Answer:
Lowering the temperature typically reduces the significance of the decrease in entropy. That makes the Gibbs Free energy of the reaction more negative. As a result, the reaction becomes more favorable overall.
Explanation:
In an addition reaction there's a decrease in the number of particles. Consider the hydrogenation of ethene as an example.
[tex]\rm H_2C\text{=}CH_2\; (g) + H_2\; (g) \stackrel{\text{Ni}^\ast}{\to} H_3C\text{-}CH_3\; (g)[/tex].
When [tex]\rm H_2[/tex] is added to [tex]\rm H_2C\text{=}CH_2[/tex] (ethene) under heat and with the presence of a catalyst, [tex]\rm H_3C\text{-}CH3[/tex] (ethane) would be produced.
Note that on the left-hand side of the equation, there are two gaseous molecules. However, on the right-hand side there's only one gaseous molecule. That's a significant decrease in entropy. In other words, [tex]\Delta S < 0[/tex].
The equation for the change in Gibbs Free Energy for a particular reaction is:
[tex]\Delta G = \Delta H + (\underbrace{- T \, \Delta S}_{\text{entropy}\atop \text{term}})[/tex].
For a particular reaction, the more negative [tex]\Delta G[/tex] is, the more spontaneous ("favorable") the reaction would be.
Since typically [tex]\Delta S < 0[/tex] for addition reactions, the "entropy term" of it would be positive. That's not very helpful if the reaction needs to be favorable.
[tex]T[/tex] (absolute temperature) is always nonnegative. However, lowering the temperature could help bring the value of
Calcium hypochlorite (Ca(OCl)2, MW = 142.983 g/mol) is often used as the source of the hypochlorite ion (OCl–, MW = 51.452 g/mol) in solutions used for water treatment. A student must prepare 50.0 mL of a 55.0 ppm OCl– solution from solid Ca(OCl)2, which has a purity of 94.0%.
A) Calculate the mass of the impure reagent required to prepare the solution. Assume the density of the solution is 1.00 g/mL.
B) Which of the following methods would work best to accurately prepare 50.0 mL of the 55.0 ppm OCl– solution?
Use an analytical balance to weigh out an amount of reagent larger than what was determined in part A. In a volumetric flask, combine the reagent and water to create a solution with a concentration greater than 55.0 ppm. Use a pipet to transfer a portion of the concentrated solution to a separate 50.0 mL volumetric flask and dilute to the line with water to create a solution with a concentration of 55.0 ppm.
Use an analytical balance to weigh out an amount of reagent larger than what was determined in part A. In a volumetric flask, combine the reagent and water to create a solution with a concentration greater than 55.0 ppm. Use a graduated cylinder to transfer a portion of the more concentrated solution to a 50.0 mL volumetric flask and dilute to the line with water to create a solution with a concentration of 55.0 ppm.
Use an analytical balance to weigh out the amount of reagent determined in part A. Transfer the reagent to a 50.0 mL volumetric flask, add water to the line, and mix thoroughly.
Use an analytical balance to weigh out the amount of reagent determined in part A. Then use a graduated cylinder to measure out 50.0 mL of water. Combine the reagent and water in a 50.0 mL volumetric flask and mix it thoroughly.
Answer:
the mass of the impure reagent required to prepare the solution is approximately 0.41875 g.
Explanation:
A) To calculate the mass of the impure reagent required to prepare the solution, we can use the formula:
mass = volume x concentration x molar mass
First, let's calculate the moles of OCl- required:
moles = volume x concentration
Given that the volume is 50.0 mL and the concentration is 55.0 ppm (parts per million), we convert the concentration to a decimal by dividing it by 1,000,000:
55.0 ppm = 55.0 / 1,000,000 = 0.000055
moles = 50.0 mL x 0.000055 = 0.00275 mol
Now, let's calculate the mass of Ca(OCl)2 needed:
mass = moles x molar mass
Given that the molar mass of Ca(OCl)2 is 142.983 g/mol, we have:
mass = 0.00275 mol x 142.983 g/mol = 0.39395 g
However, the Ca(OCl)2 is impure, with a purity of 94.0%. To find the mass of the impure reagent, we divide the mass by the purity percentage and multiply by 100:
mass of impure reagent = (0.39395 g / 94.0%) x 100 = 0.41875 g
Therefore, the mass of the impure reagent required to prepare the solution is approximately 0.41875 g.
B) The best method to accurately prepare 50.0 mL of the 55.0 ppm OCl- solution is:
- Use an analytical balance to weigh out the amount of reagent determined in part A. Transfer the reagent to a 50.0 mL volumetric flask, add water to the line, and mix thoroughly.
This method ensures that the desired concentration is achieved by precisely measuring the required mass of the impure reagent and diluting it to the desired volume in a volumetric flask. The use of an analytical balance and a volumetric flask helps to ensure accuracy and precision in the preparation of the solution.
The other methods mentioned (using a pipet or graduated cylinder) may not provide the same level of accuracy and precision as using a volumetric flask, which is specifically designed for precise volume measurements.
A binary compound created by reaction of nitrogen and an unknown element E contains 30.46% N and 69.54% E by mass. If the formula of the compound is N2E4, calculate the atomic mass of E.
Answer:
15.95
Explanation:
This question is a modification of the calculation of the empirical formula of a compound given its percent composition and atomic weights of the elements in the compound.
Here we are given the formula and the percent composition, so we know that there are 4 atoms of E per 2 atoms of N so lets solve using the information given.
In 100 grams of the binary compound we have
30.46 g N
69.54 g E
The number of moles is the mass divided by atomic weight:
mol N = 30.46 g / A.W N = 30.46 g / 14.00 g/mol = 2.18 mol N
mol E = 65.54 g / A.W E
Thus,
4 mol E/ 2 mol N = ( 69.54 g/ A.W E ) / 2.18
2 A.E = 65.54 g / 2.18 ⇒ A.W E = 69.54 g / ( 2 x 2.18 ) = 15.94 g
So the A.W is 15.94 g/mol which is close the atomic weight of O.
Rank the following compounds in order of decreasing acid strength using periodic trends. Rank the acids from strongest to weakest. To rank items as equivalent, overlap them.
H2Se
HBr
H2O
HI
Explanation:
It is known that acidic strength of hydrides of same group tends to increase when we move from top to bottom in a group. On the other hand, acidic strength of hydrides of same period elements increases when we move from left to right in a period.
As both bromine and iodine belongs to the same group. Also, selenium and oxygen are same group elements. Therefore, their acidic strength increases on moving down the group.
Therefore, we can conclude that acidic strength of given compounds from strongest to weakest is as follows.
HI > HBr > [tex]H_{2}Se[/tex] > [tex]H_{2}O[/tex]
To rank the acids in decreasing acid strength using periodic trends, consider the size, electronegativity, and presence of lone pairs of electrons. HI is the strongest acid, followed by HBr, H2O, and H2Se.
Explanation:To rank the acids in order of decreasing acid strength using periodic trends, we need to consider the size and electronegativity of the atoms. The larger the atom, the weaker the acid, and the more electronegative the atom, the stronger the acid. Additionally, we can consider the presence of lone pairs of electrons, as they increase the acidity.
HI - Iodine (I) is larger and less electronegative than the other halogens. It also has a lone pair of electrons, making it the strongest acid.HBr - Bromine (Br) is larger and less electronegative than chlorine (Cl), and it also has a lone pair of electrons. It is the second strongest acid.H2O - Oxygen (O) is smaller and more electronegative than the halogens. It does not have a lone pair of electrons, making it a weaker acid than the halogens.H2Se - Selenium (Se) is larger and less electronegative than sulfur (S). However, it does not have a lone pair of electrons, making it the weakest acid.Learn more about Periodic trends here:https://brainly.com/question/32813617
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According to the following reaction, how many grams of water are produced in the complete reaction of grams of sulfuric acid?
This question requires the mass of sulfuric acid and a balance equation. The complete question is given below
Question:
According to the following reaction, how many grams of water are produced in the complete reaction of 24.1 grams of sulfuric acid?
H₂SO₄(aq) + Zn(OH)₂(s) ---------> ZnSO₄(aq) + 2H₂O(l)
Answer:
8.85 grams of water is obtained from 24.1 grams of sulfuric acid according to the given reaction
Explanation:
In this problem, the mass of water can be determined by using the balanced chemical equation.
Step 1: Write all data
Molar mass of water = 18 g
Molar mass of sulfuric acid = 98 g
Given mass of sulfuric acid = 24.1 g
Mass of water from reaction = ?
Step 2: Write statement for conversion
Given equation shows that
1 mole of H₂SO₄ gives 2 moles of water
Step 3: Convert moles into molar mass
Convert the moles into molar mass so, the statement becomes,
98 g of H₂SO₄ gives (2)(18) g of water
1 g of H₂SO₄ gives (2)(18)/98 g of water
Step 4: Use given data
24.1 g of H₂SO₄ gives (2)(18)(24.1)/98 g of water
amount of water = 8.85 g
Indicate the number of unpaired electrons in He. Express your answer as an integer.
Answer : The number of unpaired electrons in helium is, 0
Explanation :
As we are given that the element is helium. The symbol of helium element is, (He).
The atomic number of helium element is, 2 that means it has 2 number of electrons.
Electronic configuration : It is defined as the representation of electrons around the nucleus of an atom.
Number of electrons in an atom are determined by the electronic configuration.
The electronic configuration of helium element is: [tex]1s^2[/tex]
There is no unpaired electrons in helium element.
Hence, the number of unpaired electrons in helium is, 0
The number of unpaired electrons in He is 0.
Helium (He) is an element with an atomic number of 2, which means it has 2 protons in its nucleus and, in its neutral state, 2 electrons in its electron configuration. The electron configuration of helium follows the principles of quantum mechanics, where electrons fill the lowest energy levels first according to the Pauli exclusion principle and Hund's rule.
The first energy level (n=1) can hold up to 2 electrons, and for helium, this energy level is filled with 2 electrons. These electrons occupy the 1s orbital, which is the only orbital available at this energy level. According to the Pauli exclusion principle, no two electrons can have the same set of quantum numbers, so one electron fills the 1s orbital with a spin of +1/2 (up), and the other fills it with a spin of -1/2 (down). This results in both electrons being paired with opposite spins, canceling each other out.
Since all electrons in helium are paired, there are no unpaired electrons. Therefore, the number of unpaired electrons in a helium atom is 0. This makes helium a noble gas, which is characterized by having a full valence shell and being chemically inert due to the lack of unpaired electrons available for chemical bonding.
Ammonia gas combines with hydrogen chloride gas, forming solid ammonium chloride.
a. Write a balanced chemical equation for the reaction.
b. In a reaction mixture of 3.0 g NH3 and 5.0 g HCl, which is the limiting reactant?
c. How many grams of NH4Cl could form from the reaction mixture in part (b)?
d. How much of which reactant is left over in the reaction mixture in part b?
Answer:
a. The balanced chemical equation for the reaction
[tex]NH_3(g)+HCl(g)\rightarrow NH_4Cl(g)[/tex]
b. HCl is the limiting reagent.
c. 7.3295 grams of ammonium chloride could form from the reaction mixture in part (b).
d. 0.6715 grams of ammonia is left over in the reaction mixture in part b.
Explanation:
a. The balanced chemical equation for the reaction
[tex]NH_3(g)+HCl(g)\rightarrow NH_4Cl(g)[/tex]
b.
Moles of ammonia =[tex]\frac{3.0 g}{17 g/mol}=0.1765 mol[/tex]
Moles of HCl = [tex]\frac{5.0 g}{36.5 g/mol}=0.1370 mol[/tex]
According to reaction, 1 mole of HCl reacts with 1 mole of ammonia . Then 0.1370 moles of HCl will react with :
[tex]\frac{1}{1}\times 0.1370 mol=0.1370 mol[/tex] of ammonia
Hence, HCl is the limiting reagent.
c.
Since, HCl is a limiting reagent amount of ammonium chloride will depend upon moles of HCl.
According to reaction, 1 mole of HCl gives with 1 mole of ammonium chloride . Then 0.1370 moles of HCl will give :
[tex]\frac{1}{1}\times 0.1370 mol=0.1370 mol[/tex] of ammonium chloride
Mass of 0.1370 moles of ammonium chloride :
53.5 g/mol × 0.1370 mol = 7.3295 g
7.3295 grams of ammonium chloride could form from the reaction mixture in part (b).
d.
Moles of ammonia =[tex]\frac{3.0 g}{17 g/mol}=0.1765 mol[/tex]
HCl is a limiting reagent and ammonia is an excessive reagent.
According to reaction, 1 mole of HCl reacts with 1 mole of ammonia . Then 0.1370 moles of HCl will react with :
[tex]\frac{1}{1}\times 0.1370 mol=0.1370 mol[/tex] of ammonia
Moles of Ammonia reacted = 0.1370 mol
Moles of ammonia left unreacted = 0.1765 mol - 0.1370 mol = 0.0395 mol
Mass of 0.0395 moles of ammonia :
0.0395 mol × 17 g/mol = 0.6715 g
0.6715 grams of ammonia is left over in the reaction mixture in part b.
Identify the predominant intermolecular forces in each of these substances.
Note: If you answer any part of this question incorrectly, a single red X will appear indicating that one or more of the phrases are sorted incorrectly.
1.
H
2
O
2.
C
a
C
l
2
3.
C
H
3
C
H
(
C
H
3
)
O
H
4.
C
H
4
5.
N
H
3
Answer:
1. Hydrogen bond
2. Ion-ion
3. Hydrogen bond
4. London force
5. Hydrogen bond
Explanation:
Intermolecular forces are the ones that keep the molecules, or ions, together in the substance. If the substance is metal then, metallic forces are involved in it; if it's an ionic compound, then ion-ion forces are involved in it, and, if the compound is a molecule, it may have 3 different types of forces.
The force depends on the polarity of the molecule, which is given by the measure of the dipole moment. If the dipole moment is 0, the molecule is nonpolar, and if it's different from 0, it's polar. The polarity is given by the dipole moment, which is the vector sum of the dipoles of the bonds. The dipole is the difference of electronegativity between the elements of the bond.
The nonpolar molecules have a London force, which is the weakest force. The polar molecules have dipole-dipole forces, and, if the dipole is formed by hydrogen and a high electronegative element (N, O, or F), the dipole-dipole is extremely strong and it's called a hydrogen bond.
1. The water is a polar molecule because it has angular geometry, so, the dipole of the O-H bond is not canceled. Because it has the bond between H and O, the force is hydrogen bond.
2. CaCl2 is an ionic compound formed by the ions Ca+2 and Cl-, so the force is ion-ion.
3. CH3CH(CH3)OH has a polarity at the O-H bond, so as the water it has hydrogen bond.
4. CH4 has a tetrahedral geometry, and the dipole of the C-H bonds are canceled, so it's a nonpolar molecule with the London force.
5. NH3 has a pyramid trigonal geometry, and so the dipole of N-H is not canceled, so it's polar and has the bond of H with N, so, it has hydrogen bond force.
The substances H2O, CaCl2, CH3CH(CH3)OH, CH4, NH3 predominantly have hydrogen bonding, ionic bonding, hydrogen bonding, London dispersion forces, and hydrogen bonding & dipole-dipole interaction as their intermolecular forces respectively.
Explanation:The predominant intermolecular forces in these substances are as follows:
H2O: This molecule has dipole-dipole interaction, and hydrogen bonding which is a special type of dipole-dipole interaction.CaCl2: For ionic substances like this, the main intermolecular force is ionic bonding.CH3CH(CH3)OH: This molecule, a type of alcohol, experiences hydrogen bonding and dipole-dipole interaction due to the polar hydroxyl group (OH).CH4: This molecule is nonpolar, so the main intermolecular force is London dispersion forces.NH3: This molecule experiences dipole-dipole interaction and hydrogen bonding due to the polar nature of the molecule and the presence of hydrogen.Learn more about intermolecular forces here:https://brainly.com/question/9328418
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Calculate the mass of MgCO3 precipitated by mixing 10.00 mL of a 0.200 M Na2CO3 solution with 5.00 mL of a 0.0450 M Mg(NO3)2 solution.
The mass of MgCO₃ is 1.9g.
The balanced chemical reaction is shown below
Na₂CO₃ + Mg(NO₃)₂ ⇒ 2 NaNO₃ + MgCO₃
0.200 M 0.0450 M ?
10.0 5.00 mL ?
Since the volume and concentration of Mg(NO₃)₂ and Na₂CO₃ is given , we can calculate the number of moles for each of them and then determine the limiting reagent.
Convert the volume of Mg(NO₃)₂and Na₂CO₃ to liters:
5.00 mL x ( 1 L/1000 mL ) = 5.00 x 10⁻³ L
10.00 mL x ( 1L/ 1000 mL ) = 1.000 x 10 ⁻² L
Number of mol Mg(NO₃)₂ = ( 0.0450 mol /L ) x 5.00 x 10⁻³ L
= 2.25 x 10⁻⁴ mol Mg(NO₃)₂
Number of mol Na₂CO₃ = ( 0.200 mol / L ) x 1 x 10⁻² L
= 2.000 x 10⁻³ mol Na₂CO₃
Limiting reagent
= 2.25 x 10⁻⁴ mol Mg(NO₃)₂ x ( 1 mol Na₂CO₃ / mol Mg(NO₃)₂ )
= 2.25 x 10⁻⁴ mol Na₂CO₃ required .
Limiting reagent is Mg(NO₃)₂ since 2.25 x 10⁻⁴ mol Na₂CO₃ is required to
react completely with 2.25 x 10⁻⁴Mg(NO₃)₂, and there's an excess.
Number of mole of MgCO₃ produced
= 2.25 x 10⁻⁴ mol Mg(NO₃)₂ x ( 1 mol MgCO₃ / 1 mol Mg(NO₃)₂ )
= 2.25 x 10⁻⁴ mol MgCO₃
Mole = mass/molar mass
Mass= Mole × molar mass
2.25 x 10⁻⁴ mol MgCO₃ x 84.31 g/mol = 1.90 g
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To calculate the mass of MgCO3 precipitated, determine the limiting reactant and use stoichiometry to find the moles and mass of MgCO3.
Explanation:To calculate the mass of MgCO3 precipitated, we need to determine the limiting reactant in the reaction between Na2CO3 and Mg(NO3)2. First, calculate the moles of Na2CO3 and Mg(NO3)2 using their respective concentrations and volumes. Then, compare the moles of each reactant to determine which is limiting. Use the stoichiometry of the balanced equation to find the moles of MgCO3 produced. Finally, calculate the mass of MgCO3 using its molar mass.
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calculate the change in internal energy for a system taht is giving off 25.000 kL of heat and is changing from 18.00 L to 15.00 L?
This is an incomplete question. The complete question is :
Calculate the change in internal energy (ΔE) for a system that is giving off 25.000 kJ of heat and is changing from 18.00 L to 15.00 L in volume at 1.50 atm pressure.
Answer: The change in internal energy for a system is -24544 Joules
Explanation:
According to first law of thermodynamics:
[tex]\Delta E=q+w[/tex]
[tex]\Delta E[/tex]=Change in internal energy
q = heat absorbed or released
w = work done or by the system
w = work done by the system=[tex]-P\Delta V[/tex] {Work done on the system as the final volume is lesser than initial volume and is positive}
w =[tex]-1.50atm\times (15.00-18.00)L=4.50Latm=456Joules[/tex] {1Latm=101.3J}
q = -25.000 kJ =[tex]-25.000\times 10^3J[/tex] {Heat released by the system is negative}
[tex]\Delta E=456J+(-25000)=-24544J[/tex]
Thus change in internal energy for a system is -24544 Joules
Calculate the pH for the following weak acid.
A solution of HCOOH has 0.15M HCOOH at equilibrium. The Ka for HCOOH is 1.8×10−4. What is the pH of this solution at equilibrium?
Answer:
The answer to your question is pH = 2.28
Explanation:
Chemical reaction
HCOOH(ac) + H₂O ⇔ H₃O⁺ + HCOO⁻
I 0.15 -- 0 0
C - x -- +x +x
F 0.15 - x -- x x
Write the equation of equilibrium
[tex]ka = \frac{[H3O][HCOO]}{[HCOOH]}[/tex]
Substitution
[tex]1.8 x 10^{-4} = \frac{[x][x]}{0.15 - x}[/tex]
But 0.15 - x ≈ 0.15
[tex]1.8 x 10^{-4} = \frac{x^{2} }{0.15}[/tex]
Solve for x
x² = (0.15)(1.8 x 10⁻⁴)
Simplification
x² = 0.000027
Result
x = 0.0052
pH = -log [H₃O⁺]
Substitution
pH = - log [0.0052]
Simplification and result
pH = 2.28
The pH of a 0.15M solution of HCOOH with a Ka of 1.8×10⁻´ is calculated using an ICE table, which leads to an equilibrium pH of approximately 2.78.
To calculate the pH of a 0.15M solution of HCOOH with a Ka of 1.8×10⁻´, we can set up an ICE (Initial, Change, Equilibrium) table to find the equilibrium concentrations of the species in solution:
Initial: [HCOOH] = 0.15 M, [H⁺] = 0 M, [HCOO⁻] = 0 MChange: [HCOOH] decreases by x, [H⁺] increases by x, [HCOO⁻] increases by xEquilibrium: [HCOOH] = 0.15 - x, [H⁺] = x, [HCOO⁻] = xApplying the expression for Ka, we have:
Ka = 1.8×10⁻´ = (x)(x) / (0.15 - x)
Assuming x is small and can be ignored in the denominator, we simplify to:
Ka = x² / 0.15
Solving for x (which represents [H⁺]), we find that x is approximately equal to the square root of (Ka × [HCOOH]), thus:
[H⁺] = √(1.8×10⁻´ × 0.15) ≈ 1.65×10⁻²
To find the pH, we take the negative logarithm of [H⁺]:
pH = -log(1.65×10⁻²) ≈ 2.78
The pH of the solution at equilibrium is approximately 2.78.
A 255-mL flask contains pure helium at a pressure of 746 torr . A second flask with a volume of 475 mL contains pure argon at a pressure of 726 torr . If the two flasks are connected through a stopcock and the stopcock is opened, what is the partial pressure of helium? If the two flasks are connected through a stopcock and the stopcock is opened, what is the partial pressure of argon?
If the two flasks are connected through a stopcock and the stopcock is opened, what is the total pressure?
Answer:
pHe = 261 torr
pAr = 472 torr
P = 733 torr
Explanation:
The final volume for both gases is the same: V₂ = 255 mL + 475 mL = 730 mL
We can find the final partial pressure of each gas using Boyle's law.
He
P₁ × V₁ = P₂ × V₂
P₂ = P₁ × V₁ / V₂
P₂ = 746 torr × 255 mL / 730 mL = 261 torr
Ar
P₁ × V₁ = P₂ × V₂
P₂ = P₁ × V₁ / V₂
P₂ = 726 torr × 475 mL / 730 mL = 472 torr
The total pressure is the sum of the partial pressures.
P = 261 torr + 472 torr = 733 torr
Scoring Scheme: 3-3-2-1 Part I. For each trial, enter the amount of heat gained by the cool water, qcool water. The specific heat of water is 4.184 J/goC. Report your answer to 4 digits. Note: You should always carry 1 or 2 extra digits beyond the number of significant figures until your final calculation.
The question is incomplete, complete question is:
For each trial, enter the amount of heat lost by the chemical system, qrxn.
Hints: The specific heat of water is 4.184 J/g°C. Be careful of your algebraic sign here and remember that the change in temperature is equal to the final temperature minus the initial temperature.
qrxn = - (qwater + qcalorimeter).
The heat gained by the calorimeter water, qwater, depends on the mass of water, the specific heat of water, Cpand ΔT, while the heat gained by the calorimeter, qcalorimeter, depends on the heat capacity, C and ΔT.
DATA : 1, 2, 3
[tex]T_i(^oC)[/tex] 24.2, 24.0 , 23.2
[tex]T_f(^oC)[/tex] 38.2, 37.8 , 36.6
Mass (g) 70.001 , 70.008 , 70.271
Explanation:
[tex]Q=mc\Delta T[/tex]
Where:
Q = heat absorbed or heat lost
c = specific heat of substance
m = Mass of the substance
ΔT = change in temperature of the substance
Trail 1
Heat absorbed by the water = [tex]Q_1[/tex]
Mass of water ,m,= 70.001 g
Specific heat water = c = 4.184 J/g°C
ΔT = [tex]T_f-T_i=38.2 ^oC - 24.2^oC = 14^oC[/tex]
[tex]Q_2=70.001 g\times 4.184 J/g^oC\times 14^oC=4,100.38 J[/tex]
Heat absorbed by the water is 4,100.38 J.
Trail 2
Heat absorbed by the water = [tex]Q_2[/tex]
Mass of water ,m,= 70.008 g
Specific heat water = c = 4.184 J/g°C
ΔT = [tex]T_f-T_i=37.8 ^oC - 24.0^oC = 14.2^oC[/tex]
[tex]Q_2=70.008 g\times 4.184 J/g^oC\times 14.2^oC=4,042.21 J[/tex]
Heat absorbed by the water is 4,042.21 J.
Trail 3
Heat absorbed by the water = [tex]Q_3[/tex]
Mass of water ,m,= 70.271 g
Specific heat water = c = 4.184 J/g°C
ΔT = [tex]T_f-T_i=36.6^oC - 23.2^oC = 13.4^oC[/tex]
[tex]Q_3=70.271 g\times 4.184 J/g^oC\times 13.18^oC=3939.78 J[/tex]
Heat absorbed by the water is 3,939.78 J.
One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titrating a sample of silver nitrate solution. Any chloride anions in solution will combine with the silver cations to produce bright white silver chloride precipitate.
Suppose an EPA chemist tests a Image for One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titratin sample of groundwater known to be contaminated with nickel(II) chloride, which would react with silver nitrate solution like this:
Image for One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titratin(aq) Image for One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titratinImage for One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titratin(aq) Image for One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titratinImage for One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titratin(s) Image for One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titratinImage for One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titratin(aq)
The chemist adds Image for One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titratinM silver nitrate solution to the sample until silver chloride stops forming. She then washes, dries, and weighs the precipitate. She finds she has collected Image for One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titratin of silver chloride.
Calculate the concentration of nickel(II) chloride contaminant in the original groundwater sample. Round your answer to significant digits.
Answer:
5.0x10⁻⁵ M
Explanation:
It seems the question is incomplete, however this is the data that has been found in a web search:
" One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titrating a sample of silver nitrate solution. Any chloride anions in solution will combine with the silver cations to produce bright white silver chloride precipitate. Suppose a EPA chemist tests a 250 mL sample of groundwater known to be contaminated with nickel(II) chloride, which would react with silver nitrate solution like this:
NiCl₂ + 2AgNO₃ → 2AgCl + Ni(NO₃)₂
The chemist adds 50 mM silver nitrate solution to the sample until silver chloride stops forming. She then washes, dries, and weighs the precipitate. She finds she has collected 3.6 mg of silver chloride. Calculate the concentration of nickel(II) chloride contaminant in the original groundwater sample. Round your answer to 2 significant digits. "
Keep in mind that while the process is the same, if the values in your question are different, then your answer will be different as well.
First we calculate the moles of nickel chloride found in the 250 mL sample:
3.6 mg AgCl ÷ 143.32 mg/mmol * [tex]\frac{1mmolNiCl_{2}}{2mmolAgCl}[/tex] = 0.0126 mmol NiCl₂Now we divide the moles by the volume to calculate the molarity:
0.0126 mmol / 250 mL = 5.0x10⁻⁵MDensity is an intensive physical property that relates the mass of an object to its volume. Density, which is simply the mass of an object divided by its volume, is expressed in the SI derived unit g/mLg/mL for a liquid or g/cm3g/cm3 for a solid. Most substances expand or contract when heated or cooled, so the density values for a substance are temperature dependent. A particular brand of gasoline has a density of 0.737 g/mL at 25 ?C. How many grams of this gasoline would fill a 14.6gal tank?
Answer: 40731.8 grams of this gasoline would fill a 14.6gal tank
Explanation:
Density is defined as the mass contained per unit volume.
[tex]Density=\frac{mass}{Volume}[/tex]
Given : Mass of gasoline = ?
Density of the gasoline = [tex]0.737g/ml[/tex]
Volume of the gasoline = 14.6gal = 55267.01 ml (1gal=3785.41ml)
Putting in the values we get:
[tex]0.737g/ml=\frac{mass}{55267.01ml}[/tex]
[tex]{\text {mass of gasoline}}=40731.8g[/tex]
Thus 40731.8 grams of this gasoline would fill a 14.6gal tank
During studies of the reaction below, 2 N2H4(l) + N2O4(l) ? 3 N2(g) + 4 H2O(g) a chemical engineer measured a less-than-expected yield of N2 and discovered that the following side reaction occurs. N2H4(l) + 2 N2O4(l) ? 6 NO(g) + 2 H2O(g) In one experiment, 11.5 g of NO formed when 112.4 g of each reactant was used. What is the highest percent yield of N2 that can be expected?
The highest percent yield of N2 can be calculated by comparing the moles of N2 formed in the side reaction to the moles of N2 formed in the main reaction. In this case, the highest percent yield of N2 is 200%.
Explanation:The highest percent yield of N2 that can be expected can be calculated by comparing the moles of N2 formed in the side reaction to the moles of N2 formed in the main reaction. Let's calculate:
In the side reaction, 2 moles of N2O4 produce 6 moles of NO. In this experiment, 112.4 g of N2O4 was used. So, the moles of N2O4 used is:
moles of N2O4 = mass of N2O4 / molar mass of N2O4
moles of N2O4 = 112.4 g / (92.01 g/mol) = 1.22 mol
According to the stoichiometry, 1 mole of N2 is produced for every 2 moles of N2O4 used. Therefore, the moles of N2 formed in the side reaction is:
moles of N2 = 1.22 mol / 2 = 0.61 mol
In the main reaction, 2 moles of N2H4 produce 3 moles of N2. Since the amount of N2H4 used is the same as the amount of N2O4 used (112.4 g), the moles of N2 formed in the main reaction is:
moles of N2 = 1.22 mol
Now we can calculate the highest percent yield of N2:
percent yield = (moles of N2 formed in the main reaction / moles of N2 formed in the side reaction) * 100
percent yield = (1.22 mol / 0.61 mol) * 100 = 200%
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Two members of the Boron family owe their names to bright lines in their emission spectra. Indium has a bright indigo blue line (451.1nm) and Thallium has a bright green line (535.0 nm).
What are the energies of these two spectra lines?
Answer:
E Indium = 4.407 x 10⁻¹⁹ J
E Thallium = 3.716 x 10⁻¹⁹ J
Explanation:
From Planck´s equation for a given wavelength, the energy is given by.
E = h ( c/ λ )
h: Planck´s constant, 6.626 x 10⁻³⁴ J·s
c: light speed, 3 x 10⁸ m/s
λ: wavelength in m
We will need first to convert the given wavelengths to m:
451.1 nm x ( 1 m/10⁹ nm ) = 4.511 x 10⁻⁷ m
535.0 nm x ( 1 m/10⁹ nm ) = 5.350 x 10⁻⁷ m
E Indium = 6.626 x 10⁻³⁴ J·s x 3x 10 ⁸ m/s/ 4.511 x 10⁻⁷ m = 4.407 x 10⁻¹⁹ J
E Thallium = 6.626 x 10⁻³⁴ J·s x 3x 10 ⁸ m/s/ 5.350 x 10⁻⁷ m = 3.716 x 10⁻¹⁹ J
Calculate the longest and the shortest wavelength observed in the Balmer series of the H atom spectrum
Answer:
The longest wavelength observed in the Balmer series of the H atom spectrum is 656.3 nm.
The shortest wavelength observed in the Balmer series of the H atom spectrum is 364.6 nm.
Explanation:
Using Rydberg's Equation:
[tex]\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )[/tex]
Where,
[tex]\lambda[/tex] = Wavelength of radiation
[tex]R_H[/tex] = Rydberg's Constant
[tex]n_f[/tex] = Higher energy level
[tex]n_i[/tex]= Lower energy level
[tex]\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )[/tex]
For wavelength to be longest, energy would be minimum, i.e the electron will jump from third level to second level :
[tex]n_f[/tex] = Higher energy level = [tex]3[/tex]
[tex]n_i[/tex]= Lower energy level = 2 (Balmer series)
Putting the values, in above equation, we get
[tex]\frac{1}{\lambda_{Balmer}}=R_H\left(\frac{1}{2^2}-\frac{1}{3^2} \right )[/tex]
[tex]\lambda_{Balmer}=\frac{36}{5R_H}[/tex]
[tex]\lambda_{Balmer}=\frac{36}{5\times 1.097\times 10^7 m^-1}}=6.563\times 10^{-7} m=656.3 nm[/tex]
[tex]1 m =10^9 nm[/tex]
The longest wavelength observed in the Balmer series of the H atom spectrum is 656.3 nm.
For wavelength to be shortest, energy would be maximum, i.e the electron will from infinite level to second level. :
[tex]n_f[/tex] = Higher energy level = [tex]\infty[/tex]
[tex]n_i[/tex]= Lower energy level = 2 (Balmer series)
Putting the values, in above equation, we get
[tex]\frac{1}{\lambda_{Balmer}}=R_H\left(\frac{1}{2^2}-\frac{1}{\infty^2} \right )[/tex]
[tex]\lambda_{Balmer}=\frac{4}{R_H}[/tex]
[tex]=\frac{4}{1.097\times 10^7 m^-1}=3.646\times 10^{-7} m=364.6 nm[/tex]
The shortest wavelength observed in the Balmer series of the H atom spectrum is 364.6 nm.
What is the pH of a 75.0 mL solution that is 0.047 M in weak base and 0.057 M in the conjugate weak acid ( K a = 7.2 × 10 − 8 ) ?
To calculate the pH of a solution containing a weak base and its conjugate weak acid, use the Henderson-Hasselbalch equation.
Explanation:To calculate the pH of a solution that contains a weak base and its conjugate acid, we need to apply the Henderson-Hasselbalch equation.
The Henderson-Hasselbalch equation is given by: pH = pKa + log([A-]/[HA]), where pKa is the negative logarithm of the acid dissociation constant (Ka), [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.
In this case, the weak base has a concentration of 0.047 M and the conjugate weak acid has a concentration of 0.057 M. The pKa value of the acid is 7.2 x 10^-8. Plugging these values into the Henderson-Hasselbalch equation, we can calculate the pH of the solution.
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The pH of a 75.0 mL solution that is 0.047 M in weak base and 0.057 M in the conjugate weak acid ( K a = 7.2 × 10⁻⁸) will be approximately 7.06.
To determine the pH of a solution that contains both a weak base and its conjugate acid, we can use the Henderson-Hasselbalch equation. The equation is:
pH = pKa + log([A⁻]/[HA])
Where:
Given that the Ka for the weak acid is 7.2 x 10⁻⁸, the pKa is:
pKa = -log(Ka) = -log(7.2 x 10⁻⁸)
The concentration of the weak base ([A⁻]) is given as 0.047 M, and the concentration of the conjugate weak acid ([HA]) is 0.057 M.
Substituting into the Henderson-Hasselbalch equation:
pH = pKa + log(0.047/0.057)
Performing the calculations:
pH = 7.14 + log(0.825)
pH 7.14 - 0.084 = 7.06
Therefore, the pH of the solution is approximately 7.06.
A binary compound created by reaction of an unknown element E and oxygen contains 56.05% E and 43.95% O by mass. If the formula of the compound is E2O5, calculate the atomic mass of E.
Answer:The atomic mass of E is 51
Explanation:Please see attachment for explanation
A chemist determines by measurements that .070 moles of hydrogen gas participate in a chemical reaction. Calculate the mass of hydrogen gas that participates. Round your answer to significant digits.
Answer: 0.14g
Explanation:
Mass conc. = n x molar Mass
Mass conc. = 0.070 x 2 = 0.14g
Answer:
The mass of hydrogen gas that participates is 0.14 grams
Explanation:
Step 1: Data given
Moles of H2 = 0.070 moles
Molar mass H2 = 2.02 g/mol
Step 2: Calculate the mass of hydrogen gas
Mass of H2 = moles H2 * molar mass H2
Mass of H2 = 0.070 moles * 2.02 g/mol
Mass of H2 = 0.14 grams
The mass of hydrogen gas that participates is 0.14 grams
It takes 261 kJ/mol to eject electrons from a certain metal surface. What is the longest wavelength of light (nm) that can be used to eject electrons from the surface of this metal via the photoelectric effect?
Answer: 459 nm
Explanation:
The relation between energy and wavelength of light is given by Planck's equation, which is:
[tex]E=\frac{Nhc}{\lambda}[/tex]
where,
E = energy of the light = [tex]261 kJ=261000J[/tex] (1kJ=1000J)
N= avogadro's number = [tex]6.023\times 10^{23}[/tex]
h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]
c = speed of light = [tex]3\times 10^8m/s[/tex]
[tex]\lambda[/tex] = wavelength of light = ?
Putting the values in the equation:
[tex]261000J=\frac{6.023\times 10^{23}\times 6.626\times 10^{-34}Js\times 3\times 10^8m/s}{\lambda}[/tex]
[tex]\lambda=4.587\times 10^{-7}m=459nm[/tex] [tex]1nm=10^{-9}m[/tex]
Thus the longest wavelength of light that can be used to eject electrons from the surface of this metal via the photoelectric effect is 459 nm
When the equation below is correctly balanced, what are the coefficients of Pb and H20, respectively? Pb + H2O + O2 => Pb(OH)2 A. 1,2 B. 2,1 C. 2,2 D. 1,1 E. none of these
Answer:
C
Explanation:
Pb + H2O + O2 => Pb(OH)2
Number of oxygen atom on left = 3
Number of oxygen atom on the right = 2
To balance number of oxygen
Change the coefficient of Pb and H2O to two each
2Pb + 2H2O + O2 => 2Pb(OH)2
The coefficient of Pb and H2O is 2,2
The correct coefficients for Pb and H₂O when balancing the equation Pb + H₂O + O₂ ⇒ Pb(OH)₂ are 1 and 2, respectively; therefore, the answer is A. 1,2. This ensures that the chemical equation is fully balanced with the correct number of atoms for each element on both sides.
Explanation:When balancing the chemical equation Pb + H₂O + O₂ ⇒ Pb(OH)₂, the correct coefficients for Pb and H₂O, respectively, are 1 and 2. Balancing chemical equations requires that the same number of atoms for each element be present on both sides of the equation. Starting with lead (Pb), we see that one Pb atom is needed on both sides, so we keep the coefficient for Pb as 1.
Next, we balance the hydrogen atoms. There are two hydrogens in each hydroxide ion, and since there are two hydroxide ions in Pb(OH)₂, this means there are a total of four hydrogen atoms in the product. To balance this, we need 2 molecules of H₂O (since each molecule of water contains two hydrogen atoms, giving us the four we need), making the coefficient for H₂O equal to 2.
Lastly, the oxygen atoms will balance themselves once H and Pb are balanced. With 2 H₂O molecules, we have 2 oxygen atoms, and the Pb(OH)₂ molecule has 2 oxygen atoms as well, which means we don't need any additional O₂ for balance. The balanced equation is 1 Pb + 2 H₂O + O₂ ⇒ 1 Pb(OH)₂. Therefore, the correct answer is A. 1,2.
Which of the following membrane ion channels open and close in response to changes in the membrane potential?
voltage-gated channels
Answer: voltage-gated channels
Explanation:
voltage-gated Ion channels: They are proteins of multi-subunit complexes that reacts with confrontations changes that occurs in the members potential which results in the opening and closing of the pore. They mainly help to generate electrical signals within the cell membrane. The locations of the voltage-gated ion channels includes in the cell body of the neurons, on the dendrites, on the axon hillock, on the node of Ranvier, etc. Majorly, when ion channels open, it is a form of response to stimuli.
Voltage-gated channels are the membrane ion channels that open in response to changes in membrane potential, cycling through closed, open, or inactivated states depending on the cell's electrical charge.
Explanation:The membrane ion channels that open and close in response to changes in the membrane potential are known as voltage-gated channels. These channels can exist in one of three states - closed, open, or inactivated. Initially, voltage-gated ion channels are closed, but they open in response to a change in the electrical potential of the cell membrane. Specifically, when the inner membrane voltage becomes less negative in comparison to the outside, the channels sense the change through amino acids in their structure that are sensitive to charge, prompting the pore to open and allow ions to move across the membrane. Once activated, these channels will then become inactivated for a brief period, during which time they cannot open in response to a signal even if the membrane potential changes.
What is the concentration in ppm of a 3.7 L solution (d = 1.00 g/mL) containing 4.21 × 10-7 kg of the pesticide DDT?
The concentration of the pesticide DDT in the 3.7 L solution is approximately 113.51 ppm.
The mass of the solute (DDT) must be determined to calculate the concentration in parts per million (ppm).
So, the given quantities are:
Volume of solution (V) = 3.7 L
Density of solution (d) = 1.00 g/mL
Mass of DDT (m) = 4.21 × 10^-7 kg
Let's start by using the volume and density provided to determine the mass of the solution:
Mass of solution = Volume × Density
Mass of solution = 3.7 L × 1.00 g/mL = 3.7 kg
Now, we'll convert the mass of DDT to grams:
1 kg = 1000 g
Mass of DDT in grams = 4.21 × [tex]10^-^7[/tex] kg × 1000 g/kg = 4.21 × [tex]10^-^4[/tex] g
Now we can calculate the concentration in ppm:
Concentration (ppm) = (Mass of solute / Mass of solution) × 10^6
Concentration (ppm) = (4.21 × [tex]10^-^4[/tex] g / 3.7 kg) × [tex]10^-^6[/tex] ≈ 113.51 ppm
Therefore, the concentration of the pesticide DDT in the 3.7 L solution is approximately 113.51 ppm.
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The concentration of DDT in the solution is 0.1137 ppm. This calculation involved converting mass to grams, understanding the density to find out the total mass of the solution, and then using the formula for concentration in ppm.
Explanation:To calculate the concentration of DDT in ppm (parts per million), you firstly need to convert the given mass of DDT into grams as ppm is typically expressed in terms of mass (g). We know that 1kg is equal to 1000g, so 4.21 x 10^-7 kg corresponding to 4.21 x 10^-4 g. We also know that the density of the solution is given to be 1g/mL, a 3.7L solution is therefore 3700 mL. So, the total mass of the solution will be 1.00 g/mL x 3700 mL = 3700 g.
Now, to get the concentration in ppm, we use the formula:
Concentration (ppm) = (mass of solute/mass of solution) x 10^6
Plugging in the values, we have:
Concentration (ppm) =
[tex](4.21 * 10^-^4 g / 3700 g) * 1,000,000 = 0.1137[/tex]Learn more about Concentration in ppm here:
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The metallic radius of a lithium atom is 152 pm. What is the volume of a lithium atom in cubic meters?
Answer:
Volume of lithium atom is found to be 1.47 X 10⁻²⁹ m³
Explanation:
Let us consider the volume of atom as a sphere (but it is little complex than that). This volume is mathematically expressed as,
[tex]V=\frac{4}{3}\pi R^{3}[/tex]----------------------------------------------------------------------------------------(Eq. 1)
Here, R is the radius of lithium atom. The radius is given in picometers, so firstly let us convert it into meters
[tex]R=(152pm )(1X10^{-12}\frac{m}{pm})[/tex]
[tex]R = 1.52 X 10^{-10}m[/tex]
placing this value in Eq.1 the required result is achieved
[tex]V=\frac{4}{3}\pi {1.52X10^{-10}}^{3}[/tex]
V= 1.47 X 10⁻²⁹ m³
The volume of the lithium atom is [tex]1.47 \times 10^{-29} \ m^3[/tex].
The given parameters:
Radius of the Lithium atom, V = 152 pmThe volume of the lithium atom is calculated from the volume of a sphere as follows;
[tex]V = \frac{4}{3} \pi r^3[/tex]
where;
r is the radius of the sphere in meters[tex]\\\\V = \frac{4}{3} \times \pi \times (152 \times 10^{-12})^3\\\\V = 1.47 \times 10^{-29} \ m^3[/tex]
Thus, the volume of the lithium atom is [tex]1.47 \times 10^{-29} \ m^3[/tex]
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Gluconeogenesis must use "bypass reactions" to circumvent three reactions in glycolysis that are highly exergonic and essentially irreversible. Which of the following correctly matches the glycolytic reaction with the gluconeogenic enzyme used in the corresponding bypass reaction?
(A) Fructose 6-phosphate -> fructose 1,6-bisphosphate; phosphofructokinase-2
(B) 2-phosphoglycerate -> phosphoenolpyruvate; phosphoglycerate kinase
(C) Phosphoenolpyruvate -> pyruvate; pyruvate kinase
(D) Glucose -> glucose 6-phosphate; glucose 6-phosphatase
(E) Glyceraldehyde 3-phosphate -> 1,3-bisphosphoglycerate; phosphoenolcarboxykinase
Answer: A & C
Explanation:
Hexokinase is also one of the enzymes.
(A) Fructose 6-phosphate -> fructose 1,6-bisphosphate; phosphofructokinase-2
(C) Phosphoenolpyruvate -> pyruvate; pyruvate kinase
In gluconeogenesis, bypass reactions circumvent irreversible steps in glycolysis. The correct matching of the glycolytic reaction and the gluconeogenic enzyme is glucose to glucose 6-phosphate via glucose 6-phosphatase.
Explanation:The process known as gluconeogenesis uses what are known as bypass reactions to circumvent three different steps in the glycolytic reaction process that are essentially irreversible. These bypass reactions occur via the use of different enzymes.
Options Analysis(A) Fructose 6-phosphate -> fructose 1,6-bisphosphate is incorrect; the conversion of fructose 6-phosphate to fructose 1,6-bisphosphate is catalyzed by phosphofructokinase-1, not phosphofructokinase-2.(B) 2-phosphoglycerate -> phosphoenolpyruvate is incorrect. This reaction is catalyzed by enolase, not phosphoglycerate kinase.(C) Phosphoenolpyruvate -> pyruvate is incorrect. In gluconeogenesis, the reaction of converting pyruvate back to phosphoenolpyruvate is aided by the enzyme pyruvate carboxylase, not pyruvate kinase.(D) Glucose -> glucose 6-phosphate; glucose 6-phosphatase is correct. In gluconeogenesis, the conversion of glucose 6-phosphate back to glucose is catalyzed by the enzyme glucose 6-phosphatase, which is absent in glycolysis.(E) Glyceraldehyde 3-phosphate -> 1,3-bisphosphoglycerate is incorrect. The conversion of glyceraldehyde 3-phosphate to 1,3-bisphosphoglycerate is catalyzed by glyceraldehyde 3-phosphate dehydrogenase, not phosphoenolcarboxykinase.Learn more about Gluconeogenesis here:https://brainly.com/question/34723403
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A solution contains 3.1 mM Zn(NO3)2 and 4.2 mM Ca(NO3)2. The p-function for Zn2+ is _____, and the p-function for NO3- is _____.
Answer: The p-function of [tex]Zn^{2+}[/tex] and [tex]NO_3^{-}[/tex] ions are 2.51 and 2.14 respectively.
Explanation:
p-function is defined as the negative logarithm of any concentration.
We are given:
Millimolar concentration of zinc nitrate = 3.1 mM
Millimolar concentration of calcium nitrate = 4.2 mM
Converting this into molar concentration, we use the conversion factor:
1 M = 1000 mM
Concentration of zinc nitrate = 0.0031 M = 0.0031 mol/L1 mole of zinc nitrate produces 1 mole of zinc ions and 2 moles of nitrate ions
Concentration of zinc ions = 0.0031 M
Concentration of nitrate ions in zinc nitrate, [tex]M_1=(2\times 0.0031)=0.0062M[/tex]
Concentration of calcium nitrate = 0.0042 M = 0.0042 mol/L1 mole of calcium nitrate produces 1 mole of calcium ions and 2 moles of nitrate ions
Concentration of calcium ions = 0.0042 M
Concentration of nitrate ions in calcium nitrate, [tex]M_2=(2\times 0.0042)=0.0084M[/tex]
To calculate the concentration of nitrate ions in the solution, we use the equation:
[tex]M=\frac{M_1V_1+M_2V_2}{V_1+V_2}[/tex]
Putting values in above equation, we get:
[tex]M=\frac{(0.0062\times 1)+(0.0084\times 1)}{1+1}\\\\M=0.0073M[/tex]
Calculating the p-function of zinc ions and nitrate ions in the solution:
For zinc ions:[tex]\text{p-function of }Zn^{2+}\text{ ions}=-\log[Zn^{2+}][/tex]
[tex]\text{p-function of }Zn^{2+}\text{ ions}=-\log(0.0031)\\\\\text{p-function of }Zn^{2+}\text{ ions}=2.51[/tex]
For nitrate ions:[tex]\text{p-function of }NO_3^{-}\text{ ions}=-\log[NO_3^{-}][/tex]
[tex]\text{p-function of }NO_3^{-}\text{ ions}=-\log(0.0073)\\\\\text{p-function of }NO_3^{-}\text{ ions}=2.14[/tex]
Hence, the p-function of [tex]Zn^{2+}[/tex] and [tex]NO_3^{-}[/tex] ions are 2.51 and 2.14 respectively.
The p-function values for Zn2+ and NO3- are 9.6 and 15.5, respectively.
The p-function for Zn2+ is 9.6, and the p-function for NO3- is 15.5.
Identify each process as endothermic or exothermic and indicate the sign of H . (a) an ice cube melting (b) nail polish remover quickly evaporating after it is accidentally spilled on the skin (c) gasoline burning within the cylinder of an automobile engine
Final answer:
Water boiling is endothermic, gasoline burning is exothermic, and ice forming is exothermic.
Explanation:
a. Water boiling is an endothermic process because heat is required to convert water from the liquid to the gaseous phase. The reaction absorbs energy from the surroundings, resulting in a decrease in temperature.
b. Gasoline burning is an exothermic process because it releases heat into the surroundings. The combustion of gasoline produces energy in the form of heat and light.
c. Ice forming on a pond is an exothermic process. When liquid water freezes, it releases heat into the surroundings, resulting in a decrease in temperature.
The answer is as follows: (a) Endothermic, H > 0
(b) Endothermic, H > 0
(c) Exothermic, H < 0
Explanation and logic of the
(a) When an ice cube melts, it absorbs heat from its surroundings to break the hydrogen bonds between water molecules, which requires energy. This process is endothermic, meaning it absorbs heat from the surroundings, and the enthalpy change (H) is positive because the system is gaining energy.
(b) Nail polish remover evaporating quickly after being spilled also involves an endothermic process. The liquid absorbs heat from the skin, causing it to change state from liquid to gas. This phase change requires energy, which is why the skin feels cooler as the nail polish remover evaporates. The enthalpy change (H) is positive because the system is gaining energy from the surroundings.
(c) Gasoline burning within the cylinder of an automobile engine is an exothermic reaction. When gasoline combusts, it releases a significant amount of energy in the form of heat and light. This energy is transferred to the surroundings, and the enthalpy change (H) is negative because the system is losing energy. This is characteristic of exothermic reactions, where the energy released is greater than the energy absorbed.
The density of mercury (the element) is 13.69 g/cm3. The density of Mercury (the planet) is 5.43 g/cm3. If Mercury (the planet) was made of mercury (the element) and had the same mass, what would its diameter be?
Answer:
The diameter of Mercury would be 0.735 times smaller than the actual diameter i.e., it would become 3590 km from 4879 km
Explanation:
For the sake of simplicity, let us consider Mercury spherical and uniform. The provided data shows that if the Mercury planet is purely composed of mercury then it will require less space as compared to the actual planet, due to the higher density of the mercury. The ratio of the two densities shows that the volume of Mercury would reduce 2.52 times. From the change in volume we can determine the change in diameter by taking the ratio of initial and final states, shown in the following derivation:
[tex]V=(\frac{4}{3})(\pi )(r^{3})[/tex],
Here, V is the volume and r is the radius of the Mercury.
From the ratio of initial and final states, we obtain the following relation
[tex]\frac{V_{2} }{V_{1} } = \frac{r_{2} ^{3}}{r_{1} ^{3}}[/tex]
as,
[tex]V_{2}=\frac{V_{1}}{2.52}[/tex]
so,
[tex]\frac{1}{2.52} = \frac{r_{2} ^{3}}{r_{1} ^{3}}[/tex]
[tex]r_{2}^{3}=\frac{r_{1}^{3}}{2.52}[/tex]
[tex]r_{2}= 0.735.r_{1}[/tex]
The factor will remain same for the diameter i.e.,
[tex]D_{2}= 0.735 D_{1}[/tex]
For absolute diameter of Mercury:
Actual diameter of Mercury = 4879.4 km
Hence, the diameter of Mercury, if it was made of mercury would become 3590 km.
P.S.: The element mercury first letter is kept small while the other is capped
chemist adds of a M iron(III) bromide solution to a reaction flask. Calculate the mass in grams of iron(III) bromide the chemist has added to the flask. Round your answer to significant digits
The question is incomplete, here is the complete question:
A chemist adds 370.0 mL of a 2.25 M iron(III) bromide solution to a reaction flask. Calculate the mass in grams of iron(III) bromide the chemist has added to the flask. Round your answer to 3 significant digits
Answer: The mass of iron (III) bromide is 246. grams
Explanation:
To calculate the mass of solute, we use the equation used to calculate the molarity of solution:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]
We are given:
Molarity of solution = 2.25 M
Molar mass of iron (III) bromide = 295.6 g/mol
Volume of solution = 370.0 mL
Putting values in above equation, we get:
[tex]2.25M=\frac{\text{Mass of solute}\times 1000}{295.6\times 370.0}\\\\\text{Mass of solute}=\frac{2.25\times 295.6\times 370.0}{1000}=246.1g[/tex]
Hence, the mass of iron (III) bromide is 246. grams