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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PLEASE HELP NOW
Caffeine is a weak base with a b of 4.1×10^-4 Calculate the initial molar concentration of a solution of caffeine if the pH is 10.94.
Answer:430 mg/L = 0.43g/L
Explanation:
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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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.
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?
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
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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0.250 mol of KNO3 in 0.835 L of solution
0.30M is the molarity of the given solution with 0.250 mol and 0.835 L volume of solution
The amount of moles of solute that exist in a specific number of litres of the solution, or moles per litre of a solution, is known as molar concentration or molarity. Solvent, and 'solution' to make it simpler to comprehend the principles that will follow. A homogenous combination with any number of solutes in it is referred to as a solution.
Molarity = number of moles/ volume of solution in liter
Molarity = 0.250/ 0.835
= 0.30M
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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
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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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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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)
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
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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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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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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Potential in a different kind of cell.
A typical mammalian cell at 37
∘
C, with only potassium channels open, will have the following equilibrium:
K+ (intracellular) ⇌ K+ (extracellular),
with an intracellular concentration of 150 mM K+, and 4.0 mM K+ in the extracellular fluid.
What is the potential, in volts, across this cell membrane? Note: in this case, n = the charge on the ion, and Eo for a concentration cell = 0.00 V. explain please
The potential across this cell membrane with only potassium channels open is -0.082 V, which means that the inside of the cell is negatively charged relative to the outside.
The potential across a cell membrane can be calculated using the Nernst equation:
E = (RT/zF) ln([ion]out/[ion]in)
E = potential in volts, R= gas constant (8.314 J/mol*K), T= temperature in Kelvin, z = charge on the ion, F= Faraday constant (96,485 C/mol), and [ion]out and [ion]in are the concentrations of the ion outside and inside the cell, respectively.
K+ (intracellular) ⇌ K+ (extracellular)
The charge on potassium ions is +1, so z = 1.
The temperature is 37°C or 310 K.
The concentrations of potassium ions are [K+]in = 150 mM and [K+]out = 4.0 mM.
Substituting these values into the Nernst equation,
E = (RT/zF) ln([K+]out/[K+]in)
E = (8.314 J/mol.K × 310 K)/(1 × 96,485 C/mol) ln(4.0 mM/150 mM)
E = -0.082 V
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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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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 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."--
How much energy is released when 73 grams of water cools from 72 degrees Celsius to
30 degrees Celsius?
what is the name for CH3-CH2-C(O)-OCH3
Answer:
The name for CH3-CH2-C(O)-OCH3 is ethyl methanoate.
Explanation:
a 70 piece of metal at 120 C is dropped into a kilometer with 150 g of 30C water the final temperature of the water and little changes 35c what is the specific heat of the metal
Answer:
cm=0.385 J
cm = (0.15 kg x 4.18 J/gCx (35C - 30C)) / (70 pieces x 0.1 kg/piece x (120C - 35C))
The answer is:
cm = 0.385 J/gc
Therefore, the specific heat of the metal is 0.385
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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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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A certain flexible weather balloon contains 3.1 L of helium gas. Initially, the balloon is in WP at 8500ft, where the temperature is 23.8oC and the barometric pressure is 564.8 torr. The balloon then is taken to the top of Pike’s Peak at an altitude of 14,100ft, where the pressure is 400 torr and the temperature is 6.9oC. What is the new volume of the balloon at the top of Pikes Peak?
The concept combined gas law is used here to determine the new volume of the balloon. This law relate one thermodynamic variable to another holding everything else constant. The new volume is 4.12 L.
The combination of Boyles law, Charles's law and Avogadro's law gives the combined gas law. This law states that the product of pressure volume and temperature of a system remains constant.
The equation is:
P₁V₁ / T₁ = P₂V₂ / T₂
V₂ = P₁V₁T₂ / P₂T₁
0.7455 × 3.1 × 279.9 / 0.528 × 296.8 = 4.12 L
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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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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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This molecule undergoes an E1 mechanism when stirred in water.
All the 3 chemical species are drawn in the images below/
What is E1 mechanism when stirred in waterThe E1 reaction mechanism instigates a variant of elimination reactions. It materializes in the vicinity of strong acids or bases and it initiates by eliminating a leaving group from the substrate, consequently creating an intermediate carbocation. Once completed, the mechanism eliminates a proton from a neighborly carbon, initiating the construction process of a double bond.
However, performing an E1 reaction in water may yield unexpected results due to water's nucleophilic nature, catalyzing sneaky attacks on the carbocation intermediates, leading to dissimilar products than initially intended. Furthermore, reactions performed with aqueous media cause other side-products thanks to hydrolysis mechanisms that emerge, making them undesirable.
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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.
In Activity 2, you tested various compounds for chemical changes. (Barium nitrate, sodium hydroxide, sodium hydrogen carbonate, copper (II) sulfate, potassium iodide, silver nitrate, iron (III) nitrate, and hydrochloric acid.) Write the chemical formulas for each of the reactants.
The chemical formulas are as follows:
Barium nitrate: Ba(NO₃)₂Sodium hydroxide: NaOHSodium hydrogen carbonate: NaHCO₃Copper (II) sulfate: CuSO₄Potassium iodide: KISilver nitrate: AgNO₃Iron (III) nitrate: Fe(NO₃)₃Hydrochloric acid: HClChemical formulas are shorthand notations used to represent the composition of a substance. In this case, the reactants used in Activity 2 are listed with their chemical formulas.
Barium nitrate is represented by the chemical formula Ba(NO₃)₂, which shows that it contains one barium ion (Ba²⁺) and two nitrate ions (NO₃⁻).
Sodium hydroxide is represented by the chemical formula NaOH, which shows that it contains one sodium ion (Na⁺) and one hydroxide ion (OH⁻).
Sodium hydrogen carbonate is represented by the chemical formula NaHCO₃, which shows that it contains one sodium ion (Na⁺), one hydrogen ion (H⁺), one carbonate ion (CO₃²⁻) and one hydrogen carbonate ion (HCO₃⁻).
Similarly, other reactants are represented by their respective chemical formulas.
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