Fifteen kg of iron (lll) oxide was used in a reaction to produce iron.calculate the mass of iron produced in this reaction
10.50 kg of iron will be produced from 15 kg of iron (III) oxide.
What is mass?The amount of matter in an item is measured by its mass, which is a fundamental physical quantity. It is a scalar amount that is measured in kilograms (kg) or grams (g). Regardless of an object's location or the force pressing against it, its mass always remains constant.
How do you determine it?Iron (III) oxide and elemental iron react chemically in the following balanced chemical equation:
2 Fe2O3+ 3 C = 4 Fe + 3 CO2
Due to the reaction between 2 moles of Fe2O3 and 4 moles of Fe, the mole ratio of Fe2O3 to Fe is either 2:4 or 1:2.
The amount of iron created from 15 kg of Fe2O3 can be calculated using this mole ratio:
Fe2O3 = 2 moles of Fe per mole.
Molecular weight of Fe2O3 is 159.69 g/mol.
Fe2O3 has a mass of 15 kg and a density of 15,000 g/mol, or 94.00 moles.
We can figure out how many moles of Fe were produced using the mole ratio of 1:2:
We can figure out how many moles of Fe were produced using the mole ratio of 1:2:
2 moles of Fe for each mole of Fe2O3 94.00 moles of Fe2O3 multiplied by (2 moles of Fe for each mole of Fe2O3) results in 188.00 moles of Fe.
The molar mass of Fe can then be used to convert the moles of iron to mass as follows:
Fe's molecular weight is 55.85 g/mol.
188.00 moles of Fe produced at a rate of 55.85 g/mol result in a mass of 10,499.80 g or 10.50 kg.
Hence, 10.50 kg of iron will be produced from 15 kg of iron (III) oxide.
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CAN SOMEONE HELP WITH THE HESS'S LAW?
WHAT DO YOU THINK IS THE LARGEST SOURCE OF ERROR IN THE EXPERIMENT OF HESS'S LAW? EXPLAIN THE ANSWER
The largest source of error in the experiment of Hess's Law is human error.
What is experiment?An experiment is a procedure carried out to test a hypothesis, or an educated guess, in order to support or reject it. Experiments are conducted in order to explore new phenomena, or to verify and validate existing theories or laws. Experiments provide insight into cause-and-effect by demonstrating what outcome occurs when a particular factor is manipulated. Experiments vary greatly in goal and scale, but all share the same basic structure. They begin with a hypothesis, or an educated guess, that is then tested with an experiment. Results are measured and analyzed and conclusions are drawn, based on the results of the experiment.
This includes miscalculations when measuring, recording and interpreting data, making mistakes when setting up the experiment, or misreading the results. Also, inaccurate or faulty equipment used in the experiment can lead to errors. Another potential source of error is the presence of impurities in the reactants, which can alter the reaction enthalpies and thus the results of the experiment.
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Effects of Human Activity Wet Lab
- i just need a bit of help, like examples or something because i have no clue what to do or where to start ( EDGE 2023 8TH GRD)
(I WILL BRAINLIST ONCE I KNOW HOW)
Answer:
I don’t have any specific examples of wet labs that study the effects of human activity on wetlands. However, some common human activities that can affect wetlands include the construction of buildings and roads, agricultural land use, and overfishing. Wet labs can be used to study the impact of these activities on the physical, chemical, and biological processes in wetland ecosystems.
How does activity on the Sun affect human technology on Earth and in the rest of the solar system? (Select all that apply)
A. Coronal holes reduce the amount of material leaving the Sun making it a great opportunity to launch new satellites.
B. Solar flares and coronal mass ejections can disrupt communications, damage satellites and can cause power outages on Earth.
C. During high periods of activity, the higher amount of heat radiating from the Sun will melt satellites, disrupting communications.
D. Coronal holes allow more of the Sun’s material to flow out into space.
Answer: The answer is B,C,D
Explanation: Solar activity can affect satellite orbits, communication satellites, and the local power grids. It can also impact our spacecraft throughout the solar system, especially orbiters or landers on surfaces without an atmosphere.
therefore, Solar flares and coronal mass ejections can disrupt communications, damage satellites and can cause power outages on Earth.
During high periods of activity, the higher amount of heat radiating from the Sun will melt satellites, disrupting communications.
Coronal holes allow more of the Sun’s material to flow out into space.
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The options that apply are:
B. Solar flares and coronal mass ejections can disrupt communications, damage satellites, and can cause power outages on Earth.
D. Coronal holes allow more of the Sun’s material to flow out into space.
Coronal holes, mentioned in option D, allow more of the Sun's material to flow out into space. This primarily affects the solar wind, which consists of charged particles emitted by the Sun. The increased solar wind from coronal holes can impact spacecraft and satellites throughout the solar system.
Solar flares and coronal mass ejections, mentioned in option B, are intense bursts of energy and matter from the Sun. These events can release a large number of charged particles and electromagnetic radiation. When directed towards Earth, they can interfere with satellite communications, damage sensitive electronic equipment, and disrupt power grids, potentially causing power outages.
Option A, stating that coronal holes reduce the amount of material leaving the Sun, is incorrect. Coronal holes allow more material, particularly the solar wind, to flow out into space, as mentioned in option D.
Option C, suggesting that the higher amount of heat radiating from the Sun during high activity melts satellites, disrupting communications, is not accurate. While solar activity can increase radiation levels in space, it does not typically lead to satellite melting or disruption of communications due to excess heat.
Therefore, the correct options are B and D. Solar flares and coronal mass ejections can disrupt technology on Earth, and coronal holes allow more material from the Sun to flow into space.
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Calculate the potential of a zinc/silver cell. The zinc electrode is in 0.300 M zinc nitrate solution. The silver electrode is in a 0.600 M silver nitrate solution.
The zinc/silver cell has a potential of 1.23 V.
What is a few half cells' standard electrode potential?In order to obtain a half-unique cell's reduction potential, a standard electrode potential becomes necessary. It is measured with a reference electrode called the standard hydrogen electrode (abbreviated to SHE).
Zinc(s) → Zinc2+(aq) + 2e- E° = -0.76 V
Silver+(aq) + e- → Silver(s) E° = +0.80 V
We employ the following formula to get the cell potential:
Ecell = E°cell - (0.0592 V/n) log(Q)
where n is the number of electrons transported in the balanced equation, Q is the reaction quotient, and E°cell is the standard cell potential. For a concentration cell like this, n =
Q = [Zinc2+]/[Silver+]
Plugging in the values:
n = 2 (because 2 electrons are exchanged in each half-reaction) (since 2 electrons are transferred in each half-reaction)
E°cell = E°cathode - E°anode = +0.80 V - (-0.76 V) = +1.56 V
[Zinc2+] = 0.300 M
[Silver+] = 0.600 M
Q = [Zinc2+]/[Silver+] = (0.300 M)/(0.600 M) = 0.500
Substituting all the values into the formula:
Ecell = +1.56 V - (0.0592 V/2) log(0.500) = +1.23 V
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A beaker with 2.00×102 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M . A student adds 6.90 mL of a 0.300 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.
Answer: The pH will decrease by 0.109 units.
Explanation:
To solve this problem, we will need to use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
where pH is the initial pH of the buffer, pKa is the acid dissociation constant of acetic acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.
First, we need to find the initial concentrations of [A-] and [HA] in the buffer. Since the total molarity of acid and conjugate base is 0.100 M and we know the volume of the buffer, we can use the following equation:
moles of acid = moles of conjugate base
0.100 M x 2.00x10^-2 L = [HA] x 2.00x10^-2 L
[HA] = 0.100 M
Since we know the pH of the buffer, we can use the following equation to find the concentration of the conjugate base:
pH = pKa + log([A-]/[HA])
5.000 = 4.740 + log([A-]/0.100)
[A-]/[HA] = 10^(5.000-4.740)
[A-]/[HA] = 1.995
[A-] = 1.995 x 0.100 M = 0.1995 M
Now, we need to find the new concentrations of [A-] and [HA] after the addition of HCl. Since the volume of the buffer is now 2.069x10^-2 L (2.00x10^-2 L + 6.90x10^-3 L), we can use the following equation:
moles of acid + moles of HCl = moles of conjugate base
0.100 M x 2.00x10^-2 L + 0.300 M x 6.90x10^-3 L = [HA] x 2.069x10^-2 L
[HA] = 0.1295 M
The concentration of the conjugate base can be found using the equation:
[A-]/[HA] = 10^(pH-pKa)
1.891 = 10^(pH-4.740)
pH-4.740 = log(1.891)
pH = log(1.891) + 4.740
pH = 5.000 - 0.109
Therefore, the pH will decrease by 0.109 units.
Balance [tex]Fe + O_{2} = Fe_{2}O[/tex] and calculate how many molecules [tex]Fe_{2}O[/tex] has if given 3.5 g Fe or 2.28 g [tex]O_{2}[/tex] (do both).
Balancing equation [tex]Fe + O_2 = Fe_2O: 4 Fe + 3 O_2 = 2 Fe_2O_3[/tex]
The number of molecules of [tex]Fe_2O[/tex] produced from 3.5 g of Fe:
[tex]1.88 * 10^{22} molecules[/tex] and from 2.28 g [tex]O_2[/tex] is [tex]2.86 * 10^{22[/tex] molecules.
We need to first convert mass of Fe to moles, using its molar mass 55.85 g/mol:
moles of Fe [tex]= 3.5 g / 55.85 g/ mol = 0.0626\ mol[/tex]
We can see that 2 moles of [tex]Fe_2O[/tex] are produced for every 4 moles of Fe, or equivalently, 1 mole [tex]Fe_2O[/tex] is produced for every 2 moles Fe.
Therefore, the number of moles of [tex]Fe_2O[/tex] produced from 0.0626 mol of Fe is:
moles of [tex]Fe_2O[/tex] = 0.0626 mol / 2 = 0.0313 mol
We can use Avogadro's number:
[tex]1 mol = 6.022 * 10^{23}\ particles[/tex]
number of molecules [tex]Fe_2O[/tex] =[tex]0.0313\ mol * 6.022 * 10^{23[/tex]
molecules/mol = [tex]1.88 * 10^{22[/tex] molecules
Converting mass of [tex]O_2[/tex] to moles:
moles of [tex]O_2[/tex] =[tex]2.28 g / 32.00 g/mol = 0.0713 mol[/tex]
Number of moles of [tex]Fe_2O[/tex] produced from 0.0713 mol of [tex]O_2[/tex] is:
moles of [tex]Fe_2O[/tex] =[tex]2/3 * 0.0713\ mol = 0.0475\ mol[/tex]
number of molecules of [tex]Fe_2O[/tex] = [tex]0.0475\ mol * 6.022 * 10^{23[/tex]
molecules =[tex]2.86 * 10^{22} molecules[/tex]
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6) Apply the rules for canceling complex units to simplify and find the units in an answer:
a)
=
atm / mol • (atm •L/
mol • K)
b) (mol • (atm•L / mol • K) • K ) / L
Cancelling units is also called the unit conversion or dimensional analysis. This is because we can use conversion facts along with multiplication to convert from one unit to another.
When dealing with scientific measurements in science, we can sometimes multiply a specific measurement by another measurement and units will cancel out. This is called cancelling units. Unit cancellation is just a method of converting numbers to different units.
Here the given units can be simplified as:
a) atm / mol × (atm ×L/mol × K)
cancel the mole by mole
atm / (atm ×L/ K)
b) (mol × (atm×L / mol × K) × K ) / L
cancel the mole by mole
(atm×L / K) × K ) / L
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Using your knowledge of the Brønsted-Lowry theory of acids and bases,
write equations for the following acid-base reactions and indicate each
conjugate acid-base pair:
What is matter? a. A change of state b. Anything that has mass and takes up space c. Both a and b d. None of the above
Answer: B
Explanation: Anything that has mass and takes up space
Q143.Metallic molybdenum can be produced from the mineral molybdenite, MoS2. The mineral is first oxidized in air to molybdenum trioxide and sulfur dioxide. trioxide is then reduced to metallic molybdenum using hydrogen gas. The balanced equations are MoS2 1s2 1 7 2O2 1g2 h MoO3 1s2 1 2SO2 1g2 MoO3 1s2 1 3H2 1g2 h Mo1s2 1 3H2O1l2 Calculate the volumes of air and hydrogen gas at 178C and 1.00 atm that are necessary to produce 1.00 3 103 kg pure molyb denum from MoS2. Assume air contains 21% oxygen by volume, and assume 100% yield for each reaction.
250ml is the volumes of air and hydrogen gas at 178°C and 1.00 atm that are necessary to produce 1.00 3 103 kg pure molyb denum from MoS2.
A measurement of three-dimensional space is volume. It is frequently expressed quantitatively using SI-derived units, like the cubic metre or litre, or different imperial or US-standard units, including the gallon, quart and cubic inch. Volume and length (cubed) have a symbiotic relationship.
The volume much a container is typically thought of as its capacity, not as the amount of space it takes up. In other words, the volume is the quantity of fluid (liquid or gas) that the container may hold.
MoS[tex]_2[/tex] (g) + 7/2 O[tex]_2[/tex] (g) -------> MoO[tex]_3[/tex] (s) + 2SO[tex]_2[/tex] (g)
MoO[tex]_3[/tex] (s) + 3H[tex]_2[/tex] (g) ------------> Mo (s) + 3H[tex]_2[/tex]O (l)
P×V = n×R×T
moles = 1000/ 95.96=10.6
1×V =10.6×0.821×290
V= 250ml
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250ml is the volumes of air and hydrogen gas at 178°C and 1.00 atm that are necessary to produce 1.00 3 103 kg pure molyb denum from MoS2.
A measurement of three-dimensional space is volume. It is frequently expressed quantitatively using SI-derived units, like the cubic metre or litre, or different imperial or US-standard units, including the gallon, quart and cubic inch. Volume and length (cubed) have a symbiotic relationship.
The volume much a container is typically thought of as its capacity, not as the amount of space it takes up. In other words, the volume is the quantity of fluid (liquid or gas) that the container may hold.
MoS[tex]_2[/tex] (g) + 7/2 O[tex]_2[/tex] (g) -------> MoO[tex]_3[/tex] (s) + 2SO[tex]_2[/tex] (g)
MoO[tex]_3[/tex] (s) + 3H[tex]_2[/tex] (g) ------------> Mo (s) + 3H[tex]_2[/tex]O (l)
P×V = n×R×T
moles = 1000/ 95.96=10.6
1×V =10.6×0.821×290
V= 250ml
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How many moles of PCl5 can be produced from 25.0 g of P4 (and excess Cl2 )?
Answer: The balanced equation for the reaction between P4 and Cl2 to form PCl5 is:
P4 + 10Cl2 → 4PCl5
The molar mass of P4 is 123.9 g/mol, which means that 25.0 g of P4 is equal to:
25.0 g P4 x (1 mol P4 / 123.9 g P4) = 0.202 mol P4
According to the balanced equation, 1 mol of P4 reacts with 10 mol of Cl2 to produce 4 mol of PCl5. Therefore, the number of moles of PCl5 that can be produced from 0.202 mol of P4 is:
0.202 mol P4 x (4 mol PCl5 / 1 mol P4) = 0.808 mol PCl5
Therefore, 0.808 moles of PCl5 can be produced from 25.0 g of P4 (assuming excess Cl2).
Identify and discuss one common way of deciding that a reaction has occurred, and give an example.
Answer: Some signs that could be an indicator of a chemical reaction occurring are changes in color and temperature, the formation of gas, and precipitate.
Explanation: There is no single way to conclude that a reaction has occurred. Some reactions show changes in temperature and color, while others may form ppt. It varies for each and every reaction.
E.g. This reaction of Copper Sulphate with Sodium Hydroxide gives Copper Hydroxide ppt.
CuSO4 + 2 NaOH ------> Na2SO4 + Cu(OH)2 (ppt.)
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The reaction below can be classified as: 2NaC1(s)- 2Na(s)+C12(g)
Answer:
decomposition reaction
Explanation:
The given chemical reaction, 2NaCl(s) → 2Na(s) + Cl2(g), represents the decomposition of sodium chloride (NaCl) into its constituent elements, sodium (Na) and chlorine (Cl2).
This reaction can be classified as a decomposition reaction, which is a type of chemical reaction where a single compound breaks down into two or more simpler substances. In this case, the reactant NaCl decomposes into Na and Cl2 as the products.
The reaction is facilitated by heating or passing an electric current through the NaCl to break the bonds between the Na and Cl atoms.
Hope this helps!
Yolanda is focusing on eccentric contractions during her workout today. How is this MOST likely demonstrated in her workout?
Question 7 of 10
Komal found that her vial of isopropyl alcohol showed a much better surface
tension bubble shape (a higher bubble) than her vial of water. Her peer group
suggested some experimental errors that may have caused this to happen.
Which three experimental errors are most likely to have occurred?
A. Someone might have jiggled the table and made the water
surface tension bubble spill over.
B. Komal could have mixed up the labels on the vials.
C. Someone might have jiggled the table and made the isopropyl
alcohol surface tension bubble spill over.
D. The water could actually have been saltwater instead of pure
water.
The three experimental error that most likely would have occurred are Someone might have jiggled the table and made the water surface tension bubble spill over, Komal could have mixed up the labels on the vials and Someone might have jiggled the table and made the isopropyl alcohol surface tension bubble spill over. The correct option to this question are A,B and C.
What are the best practices for handling liquid chemicals in a chemistry lab without contaminating them?All chemical labels must be included on every container. Wear a lab coat and use safety eye gear to keep your eyes protected. Keep chemicals away from your intentional nose, mouth, and tongue. Stay away from your hands, face, clothes, and shoes when using chemicals, and always avoid direct contact.While working with chemicals, keep your hands off of your face, eyes, mouth, and body. Never enter the laboratory or chemical storage area with open or closed food or beverages. Never eat or drink out of a lab-glass container. In the storage or lab areas, avoid applying cosmetics.For more information on experimental error kindly visit to
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Titration of 20.0 mL of an NaOH solution required 9.0 mL of a 0.30 M KNO3 solution. What is the morality of the NaOH solution?
nitric acid is electrolysed using graphite, what are the ions ?
Answer:
When nitric acid (HNO3) is electrolyzed using graphite electrodes, the following ions are present:
Nitrate ions (NO3-): These are negatively charged ions that are formed from the dissociation of nitric acid. During electrolysis, they will move towards the positively charged anode.
Hydrogen ions (H+): These positively charged ions are also formed from the dissociation of nitric acid. During electrolysis, they will move towards the negatively charged cathode.
Water molecules (H2O): Small amounts of water may also be present in the nitric acid solution, and they will also be involved in the electrolysis process. The water molecules can be oxidized at the anode to form oxygen gas and positively charged hydrogen ions (H+).
230 Calories = ______ Joules
Answer:
962.32 or 962320 if it's 4184j per cal
Explanation:
1cal = 4.184j
so 230cal is 962.32Question 7 of 10
The bonds of the products store 22 kJ more energy than the bonds of the
reactants. How is energy conserved during this reaction?
A. The reaction creates 22 kJ of energy when bonds form.
B. The surroundings absorb 22 kJ of energy from the reaction
system.
C. The reaction system absorbs 22 kJ of energy from the
surroundings.
D. The reaction uses up 22 kJ of energy when bonds break
The total energy of the system remains constant, and energy is conserved. Option A.
Conservation of energyAccording to the law of conservation of energy, energy cannot be created or destroyed, it can only be transformed from one form to another.
In this case, the excess 22 kJ of energy is stored in the chemical bonds of the products when they are formed. The energy comes from the chemical bonds of the reactants, which break during the reaction and release energy.
Therefore, the total energy of the system remains constant, and energy is conserved.
Option B is incorrect because the surroundings cannot absorb energy from the reaction system without an external source of energy.Option C is incorrect because the reaction system cannot absorb energy from the surroundings without an external source of energy.Option D is incorrect because the breaking of bonds requires energy to be inputted, not the energy to be used up.More on bond energy can be found here: https://brainly.com/question/26141360
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What is the approximate temperature of each of these regions?
1.photosphere:
A. 4500K to 6800K
B. 10^4 K to 10^6 K
C. 10^4 K
D. A few million K
2. Chromosphere:
A. 4500K to 6800K
B. 10^4 K to 10^6 K
C. 10^4 K
D. A few million K
3. Corona:
A. 4500K to 6800K
B. 10^4 K to 10^6 K
C. 10^4 K
D. A few million K
the approximate temperature of each of these regions are:
Photosphere: 4500K to 6800K
Chromosphere: 10⁴ K to 10⁶K
Corona: A few million k
How to solve the question?
The photosphere is the visible surface of the sun, and its temperature ranges from approximately 4500K to 6800K (Kelvin). This temperature range is where the majority of the sun's light and radiation is emitted, and it is also where sunspots and solar flares occur. The photosphere is also where the sun's magnetic field lines emerge and interact with the surrounding plasma.
The chromosphere is the region of the sun that lies just above the photosphere. Its temperature ranges from approximately 10^4 K to 10^6 K, making it hotter than the photosphere. This region is visible during a solar eclipse as a red or pink ring around the sun. The chromosphere is also where solar prominences and flares occur, which are large eruptions of hot plasma from the sun's surface.
The corona is the outermost layer of the sun's atmosphere, and its temperature is several million Kelvin (a few million K). The corona is much hotter than the layers below it, despite being farther away from the sun's core. The corona is visible during a total solar eclipse as a white halo around the sun. The corona is also where the solar wind originates, which is a stream of charged particles that flows out into space and can affect the Earth's magnetosphere.
In summary, the approximate temperature of each of these regions are:
Photosphere: 4500K to 6800K
Chromosphere: 10⁴ K to 10⁶ K
Corona: A few million K
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I NEED HELP! CHEMISTRY! PLEASE HELP ME!
Procedures:
Measure the mass of 2.5 mL of Lye (or ½ teaspoon of powdered lye)
If your scale is not accurate you can find the mass of a larger volume such as 25 mL and divide by 10.
Add 100.0 mL (½ cup) of vinegar to the calorimeter.
Measure the temperature of the vinegar (initial temperature)
Add 2.5 mL of Lye to the vinegar and record the highest temperature reached (final temperature).
Calculate the heat of this reaction, ∆H, using q = mc∆T (use the calculation steps below). ∆H = q
Determine the ∆H/mole of NaOH.
Your observations include the pertinent data and calculations from your experiment.
Record your observations here:
Specific heat of vinegar (c): 4.1 J/g oC
Density of vinegar: 0.99 g/mL
Mass of 2.5 mL (or ½ teaspoon) of lye:
Volume of vinegar:
Initial temperature of vinegar:
Final temperature of vinegar:
Energy gained by the contents of the calorimeter (q) = heat of the reaction (∆H)
q = mc∆T
Mass of 100.0 mL of vinegar (multiply by density of vinegar):
Mass of contents (vinegar + lye); this is m in our equation:
Change in temperature (Final – Initial), this is ∆T in our equation:
Energy gained by the calorimeter contents (this is q):
Moles of NaOH (assumes all of the lye is NaOH):
Value of ∆H in J/mole ( q / moles of NaOH):
To calculate the value of H in J/mole: H value in J/mole (q / moles NaOH): [q] / ([mass of 2.5 mL (or 12 teaspoon) lye] / 40.0 g/mol).
How to determine specific heat capacity?Based on the given procedures, record the following information:
Mass of 2.5 mL (or ½ teaspoon) of lye: [missing, needs to be measured]
Volume of vinegar: 100.0 mL
Initial temperature of vinegar: [measurement needed]
Final temperature of vinegar: [measurement needed]
Specific heat of vinegar (c): 4.1 J/g oC
Density of vinegar: 0.99 g/mL
To calculate the mass of 2.5 mL of lye, to measure it using a scale. Once we have that measurement, we can proceed with the calculations:
Mass of 2.5 mL (or ½ teaspoon) of lye: [measurement needed]
Volume of vinegar: 100.0 mL
Initial temperature of vinegar: [measurement needed]
Final temperature of vinegar: [measurement needed]
Specific heat of vinegar (c): 4.1 J/g oC
Density of vinegar: 0.99 g/mL
To calculate the energy gained by the contents of the calorimeter (q), we can use the formula q = mc∆T, where q is the energy gained, m is the mass of the contents (vinegar + lye), c is the specific heat capacity of vinegar, and ∆T is the change in temperature (final temperature - initial temperature):
Mass of 100.0 mL of vinegar (multiply by density of vinegar): 99.0 g
Mass of contents (vinegar + lye); this is m in our equation: 99.0 g + [mass of 2.5 mL (or ½ teaspoon) of lye]
Change in temperature (Final – Initial), this is ∆T in our equation: [Final temperature] - [Initial temperature]
Energy gained by the calorimeter contents (this is q): q = (99.0 g + [mass of 2.5 mL (or ½ teaspoon) of lye]) x 4.1 J/g oC x ([Final temperature] - [Initial temperature])
To calculate the moles of NaOH, we need to convert the mass of lye used to moles using the molar mass of NaOH (40.0 g/mol):
Moles of NaOH (assumes all of the lye is NaOH): [mass of 2.5 mL (or ½ teaspoon) of lye] / 40.0 g/mol
Finally, to determine the value of ∆H in J/mole, divide the energy gained by the calorimeter contents (q) by the moles of NaOH:
Value of ∆H in J/mole (q / moles of NaOH): [q] / ([mass of 2.5 mL (or ½ teaspoon) of lye] / 40.0 g/mol)
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120cm3 of a gas at 25°c exerts a pressure of 750mmHg. calculate its pressure if its volume increased to 150cm3 at 40°c.
Answer:
P2 = 1125 mmHg
Explanation:
Gas Pressure Calculation
To solve this problem, we need to use the combined gas law, which relates the pressure, volume, and temperature of a gas under different conditions. The combined gas law is given by:
(P1V1)/T1 = (P2V2)/T2
where P1, V1, and T1 are the initial pressure, volume, and temperature of the gas, and P2, V2, and T2 are the final pressure, volume, and temperature.
Let's start by calculating the initial conditions:
P1 = 750 mmHg
V1 = 120 cm^3
T1 = 25°C + 273.15 = 298.15 K (temperature in Kelvin)
Now we can plug in these values and solve for P2:
(P1V1)/T1 = (P2V2)/T2
(750 mmHg x 120 cm^3) / 298.15 K = (P2 x 150 cm^3) / (40°C + 273.15)
Simplifying this equation, we get:
P2 = (750 mmHg x 120 cm^3 x (40°C + 273.15)) / (298.15 K x 150 cm^3)
P2 = 1125 mmHg
Therefore, the pressure of the gas would increase to 1125 mmHg if its volume increased to 150 cm^3 at 40°C.
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In the first order "A -> products" reaction, initially [A] = 0.816 M, after 16.0 minutes it is [A] = 0.632 M. rate constant , what is the value of k ?
The first order reaction is defined by the rate law:
rate = k[A]
where k is the rate constant and [A] is the concentration of the reactant.
If we integrate this rate law, we get:
ln([A]_t/[A]_0) = -kt
where [A]_t is the concentration of A at time t, [A]_0 is the initial concentration of A, k is the rate constant, and t is time.
We can use this equation to solve for k given the initial and final concentrations of A and the time interval.
In this case, we have:
[A]_t = 0.632 M
[A]_0 = 0.816 M
t = 16.0 min
Substituting these values into the equation above, we get:
ln(0.632/0.816) = -k(16.0)
Solving for k, we get:
k = (1/16.0) * ln(0.816/0.632) = 0.0316 min^-1
Therefore, the value of the rate constant k for this first order reaction is 0.0316 min^-1.
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750mL of water (initially at 20°C) is mixed with an unknown mass of iron (initially at 120°C). When thermal equilibrium is reached, the system has a temperature of 30°C. Find the mass of the iron. Heat capacity of iron is 0.46 J/g°C.
The mass of the iron is 17.64 kg.
To solve this problem, we need to use the principle of heat transfer, which states that the total heat gained by one object must be equal to the total heat lost by the other object when they are in thermal equilibrium.
Let's start by calculating the heat lost by the iron and the heat gained by the water. We can use the following formula:
Q = m×c×ΔT
where Q is the heat transferred, m is the mass of the object, c is its specific heat capacity, and ΔT is the temperature change.
Heat lost by iron = Heat gained by the water
[tex]m_{iron}[/tex] x [tex]c_{iron}[/tex] x ( [tex]T_{f}[/tex] - [tex]T_{i}[/tex] ) = [tex]m_{water}[/tex] x [tex]c_{water}[/tex] x ( [tex]T_{f}[/tex] - [tex]T_{i}[/tex] )
where [tex]T_{f}[/tex] is the final temperature of the system (30°C), [tex]T_{i}[/tex] is the initial temperature of each substance (120°C for the iron and 20°C for the water), [tex]c_{iron}[/tex] and [tex]c_{water}[/tex] are the specific heat capacities of iron and water, respectively.
We need to solve for [tex]m_{iron}[/tex], which is the unknown mass of iron. Rearranging the equation, we get:
[tex]m_{iron}[/tex] = ([tex]m_{water}[/tex] x [tex]c_{water}[/tex] x ([tex]T_{f}[/tex] - [tex]T_{i}[/tex] )) / ([tex]c_{iron}[/tex] x ([tex]T_{i}[/tex] - [tex]T_{f}[/tex] ))
Substituting the given values, we get:
[tex]m_{iron}[/tex]= (750g x 4.18 J/g°C x (30°C - 20°C)) / (0.45 J/g°C x (120°C - 30°C))
[tex]m_{iron}[/tex]= 17,640 g = 17.64 kg (to two decimal places)
Therefore, the mass of the iron is 17.64 kg.
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(15 pts!) Match the following items with the correct descriptions.
1. releases hydrogen ions
two
2. accepts hydrogen ions
eight
3. poor conductor of electricity
base
4. shiny and lustrous
acid
5. maximum number of electrons in first shell
nonmetal
6. maximum number of electrons in second shell
metal
A certain mass of nitrogen gas occupies a volume of 8.52 L at a pressure of 5.06 atm. At what pressure will the volume of this
sample be 10.90 L? Assume constant temperature and ideal behavior.
= P =
atm
At a volume of 10.90 L, the pressure of the nitrogen gas would be approximately 3.95 atm.
To solve this problem
Using the ideal gas law, we have:
PV = nRT
Where
P is pressureV is volume n is the number of moles of gas R is the universal gas constantT is the temperature in KelvinAssuming constant temperature and ideal behavior, we can set up a proportion:
(P1)(V1) = (P2)(V2)
Where
P1 is the initial pressureV1 is the initial volumeP2 is the final pressureV2 is the final volumeSubstituting the given values, we get:
(5.06 atm)(8.52 L) = (P2)(10.90 L)
Solving for P2, we get:
P2 = (5.06 atm)(8.52 L) / (10.90 L)
P2 = 3.95 atm
Therefore, at a volume of 10.90 L, the pressure of the nitrogen gas would be approximately 3.95 atm.
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Conditions required to separate chromatograms in dyes
Answer:
All substances should be equally soluble (or equally insoluble) in both.
Specific Heat of Metals Lab 5. What assumptions do we make in this lab about heat transfers? 6. What happens if the assumptions you made in the above question are false? 7. What are some sources of error and how did they affect your data and results? 8. Which metal has the highest specific heat capacity? What does that mean about th metal?
Assumptions made in the lab about heat transfers (Question 5):
1. The specific heat capacity of the metal being tested is constant and does not change with temperature.
2. The heat transfer between the metal sample and the surroundings is only due to conduction, and other modes of heat transfer such as convection or radiation are negligible.
3. The heat lost or gained by the metal sample is fully absorbed or released by the water in the calorimeter and the calorimeter itself has negligible heat capacity.
4. There is no heat exchange with the surrounding environment, and the calorimeter is perfectly insulated.
Consequences of false assumptions (Question 6):
If any of the above assumptions are false, it could affect the accuracy of the results obtained in the lab. For example:
1. If the specific heat capacity of the metal being tested changes with temperature, it could lead to inaccurate measurements of heat transfer and calculated specific heat capacity values.
2. If convection or radiation is not negligible, it could result in additional heat transfer between the metal sample and the surroundings, leading to inaccurate results.
3. If the calorimeter has significant heat capacity or there is heat exchange with the surrounding environment, it could affect the measurement of heat transfer and calculated specific heat capacity values.
Sources of error and their effects on data (Question 7):
1. Measurement errors: Errors in measuring the initial and final temperatures of the metal sample and water, or the mass of the metal sample and water, could affect the accuracy of calculated specific heat capacity values.
2. Heat loss or gain to the environment: If there is heat exchange with the surrounding environment during the experiment, it could result in inaccurate measurements of heat transfer and specific heat capacity values.
3. Assumption violations: If any of the assumptions mentioned earlier are not valid, it could affect the accuracy of the results obtained in the lab.
Metal with highest specific heat capacity (Question 8):
The metal with the highest specific heat capacity would be the one that requires the most amount of heat energy to raise its temperature by a given amount compared to the other metals tested in the lab. The metal with the highest specific heat capacity would have a larger value of specific heat capacity, indicating that it can absorb more heat energy per unit mass per unit temperature change compared to the other metals. This means that the metal with the highest specific heat capacity can store more heat energy without experiencing a significant increase in temperature, and it has a greater ability to resist changes in temperature when heat is added or removed.
How many aluminum atoms are in 25g of aluminum
Answer: 5.58x10^23 atoms of Al
Explanation:
Here's the math:
I converted grams to mols then mols to atoms.