in the 1800s. The science also began to develop a more theoretical foundation. It was in 1807, John Dalton put forth his atomic theory.
It was not until the era of the ancient Greeks that we have any record of how people explained the chemical changes they observed and used. At that time natural objects were thought to consist of only four basic elements like earth, air, fire and water.
It was in the fourth century BC, two Greek philosophers Democritus and Leucippus suggested that matter was not infinitely divisible into smaller particles but instead consists of the fundamental particles called the atoms. Chemistry took its present form in the 18th century.
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An atom has 21 electrons, 17 protons and 18 neutrons. What is the total charge of the atom?
Explanation:
Only protons and electrons contribute to the charge of an atom ...NEUTRons are NEUTRal...
Protons are + charge
Electrons are - charge
21 negative charges added to 17 positive charges results in
- 4 charge
100 mL of 3 M HCl can be neutralized with exactly 200 mL of 1.5 M NaOH. If this reaction is done in a coffee cup calorimeter, what temperature change would you expect to observe?
100 mL of 3 M HCl can be neutralized with exactly 200 mL of 1.5 M NaOH. If this reaction is done in a coffee cup calorimeter, temperature would increase.
A chemical reaction known as neutralisation occurs when an acid and a base quantitatively react with one another. Alternate spellings include Neutralisation. The pH in the neutralised solution is determined by the acidity of the reactant.
Have you ever overindulged in spicy food and felt your stomach start to burn? This results from the stomach producing acid. The use of an antacid, which counteracts the effects of acid, can solve this issue; this process is known as a neutralisation response. 100 mL of 3 M HCl can be neutralized with exactly 200 mL of 1.5 M NaOH. If this reaction is done in a coffee cup calorimeter, temperature would increase.
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Using this balanced equation:
NaHCO3 + CH3COOH H2O + CO2 + NaC2H3O2
In an experiment , the following mass measured 3.0 grams, 5.5 grams and 7.0 grams of
sodium bicarbonate is mixed with 0.5 mol of acetic acid (vinegar) to form carbon dioxide as
a product formed
The stoichiometric ratio of NaHCO₃ to CH₃COOH, which is determined by the balancing equation, is 1:1, meaning that 1 mole of NaHCO₃ reacts with 1 mole of CH₃COOH.
What is the balanced equation between vinegar's acetic acid CH₃COOH and sodium bicarbonate NaHCO₃?Acetic acid (CH₃COOH), a component of vinegar, interacts with sodium bicarbonate (NaHCO₃) in baking soda to produce water, carbon dioxide gas, and sodium acetate.
When NaHCO₃ and HC₂H₃O₂ are joined in a closed system, what do you suppose would happen?CO₂(g) + H₂O(l) + NaC₂H₃O₂ = NaHCO₃(aq) + HC₂H₃O₂(aq) when yellow bubbles appear, an acidic gas H₂CO₃ has generated as a result of the reaction between CO₂ and water. The salt in the solution (NaC₂H₃O₂) is basic since the solution is red in colour.
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A sample of air from a factory smokestack measured at 35 °C contained SO₃ at a partial pressure of 8.75 torr. What mass, in g, of SO₃ is in 1.00 L of the air sample?
The mass, in grams, of SO₃ in 1.00 L of the air sample is 27.48 g.
To determine the mass of SO₃ in 1.00 L of an air sample, we need to use the ideal gas law:
PV = nRT
where P is the partial pressure of SO₃, V is the volume of the sample (1.00 L), n is the number of moles of SO₃, R is the gas constant, and T is the temperature in Kelvin (308 K)
n = PV/RT
n = (8.75 torr) × (1.00 L) ÷ (0.0821 L·atm/mol·K) × (308 K)
n = 0.343 mol
To convert this to mass, we need to use the molar mass of SO₃, which is 80.06 g/mol. Therefore, the mass of SO₃ in 1.00 L of the air sample is:
mass = n × molar mass
mass = 0.343 mol × 80.06 g/mol
mass = 27.48 g
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Suppose that a certain quantity of methane occupies a volume of 0.138 L under a pressure of 300 atm at 200 °C, and the volume required at 600 atm at 0 °C. For 300 atm and at 200 °C, Z=1.067, while for 600 atm at 0 °C, Z=1.367.
Answer:
Therefore, the volume required at 600 atm and 0 °C is 0.319 L.
Explanation:
We can use the ideal gas law to solve for the number of moles of methane present, assuming ideal gas behavior at both conditions:
PV = nRT
At 300 atm and 200 °C:
n = PV/RT = (300 atm * 0.138 L) / [(0.08206 L atm mol^-1 K^-1) * (200 + 273.15) K * 1.067]
n = 2.451 mol
At 600 atm and 0 °C:
n = PV/RT = (600 atm * V2) / [(0.08206 L atm mol^-1 K^-1) * (273.15 K) * 1.367]
n = 7.682 V2
Since the number of moles of methane must be the same at both conditions:
2.451 mol = 7.682 V2
Solving for V2:
V2 = 0.319 L
Therefore, the volume required at 600 atm and 0 °C is 0.319 L.
What is the molarity of a solution containing 65 grams of KCl if there is
8.3 L of solution?
Answer: The molarity of the solution is 0.11 M.
Explanation: You first need to convert 65g of KCl to moles of KCl. When you do that you will get 0.871898055 moles of KCl, round anykind of answers you get at the end.
The second step is to use the molarity formula to find the molarity of the solution. Molarity = moles/liters
Molarity = 0.871898055 moles of KCl / 8.3 Liters of solution
M = 0.11
Make sure that your answer always has the correct number of significant figures. In the question both numbers given to you have 2 sig figs, therefore your final answer also needs to have 2 sig figs.
Convert 231 μm3 to cm3 .
To convert micrometers cubed (μm^3) to cubic centimeters (cm^3), we need to divide by 10^12 (since there are 10^12 μm^3 in 1 cm^3).
So, to convert 231 μm^3 to cm^3, we can use the following formula:
231 μm^3 ÷ (10^12 μm^3/cm^3) = 2.31 x 10^-10 cm^3
Therefore, 231 μm^3 is equal to 2.31 x 10^-10 cm^3.
:)
To convert micrometers cubed (μm^3) to cubic centimeters (cm^3), we need to divide by 10^12 (since there are 10^12 μm^3 in 1 cm^3).
So, to convert 231 μm^3 to cm^3, we can use the following formula:
231 μm^3 ÷ (10^12 μm^3/cm^3) = 2.31 x 10^-10 cm^3
Therefore, 231 μm^3 is equal to 2.31 x 10^-10 cm^3.
:)
what is the force that exists between atoms that are in different molecules
Intermolecular force connects atoms in different molecules. Intermolecular forces determine physical qualities including boiling temperature, melting point, viscosity, and surface tension.
London dispersion, dipole-dipole, and hydrogen bonding are intermolecular forces. London dispersion forces, caused by electron movement-induced dipoles, are the smallest intermolecular interactions between all atoms and molecules.
Polar molecules have greater dipole-dipole interactions than London dispersion forces. Hydrogen bonding is a specific dipole-dipole interaction between a hydrogen atom bound to an electronegative atom (N, O, or F) in one molecule and an electronegative atom in another molecule.
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What would you expect the effect of pH to have on the surface charge of aqueous solutions of gold nanocrystals capped with alkanethiols bearing either terminal carboxylic acid or amine groups? What would you expect to happen on mixing these two kinds of nanocrystals at different pH?
Depending on the pH of the solution, the surface charge of each nanocrystal, and the mixing of these two types of nanocrystals, complicated interactions may occur.
What feature of nanoparticles is most significant?Friction is the most significant characteristic of nano metals. One of the many characteristics that make nanomaterials special is their small size. Nanomaterials can have a size up to a thousand times smaller than a human hair. The ratio of a nanoparticle's surface area to volume is very high.
What occurs when different pH levels of gold nanocrystals with carboxylic acid- and amine-capped surfaces are combined?Because of their distinct surface charges, these two different kinds of nanocrystals may exhibit electrostatic attraction or repulsion when combined in different pH environments. Positively charged nanocrystals with carboxylic acid caps may draw negatively charged ones with amine caps when the pH is low. The carboxylic acid-capped nanocrystals have a greater negatively charged surface charge when the pH rises, which may cause them to reject one another. The amine-capped nanocrystals, on the other hand, have a higher positive change in surface charge, which could cause them to attract.
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What is the cell potential for the reaction
Mg(s)+Fe^2+(aq)→Mg^2+(aq)+Fe(s)
at 89 degrees celsius when [Fe2+] = 2.80 M
and [Mg2+] = 0.110 M
solve and show steps. I have tried 2.86 and 6.2653 and both were not correct
show answer in 3 sig figs
The cell potential for the reaction Mg(s)+Fe2+(aq)Mg2+(aq)+Fe(s) is 2.87 V at 89 degrees Celsius.
The Nernst equation can be used to compute this: Where E° cell is the standard cell potential, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transported, F is the Faraday constant, and Q is the reaction quotient, E cell is defined as E° cell - (RT/nF)lnQ.
E° cell = -2.37 V, R = 8.314 J/K-mol, T = 362 K, n = 2, F = 96485 C/mol, and Q = [Mg2+]/[Fe2+] = 0.110/2.80 = 0.039 in this instance. The Nernst equation using these values as input produces the result E cell = -2.37 - (8.314*362/2*96485)ln(0.039) = 2.87 V.
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How many grams of PbBr2 will precipitate when excess FeBr2 solution is added to 58.9 mL of 0.505 M Pb(NO3)2 solution?
when a sufficient [tex]FeBr_{2[/tex] solution is introduced to 58.9 mL of 0.505 M [tex]Pb(NO_3)_2[/tex] solution, 5.45 grammes of [tex]PbBr_2[/tex] will precipitate.
Calculation-We can figure out the amount of lead(II) bromide (PbBr2) that will precipitate by using the stoichiometry of the reaction between lead(II) nitrate [tex]Pb(NO_3)_2[/tex] and iron(II) bromide [tex](FeBr_2).[/tex]
The reaction's chemically balanced equation is as follows:
[tex]PbBr_2(s) + 2 Fe(NO_2)2 = Pb(NO_2)2(aq) + 2 FeBr_2 (aq)(aq)[/tex]
According to the equation, two moles [tex]FeBr_2[/tex] react with one mole of [tex]Pb(NO_3)_2[/tex] to create one mole [tex]FeBr_2[/tex]. The amount [tex]PbBr_2[/tex] that will precipitate will therefore be equal to half of the amount of [tex]Pb(NO_3)_2[/tex] that is present in the solution.
We can use the following formula to get the amount of[tex]Pb(NO_3)_2[/tex] in moles:
moles = volume x concentration
where volume is given as 58.9 mL, which we will divide by 1000 to get litres, and concentration is given as 0.505 M:
[tex]moles of Pb(NO3)2 = 0.505 M x 0.0589 L = 0.0297 moles[/tex]
Therefore, the amount of PbBr2 that will precipitate is:
[tex]PbBr2 moles =0.0297 moles / 2 moles, or 0.01485 moles.[/tex]
Finally, using the molar mass of [tex]PbBr_2[/tex], which is 367.01 g/mol, we can convert the number of moles [tex]PbBr_2[/tex] to grammes:
[tex]mass of PbBr2 = 0.01485 moles x 367.01 g/mol = 5.45 g[/tex]
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state and explainvtwo conditions under which the intesity of the brown colour of the equlibrium mixture can be increased
a)condition1
b)condition2
What materials formed the solar system?
Answer:
The solar system formed from a cloud of gas and dust, known as the solar nebula, that collapsed under its own gravity about 4.6 billion years ago. As the solar nebula collapsed, it formed a spinning disk of gas and dust, with most of the material concentrated at the center. Over time, the material in the disk began to clump together due to gravity, forming small rocky and icy planetesimals, which collided and merged to form larger bodies. These larger bodies eventually became the planets, moons, asteroids, and other objects that make up the solar system today. The exact composition of the solar nebula and the materials that formed the solar system varied depending on the distance from the Sun and the temperature and pressure conditions in different regions of the nebula. Generally, the inner solar system (including the Sun, Mercury, Venus, Earth, and Mars) is composed mostly of rocky materials, while the outer solar system (including the gas giants Jupiter, Saturn, Uranus, and Neptune, as well as the dwarf planet Pluto) is composed mostly of ices, such as water, ammonia, and methane, as well as gases like hydrogen and helium.
short term answer: dust and gas.
Explanation: learned it in 5th grade science class, bye have a great day !
What is molecular weight of a substance given that 1.22g of the sample was vaporised in 100ml flask at 45°C and 687mmHg.
To calculate the molecular weight of a substance from its vapor density, we can use the following formula:
Molecular weight = (RT/P) x d
where:
R is the gas constant (0.0821 L atm/mol K)
T is the temperature in Kelvin
P is the pressure in atm
d is the vapor density (in g/L) of the substance
First, we need to calculate the vapor density of the substance using the given information. We can use the ideal gas law to find the number of moles of the substance in the flask:
PV = nRT
where:
P is the pressure (687 mmHg = 0.903 atm)
V is the volume (100 mL = 0.1 L)
n is the number of moles of gas
R is the gas constant (0.0821 L atm/mol K)
T is the temperature in Kelvin (45°C = 318 K)
Solving for n, we get:
n = PV/RT = (0.903 atm)(0.1 L)/(0.0821 L atm/mol K)(318 K) = 0.00372 mol
Next, we can use the mass and volume information to find the density of the substance:
density = mass/volume = 1.22 g/0.1 L = 12.2 g/L
Since the vapor density is half of the density of the substance in the liquid state, we can calculate the vapor density:
vapor density = density/2 = 6.1 g/L
Finally, we can use the formula above to find the molecular weight:
Molecular weight = (RT/P) x d
Molecular weight = (0.0821 L atm/mol K)(318 K)/(0.903 atm) x 6.1 g/L
Molecular weight = 92.2 g/mol
Therefore, the molecular weight of the substance is 92.2 g/mol.
Calculate the cell potential for the following reaction that takes place in an electrochemical cell at 25°C.
Cu(s) | Cu2+(aq, 0.0032 M) || Cu2+(aq, 4.48 M) | Cu(s)
Calculate the cell potential for the following reaction that takes place in an electrochemical cell at 25°C.
Cu(s) | Cu2+(aq, 0.0032 M) || Cu2+(aq, 4.48 M) | Cu(s)
+0.186 V
0.00 V
+0.093 V
+0.34 V
+0.052 V
Okay, let's solve this step-by-step:
1) The standard reduction potentials for Cu2+/Cu are: E°Cu2+/Cu = +0.34 V.
This is the reduction potential when [Cu2+] = 1 M and [Cu] = 1 M.
2) The actual reduction potential (Ered) depends on the concentrations of oxidized and reduced species.
Here,
[Cu2+] = 4.48 M on the right side.
[Cu2+] = 0.0032 M on the left side.
3) Ered = E° + 0.0591 log([oxidized]/[reduced]) (Nernst equation)
So for the right side:
Ered = +0.34 + 0.0591 log(4.48/1) = +0.34 + 0.186 = +0.526 V
And for the left side:
Ered = +0.34 + 0.0591 log(0.0032/1) = +0.34 - 0.093 = +0.247 V
4) The cell potential (Ecell) is the difference between the two half-cell potentials:
Ecell = +0.526 - 0.247 = +0.279 V
So the cell potential for the given reaction at 25°C is +0.279 V.
Let me know if you have any other questions!
What would the expected temperature change be (in F) if 0.5 gram sample of water released 0.0501 j of heat energy? The specific heat of liquid water 4.184 j/g
The expected temperature change would be approximately 4.32°F if 0.5-gram sample of water released 0.0501 j of heat energy.
The specific heat of a substance is the amount of heat energy required to raise the temperature of one gram of the substance by one degree Celsius. The expected temperature change can be calculated using the formula:
q = m × c × ΔT
where q is the heat energy released, m is the mass of water, c is the specific heat of water, and ΔT is the change in temperature.
Rearranging the formula to solve for ΔT, we get:
ΔT = q ÷ (m × c)
Substituting the given values, we get:
ΔT = 0.0501 J ÷ (0.5 g × 4.184 J/g°C)
ΔT ≈ 2.4°C or 4.32°F
When a substance undergoes a change in temperature, it either gains or loses heat energy.
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2) The lowest atmospheric pressure at sea level in the Western Hemisphere was recorded in 2015 during
hurricane Patricia: a pressure of 656 torr. Show unit cancelation in work.
a) What is this pressure in kilopascals?
To convert torr to kilopascals, we can use the following conversion factors:
1 torr = 1/760 atm
1 atm = 101.325 kPa
So, we can write: 656 torr × (1/760 atm) × (101.325 kPa/atm) = 87.0 kPa
Therefore, the atmospheric pressure during hurricane Patricia was 87.0 kPa.
Certainly!
In this exercise, converting a pressure value from torr to kilopascals (kPa) is required. Torr is a common unit of pressure used in physics and chemistry, while kPa is a common unit of pressure used in engineering and other disciplines.
We may utilise a conversion factor that connects torr to kPa to carry out the conversion. We must employ the following conversion factor:
0.133322 kPa per torr
As a result, 0.133322 kPa is equal to 1 torr. A pressure value in torr may be converted to a pressure value in kPa by multiplying the torr value by the conversion factor. This enables us to eliminate the torr units and obtain a value in kPa.
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A balloon has a volume of 145 mL at room temperature (25°C = 298°K). Alyssa decides to place the balloon in the freezer to see what happens. After being in the freezer for an hour, the balloon has a new volume of 35mL. What is the temperature inside the freezer?
The temperature inside the freezer is approximately -164°C.
To solve this problem, we can use the combined gas law equation:
[tex](P1V1)/T1 = (P2V2)/T2[/tex]
where P is the pressure, V is the volume, and T is the temperature of the gas.
We know that the initial volume of the balloon is 145 mL and the final volume is 35 mL. We also know that the initial temperature is 25°C or 298 K, and we need to find the final temperature.
Assuming the pressure of the gas remains constant, we can rearrange the combined gas law equation to solve for the final temperature:
[tex]T2 = (P2V2/T1)(P1/V1)[/tex]
Plugging in the values we know, we get:
[tex]T2 = (1 atm * 35 mL/298 K)(1 atm/145 mL) = 0.0808 atm/mL[/tex]
Multiplying both sides by 298 K and dividing by 0.0808 atm/mL, we get:
T2 = 109.15 K or approximately -164°C
Therefore, the temperature inside the freezer is approximately -164°C.
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[tex]a. \: Ca + Cl _{2} → CaCl _{2} \\
\\ b. \: Cl _{2}+H _{2} O+ NaOH → \\ NaCl+ H _{2}O \\ \\
c. \: \: \: H _{2} SO _{4} +CaCO _{3} → \\ CaSO _{4} +H _{2}O+CO _{2} \\ \\
d. \: \: Fe+Cu(NO _{3}) _{2} → \\ Fe(NO _{3} ) _{2}+Cu[/tex]
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Among the elements of the main group, the first ionization energy increases
from left to right across a period.
from right to left across a period.
when the atomic radius increases.
down a group.
The first ionisation energy increases over time from left to right among the major group of elements. answer is option (a).
What is Ioniztion?When an element loses its valence electron, its oxidation number increases (a process known as oxidation), and this energy loss is known as ionisation (Ei).
Earth alkaline metals, which are located immediately next to alkaline metals, have higher ionisation energies than alkaline metals because they have two valence electrons, while alkaline metals, which are located far left in the main group, have the lowest ionisation energies and are easiest to remove.
Because they contain a large number of valence electrons, nonmetals are far to the right in the main group and have the highest ionisation energy.
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The complete question is,
Among the elements of the main group, the first ionization energy increases
a. from left to right across a period.
b. from right to left across a period.
c. when the atomic radius increases.
d. down a group.
1) The last state of matter we will study is gases. Gas quantities are measured using four common
variables:
a) P for ____which is often measured in units of ____, ____, ____, and ____
b) V for___ which is often measured in units of___,___,___, and ___
c) T for _____ _____ which must be in units ____.
d) n for ____, which is often found by converting from grams of a gas.
1) The last state of matter we will study is gases.
a) P for Pressure which is often measured in units of atmospheres (atm), milliamps (mm Hg), pounds per square inch (psi), pascals (Pa)
b) V for Volume which is often measured in units of liters (L), cubic meters (m3), cubic feet (ft3), cubic inches (in3)
c) T for Temperature which must be in units Kelvin (K) or Celsius (°C).
d) n for Number of moles , which is often found by converting from grams of a gas.
determine the mass in grams of 0.75 moles of cr2(so4)3
Answer: The mass in grams is 290g.
Explanation: In order to find the mass you need to multiply the number of moles given by the molar mass of the compound.
You find the molar mass by adding all the atomic masses of each atom in the compound.
Since 0.75 has 2 significant figures, your final answer also needs to have 2 sig figs.
find [N₂ O₄] given K = 0.028, [NO₂] = 0.0.042 M N₂O₄ ← →2 NO₂
The reaction between N2O4 and is at equilibrium at this time.2 Given by K = 0.028 is NO2. The amount of NO2 in the sample is 0.042 M.
K = [NO2] is the equilibrium expression for the reaction.²/[N₂O₄] When K and [NO2] are substituted in the equilibrium expression, we obtain 0.028 = (0.042).²/[N₂O₄]
We get at [N2O4] = 0.028/0.0422 = 0.0075 M after solving for [N2O4].
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The cell diagram for the lead-acid cell that is used in automobile and truck batteries is
Pb(s)∣∣PbSO4(s)∣∣H2SO4(aq)∣∣PbO2(s),PbSO4(s)∣∣Pb(s)
The comma between PbO2(s) and PbSO4(s) denotes a heterogeneous mixture of the two solids. The right-hand lead electrode is nonreactive.
Write the balanced equation for the net cell reaction.
Look up standard potentials for the oxidation and reduction half-reactions, and then calculate the value of ∘cell
Calculate the value of Δ∘rxn
Calculate the value of cell at 25 ∘C if [H2SO4]=10.0 M
How many lead-acid cells are in a 12 Vcar battery? Round to the nearest integer.
number of lead-acid cells:
The net cell reaction is described by the equation: Pb(s) + PbO2(s) + 4H2SO4(aq) 2PbSO4(s) + 2H2O(l). 2.14 V and -0.36 V, respectively, are the standard potentials for the oxidation and reduction half-reactions. The result is that cell = 2.50 V.
The computed value of rxn is 2.50 V, which is the difference between the two standard potentials. The computed cell potential with [H2SO4] = 10.0 M at 25 C is 2.50 V. Six lead-acid cells connected in series make up a 12 V automobile battery.
Six cells are required to obtain 12 V because each cell has a voltage of 2.0 V. Consequently, there are six lead-acid cells in a 12 V automobile battery.
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Can you help me solve number 4?
The pressure of the gas in the flask, given that the temperature is reduced to -15 is 0.38 mmHg
How do i determine the pressure of the gas?From the question given above, the following data were obtained:
Initial pressure (P₁) = 338 mmHgInitial temperature (T₁) = 72 °C = 72 + 273 = 345 KInitial volume (V₁) = 0.225 LNew volume (V₂) = 150 LNew temperature (T₂) = -15 °C = -15 + 273 = 258 KNew pressure (P₂) = ?The new pressure of the gas can be obtained by using the combined gas equation as shown below:
P₁V₁ / T₁ = P₂V₂ / T₂
(0.225 × 338) / 345 = (P₂ × 150) / 258
Cross multiply
345 × 150 × P₂ = 0.225 × 338 × 258
Divide both sides by (345 × 150)
P₂ = (0.225 × 338 × 258) / (345 × 150)
P₂ = 0.38 mmHg
Thus, we can conclude the pressure of the gas is 0.38 mmHg
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