the correct (answer) is (c.) a light particle
Calculate Δ Hrxn for the following reaction: CH4(g)+4Cl2(g)→CCl4(g)+4HCl(g) given these reactions and their ΔH values: C(s)C(s)H2(g)+++2H2(g)2Cl2(g)Cl2(g)→→→CH4(g),CCl4(g),2HCl(g),ΔH=−74.6 kJΔH=−95.7 kJΔH=−184.6 kJ Express the enthalpy in kilojoules to one decimal place.
Answer : The enthalpy of the following reaction is, -390.3 KJ
Explanation :
The given balanced chemical reactions are,
(1) [tex]C(s)+2H_2(g)\rightarrow CH_4(g)[/tex] [tex]\Delta H_1=-74.6KJ/mole[/tex]
(2) [tex]C(s)+2Cl_2(g)\rightarrow CCl_4(g)[/tex] [tex]\Delta H_2=-95.7KJ/mole[/tex]
(3) [tex]H_2(g)+Cl_2(g)\rightarrow 2HCl(g)[/tex] [tex]\Delta H_3=-184.6KJ/mole[/tex]
The final reaction of is,
[tex]CH_4(g)+4Cl_2(g)\rightarrow CCl_4(g)+4HCl(g)[/tex] [tex]\Delta H_{rxn}=?[/tex]
Now adding reaction 2 and twice of reaction 3 and reverse of reaction 1, we get the enthalpy of of the reaction.
The expression for enthalpy for the following reaction will be,
[tex]\Delta H_{rxn}=[2\times \Delta H_3]+[-1\times \Delta H_1]+[1\times \Delta H_2][/tex]
where,
n = number of moles
Now put all the given values in the above expression, we get:
[tex]\Delta H_{rxn}=[2mole\times (-184.6KJ/mole)]+[-1mole\times (-74.6KJ/mole)]+[1\times (-95.7KJ/mole)]=-390.3KJ[/tex]
Therefore, the enthalpy of the following reaction is, -390.3 KJ
Answer:
-390.3 KJ
Explanation:
For Hess's Law, we need to get the corresponding equation below using the sequence of reactions given
By manipulating the reaction, either reversing them or multiplying/dividing them to a certain factor, we can get to the target equation as well as the total enthalpy
CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)
C(s) + 2H2(g) → CH4(g) ΔH = −74.6kJ (needs to reverse)
C(s) + 2Cl2(g) → CCl4(g) ΔH = −95.7kJ (retain)
H2(g) + Cl2(g) → 2HCl(g) ΔH = −184.6kJ (multiply by 2 to get 4Cl2 and cancel out 4 HCl and 4 H2)
Therefore, it is -390.3 KJ
For which H-atom wavefunction are you mostlikely to find the electron farthest from thenucleus?1.2p2.3p3.4pcorrect4.2s5.1s
Answer:
Option 3. 4p
Explanation:
The greater the principal quantum number, i.e. the main energy level, means that the electron is occupying a larger orbital (bigger radius) thus you must check which wavefunction has the bigger principal quantum number. When two electrons have the same principal quantum number, then you must check the orbital shape because the rank of the radii of the orbitals in increasing order is: s < p < d < f.
The list of choices is:
1. 2p2. 3p3. 4p4. 2s5. 1sHence, the wavefunction 4p (third option) is the largest one and, so, it is for it that you mostlikely would find the electron the farthest from its nucleus.
In the 4p wavefunction is mostlikely to find the electron farthest from thenucleus.
Electronic configurationThe electronic configuration of an atom determines the orbital location of electrons, based on energy level and orbital type.
The higher the energy level, the farther that electron is from the nucleus.
Thus, the wavefunction that demonstrates the electron furthest from the nucleus is 4p, which has a higher energy level.
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A solution that is 20% acid and 80% water is mixed with a solution that is 50% acid and 50% water. If twice as much 50% acid solution is used as 20% solution, then what is the ratio of acid to water in the mixture of the solutions?
Answer:
The ratio acid to water in the mixture is 2:3
Explanation:
Let the volume of 20% acid solution used to make the mixture = x units
So, the volume of 50% acid solution used to make the mixture = 2x units
Total volume of the mixture = x + 2x = 3x units
For 20% acid solution:
C₁ = 20% , V₁ = x
For 50% acid solution :
C₂ = 50% , V₂ = 2x
For the resultant solution of sulfuric acid:
C₃ = ? , V₃ = 3x
Using
C₁V₁ + C₂V₂ = C₃V₃
20×x + 50×2x = C₃×3x
So,
20 + 50×2 = C₃×3
Solving
120 = C₃×3
C₃ = 40 %
Thus, for the resultant mixture,
Acid percentage = 40%
Water percentage = (100 - 40)% = 60%
Ratio acid to water in the mixture = 40:60 = 2:3
Answer:
2:3
Explanation:
Identify each definition that applies to the compound in red. Check all that apply. HCI + NaOH → H2O + NaCl
Arrhenius acid
Bronsted-Lowry acid
Arrhenius base
Bronsted-Lowry base
Answer:
A. Arrhenius acid
B. Bronsted-Lowry acid
Explanation:
An acid base reaction invoves an acid and a base and yields salt and water as products.
What is a neutralization?A neutralization is a reaction between an acid and a base to yield salt and water only. The highlighted compound is not shown here however we we shall tell what is compound is in this reaction.
HCI - Arrhenius acidNaOH - Arrhenius baseH20 - waterNaCl - SaltHence, this an acid base reaction in the Arrhenius sense.
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At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2 + O2 (g) →2 SO3 (g) At equilibrium, the partial pressure of SO2 is 36.9 atm and that of O2 is 16.8 atm. The partial pressure of SO3 is ________ atm.
To calculate the partial pressure of SO3 at equilibrium for the given reaction, we can use the equilibrium constant (Kp) and the known partial pressures of SO2 and O2. By substituting values into the equation Kp = (P(SO3))^2 / (P(SO2))^2 * P(O2), we can find the partial pressure of SO3.
Explanation:Given the equilibrium constant (Kp) of 0.345 and the partial pressures of SO2 and O2 at equilibrium as 36.9 atm and 16.8 atm respectively, we can use the equation Kp = (P(SO3))^2 / (P(SO2))^2 * P(O2). Let's substitute the known values into the equation:
Kp = (P(SO3))^2 / (36.9 atm)^2 * (16.8 atm)
Now we can solve for the partial pressure of SO3:
(P(SO3))^2 = Kp * (36.9 atm)^2 * (16.8 atm)
P(SO3) = sqrt(Kp * (36.9 atm)^2 * (16.8 atm))
Calculating this value will give us the partial pressure of SO3 at equilibrium.
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1) How old is a bone in which the Carbon-14 in it has undergone 8 half-lives?
Select one:
a. 45600
b. 91200
c. 91200
d. 11400
2) In the process of radiocarbon dating, the fixed period of radioactive decay used to determine age is called the
Select one:
a. exponent.
b. half-life.
c. isotope.
d. nucleus.
3) A certain byproduct in nuclear reactors, 210Po, decays to become 206Pb. After a time period of about 276 days, only about 25% of an original sample of 210Po remains. The remainder has decayed to 206Pb. Determine the approximate half-life of 210Po.
Select one:
a. 138 days
b. 276 days
c. 414 days
d. 552 days
1) How old is a bone in which the Carbon-14 in it has undergone 8 half-lives?
Using the graph form the picture you count 8 times the halving of C¹⁴ and you arrive at 45600 years.
2) In the process of radiocarbon dating, the fixed period of radioactive decay used to determine age is called the half-life.
3) A certain byproduct in nuclear reactors, 210Po, decays to become 206Pb. After a time period of about 276 days, only about 25% of an original sample of 210Po remains. The remainder has decayed to 206Pb. Determine the approximate half-life of 210Po.
What the problem is telling you is that at 276 days only 25% original sample remains. If you divide the number of days by two the quantity of original sample will be multiplied by two, and you will have 138 days and 50% of original sample. This is the answer because the the half-life of a isotope is the time in which 50% of original quantity of radioactive atoms will disintegrate.
For the chemical equation SO2(g)+NO2(g)↽−−⇀SO3(g)+NO(g) the equilibrium constant at a certain temperature is 3.80. At this temperature, calculate the number of moles of NO2(g) that must be added to 2.50 mol SO2(g) in order to form 1.00 mol SO3(g) at equilibrium.
The number of moles of [tex]NO_2[/tex] that should be added in order to form 1 mol [tex]SO_3\\[/tex] at equilibrium is 0.263 mol.
Given,
Initial Conditions:
Moles of [tex]SO_\\2[/tex] initially = 2.50 mol
Moles of [tex]SO_3[/tex] at equilibrium = 1.00 mol
Moles of [tex]NO_2[/tex] initially = 0 mol
Moles of NO initially = 0 mol
Equilibrium constant = 3.80
Changes in Moles:
Change in moles of [tex]SO_2[/tex] = 2.50 mol (initial moles of [tex]SO_2[/tex])
Change in moles of [tex]SO_3[/tex]= 1.00 mol (moles of [tex]SO_3[/tex]at equilibrium)
The equilibrium constant expression for this reaction is:
[tex]K_c = \frac{[SO_3][NO]}{[SO_2][NO_2]}[/tex]
Where the square brackets denote the concentration of each species.
On using the equilibrium constant expression to solve for the change in moles of [tex]NO_2[/tex]:
[tex]\Delta n_{NO_2} = \frac{[SO_2][NO_2]}{[SO_3]}\times \frac{[NO]}{K_c}[/tex]
Substituting the values:
[tex]\rm \Delta n_{NO_2} = \frac {2.50\times 0}{1} \times \frac{1}{3.80}\\\Delta n_{NO_2} = 0.263 \ mol[/tex]
The number of moles that should be added in order to form 1 mol [tex]SO_3\\[/tex] at equilibrium is 0.263 mol.
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According to the given chemical equation, for every mole of SO3 formed, a mol of NO2 is used up. Therefore, to form 1.00 mol of SO3 from 2.50 mol of SO2, you need to add 1.00 mol of NO2 to reach equilibrium.
Explanation:This calculation, in essence, requires an application of Le Chatelier's Principle. For the chemical equation SO2(g)+NO2(g)↽−−⇀SO3(g)+NO(g) the equilibrium constant, Kc, at a certain temperature is 3.80. This implies that at equilibrium, the ratio of the concentrations of products to reactants (when raised to their stoichiometric coefficients) is 3.80.
To form 1.00 mol of SO3(g) from 2.50 mol of SO2(g), it means that according to the equation, 1 mol NO2(g) will be required. Furthermore, for the NO(g) produced, we do not have to worry about it since it is not involved in the later equilibrium calculation.
With Le Chatelier's Principle, when you increase the amount of one of the reactant (in this case NO2(g)), the equilibrium will shift to the right side to try and 'use up' that extra NO2 added, this will cause more SO3(g) to be produced and reach a new equilibrium state.
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Identify the following as combination, decomposition, replacement, or ion exchange reactions: ZnCO3(s) → ZnO(s) + CO2 ↑ 2 NaBr(aq) + Cl2(g) → 2 NaCl(aq) + Br2(g) 2 Al(s) + 3 Cl2(g) → 2 AlCl3(s) Ca(OH)2(aq) + H2SO4(aq) → CaSO4(aq) + 2 H2O(l) Pb(NO3)2(aq) + H2S(g) → 2 HNO3(aq) + PbS↓ C(s) + ZnO(s) → Zn(s) + CO↑
Answer:
Explanation:
ZnCO₃(s) → ZnO(s) + CO₂ ↑
Decomposition reaction:
This involves the breakdown of a compound into its component individual elements or other compounds.
2NaBr(aq) + Cl₂(g) → 2NaCl(aq) + Br₂(g)
Replacement reaction
This is a single replacement reaction in which one substance replaces the other.
2Al(s) + 3Cl₂(g) → 2AlCl₃(s)
Combination reaction:
Here, a single compound forms from two or more reacting species.
Ca(OH)₂(aq) + H₂SO₄(aq) → CaSO₄(aq) + 2H₂O(l)
Replacement reaction:
This is a double replacement reaction in which ions are exchanged to form new compounds.
Pb(NO₃)₂(aq) + H₂S(g) → 2HNO₃(aq) + PbS↓
Replacement reaction:
This is a double replacement reaction in which ions are exchanged to form new compounds.
C(s) + ZnO(s) → Zn(s) + CO↑
Replacement reaction
This is a single replacement reaction in which one substance replaces the other.
Answer:
ZnCO₃(s) → ZnO(s) + CO₂ ↑ Decomposition
2NaBr(aq) + Cl₂(g) → 2NaCl(aq) + Br₂(g) Replacement
2Al(s) + 3Cl₂(g) → 2AlCl₃(s) Combination
Ca(OH)₂(aq) + H₂SO₄(aq) → CaSO₄(aq) + 2H₂O(l) Ion Exchange.
Pb(NO₃)₂(aq) + H₂S(g) → 2HNO₃(aq) + PbS↓ Ion Exchange.
C(s) + ZnO(s) → Zn(s) + CO↑ Replacement.
Explanation:
We have a st of reaction, we need to identify them as Combination, decomposition, replacement or ion exchange reactions:
First reaction is:
ZnCO₃(s) → ZnO(s) + CO₂ ↑
This reaction involves the breakdown of the reagent into its component individual elements or other compounds. We can identify this reaction as: Decomposition reaction.
Second Reaction is:
2NaBr(aq) + Cl₂(g) → 2NaCl(aq) + Br₂(g)
This is a single replacement reaction in which Cl replaces the Br in the molecule. We can identify this reaction as: Replacement reaction.
Third reaction is:
2Al(s) + 3Cl₂(g) → 2AlCl₃(s)
Here two reagents react to form a single product. We can identify this reaction as: Combination reaction:
Fourth reaction:
Ca(OH)₂(aq) + H₂SO₄(aq) → CaSO₄(aq) + 2H₂O(l)
In this reaction, SO₄ and OH Ions exchange position. We can identify this reaction as: Ion Exchange.
Fifth reaction:
Pb(NO₃)₂(aq) + H₂S(g) → 2HNO₃(aq) + PbS↓
In this reaction, S and NO₃ Ions exchange position. We can identify this reaction as: Ion Exchange.
Sixth reaction:
C(s) + ZnO(s) → Zn(s) + CO↑
This is a single replacement reaction in which Oxygen replaces Carbon. This reaction can be identified as: Replacement reaction.
When coal is burned to generate electricity, 70% of the primary energy in the coal is transformed into unused heat. This means that the energy conversion efficiency from chemical to electrical energy is _____
Answer:
0.3
Explanation:
The energy conversion efficiency is defined as the ratio of the useful output energy being generated by an energy machine to the input energy.It is denote by η.
The input and the useful output can be electric power, chemical, mechanical power, heat, etc.
Coal is being burnt to generate energy. Let the the input energy of coal be x units.
The energy lost in unused heat = 70% of x = 0.7x units
Useful output energy = 100% - 70% = 30% of x = 0.3x units
Thus,
η = Useful Energy/ Input energy
η = 0.3x/x = 0.3
Thus,
The energy conversion efficiency from chemical to electrical energy is 0.3.
Determine the molarity of a solution formed by dissolving 97.7 g libr in enough water to yield 750.0 ml of solution.
Answer:
1.5 M.
Explanation:
Molarity (M) is defined as the no. of moles of solute dissolved in a 1.0 L of the solution.M = (no. of moles of LiBr)/(Volume of the solution (L).
∵ no. of moles of LiBr = (mass/molar mass) of LiBr = (97.7 g)/(86.845 g/mol) = 1.125 mol.
Volume of the solution = 750.0 mL = 0.75 L.
∴ M = (no. of moles of luminol)/(Volume of the solution (L) = (1.125 mol)/(0.75 L) = 1.5 M.
Hello!
Determine the molarity of a solution formed by dissolving 97.7 g LiBr in enough water to yield 750.0 ml of solution.
We have the following data:
M (Molarity) =? (in mol / L)
m1 (mass of the solute) = 97.7 g
V (solution volume) = 750 ml → V (solution volume) = 0.75 L
MM (molar mass of LiBr)
Li = 6.941 u
Br = 79.904 u
---------------------------
MM (molar mass of LiBr) = 6.941 + 79.904
MM (molar mass of LiBr) = 86.845 g/mol
Now, let's apply the data to the formula of Molarity, let's see:
[tex]M = \dfrac{m_1}{MM*V}[/tex]
[tex]M = \dfrac{97.7}{86.845*0.75}[/tex]
[tex]M = \dfrac{97.7}{65.13375}[/tex]
[tex]M = 1.49999... \to \boxed{\boxed{M \approx 1.5\:mol/L}}\:\:\:\:\:\:\bf\green{\checkmark}[/tex]
________________________
________________________
*** Another way to solve is to find the number of moles (n1) and soon after finding the molarity (M), let's see:
[tex]n_1 = \dfrac{m_1\:(g)}{MM\:(g/mol)}[/tex]
[tex]n_1 = \dfrac{97.7\:\diagup\!\!\!\!\!g}{86.845\:\diagup\!\!\!\!\!g/mol}[/tex]
[tex]n = 1.12499.. \to \boxed{n_1 \approx 1.125\:mol}[/tex]
[tex]M = \dfrac{n_1\:(mol)}{V\:(L)}[/tex]
[tex]M = \dfrac{1.125\:mol}{0.75\:L}[/tex]
[tex]\boxed{\boxed{M = 1.5\:mol/L}}\:\:\:\:\:\:\bf\blue{\checkmark}[/tex]
_____________________
[tex]\bf\purple{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}[/tex]
What condition must be met for a battery to be rechargeable? Either its anode or its cathode must generate a gas as a result of the electrochemical reaction. It must generate electricity via an acid-base reaction rather than via an oxidation-reduction reaction. The battery must be open to the outside so that it can vent any internal pressure that builds up from gases within it. The electrochemical reaction of the battery must be reversible.
Answer:
The electrochemical reaction of the battery must be reversible.
Explanation:
An electrochemical cell is device in which chemical reactions produce electricity. There are two main types of electrochemical cells:
Primary cells: In primary cells, the chemical reaction through which electric current is generated is irreversible and hence cannot be re-charged.
Secondary cells: are those in which chemical reactions through which electricity is generated is reversible. Hence they can be re-charged.
A 0.8870 g sample of a mixture of NaCl and KCl is dissolved in water, and the solution is then treated with an excess of AgNO3 to yield 2.142 g of AgCl. Calculate the percent by mass of each compound in the mixture
Answer : The percent by mass of NaCl and KCl are, 92.22 % and 7.78 % respectively.
Explanation :
As we know that when a mixture of NaCl and KCl react with excess [tex]AgNO_3[/tex] then the silver ion react with the chloride ion in both NaCl and KCl to form silver chloride.
Let the mass of NaCl be, 'x' grams and the mass of KCl will be, (0.8870 - x) grams.
The molar mass of NaCl and KCl are, 58.5 and 74.5 g/mole respectively.
First we have to calculate the moles of NaCl and KCl.
[tex]\text{Moles of }NaCl=\frac{\text{Mass of }NaCl}{\text{Molar mass of }NaCl}=\frac{xg}{58.5g/mole}=\frac{x}{58.5}moles[/tex]
[tex]\text{Moles of }KCl=\frac{\text{Mass of }KCl}{\text{Molar mass of }KCl}=\frac{(0.8870-x)g}{74.5g/mole}=\frac{(0.8870-x)}{74.5}moles[/tex]
As, each mole of NaCl and KCl gives one mole of chloride ions.
So, moles of chloride ions in NaCl = [tex]\frac{x}{58.5}moles[/tex]
Moles of chloride ions in KCl = [tex]\frac{(0.8870-x)}{74.5}moles[/tex]
The total moles of chloride ions = [tex]\frac{x}{58.5}moles+\frac{(0.8870-x)}{74.5}moles[/tex]
Now we have to calculate the moles of AgCl.
As we know that, this amount of chloride ion is same as the amount chloride ion present in the AgCl precipitate. That means,
Moles of AgCl = Moles of chloride ion = [tex]\frac{x}{58.5}moles+\frac{(0.8870-x)}{74.5}moles[/tex]
Now we have to calculate the moles of AgCl.
The molar mass of AgCl = 143.32 g/mole
[tex]\text{Moles of }AgCl=\frac{\text{Mass of }AgCl}{\text{Molar mass of }AgCl}=\frac{2.142g}{143.32g/mole}=0.0149moles[/tex]
Now we have to determine the value of 'x'.
Moles of AgCl = [tex]\frac{x}{58.5}moles+\frac{(0.8870-x)}{74.5}moles[/tex]
0.0149 mole = [tex]\frac{x}{58.5}moles+\frac{(0.8870-x)}{74.5}moles[/tex]
By solving the term, we get the value of 'x'.
[tex]x=0.818g[/tex]
The mass of NaCl = x = 0.818 g
The mass of KCl = (0.8870 - x) = 0.8870 - 0.818 = 0.069 g
Now we have to calculate the mass percent of NaCl and KCl.
[tex]\text{Mass percent of }NaCl=\frac{\text{Mass of }NaCl}{\text{Total mass of mixture}}\times 100=\frac{0.818g}{0.8870g}\times 100=92.22\%[/tex]
[tex]\text{Mass percent of }KCl=\frac{\text{Mass of }KCl}{\text{Total mass of mixture}}\times 100=\frac{0.069g}{0.8870g}\times 100=7.78\%[/tex]
Therefore, the percent by mass of NaCl and KCl are, 92.22 % and 7.78 % respectively.
The percentages of NaCl and KCl in the chemical mixture are 97.97% and 125.53% respectively, but these percentages raise issues with the experimental results as KCl exceeds 100%. Conversion from grams to moles and vice versa is essential in this calculation.
Explanation:To answer your question, let's start by figuring out the amount of NaCl and KCl in the mixture. When mixed with AgNO3, both these compounds yield AgCl. The given mass of AgCl is 2.142 g, which we can convert into moles using its molar mass (143.32 g/mol), yielding 0.0149 mol of AgCl.
The molar amounts of NaCl and KCl in the original mixture are also equal to this value because each NaCl or KCl molecule yields one AgCl molecule. Now, let's convert these molar amounts back into grams using the molar masses of NaCl (58.44 g/mol) and KCl (74.56 g/mol), yielding 0.869 g of NaCl and 1.113 g of KCl.
However, these do not add up to the original sample weight (0.887g). This can be due to experimental errors or purity issues. Nevertheless, to calculate the percentage mass of each compound in the mixture, you can use the formula: (observed mass / total mass) * 100%, hence, the percentages are: (0.869 g / 0.8870 g) * 100% = 97.97% NaCl and (1.113 g / 0.8870 g) * 100% = 125.53% KCl.
Yet, these results seem incorrect due to the KCl percentage being higher than 100%, suggesting a need for a re-check of experimental results or re-calculation.
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The scientific principle which is the basis for balancing chemical equations is
Answer:
Law of Conservation of Mass
Explanation:
The Law of Conservation of Matter forms the basis for balancing chemical equations, ensuring that the number of each element is equal on both sides of the equations. This principle also serves to describe a reaction's stoichiometry, where amounts of reactants and products in a reaction are considered.
Explanation:The scientific principle which forms the basis for balancing chemical equations is the Law of Conservation of Matter. According to this principle, the same number of each element must be represented on the reactant (input) and product (output) sides of an equation. This ensures that equations accurately reflect the reality that matter is not created or destroyed in a chemical reaction.
For instance, consider the following chemical equation: PC15 (s) + H₂O(1) →→→ POC13 (1) + 2HCl(aq). In this equation, we balance the equation by ensuring that for each of the elements involved (P, C, and H), the number of atoms of that element is equal on both sides of the equation.
Beyond simply balancing the equation, this principle also helps describe a reaction's stoichiometry, which involve the relationships between amounts of reactants and products. Coefficients from the balanced equation can be used in computations relating to reactant and product masses, molar amounts, and other quantitative properties.
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Using the phase diagram for CO2, what phase is carbon dioxide in at -60C and 1 atm pressure? A. Solid B. Gas C. At the melting point D. Liquid
Answer:
B. Gas.
Explanation:
Kindly, see the attached image of CO₂ phase diagram.Extrapolating the T from - 60.0°C and P from 1.0 atm, you will find that CO₂ is in the gas phase.So, the right choice is: B. Gas.Using the phase diagram for CO2, at -60°C and 1 atm pressure, carbon dioxide is in the solid phase. This phase is also known as dry ice.
Explanation:I understand you're asking about the phase of CO2 at specific conditions using a phase diagram. A phase diagram is a graphical representation of the state of a substance at different temperatures and pressures. In the case of CO2 at -60°C and 1 atm pressure, we refer to its phase diagram. According to the CO2 phase diagram, at -60°C and 1 atm pressure, CO2 is in the solid phase, so the correct answer is A. Solid. This phase of CO2 is also known as dry ice.
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What is the maximum number of electrons that can be contained in the first level
Your answer is:
The first level (or shell) can hold up to 2 electrons.
Hope this helps! C:
What temperature must a gas initially at 10 °c be brought to for the pressure to triple?
Answer:
I think that depends on the type of gas and the volume of the container.
How can the value of Ksp be related to the molar solubility of a compound? View Available Hint(s) How can the value of Ksp be related to the molar solubility of a compound? When a common ion is present, the solubility of the compound decreases, and this is reflected in a lower value of Ksp. The square of a compound's molar solubility equals the value of Ksp for the compound. The value of Ksp equals the concentration of the compound in a saturated solution, which can be converted to the molar solubility using the molar mass. The molar solubility can be used to calculate the concentrations of ions in solution, which in turn are used to calculate Ksp.
Answer:
The molar solubility can be used to calculate the concentrations of ions in solution, which in turn are used to calculate Ksp.
Explanation:
Consider a slightly soluble solid with formula M₃X₂. Its solubility product expression is
[tex]\begin{array}{rcccc}M_{3}X_{2}(s) & \rightleftharpoons&3M^{2+}(aq) & + & 2X^{3-}(aq)\\& & 3s & &2s\\\\K_\text{sp}& = & [3s]^{3}[2s]^{2}&= & 108s^{5}\\\\\end{array}[/tex]
Thus, the molar solubility can be used to calculate the concentrations of ions in solution, which in turn are used to calculate Ksp.
A is wrong. The solubility product constant is a constant. It does not change in the presence of a common ion.
B is wrong. It is correct only for compounds with formula MX.
C is wrong. Ksp does not equal the concentration of the compound in solution.
The value of Ksp is related to the molar solubility in that molar solubility helps calculate the ion concentrations in a solution, which are used to calculate Ksp. The presence of a common ion affects both solubility and Ksp through the common ion effect.
The value of Ksp, or the solubility product constant, is closely related to the molar solubility of a compound, which is a measure of how much of the compound can dissolve in a given amount of solvent to form a saturated solution. The connection between Ksp and molar solubility can be understood through a series of steps:
First, molar solubility is calculated by converting the solubility of a compound (often given in g/L) to moles per liter using the molar mass of the compound.Next, the dissociation equation of the compound is used to determine the concentration of each ion in the solution.Finally, these ion concentrations are used in the Ksp expression to calculate the solubility product constant.Furthermore, if a common ion is present in the solution, it affects the solubility of the compound and thus the Ksp value. This is a demonstration of the common ion effect, where the presence of a common ion decreases the solubility and Ksp of the compound according to Le Chatelier's principle.
Therefore, understanding the relationship between Ksp and molar solubility is essential for predicting the solubility of compounds and determining the possible concentrations of ions in solution.
If you start with 512 grams of aluminum and 1147 grams of copper chloride to make aluminum chloride and copper, what is the limiting reagent? 2Al + 3CuCl -> 2AlCl3 + 3Cu
First you need to calculate the number of moles of aluminium and copper chloride.
number of moles = mass / molecular weight
moles of Al = 512 / 27 = 19 moles
moles of CuCl = 1147 / 99 = 11.6 moles
From the reaction you see that:
if 2 moles of Al will react with 3 moles of CuCl
then 19 moles of Al will react with X moles of CuCl
X = (19 × 3) / 2 = 28.5 moles of CuCl, way more that 11.6 moles of CuCl wich is the quantity you have. So the copper chloride is the limiting reagent.
The octet rule pertains to:
a. only the noble gases.
b. only groups 1A and 2A.
c. elements in Groups 1A - 7A.
d. all elements in the periodic table.
Of the following elements, which could be classified as a metalloid (semimetal)? Select all that apply.
(A) Mg
(B) Ti
(C) Ge
(D) Si
(E) Rn
(F) Au
(G) Bi
(H) At
(I) Br
(J) B
(K) Eu
Answer:
(C) Ge
(D) Si
(H) At
(J) B
Hope this helps!
What volume of 24% trichloroacetic acid (tca) is needed to prepare eight 3 ounce bottles of 10% tca solution?
Answer:
295.7 mL of 24% trichloroacetic acid (tca) is needed .
Explanation:
Let the volume of 24% trichloroacetic acid solution be x
Volume of required 10% trichloroacetic acid solution =8 bottles of 3 ounces
= 24 ounces = 709.68 mL
(1 ounces = 29.57 mL)
Amount of trichloroacetic acid in 24% solution of x volume of solution will be equal to amount of trichloroacetic acid in 10% solution of volume 709.68 mL.
[tex]x\times \frac{24}{100}=709.68 mL\times \frac{10}{100}[/tex]
x = 295.7 mL
295.7 mL of 24% trichloroacetic acid (tca) is needed .
How many grams of barium sulfate can be produced from the reaction of 2.54 grams sodium sulfate and 2.54 g barium chloride? Na2SO4(aq) + BaCl2(aq) --> BaSO4(s) + 2NaCl(aq) Report your answer to 3 decimal places.
Answer: 2.796 grams
Explanation:
[tex]Na_2SO_4+BaCl_2\rightarrow 2NaCl+BaSO_4[/tex]
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
[tex]\text{Number of moles of sodium sulphate}=\frac{2.54g}{142g/mol}=0.018moles[/tex]
[tex]\text{Number of moles of barium chloride}=\frac{2.54g}{208g/mol}=0.012moles[/tex]
According to stoichiometry:
1 mole of [tex]BaCl_2[/tex] reacts with 1 mole of [tex]Na_2SO_4[/tex]
0.012 moles of [tex]BaCl_2[/tex] will react with=[tex]\frac{1}{1}\times 0.012=0.012moles[/tex] of [tex]Na_2SO_4[/tex]
Thus [tex]BaCl_2[/tex] is the limiting reagent as it limits the formation of product. [tex]Na_2SO_4[/tex] is the excess reagent as (0.018-0.012)=0.006 moles are left unused.
1 mole of [tex]BaCl_2[/tex] produces 1 mole of [tex]BaSO_4[/tex]
0.012 moles of [tex]BaCl_2[/tex] will produce=[tex]\frac{1}{1}\times 0.012=0.012moles[/tex] of [tex]BaSO_4[/tex]
Mass of [tex]BaSO_4=moles\times {\text {Molar mass}}=0.012\times 233=2.796g[/tex]
Thus 2.796 grams of [tex]BaSO_4[/tex] are produced.
Which is the part of an experiment that serves as the point of comparison for the results?
hypothesis
independent variable
constant
control
What is the percent-by-mass concentration of acetic acid (CH3COOH) in a vinegar solution that contains 51.80 g acetic acid in a 1.000−L solution? The density of this solution is 1.005 g/mL.
Final answer:
The percent-by-mass concentration of acetic acid in the vinegar solution is approximately 5.15%, calculated by dividing the acetic acid mass by the total solution mass and then multiplying by 100.
Explanation:
To find the percent-by-mass concentration of acetic acid in vinegar, the mass of acetic acid is divided by the total mass of the solution and then multiplied by 100. First, convert the solution volume to mass using the given density. The density of the solution is 1.005 g/mL, which means 1.000 L (or 1000 mL) of solution has a mass of 1005 g (1000 mL × 1.005 g/mL). The mass of acetic acid is given as 51.80 g. Thus, the percent-by-mass concentration is calculated as (51.80 g / 1005 g) × 100%.
The calculation gives a percent-by-mass concentration of approximately 5.15%. This value represents the mass of acetic acid expressed as a percentage of the total mass of the solution.
Determination of the tertiary structure of a membrane protein finds that the outer surface is composed primarily of hydrophobic residues. Which conclusion is most likely from this observation?
Answer:
The hydrophobic residues functions as a binding agent for the protein to the membrane
Explanation:
The membrane is a phospholipid bilayer that consists primarily of fatty acetyl groups. The protein has hydrophobic side chains that interacts with those fatty acetyl groups, forming a bond that will basically have the the protein anchored to the phospholipid bilayer membrane.
one method for generating chlorine gas is by reacting potassium permanganate and hydrochloric acid. how many liters of Cl2 at 40 C and a pressure of 1.05 atm can be produced by the reaction of 6.23 g KMnO4 with 45.0 ml of 6.00 m HCl?
Answer: The volume of chlorine gas produced in the reaction is 2.06 L.
Explanation:
For potassium permanganate:To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of potassium permanganate = 6.23 g
Molar mass of potassium permanganate = 158.034 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of potassium permanganate}=\frac{6.23g}{158.034g/mol}=0.039mol[/tex]
For hydrochloric acid:To calculate the moles of hydrochloric acid, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
Molarity of HCl = 6.00 M
Volume of HCl = 45.0 mL = 0.045 L (Conversion factor: 1 L = 1000 mL)
Putting values in above equation, we get:
[tex]6.00mol/L=\frac{\text{Moles of HCl}}{0.045L}\\\\\text{Moles of HCl}=0.27mol[/tex]
For the reaction of potassium permanganate and hydrochloric acid, the equation follows:[tex]2KMnO_4+16HCl\rightarrow 2MnCl_2+5Cl_2+2KCl+8H_2O[/tex]
By Stoichiometry of the reaction:
16 moles of hydrochloric acid reacts with 2 moles of potassium permanganate.
So, 0.27 moles of hydrochloric acid will react with = [tex]\frac{2}{16}\times 0.27=0.033moles[/tex] of potassium permanganate.
As, given amount of potassium permanganate is more than the required amount. So, it is considered as an excess reagent.
Thus, hydrochloric acid is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
16 moles of hydrochloric acid reacts with 5 moles of chlorine gas.
So, 0.27 moles of hydrochloric acid will react with = [tex]\frac{5}{16}\times 0.27=0.0843moles[/tex] of chlorine gas.
To calculate the volume of gas, we use the equation given by ideal gas equation:[tex]PV=nRT[/tex]
where,
P = pressure of the gas = 1.05 atm
V = Volume of gas = ? L
n = Number of moles = 0.0843 mol
R = Gas constant = [tex]0.0820\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature of the gas = [tex]40^oC=[40+273]K=313K[/tex]
Putting values in above equation, we get:
[tex]1.05atm\times V=0.0843\times 0.0820\text{ L atm }mol^{-1}K^{-1}\times 313K\\\\V=2.06L[/tex]
Hence, the volume of chlorine gas produced in the reaction is 2.06 L.
Final answer:
The question involves a stoichiometry problem in chemistry, where we calculate the volume of chlorine gas from the reaction of potassium permanganate with hydrochloric acid. We identify the limiting reactant and then use the ideal gas law to find the volume under the given conditions of temperature and pressure.
Explanation:
The student's question concerns a chemical reaction between potassium permanganate (KMnO4) and hydrochloric acid (HCl) to produce chlorine gas (Cl2). This is a stoichiometry problem where we will calculate the volume of chlorine gas generated at specific conditions using the ideal gas law. To find the number of moles of chlorine gas that can be produced, we first determine the limiting reactant. Then, using the ideal gas law (PV=nRT), we can calculate the volume at the given temperature and pressure.
First, we must find the reaction equation:
KMnO4 + 16HCl → 2KCl + 5Cl2 + 2MnCl2 + 8H2O
Next, we calculate the moles of KMnO4 and HCl. Following this, we determine the limiting reactant and use it to calculate moles of Cl2 produced. Lastly, we use the ideal gas law (PV=nRT, with R=0.0821 L·atm/(mol·K)) to find the volume of Cl2 at 40°C and 1.05 atm.
A calorimeter contains 22.0 mL of water at 14.0 ∘C . When 2.50 g of X (a substance with a molar mass of 82.0 g/mol ) is added, it dissolves via the reaction X(s)+H2O(l)→X(aq) and the temperature of the solution increases to 28.0 ∘C . Calculate the enthalpy change, ΔH, for this reaction per mole of X. Assume that the specific heat of the resulting solution is equal to that of water [4.18 J/(g⋅∘C)], that density of water is 1.00 g/mL, and that no heat is lost to the calorimeter itself, nor to the surroundings.
Answer:
The enthalpy change in the the reaction is -47.014 kJ/mol.
Explanation:
[tex]X(s)+H_2O(l)\rightarrow X(aq)[/tex]
Volume of water in calorimeter = 22.0 mL
Density of water = 1.00 g/mL
Mass of the water in calorimeter = m
[tex]m=1.00 g/mL\times 22.0 mL=22 g[/tex]
Mass of substance X = 2.50 g
Mass of the solution = M = 2.50 g + 22 g = 24.50 g
Heat released during the reaction be Q
Change in temperature =ΔT = 28.0°C - 14.0°C = 14.0°C
Specific heat of the solution is equal to that of water :
c = 4.18J/(g°C)
[tex]Q=Mc\times \Delta T[/tex]
[tex]Q=24.50 g\times 4.18 J/g ^oC\times 14.0^oC=1,433.74 J=1.433 kJ[/tex]
Heat released during the reaction is equal to the heat absorbed by the water or solution.
Heat released during the reaction =-1.433 kJ
Moles of substance X= [tex]\frac{2.50 g}{82.0 g/mol}=0.03048 mol[/tex]
The enthalpy change, ΔH, for this reaction per mole of X:
[tex]\Delta H=\frac{-1.433 kJ}{0.03048 mol}=-47.014 kJ/mol[/tex]
2.50 g of X dissolves in a calorimeter containing 22.0 mL of water, causing the temperature to increase from 14.0 °C to 28.0 °C. The enthalpy change of the reaction is -46.9 kJ/mol.
First, we will convert 22.0 mL of water to grams using its density (1.00 g/mL).
[tex]22.0 mL \times \frac{1.00g}{mL} = 22.0 g[/tex]
The solution contains 22.0 g of water and 2.50 g of X. The mass of the solution is:
[tex]m = 22.0 g + 2.50 g = 24.5 g[/tex]
We can calculate the heat absorbed by the solution (Qs) using the following expression.
[tex]Qs = c \times m \times \Delta T = \frac{4.18J}{g. \° C } \times 24.5 g \times (28.0 \° C - 14.0 \° C) = 1.43 \times 10^{3} J = 1.43 kJ[/tex]
According to the law of conservation of energy, the sum of the heat absorbed by the solution and the heat released by the reaction (Qr) is zero.
[tex]Qs + Qr = 0\\\\Qr = -Qs = -1.43 kJ[/tex]
Then, we will convert 2.50 g of X to moles using its molar mass (82.0 g/mol).
[tex]n = 2.50 g \times \frac{1mol}{82.0g} = 0.0305 mol[/tex]
Finally, we will calculate the enthalpy change, ΔH, for this reaction per mole of X using the following expression.
[tex]\Delta H = \frac{Qr}{n} = \frac{-1.43 kJ}{0.0305mol} = -46.9 kJ/mol[/tex]
2.50 g of X dissolves in a calorimeter containing 22.0 mL of water, causing the temperature to increase from 14.0 °C to 28.0 °C. The enthalpy change of the reaction is -46.9 kJ/mol.
Learn more: https://brainly.com/question/16104165
Select all that apply: The melting points of organic compounds are usually lower than those of inorganic compounds. The boiling points of organic compounds are usually lower than those of inorganic compounds. The flammability of organic compounds is usually lower than that of inorganic compounds.
Answer:
The melting points of organic compounds are usually lower than those of inorganic compounds.
The boiling points of organic compounds are usually lower than those of inorganic compounds.
Explanation:
Organic compounds are compounds of carbon with exception of simple ones like oxides, carbides and carbonates. Inorganic compounds are other compounds excluding organic compounds.
Inorganic compounds due to the nature of their bonds typically have higher melting and boiling points compared to organic compounds.
Organic compounds are also highly flammable and would easily burn compared to inorganic ones.
Use dimensional analysis to convert 14.5mi/hr to km/s
Answer:
0.006 48 km/s
Explanation:
1. Convert miles to kilometres
14.5 mi × (1.609 km/1 mi) = 23.33 km
2. Convert hours to seconds
1 h × (60 min/1h) × (60 s/1 min) = 3600 s
3. Divide the distance by the time
14.5 mi/1 h = 23.3 km/3600 s = 0.006 48 km/s
Which of the following is a valid conversion factor?
1m/10 cm
10 hm/1 m
1 cm/1L
1 kg/1000 g
Answer:
1kg/1000g so D.
Explanation:
Answer: The correct answer is 1 kg = 1000 g
Explanation:
Unit is defined as the quantity that is used as a standard for measurement.
Meter, Centimeters, hectometers are the units which is used to express length of a substance.
Liter is the unit which is used to express volume of a substance.
Kilogram, grams are the units which is used to express mass of a substance.
All the units of a particular parameter are inter changeable.
For the given options:
Option 1: 1 m/10 cmThe conversion of meter to centimeters is:
1 m = 100 cm
Thus, this is not a valid conversion factor.
Option 2: 10 hm/1 mThe conversion of meter to hectometers is:
1 m = 100 hm
Thus, this is not a valid conversion factor.
Option 3: [tex]1cm^3/1L[/tex]The conversion of cubic centimeter to liters is:
[tex]1L=1000cm^3[/tex]
Thus, this is not a valid conversion factor.
Option 4: 1 kg/1000 gThe conversion of kilograms to grams is:
1 kg = 1000 g
Thus, this is a valid conversion factor.
Hence, the correct answer is 1 kg = 1000 g