In 100 mL of 0.12 M Ca3(PO4)2 solution, there are 0.036 moles of calcium ions and 0.024 moles of phosphate ions.
To find the number of moles of each ion present in 100 mL of 0.12 M Ca3(PO4)2 solution, we first need to determine the formula weight of Ca3(PO4)2.
Ca3(PO4)2 contains three calcium ions (Ca2+) and two phosphate ions (PO43-).
The atomic weight of calcium (Ca) is 40.08 g/mol, and there are three calcium ions in Ca3(PO4)2, so the total weight of calcium in Ca3(PO4)2 is 3 x 40.08 = 120.24 g/mol.
The atomic weight of phosphorus (P) is 30.97 g/mol, and the atomic weight of oxygen (O) is 16.00 g/mol.
The molecular weight of the phosphate ion (PO43-) is therefore 30.97 + (4 x 16.00) = 94.97 g/mol. Since there are two phosphate ions in Ca3(PO4)2, the total weight of phosphate in Ca3(PO4)2 is 2 x 94.97 = 189.94 g/mol.
The total formula weight of Ca3(PO4)2 is therefore 120.24 + 189.94 = 310.18 g/mol.
To find the number of moles of each ion present in 100 mL of 0.12 M Ca3(PO4)2 solution, we can use the formula:
moles = concentration x volume
For calcium ions:
moles of Ca2+ = concentration of Ca3(PO4)2 x (3 moles of Ca2+ / 1 mole of Ca3(PO4)2) x volume
moles of Ca2+ = 0.12 M x (3 / 1) x 0.1 L
moles of Ca2+ = 0.036 mol
For phosphate ions:
moles of PO43- = concentration of Ca3(PO4)2 x (2 moles of PO43- / 1 mole of Ca3(PO4)2) x volume
moles of PO43- = 0.12 M x (2 / 1) x 0.1 L
moles of PO43- = 0.024 mol
Therefore, in 100 mL of 0.12 M Ca3(PO4)2 solution, there are 0.036 moles of calcium ions and 0.024 moles of phosphate ions.
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The unit you work with is leaving a forward area rearming/refueling point and has unused ammunition. The ammunition should be
It is necessary to account for all unused ammunition and return it to a safe storage place or to the concerned authorities for disposal.
What is Ammunition?Bullets, shells and explosives are examples of physical objects that serve as ammunition to project force at a target. These objects are intended to be fired from a weapon such as a gun, rifle, or artillery piece and may be made of a variety of materials such as metal, plastic, or composite materials.
Depending on the weapon and the purpose of the attack, such as whether it is intended for training, target shooting, hunting or fighting, the ammo used will vary. Governments around the world have strict regulations governing ammunition, to ensure its safe handling, movement and application.
Therefore, it is necessary to account for all unused ammunition and return it to a safe storage place or to the concerned authorities for disposal.
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a reaction between 1.7 moles of zinc iodide and excess sodium
Answer: I got you fam
Explanation:
The formula is Na2CO3 + ZnI2 → 2NaI + ZnCO3
Or -5.19
Given the thermochemical equation:
4 AlCl3 (s) + 3 O2 (g) ⇒ 2 Al2 03 (s) + 6 Cl2 (g) ; ΔH = -529 kJ
Find ΔH for the following reaction:
1) 3 Al2O3 (s) + Cl2 (g) ⇒ 2/3 AlCl3 (s) + 1/2 O2 (g) ΔH= ?kJ
2) 88.2 kJ
b) 264.5 kJ
c) 529 kJ
d) 176.3 kJ
e) - 176.3 kJ
A thermochemical equation can be written by expressing the heat evolved or absorbed in terms of the enthalpy change ΔH. Here ΔH for the following reaction +88.2 kJ. The correct option is A.
A chemical equation which indicates the heat change occuring during the reaction is defined as the thermochemical equation. In thermochemical equations, physical states of the reactants and products should be specified.
Here the given reaction 4 AlCl₃ (s) + 3 O₂ (g) ⇒ 2 Al₂O₃ (s) + 6 Cl₂ (g) is reversed as 1 /3 Al₂O₃ (s) + Cl₂ (g) ⇒ 2/3 AlCl₃ (s) + 1/2 O₂ (g) and multiplied by 1/6.
So the new enthalpy is +88.16 ≈ 88.2 kJ
Thus the correct option is A.
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A soft drink contains 33g of sugar in 349g of H2O. A soft drink contains 33g of sugar in 349g of H2O. What is the concentration of sugar in the soft drink in mass percent?
A soft drink contains 33g of sugar in 349g of H[tex]_2[/tex]O. 14.3% is the concentration of sugar in the soft drink in mass percent.
One approach to indicate the concentration of any dissolved component in a solution is by mass percentage. Mass percentage is the ratio of the total weight of a compound in a solution to the overall mass of the solution, expressed in percentages.
In order to express the mass percent of a solution, the grammes of solute are divided by the grammes of solution, and the result is multiplied by 100. As long as you use a comparable number for both the component and solute mass.
Mass percent = (mass of solute/mass of solute+ mass of solvent)×100
= ( 33/ 33+ 349)×100
=14.3%
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Assuming that ground water flow does follow the contours of the land, is it possible that there are two sources of contamination? What would you expect to find if all three companies had leaking storage tanks and were actual sources of contamination?o
If ground water flow follows the contours of the land, it is possible that there are two sources of contamination.
This can occur if there are multiple locations of contamination that are not connected through the flow of groundwater. For example, if two companies had leaking storage tanks on opposite sides of a hill, the contaminants from each site could flow in different directions and not mix with each other.
If all three companies had leaking storage tanks and were actual sources of contamination, we would expect to find that the groundwater near each company contained contaminants associated with that company's stored chemicals. The contaminants may be different for each company, depending on the type of chemicals stored.
However, if the contamination has been ongoing for a long time, the chemicals may have mixed together in the groundwater, making it difficult to identify the specific source of contamination for each chemical. In this case, further investigation would be needed to determine the specific sources and extent of contamination from each company.
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A 2.26 gg lead weight, initially at 11.1 ∘C∘C, is submerged in 7.45 gg of water at 52.2 ∘C∘C in an insulated container.
Answer:
The temperature of the water and lead weight is 31.0°C.
Explanation:
To solve this problem, we can use the formula:
q = mcΔT
where q is the heat transferred, m is the mass, c is the specific heat, and ΔT is the change in temperature.
First, let's calculate the heat transferred from the water to the lead weight:
q1 = mcΔT = (7.45 g)(4.184 J/g°C)(52.2°C - T) where T is the final temperature of the water and lead weight
q2 = mcΔT = (2.26 g)(0.128 J/g°C)(T - 11.1°C)
Since the container is insulated, the heat transferred from the water to the lead weight is equal to the heat transferred from the lead weight to the water:
q1 = q2
(7.45 g)(4.184 J/g°C)(52.2°C - T) = (2.26 g)(0.128 J/g°C)(T - 11.1°C)
Simplifying and solving for T:
T = 31.0°C
Therefore, the final temperature of the water and lead weight is 31.0°C.
Balance the following reaction by typing in the correct coefficients in front of each reactant and product.
H3PO4(s) -
-->
H₂(g) +
P(s) +
O₂(g)
A mixture of gases contains 10.25g of F2, 2.83g of H2, and 5.95g of CO2. If the total pressure of the mixture is 2.75 atm, what is the partial pressure of each component?
1. Calculate the number of grams of Al in 371 g of Al2O3.
2. Urea [(NH2)2 CO] is used for fertilizer and many other things. Calculate the mass of N, C, O, and H atoms in 1.68 x 10^4 g of urea.
For the following calculations:
371 g of Al₂O₃ contains 196.53 g of Al.
Urea contains 7847.55 g N, 3362.69 g C, 152.18 g H, and 4472.57 g O.
How to calculate contents?1. To calculate the number of grams of Al in 371 g of Al₂O₃, calculate the molar mass of Al₂O₃ and then use stoichiometry to find the mass of Al.
Molar mass of Al₂O₃ = (2 x atomic mass of Al) + (3 x atomic mass of O)
= (2 x 26.98 g/mol) + (3 x 16.00 g/mol)
= 101.96 g/mol
Using stoichiometry to find the mass of Al:
1 mol Al₂O₃ contains 2 mol Al
So, 101.96 g Al₂O₃ contains (2 x 26.98) = 53.96 g Al
Therefore, 371 g of Al₂O₃ contains (53.96/101.96 x 371) = 196.53 g of Al.
2. To calculate the mass of N, C, O, and H atoms in 1.68 x 10⁴ g of urea:
Molar mass of urea = (2 x atomic mass of N) + (1 x atomic mass of C) + (3 x atomic mass of H) + (1 x atomic mass of O)
= (2 x 14.01 g/mol) + (1 x 12.01 g/mol) + (3 x 1.01 g/mol) + (1 x 16.00 g/mol)
= 60.06 g/mol
Using stoichiometry to find the mass of each element:
1 mol urea contains 2 mol N, 1 mol C, 3 mol H, and 1 mol O
So, 60.06 g urea contains 2 x (14.01 g N) = 28.02 g N,
1 x (12.01 g C) = 12.01 g C,
3 x (1.01 g H) = 3.03 g H, and
1 x (16.00 g O) = 16.00 g O.
Therefore, 1.68 x 10⁴ g of urea contains:
(28.02/60.06 x 1.68 x 10⁴) = 7847.55 g N,
(12.01/60.06 x 1.68 x 10⁴) = 3362.69 g C,
(3.03/60.06 x 1.68 x 10^4) = 152.18 g H, and
(16.00/60.06 x 1.68 x 10⁴) = 4472.57 g O.
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What volume in milliliters of a 1.00 M solution of sodium hydroxide is required to
make 125 mL of a 0.0600 M solution?
7.50 mL
12.5 mL
16.7 mL
208 mL
How much energy is released when 73 grams of water cools from 72 degrees Celsius to
30 degrees Celsius?
The Galapagos Islands are a series of islands near the coast of South America. The finches (a type of bird) on the Galapagos look similar to the finches in South America, but the birds on each island have some variations, including unique beak structures. Explain what differences on the islands would have led to this beak structure variation.
This archipelago and its enormous marine reserve are referred to be a "living museum and showcase of evolution" for their singularity.
Thus, The 127 islands, islets, and rocks that make up the Galapagos archipelago, 19 of which are major and 4 of which are inhabited, are located roughly 1,000 kilometres from the Ecuadorian mainland. In 1959, national parks were established on 7,665,100 ha, or 97% of the entire emergent surface.
On the remaining three percent of the islands, only human habitation is permitted in designated rural and urban zones (the fifth island only includes an airport, a tourist dock, a fuel containment system, and military installations).
The Galapagos Marine Reserve, one of the largest marine reserves in the world, was established in 1986 and expanded to its current area in 1998. It surrounds the islands. Inland waters of the archipelago are also a part of the marine reserve.
Thus, This archipelago and its enormous marine reserve are referred to be a "living museum and showcase of evolution" for their singularity.
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Does anyone have the Labs: Acid and Bases, lab report. I will give brainliest.
A variety of lichen was used to create the natural acid-base indicator known as litmus.
Since we can't taste everything to determine if it constitutes an acid or a base, litmus paper is essentially an indicator that is used to differentiate acids from bases. It is also known as an acid-base indicator since it can detect the presence of either a base or an acid in a solution.
Litmus is combined with wood cellulose paper to create a litmus paper. The lichen plants that are used to make the purple dye known as litmus are classified as members of the Thallophyta division. A variety of lichen was used to create the natural acid-base indicator known as litmus. You can determine whether a variety of solutions are bases or acids by testing them with red and blue coloured litmus paper.
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why is often difficult to identify a highly weathered mineral
Weathering changes the chemical and physical nature of an element that is why it is often difficult to identify a highly weathered mineral.
The breakdown and alteration of rocks and minerals at or near the Earth's surface as a result of exposure to various weathering agents, such as water, wind and temperature changes is known as weathering.
Minerals can undergo physical changes as a result of weathering, such as being broken up into smaller pieces or having their color and texture altered. Additionally, it may result in chemical alterations such as the removal or addition of specific chemical components.
This may lead to the creation of brand-new minerals or the modification of already existing minerals into new ones. Highly weathered minerals might not still possess the same physical and chemical characteristics as their unweathered counterparts.
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What mass in grams of oxygen gas are produced when 2.43 x 10-4 g of KCIO,
are completely reacted according to the following chemical equation:
2 KCIO₂ (s) → 2 KCI (s) + 3 0₂(g)
Calculate the volume of barium hydroxide (0.1177 M) necessary to react with 25.00 mL of phosphoric acid (0.1002 M)
The concept molarity is an important method which is used to calculate the concentration of a solution. It is mainly employed to calculate the concentration of a binary solution. Here the volume of barium hydroxide is 21.28 mL.
Molarity of a solution is defined as the number of moles of the solute present per litre of the solution. It is represented as 'M' and its unit is mol/L.
The equation connecting molarity and volume of two solution is given as:
M₁V₁ = M₂V₂
V₁ = M₂V₂ / M₁
0.1002 × 25.00 / 0.1177 = 21.28 mL
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The gas pressure in a can is 2.5 atm at 25 °C. Assuming that the gas obeys the ideal-gas equation, what is the pressure (in atm) when the can is heated to 525 °C?
The concept combined gas law is used here to determine the new pressure of the gas. This law states that the ratio between the product of pressure-volume and temperature of a system remains constant.
The combined gas law is the combination of Boyle's law, Charles's law and the Avogadro's law. These laws relate one thermodynamic variable to another holding everything else constant.
Here volume is constant, so the equation is:
P₁ / T₁ = P₂ / T₂
T₁ = 298 K
T₂ = 798 K
Pressure is:
P₂ = P₁ T₂/T₁
P₂= 2.5 × 798 / 298
P₂ = 6.69 atm
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PLEASE HELP ASAP
(50 POINTS)
You have 400,000 atoms of a radioactive substance. After 2 half-lives have past, how
many atoms remain?
Remember that you cannot have a fraction of an atom, so round the answer to the
nearest whole number.
Answer:
If 2 half-lives have passed, it means that the radioactive substance has decayed twice, so the number of remaining atoms would be:
1st half-life: 400,000 / 2 = 200,000 atoms remaining
2nd half-life: 200,000 / 2 = 100,000 atoms remaining
Therefore, after 2 half-lives have passed, 100,000 atoms would remain, rounded to the nearest whole number
Explanation:
IF U NEED FURTHER HELP AND WANT TO BE FRIEND , SN AP = m_oonlight781
0.152 mol of sucrose in 602 mL of solution
Answer:
To convert from moles to molarity, we need to divide the number of moles of solute by the volume of solution in liters.
First, we need to convert the volume of solution from milliliters to liters:
602 mL = 0.602 L
Now, we can calculate the molarity:
Molarity = moles of solute / liters of solution
Molarity = 0.152 mol / 0.602 L
Molarity = 0.252 M
Therefore, the molarity of the sucrose solution is 0.252 M.
Explanation:
Find the critical value(s) and rejection region(s) for the indicated t-test, level of significance , aand sample size n.
right-tailed test, a= 0.10, n=9 The critical value is
The critical value of the indicated t-test is 1.397
t > 1.397 is the rejection region.
How to find critical value and rejection region?A t-distribution table or calculator is required to calculate the critical value for a right-tailed t-test with a significance level of 0.10 and a sample size of 9. The critical value is 1.397 when using a t-distribution table with 8 degrees of freedom (n - 1 = 9 - 1 = 8) and a significance level of 0.10.
t > 1.397 is the rejection zone for a right-tailed t-test with a significance level of 0.10 and a sample size of 9. In other words, if the estimated t-value is bigger than 1.397, the null hypothesis may be rejected.
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A piece of metal with a mass of 32.8 g is heated to 100.5 C and dropped into 138.2 g of water at 20.0 C. the final temperature of the system is 30.2 C. What is the specific heat capacity of the metal
Answer:
To solve this problem, we can use the equation:
Q = m * c * ΔT
where Q is the heat transferred, m is the mass of the substance, c is its specific heat capacity, and ΔT is the change in temperature.
In this case, we know that the heat lost by the metal is equal to the heat gained by the water:
Q lost = Q gained
We can calculate the heat lost by the metal using the equation:
Q lost = m * c * ΔT
where m is the mass of the metal, c is the specific heat capacity of the metal (which we are trying to find), and ΔT is the change in temperature of the metal (100.5 C - 30.2 C = 70.3 C).
We can calculate the heat gained by the water using the equation:
Q gained = m * c * ΔT
where m is the mass of the water and ΔT is the change in temperature of the water (30.2 C - 20.0 C = 10.2 C).
Setting the two equations equal to each other, we get:
m * c * ΔT (metal) = m * c * ΔT (water)
Simplifying, we get:
c (metal) = (m * c * ΔT (water)) / (m * ΔT (metal))
Plugging in the values we know:
m (metal) = 32.8 g
ΔT (metal) = 70.3 C
m (water) = 138.2 g
ΔT (water) = 10.2 C
c (metal) = (138.2 g * 4.184 J/g·K * 10.2 C) / (32.8 g * 70.3 C)
c (metal) = 0.192 J/g·K
Therefore, the specific heat capacity of the metal is 0.192 J/g·K.
how many grams of MgCl2 are contained in 0.50 L kc a 1.5 m solution?
0.50 L of a 1.5 m solution contains 71.4 grams of MgCl2.
To solve this problemThe equation moles of solute = molarity x volume (in liters) can be used.
When converting to grams, we may use the molar mass of the MgCl2 to determine how many moles there are in the solution.
MgCl2 has a molar mass of roughly 95.21 g/mol.
The volume must first be changed from liters to milliliters:
0.50 L = 500 mL
Next, we may determine how many moles of MgCl2 are present in the solution:
moles of MgCl2 = molarity x volume (in liters)
moles of MgCl2 = 1.5 mol/L x 0.50 L
moles of MgCl2 = 0.75 moles
Finally, we can figure out how much MgCl2 is present in the solution:
mass of MgCl2 = moles of MgCl2 x molar mass
mass of MgCl2 = 0.75 moles x 95.21 g/mol
mass of MgCl2 = 71.4 grams
Therefore, 0.50 L of a 1.5 m solution contains 71.4 grams of MgCl2.
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Look back at parts A and B to compare the properties of the unknown elements with the properties of the known
elements. Based on these properties, match each unknown element to its group in the periodic table.
Drag each tile to the correct box.
Tiles
element 1 element 2
Pairs
group 1
group 2
group 11
group 14
group 17
group 18
element 3
element 4
element 5
element 6
Based on the properties of elements, elements can be arranged into groups in the periodic table as follows:
Group 1 to 3 - metals
Group 14 - non-metals, metalloids, and metals
Group 15 to 18 - non-metals
What are groups and periods in the periodic table?Groups are the names given to the periodic table's columns. In the table, individuals who belong to the same group make bonds of the same kind and have an equal number of electrons in their atoms' outermost shells.
Periods are the horizontal rows found in the periodic table.
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I just need # 6,8 and 12 pls
6. 1 [tex]SO_{3}[/tex] + 1 [tex]H_{2}O[/tex] --> 1 [tex]H_{2} SO_{4}[/tex]
8. 1 [tex]K_{2}O[/tex] + 1 [tex]H_{2}O[/tex] --> 2 KOH
12. 1 [tex]CdSO_{4}[/tex] + 1 [tex]H_{2} S[/tex] --> 1 CdS + 1 [tex]H_{2} SO_{4}[/tex]
Please help asp and don 't just put random answers please
The value of Tan P as fraction simplest form is 15 / 29
How do determine the value of tan P?The following data were obtained from the question:
Angle θ = POpposite = 15Adjacent = 29Tan P =?Tan θ ratio is express as:
Tan θ = Opposite / Adjacent
Inputting the various parameters obtained from the question, we can obtain Tan P as shown below:
Tan P = Opposite / Adjacent
Tan P = 15 / 29
Thus, from the above calculation, we can conclude that the value of tan P in it's lowest fraction is 15 / 29
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The diagram shows sound and light waves from an emergency vehicle traveling toward a brick wall. The brick wall has both smooth and rough surfaces.
Select the correct answer from each drop-down menu to complete the sentences about how each wave is affected by the brick wall.
The sound waves from the siren will
the smooth surface of the wall. The light waves from the emergency vehicle will
the smooth surface of the wall. Rougher sections of the wall surface will cause the
from the emergency vehicle to scatter.
The sound waves from the siren will reflect off the smooth surface of the wall. The light waves from the emergency vehicle will reflect off the smooth surface of the wall. Rougher sections of the wall surface will cause the light waves from the emergency vehicle to scatter.
When sound waves hit a smooth surface, they reflect off the surface in a predictable way called the law of reflection. So, the sound waves from the siren will reflect off the smooth surface of the wall.
Similarly, light waves also follow the law of reflection when they hit a smooth surface. Therefore, the light waves from the emergency vehicle will also reflect off the smooth surface of the wall.
However, when light waves encounter a rough surface, they scatter in all directions due to the irregularities on the surface. Therefore, rougher sections of the wall surface will cause the light waves from the emergency vehicle to scatter.
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Calculate the pH of the three solutions described below. (25 pts)
a) 6.0 x 10-3 M HClO4
b) .0009 M Sr(OH)2
c) The solution made by mixing 25.00 mL of 0.121 M HCl + 30.00 mL of 0.100 M KOH
a) HClO₄ is a strong acid, and in water, it will dissociate completely into H+ and ClO₄⁻. The concentration of H⁺ in this solution is therefore equal to the concentration of HClO₄:
[H⁺] = 6.0 x 10⁻³ M
Using the definition of pH, we have:
pH = -log[H⁺]
pH = -log(6.0 x 10⁻³)
pH = 2.22
b) Sr(OH)₂ will dissociate completely into Sr²⁺ and 2OH⁻. The concentration of OH⁻ in this solution is therefore twice the concentration of Sr(OH)₂:
[OH-] = 2 x 0.0009 M
[OH-] = 0.0018 M
Using the definition of pH, we have:
pOH = -log[OH-]
pOH = -log(0.0018)
pOH = 2.74
Since pH + pOH = 14, we have:
pH = 14 - pOH
pH = 14 - 2.74
pH = 11.26
c) Solution made by mixing 25.00 mL of 0.121 M HCl and 30.00 mL of 0.100 M KOH
HCl + KOH → H2O + KCl
The moles of HCl in 25.00 mL of 0.121 M HCl are:
moles HCl = concentration x volume
moles HCl = 0.121 mol/L x 0.025 L
moles HCl = 0.003025 mol
The concentration of HCl in the final solution is:
[HCl] = moles HCl / total volume of solution
[HCl] = 0.003025 mol / (25.00 mL + 30.00 mL)
[HCl] = 0.003025 mol / 0.055 L
[HCl] = 0.055 M
Similarly, the moles of KOH in 30.00 mL of 0.100 M KOH is:
moles KOH = concentration x volume
moles KOH = 0.100 mol/L x 0.030 L
moles KOH = 0.003 mol
The concentration of OH⁻ in the final solution is:
[OH⁻] = moles OH⁻ / total volume of solution
[OH⁻] = 0.006 mol / 0.055 L
[OH⁻] = 0.109 M
Using the definition of pH, we have:
pH = 14 - pOH
pH = 14 - (-log[OH⁻])
pH = 14 - (-log(0.109))
pH = 12.06
Thus, the pH of (a) is 2.22, (b) is 11.26, and (c) is 12.06.
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A student used a video camera to record another student dropping a marble through water in a graduated cylinder. The students watched the video in slow motion and made the observations shown below. During which part or parts of the marble’s fall did the marble experience unbalanced forces?
Parts B and C of the marble's fall did the marble experience unbalanced forces. Option 4 is correct.
A force is a push or pull (interaction) which changes the momentum of an object, either stationary or in motion when unopposed. All objects experience different forces depending on their environment. When immersed in fluids, unbalanced forces of one upward moving force tends to cancel the gravity force moving downward on a sinking object causing deceleration to a constant sinking speed.
This upward moving force is called as Buoyant force. This is where at part A, the object will experiences a balanced force of gravity which accelerates due to the absence of an opposing force acting upwards on the object. At part B, the speed of the sinking object decreases due to an unbalanced force that cancels the acceleration by the buoyant force. Once the sinking object’s acceleration is cancelled, its sinking speed turns constant at part C.
Hence, 4. is the correct option.
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--The given question is incomplete, the complete question is
"A student used a video camera to record another student dropping a marble through water in a graduated cylinder. The students watched the video in slow motion and made the observations shown below. During which part or parts of the marble's fall did the marble experience unbalanced forces? (1) Part A only (2) Parts A and B only (3) Part C only (4) Parts B and C only."--
Suppose that 0.95 g of water condenses on a 75.0 g block of iron that is initially at 22 °C. If the heat released during condensation goes only to warming the iron block, what is the final tempera- ture (in Celsius) of the iron block? (Assume a constant enthalpy ofvaporizationforwaterof44.0kJmol-1.)
Answer:
The temperature of the iron block is 68.5°C.
Explanation:
The heat released during condensation of water is used to warm the iron block:
q = m_H2O * ΔH_vap = m_fe * c_fe * ΔT
where q is the heat released, m_H2O is the mass of water condensed, ΔH_vap is the enthalpy of vaporization for water, m_fe is the mass of iron, c_fe is the specific heat capacity of iron, and ΔT is the change in temperature of the iron block.
Rearranging the equation gives:
ΔT = (m_H2O * ΔH_vap) / (m_fe * c_fe)
Substituting the given values gives:
ΔT = (0.95 g * 44.0 kJ/mol) / (75.0 g * 0.449 J/(g°C))
ΔT = 46.5°C
Therefore, the final temperature of the iron block is:
T_f = T_i + ΔT = 22°C + 46.5°C = 68.5°C.
The final temperature of the iron block is 68.5°C.
PROBLEM 19.12 Draw the structure of a triacylglycerol that fits each description: a. a saturated triacylglycerol formed from three 12-carbon fatty acids b. an unsaturated triacylglycerol that contains three cis double bonds c. a trans triacylglycerol that contains a trans double bond in each hydrocarbon chain