This is an incomplete question, here is a complete question.
Determine whether each of the following electron configurations is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choices.
(a) [tex]1s^22s^22p^63s^23p^5[/tex]
(b) [tex]1s^22s^22p^63s^23p^63d^74s^2[/tex]
(c) [tex]1s^22s^22p^63s^23p^63d^{10}4s^24p^6[/tex]
(d) [tex]1s^22s^22p^63s^23p^63d^4s^1[/tex]
Answer :
(a) [tex]1s^22s^22p^63s^23p^5[/tex] → Halogen
(b) [tex]1s^22s^22p^63s^23p^63d^74s^2[/tex] → Transition metal
(c) [tex]1s^22s^22p^63s^23p^63d^{10}4s^24p^6[/tex] → Transition metal
(d) [tex]1s^22s^22p^63s^23p^63d^4s^1[/tex] → Transition metal
Explanation :
Inert gas : These are the gases which lie in group 18.
Their general electronic configuration is: [tex]ns^2np^6[/tex] where n is the outermost shell.
Halogen : These are the elements which lie in group 17.
Their general electronic configuration is: [tex]ns^2np^5[/tex] where n is the outermost shell.
An alkali metal : These are the elements which lie in group 1.
Their general electronic configuration is: [tex]ns^1[/tex] where n is the outermost shell.
An alkaline earth metal : These are the elements which lie in group 2.
Their general electronic configuration is: [tex]ns^2[/tex] where n is the outermost shell.
Transition elements : They are the elements which lie between 's' and 'p' block elements. These are the elements which lie in group 3 to 12. The valence electrons of these elements enter d-orbital.
Their general electronic configuration is: [tex](n-1)d^{1-10}ns^{0-2}[/tex] where n is the outermost shell.
(a) [tex]1s^22s^22p^63s^23p^5[/tex]
The element having this electronic configuration belongs to the halogen family.
(b) [tex]1s^22s^22p^63s^23p^63d^74s^2[/tex]
The element having this electronic configuration belongs to the transition family.
(c) [tex]1s^22s^22p^63s^23p^63d^{10}4s^24p^6[/tex]
The element having this electronic configuration belongs to the transition family.
(d) [tex]1s^22s^22p^63s^23p^63d^4s^1[/tex]
The element having this electronic configuration belongs to the transition family.
1. [tex][Ar]4s^2 3d^10 4p^6[/tex] is an inert gas (krypton, Kr).
2. [tex][Kr]5s^2 4d^10 5p^2[/tex] is a transition metal (zinc, Zn).
3. [tex][He]2s^1[/tex] is an alkali metal (lithium, Li).
4. [tex][Ne]3s^2 3p^5[/tex] is a halogen (chlorine, Cl).
5. [tex][Ar]4s^2[/tex] is an alkaline earth metal (calcium, Ca).
6. [tex][Xe]6s^2 4f^14 5d^10 6p^3[/tex] is not a transition metal but a post-transition metal (bismuth, Bi).
To determine the type of element based on its electron configuration, one must understand the characteristic configurations of the different groups of elements in the periodic table. Here are some general rules:
- Inert gases (also known as noble gases) have completely filled s and p subshells in their valence shell. Their configurations typically end in [tex]ns^2 np^6[/tex], except for helium, which has [tex]1s^2[/tex].
- Halogens have the [tex]ns^2 np^5[/tex] configuration in their valence shell, with the exception of helium, which has [tex]1s^2[/tex].
- Alkali metals have a single electron in their outermost s subshell, denoted as [tex]ns^1[/tex].
- Alkaline earth metals have two electrons in their outermost s subshell, denoted as [tex]ns^2[/tex].
- Transition metals have incompletely filled d subshells in their penultimate energy level. Their configurations typically start with [tex](n-1)d^1 to (n-1)d^10[/tex] in the valence shell.
Let's analyze the given electron configurations:
1. [tex][Ar]4s^2 3d^10 4p^6[/tex]
- This configuration has a filled 4s subshell, a filled 3d subshell (10 electrons), and a filled 4p subshell. It corresponds to the configuration of an inert gas, specifically krypton (Kr), which has the full configuration [tex][Ar]4s^2 3d^10 4p^6[/tex].
2. [tex][Kr]5s^2 4d^10 5p^2[/tex]
- This configuration has a filled 5s subshell and a filled 4d subshell, but the 5p subshell is not filled (only 2 electrons out of a possible 6). This does not correspond to an inert gas, a halogen, an alkali metal, or an alkaline earth metal. It is a configuration of an element in group 12, which is a transition metal. Specifically, this is the configuration for zinc (Zn).
3. [tex][He]2s^1[/tex]
- This configuration has a single electron in the 2s subshell, which is characteristic of an alkali metal. Specifically, this is the configuration for lithium (Li).
4. [tex][Ne]3s^2 3p^5[/tex]
- This configuration has a filled 3s subshell and the 3p subshell is missing one electron to be filled (5 electrons out of a possible 6). This is the configuration of a halogen, specifically chlorine (Cl).
5. [tex][Ar]4s^2[/tex]
- This configuration has a filled 4s subshell, which is characteristic of an alkaline earth metal. Specifically, this is the configuration for calcium (Ca).
6. [tex][Xe]6s^2 4f^14 5d^10 6p^3[/tex]
- This configuration has a filled 6s subshell, a filled 4f subshell (14 electrons), and a filled 5d subshell (10 electrons), but the 6p subshell is not filled (only 3 electrons out of a possible 6). This does not correspond to an inert gas, a halogen, an alkali metal, or an alkaline earth metal. It is the configuration of an element in group 13, which is not a transition metal but a post-transition metal. Specifically, this is the configuration for bismuth (Bi).
In summary:
1. [tex][Ar]4s^2 3d^10 4p^6[/tex] is an inert gas (krypton, Kr).
2. [tex][Kr]5s^2 4d^10 5p^2[/tex] is a transition metal (zinc, Zn).
3. [tex][He]2s^1[/tex] is an alkali metal (lithium, Li).
4. [tex][Ne]3s^2 3p^5[/tex] is a halogen (chlorine, Cl).
5. [tex][Ar]4s^2[/tex] is an alkaline earth metal (calcium, Ca).
6.[tex][Xe]6s^2 4f^14 5d^10 6p^3[/tex] is not a transition metal but a post-transition metal (bismuth, Bi).
127 grams of copper metal react with 32 grams of oxygen gas to form a new compound, with no copper or oxygen left over. What is the mass of the new compound in grams?
Answer: The mass of the new compound formed will be 159 grams.
Explanation:
Synthesis reaction is defined as the reaction in which smaller substances combine in their elemental state to form a larger substance.
[tex]A+B\rightarrow AB[/tex]
Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.
This also means that total mass on the reactant side must be equal to the total mass on the product side.
The chemical equation for the synthesis of copper oxide follows:
[tex]2Cu+O_2\rightarrow 2CuO[/tex]
Let the mass of new compound (copper oxide) formed be 'x' grams
We are given:
Mass of copper metal = 127 grams
Mass of oxygen gas = 32 grams
Total mass on reactant side = 127 + 32 = 159 g
Total mass on product side = x
So, by applying law of conservation of mass, we get:
x = 159 g
Hence, the mass of the new compound formed will be 159 grams.
A mineral sample is obtained from a region of the country that has high arsenic contamination. An elemental analysis yields the following elemental composition:
Element Atomic Weight (g/mol) Percent Composition
Ca 40.078 22.3%
As 74.9216 41.6%
O 15.9994 35.6%
H 1.00794 60%
What is the empirical formula of this mineral?
Answer:
CaAsHO₄
Explanation:
The data has a mistake in one of the values there. I believe the mistake is on the hydrogen. So, I'm going to assume the value of Hydrogen is 0.6%, so the total percent composition would be 100.1% (Something better). All you have to do is replace the correct value of H (or the value with the mistaken option) and do the same procedure.
Now, to calculate the empirical formula, we can do this in three steps.
Step 1. Calculate the amount in moles of each element.
In these case, we just divide the percent composition with the molar mass of each one of them:
Ca: 22.3 / 40.078 = 0.5564
As: 41.6 / 74.9216 = 0.5552
O: 35.6 / 15.9994 = 2.2251
H: 0.6 / 1.00794 = 0.5953
Now that we have done this, let's calculate the ratio of mole of each of them. This is doing dividing the smallest number of mole between each of the moles there. In this case, the moles of As are the smallest so:
Ca: 0.5564/0.5552 = 1.0022
As: 0.5552/0.5552 = 1
O: 2.2251/0.5552 = 4.0077
H: 0.5953/0.5552 = 1.0722
Now, we round those numbers, and that will give us the number of atoms of each element in the empirical formula
Step 3. Write the empirical formula with the rounded numbers obtained
In this case we will have:
Ca: 1
As: 1
O: 4
H: 1
The empirical formula would have to be:
CaAsHO₄
Write the rate law for the following elementary reaction:
2N2O5(g) - 2N204 Use k1 to stand for the rate constant.
Answer:
Rate law = k1 [N2O5]^2
Explanation:
Rate law only cares about reactants and rate law can only be determined experimentally or by using the coefficients of reactants in elementary reactions
If 2A---->B is an elementary reaction... rate law for this is rate=k[A]^2
So if we look at your question...2N2O5---->2N2O4 + 02 (I think you are missing O2 in question because otherwise equation is unbalanced)
Rate law = k1 [N2O5]^2
The rate law for the reaction 2N2O5(g) → 2N2O4(g) is rate = k1[N2O5]^2, indicating a second-order dependence on [N2O5]. For the reaction H2(g) + 2NO(g) → N2O(g) + H2O(g), the rate law is rate = k[NO]^2[H2], with the reaction being second order in NO and first order in H2.
Explanation:The rate law for an elementary reaction can be written directly from the stoichiometry of the reaction. Considering the provided elementary reaction 2N2O5(g) → 2N2O4(g), the rate law would be rate = k1[N2O5]^2, which indicates that the reaction is second order with respect to N2O5.
In general, for an elementary reaction like H2(g) + 2NO(g) → N2O(g) + H2O(g), the rate law given is rate = k[NO]^2[H2], which means that the reaction is second order with respect to NO (order of 2) and first order with respect to H2 (order of 1), making the overall order of the reaction three.
Study this chemical reaction: FeSO4 (aq) + Zn (s) --> Fe (s) + ZnSO4 (aq) Then, write balanced half-reactions describing the oxidation and reduction that happen in this reaction.
oxidation:
reduction:
Answer: The oxidation and reduction half reactions are written below.
Explanation:
Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.
[tex]X\rightarrow X^{n+}+ne^-[/tex]
Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.
[tex]X^{n+}+ne^-\rightarrow X[/tex]
For the given chemical reaction:
[tex]FeSO_4(aq.)+Zn(s)\rightarrow Fe(s)+ZnSO_4(aq.)[/tex]
The half reactions for the given reaction follows:
Oxidation half reaction: [tex]Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-[/tex]
Reduction half reaction: [tex]Fe^{2+}(aq.)+2e^-\rightarrow Fe(s)[/tex]
Hence, the oxidation and reduction half reactions are written above.
The balanced half-reaction for oxidation in the reaction is Zn (s) -> Zn2+ (aq) + 2e-, signifying Zinc is being oxidised. The reduction half-reaction is Fe2+ (aq) + 2e- -> Fe (s) demonstrating Iron being reduced.
Explanation:In this chemical reaction:FeSO4 (aq) + Zn (s) --> Fe (s) + ZnSO4 (aq) two processes occur simultaneously; oxidation and reduction. Oxidation involves loss of electrons, and in this case, it is the Zinc metal that gets oxidized. It unites with Sulphate forming ZnSO4 and in the process gains a positive charge by losing 2 electrons.
The balanced half-reaction for oxidation is: Zn (s) --> Zn2+ (aq) + 2e-
Reduction on the other hand involves gain of electrons. Iron in this reaction gains electrons to form metallic iron (Fe). This can be represented by the following balanced half-reaction:
Fe2+ (aq) + 2e- --> Fe (s).
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Calculate the radius of a platinum atom in cm, given that Pt has an FCC crystal structure, a density of 21.4 g/cm3, and an atomic weight of 195.08 g/mol.
Answer:
Radius of platinum atom is [tex]1,38\times10^{-8} \ cm.[/tex]
Explanation:
We know, Density in solid state is also given by , [tex]\rho=\dfrac{Z\times M}{N_A\times a^3}[/tex]
Here, Z= total no of atom or molecule per unit cell. ( Z = 4 for FCC structure )
M = atomic weight = 195.08 gm/mol.
N_A= Avogadro's number = [tex]6.023\times 10^{23} .[/tex]
a= size of side of cube.
Also for, FCC crystal [tex]a=\sqrt2\times 2r[/tex] ....1
Since,[tex]\rho=\dfrac{Z\times M}{N_A\times a^3}\\[/tex]
Therefore, [tex]a^3=\dfrac{Z\times M}{N_A\times \rho}[/tex]
Putting all values , [tex]a=3.9\times 10^{-8} \ cm.[/tex]
Putting value of a in equation 1 we get,
[tex]r=\dfrac{a}{\sqrt2\times2}=1.38\times 10^{-8} \ cm.[/tex]
Hence, this is the required solution.
The radius of a platinum atom can be found using the density, atomic weight, and the nature of the crystal structure. The calculation involves finding the volume of an atom from the atomic weight, density and Avogadro's number and converting this volume to a radius by considering the atom as a sphere.
Explanation:The radius of a platinum atom can be calculated using the density, atomic weight, and the information about the crystal structure (in this case, face-centered cubic or FCC). Platinum (Pt) has a density (d) of 21.4 g/cm3 and an atomic weight (AW) of 195.08 g/mol.
First, let's calculate the number of atoms (n) in the unit cell of an FCC structure. An FCC unit cell has 4 atoms. Avogadro's number (Na) is 6.022 x 1023 mol-1.
The volume (V) of an atom can be calculated as follows: V = AW / (d * n * Na).Now we will calculate the radius (r) from the volume, considering the atom as a sphere: r = ((3*V) / (4*π))1/3.Keep in mind that the values have to be converted to cm in the end.
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Consider a mixture of four proteins with various molecular weights. A histone molecule weighs 15 kDa, 15 kDa, a p53 molecule weighs 53 kDa, 53 kDa, an actin molecule weighs 42 kDa, 42 kDa, and an IgG molecule weighs 150 kDa. 150 kDa. Arrange the molecules in order of their elution from a gel filtration column.
Answer:
IgGp53ActinHistoneExplanation:
In gel filtration chromatography a porous matrix is used as the stationary phase. The smaller molecules are able to diffuse into this matrix while the larger ones are not. For this reason the larger molecules will elute first, followed by the smaller ones in decreasing order of size.
In other words the largest molecule will elute first and the smallest molecule will elute last:
IgGp53ActinHistoneWhat is the mass % of ammonium chloride in a 1.73 M ammonium chloride aqueous solution at 20 °C?
The question is incomplete, here is the complete question:
What is the mass % of ammonium chloride in a 1.73 M ammonium chloride aqueous solution at 20 °C? The density of the solution is 1.0257 g/mL
Answer: The mass percent of ammonium chloride in solution is 9.03 %
Explanation:
We are given:
Molarity of ammonium chloride solution = 1.73 M
This means that 1.73 moles of ammonium chloride is present in 1 L or 1000 mL of solution.
To calculate the mass of solution, we use the equation:[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
Density of solution = 1.0257 g/mL
Volume of solution = 1000 mL
Putting values in above equation, we get:
[tex]1.0257g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.0257g/mL\times 1000mL)=1025.7g[/tex]
To calculate the number of moles, we use the equation:[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of ammonium chloride = 1.73 moles
Molar mass of ammonium chloride = 53.5 g/mol
Putting values in above equation, we get:
[tex]1.73mol=\frac{\text{Mass of ammonium chloride}}{53.5g/mol}\\\\\text{Mass of ammonium chloride}=(1.73mol\times 53.5g/mol)=92.6g[/tex]
To calculate the mass percentage of ammonium chloride in solution, we use the equation:[tex]\text{Mass percent of ammonium chloride}=\frac{\text{Mass of ammonium chloride}}{\text{Mass of solution}}\times 100[/tex]
Mass of solution = 1025.7 g
Mass of ammonium chloride = 92.6 g
Putting values in above equation, we get:
[tex]\text{Mass percent of ammonium chloride}=\frac{92.6g}{1025.7g}\times 100=9.03\%[/tex]
Hence, the mass percent of ammonium chloride in solution is 9.03 %
write out the acid dissociation reaction for hydrochloric acid. b) calculate the pH of a solution of 5.0 x 10^-4 M HCl. c) write out the acid dissociation reaction for sodium hydroxide. d) calculate the pH of a solution of 7.0 x 10^-5 M NaOH.
Answer:
The answer to your question is below
Explanation:
a) Acid dissociation reaction for HCl
HCl(g) + H₂O(l) ⇒ H⁺¹ (aq) + Cl⁻¹ (aq)
b) pH = -log [H⁺¹]
Concentration = 5 x 10⁻⁴
pH = -log [5 x 10⁻⁴]
pH = 3.3
c) Acid dissociation reaction
NaOH(s) + H₂O ⇒ Na⁺¹(aq) + OH⁻¹(aq)
d) pH
pOH = -log [7 x 10⁻⁵]
pOH = 4.2
pH = 14 - 4.2
pH = 9.8
The acid dissociation and pH for HCl and NaOH were detailed. For HCl, the reaction and solution pH were HCl -> H+ + Cl- and 3.3 respectively. For NaOH, the dissociation and solution pH were NaOH -> Na+ + OH- and 9.8 respectively.
Explanation:a) The acid dissociation reaction for hydrochloric acid (HCl) is as follows:
HCl → H+ + Cl-
b) The pH of a solution of hydrochloric acid can be calculated using the formula:
pH = -log[H+].
As HCl is a strong acid, it will completely ionize in water, meaning the concentration of H+ ions equals the concentration HCl. Therefore, the pH for a 5.0 x 10^-4 M HCl solution is -log(5.0 x 10^-4) = 3.3.
c) Sodium hydroxide is a base, not an acid, so its ionization does not involve releasing protons into the solution, but rather attracting them. It has the following dissociation reaction:
NaOH → Na+ + OH-.
d) As for the pH of a solution of 7.0 x 10^-5 M NaOH, we first need to find the pOH using the formula pOH = -log[OH-], since NaOH being a strong base completely ionizes. This equals -log(7.0 x 10^-5) = 4.2. Given that pH and pOH are related by the expression pH + pOH = 14 at 25°C, we can solve for the pH of the NaOH solution = 14 - pOH = 14 - 4.2 = 9.8.
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The frequency of stretching vibrations is correlated to the strength and stiffness of the bond between two atoms. This can be thought of as a ball-and-spring model. Using this knowledge, rank the following bonds in each part of the question by increasing frequency.
(a) Alkyne
(b) Alkane
(c) alkene
Answer:
a > c > b
Explanation:
As higher is the strength and stiffness of the bond between two atoms, more stable it is, and more difficult is to these bonds vibrate. So, the stretching vibration decreases when the strength and stiffness increases.
As more bonds are done between the atoms, more strength, and stiffness they have. So, the order of increase is:
simple bond > double bond > triple bond
And the increased frequency of vibration is:
triple bond > double bond > simple bond
An alkane is a hydrocarbon that has only simple bonds between carbons, an alkene is a hydrocarbon with one double bond between carbon, and an alkyne is a hydrocarbon with one triple bond. So, the increase in vibration of them is:
alkyne (a) > alkene (c) > alkane (b)
Answer: (A) < (C) < (B)
Ranking in order of increasing frequency.
Explanation:
An acid that dissociates to the extent of 92% in water would be termed a strong acid.
True or Fasle?
Answer: False
Explanation:
Strong acid is defined as the acid which completely dissociates when dissolved in water. They have low pH. These releases ions in their aqueous states.
[tex]HNO_3(aq.)\rightarrow H^+(aq.)+NO_3^-(aq.)[/tex]
Weak acid is defined as the acid which does not completely dissociates when dissolved in water. They have high pH. These releases ions in their aqueous states.
[tex]CH_3COOH\rightleftharpoons CH_3COO^-+H^+[/tex]
Thus as the given acid is only 92% dissociated and cannot be completely dissociated, it is termed as a weak acid.
Final answer:
The given statement," An acid that dissociates to the extent of 92% in water would be termed a strong acid." is false.
Explanation:
In chemistry, the degree to which an acid dissociates in water classifies it as either strong or weak. An acid that dissociates 100% into ions when dissolved in water is deemed a strong acid. In contrast, acids that do not reach 100% dissociation are termed weak acids. An example of a weak acid is acetic acid (CH3COOH), which has a dissociation level significantly lower than 100% when dissolved in water.
Considering the given information, an acid that dissociates to the extent of 92% in water would not meet the criteria for a strong acid, as it does not fully dissociate. Thus, even with a high dissociation percentage such as 92%, this acid would still be classified as a weak acid. It is important to note that in practical terms a 92% dissociation indicates a relatively strong acid amongst weak acids, but it is not categorized as a strong acid in the technical sense.
To what does the term 'drying a liquid' refer?
a. Removing all traces of water to form a solid.
b. The technique of removing all traces of an organic solvent from water.
c. Removing all traces of water from an organic solvent.
d. The extraction of a solid from a polar solvent.
Answer: a. Removing all traces of water to form a solid
Explanation:
A chemist makes of magnesium fluoride working solution by adding distilled water to of a stock solution of magnesium fluoride in water. Calculate the concentration of the chemist's working solution. Round your answer to significant digits.
The question is incomplete, here is the complete question:
A chemist makes 600. mL of magnesium fluoride working solution by adding distilled water to 230. mL of a stock solution of 0.00154 mol/L magnesium fluoride in water. Calculate the concentration of the chemist's working solution. Round your answer to 3 significant digits.
Answer: The concentration of chemist's working solution is [tex]5.90\times 10^{-4}M[/tex]
Explanation:
To calculate the molarity of the diluted solution (chemist's working solution), we use the equation:
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of the stock magnesium fluoride solution
[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of chemist's magnesium fluoride solution
We are given:
[tex]M_1=0.00154M\\V_1=230mL\\M_2=?M\\V_2=600mL[/tex]
Putting values in above equation, we get:
[tex]0.00154\times 230=M_2\times 600\\\\M_2=\frac{0.00154\times 230}{600}=5.90\times 10^{-4}M[/tex]
Hence, the concentration of chemist's working solution is [tex]5.90\times 10^{-4}M[/tex]
How many grams of Al were reacted with excess HCl if 2.34 L of hydrogen gas were collected at STP in the following reaction2Al + 6HCl --> 2AlCl3 + 3H2
Answer:
1.87 g of Al were reacted
Explanation:
We must apply the Ideal Gases law to solve the amount of moles of H₂ produced at STP
T = 273 K
P = 1 atm
P. V = n . R . T
1 atm . 2.34 L = n . 0.082 . 273 K
( 1 atm . 2.34 L ) / ( 0.082 . 273 K) = n
0.104 moles = n
Now we can find the moles of Al with the reaction. As ratio is 3:2 we have to make a rule of three to determine the moles we used.
3 moles of H₂ came from 2 moles of Al
0.104 moles of H₂ would come from ( 0.104 . 2 ) /3 = 0.0693 moles of Al
Finally we can know the mass of Al we used ( moles . molar mass)
0.0693 mol . 26.98 g/mol = 1.87 g
To determine the grams of Al reacted with excess HCl, we need to calculate the number of moles of H2 gas produced at STP, then use stoichiometry to convert the moles of H2 to moles of Al, and finally convert the moles of Al to grams using its molar mass.
Explanation:To determine how many grams of Al were reacted with excess HCl, we need to first calculate the number of moles of H2 gas produced at STP. From the balanced chemical equation, we can see that for every 2 moles of Al, 3 moles of H2 gas are produced. Since 1 mole of any ideal gas occupies 22.4 L at STP, we can convert the given volume of H2 gas to moles by dividing 2.34 L by 22.4 L/mol. This gives us approximately 0.1045 moles of H2 gas.
Using the stoichiometry of the reaction, we can then determine the number of moles of Al, which is the same as the number of moles of H2 gas produced. Finally, we can convert the moles of Al to grams using the molar mass of Al, which is 26.98 g/mol. Therefore, the number of grams of Al reacted would be approximately 0.1045 moles multiplied by 26.98 g/mol, resulting in approximately 2.82 grams of Al.
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A chemist adds of a M zinc nitrate solution to a reaction flask. Calculate the mass in grams of zinc nitrate the chemist has added to the flask. Round your answer to significant digits.
corrected question: A chemist adds 135mL of a 0.21M zinc nitrate solution to a reaction flask. Calculate the mass in grams of zinc nitrate the chemist has added to the flask. Round your answer to significant digits.
Answer:
5.37g
Explanation:
0.21M means ; 0.21mol/dm³
1dm³=1L , so we can say 0.21mol/L
if 0.21mol of Zinc nitrate is contained in 1L of water
x will be contained in 135mL of water
x= 0.21*135*10³/1
=0.02835moles
number of moles= mass/ molar mass
mass= number of moles *molar mas
molar mass of Zn(NO₃)₂=189.36 g/mol
mass= 0.02835 *189.36
mass=5.37g
The synthesis of dipropyl ether can be accomplished using 1-propanol. What reactants and conditions are necessary for this to occur
Answer: Reactants==>2 molecules of propanol.
Condition for Reaction to occur==> Strong acid,e.g HCl and high temperature (140°C).
Explanation:
Dipropyl ether can be sythesized commonly in two ways;
(1). Williamson ether synthesis of dipropyl ether: this chemical reaction involves the reaction of alkyl halide(propyl halide) with conjugate base of propanol.
(2). Acid catalyzed ether synthesis of dipropyl ether: this is the reaction this question is talking about. A strong acid is used in the Chemical Reaction, and , it is reaction between two(2) molecules of propanol. This is done in the presence of strong acid such as HCl and high temperature (up to 140°C).
The Reactants are the two molecules of propanol. The condition necessary for the reaction to occur is that the temperature must be high(140-143°C) and it must be in the presence of a strong acid.
In a dehydration reaction, 2 molecules lose one water molecule to combine. In presence of strong acid ([tex]\bold{H_2SO4}[/tex]) and high temperature at [tex]\bold{145^oC}[/tex] propanol can be accomplished to synthesize dipropyl ether.
Dehydration of Alcohol:
In high acidic and high-temperature conditions, alcohol dehydrates to form ether.
[tex]\bold{2R-OH \rightarrow R-O-R +H_2O}[/tex]
2 molecules of propanol will combine to form dipropyl ether in the presence of strong acid ([tex]\bold{H_2SO4}[/tex]) at [tex]\bold{145^oC}[/tex].
[tex]\bold{2CH_3 CH_2CH_2OH \rightarrow CH_3 CH_2CH_2-O-CH_2CH_2CH_3 + H_2O }[/tex]
Therefore, In presence of strong acid ([tex]\bold{H_2SO4}[/tex]) and high temperature at [tex]\bold{145^oC}[/tex]
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While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene with water vapor at elevated temperatures. A chemical engineer studying this reaction fills a flask with of ethylene gas and of water vapor. When the mixture has come to equilibrium she determines that it contains of ethylene gas and of water vapor. The engineer then adds another of water, and allows the mixture to come to equilibrium again. Calculate the pressure of ethanol after equilibrium is reached the second time. Round your answer to significant digits.
Answer:
Pressure of alcohol after reaching equilibrium the second time = 2.62atm
Explanation:
Detailed step by step with explanation is as shown in the attachment.
A mixture of xenon and oxygen gas is compressed from a volume of 99.0 L to a volume of 98.0 L, while the pressure is held constant at 69.0 atm. Calculate the work done on the gas mixture. Be sure your answer has the correct sign (positive or negative) and the correct number of significant digits
The work done on the gas mixture when it is compressed from 99.0 L to 98.0 L at a constant pressure of 69.0 atm is -69.0 atm L.
Explanation:In order to calculate the work done on a gas, we can use the equation:
Work = Pressure * Change in Volume
In this case, the pressure is held constant at 69.0 atm, and the change in volume is from 99.0 L to 98.0 L.
Using the formula, we can calculate the work as follows:
Change in Volume = Final Volume - Initial Volume = 98.0 L - 99.0 L = -1.0 LWork = Pressure * Change in Volume = 69.0 atm * -1.0 L = -69.0 atm LThe work done on the gas mixture is -69.0 atm L. The negative sign indicates that work is done on the system rather than by the system.
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In an ionic compound, the size of the ions affects the internuclear distance (the distance between the centers of adjacent ions), which affects lattice energy (a measure of the attractive force holding those ions together).
1. Based on ion sizes, arrange these compounds by their expected lattice energy. Note that many sources define lattice energies as negative values. Please arrange by magnitude and ignore the sign. Great Lateral Energy to least (weakest) lattice energy.
O Nal
O NaCl
O NaBr
O NaF
Calcium reacts with sulfur forming calcium sulfide. What is the theoretical yield (g) of CaS(s) that could be prepared from 8.54 g of Ca(s) and 2.33 g of sulfur(s)? Do not type units with your answer.
Answer:
5.242
Explanation:
Molar mass of Ca = 40.078 g/mol
Molar mass of S = 32.065 g/mol
Equation of reaction
Ca (s) + S(s) - CaS(s)
To find the limiting reagent
Mole of Ca = 8.54/40.078 (g/mol) = 0.213 mol
Mole of S = 2.33g / 32.065 (g/mol) = 0.072664 mol
From their mole ratio, sulphur is the limiting reagent
From the reaction
1 mole of calcium react with 1 mole of Sulphur to yield 1 mole of calcium sulphide
0.072664 mol of calcium will react with 0.072664 of sulphur to yield 0.072664 mole of CaS
Theoretical yield of CaS = 0.072664 × molar mass of CaS = 0.072664 × 72.143 g/mol = 5.242 g
A brick has a mass of and the Earth has a mass of . Use this information to answer the questions below. Be sure your answers have the correct number of significant digits. What is the mass of 1 mole of bricks? How many moles of bricks have a mass equal to the mass of the Earth?
Complete question
A brick has a mass of 4.0 kg and the Earth has a mass of 6.0 X 10²⁷ g. Use this information to answer the questions below. Be sure your answers have the correct number of significant digits. What is the mass of 1 mole of bricks? Round your answer to 2 significant digits. How many moles of bricks have a mass equal to the mass of the Earth? Round your answer to 2 significant digits.
Answer:
The mass of 1 mole of bricks = 24 X 10²⁶ g (2 significant digits), and
2.5 moles of bricks (2 significant digits) have a mass equal to the mass of the Earth.
Explanation:
Given mass of bricks = 4.0 kg
4.0kg of brick = 4000g
1 mole of any substance always contain 6.022 X 10²³ Avogadro's number
1 mole of bricks contains 6.022 X 10²³ bricks
Therefore, mass of 1 mole of brick = 6.022 X 10²³ X 4000g
= 4000 X 6.022 X 10²³ g
= 24088 X 10²³ g
The mass of 1 mole of bricks = 24 X 10²⁶ g (2 significant digits)
-----------------------------------------------------------------------------------------------------
⇒How many moles of bricks have a mass equal to the mass of the Earth?
Given mass of Earth = 6.0 X 10²⁷ g
24088 X 10²³ g --------> mass of 1 mole of bricks
6.0 X 10²⁷ g --------------> ?
= (6.0 X 10²⁷ g)/(24088 X 10²³ g)
= 2.4909 moles of bricks
≅ 2.5 moles of bricks ( 2 significant digits)
Therefore, 2.5 moles of bricks have a mass equal to the mass of the Earth
Answer
2.4x10^27 g
2.5 mol
An electron is confi ned to a linear region with a length of the same order as the diameter of an atom (about 100 pm). Calculate the minimum uncertainties in its position and speed.
Explanation:
The minimum uncertainty position is 100 nm.
Therefore, by Heisenberg Uncertainty principal
ΔxΔp≥h/2
Δp≥h/(2Δx)
we know that
h= [tex]1.0546\times10^{-34}[/tex] Js
Δx= [tex]100\times10^{-12} m[/tex]
therefore,
[tex]\Delta p =\frac{1.0546\times10^{-34}}{2\times12\times10^{-12}} = 5.3\times10^{-25} kgms^{-1}[/tex]
therefore,
[tex]\Delta v= \frac{\Delta p}{m} =\frac{5.3\times10^{-25}}{9.11\times10^{-31}}= 5.8\times10^{5} ms^{-1}[/tex]
The gas in a 250. mL piston experiences a change in pressure from 1.00 atm to 4.45 atm. What is the new volume (in mL) assuming the moles of gas and temperature are held constant
Answer:
56.2 mL
Explanation:
Given data
Initial volume (V₁): 250 mLInitial pressure (P₁): 1.00 atmFinal volume (V₂): ?Final pressure (P₂): 4.45 atmAssuming the gas has an ideal behavior, we can find the final volume using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁/P₂
V₂ = 1.00 atm × 250 mL/ 4.45 atm
V₂ = 56.2 mL
According to the question,
Volume,
[tex]V_1 = 250 \ mL[/tex][tex]V_2 = \ ?[/tex]Pressure,
[tex]P_1 = 1.00 \ atm[/tex][tex]P_2 = 4.45 \ atm[/tex]By using the Boyle's Law, we get
→ [tex]P_1 V_1 = P_2 V_2[/tex]
or,
→ [tex]V_2 = \frac{P_1V_1}{P_2}[/tex]
By putting the values, we get
[tex]= \frac{1.00\times 250}{4.45}[/tex]
[tex]= \frac{250}{4.45}[/tex]
[tex]= 56.2 \ mL[/tex]
Thus the above answer is right.
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Which of the following personal items are permitted to be at your workstation in the organic laboratory? Select all that apply. a. Pen or pencil b. Drinking water bottle c. I-Pad d. Lab manual e. Cell phone
Explanation:
While working in a laboratory it is necessary to be free from any kind of distraction. This is because a distraction while working in a laboratory can lead to serious consequences.
For this it is important to put all your electronic devices like mobile phones, tablets etc in your backpack. Also, you should not carry any kind of eatable, water bottle or anything into the lab.
This is because chemicals do react with food items. Hence, then food does not remain fit for consumption.
Thus, we can conclude that pen or pencil and lab manual personal items are permitted to be at your workstation in the organic laboratory.
Which of the following compounds would have the highest boiling point?
a. 1-propanol,
b. CH3CH2CH2OH ethanol,
c. CH3CH2OH dimethyl ether,
d. CH3OCH3 ethyl methyl ether,
e. CH3OCH2CH3
Answer:a. 1-propanol,
Explanation:
Boiling point depends on the relative molecular mass of the compound and the kind of intermolecular interactions present in the substance. 1-propanol has a relative molecular mass of 60 and posses strong intermolecular hydrogen bonding. The presence of hydrogen bonding is known to lead to an increase in boiling point. As a result of this, it is obvious that 1-propanol has the highest boiling point among other compounds in the list.
Final answer:
1-propanol has the highest boiling point among the mentioned compounds due to its ability to form hydrogen bonds, which significantly elevate its boiling point compared to the other options.
Explanation:
The question pertains to determining which of the listed compounds would have the highest boiling point. When comparing boiling points, factors such as molecular mass, polarity, and the presence of hydrogen bonding play crucial roles. Among the options, 1-propanol (C3H7OH) stands out because it is capable of hydrogen bonding, which significantly increases its boiling point compared to the other compounds listed, such as dimethyl ether (CH3OCH3) and ethyl methyl ether (CH3OCH2CH3), which primarily exhibit dipole-dipole interactions and London dispersion forces. Ethanol (CH3CH2OH) also exhibits hydrogen bonding but has a shorter carbon chain than 1-propanol, leading to a slightly lower boiling point.
You have a 275 mg sample of a compound that contains aluminum and a group 7A element (F, Cl, Br, I or At). This compound reacts with excess silver nitrate as follows ("X" represents the unknown element):
Complete Question:
You have a 275 mg sample of a compound that contains aluminum and a group 7A element (F, Cl, Br, I or At). This compound reacts with excess silver nitrate as follows (“X” represents the unknown element): AlX3 + 3 AgNO3 → 3 AgX + Al(NO3)3. This reaction forms 581 mg of AgX. Identify element X.
Answer:
Br
Explanation:
By the stoichiometry of the reaction, 1 mol of AlX3 will form 3 moles of AgX. If we call the molar mass of X as m, and by the periodic table the molar masses of Al is 26.982 g/mol, and of Ag is 107.87, thus the molar masses of the compounds will be:
AlX3 = 26.982 + 3m
AgX = 107.87 + m
And the mass relationship will be the number of moles multiplied by the molar mass. So, by a simple direct rule of three:
26.982 + 3m g of AlX3 ------------- 3*(107.87 + m) g of AgX
275 mg of AlX3 ----------- 581 mg
825*(107.87 + m) = 15676.542 + 1743m
88992.75 + 825m = 15676.542 + 1743m
-918m = -73316.208
m = 79.86 g/mol
So, comparing with the values of the masses of the group 7 elements, X must be Br (79.904 g/mol).
102.1 g of Aluminum nitrite and 174.3 g of ammonium chloride react to form aluminum chloride, nitrogen, and water.
How many grams of the excess reagent remains after the reaction?
Answer:
57.39 g excess Aluminum nitrite
Explanation:
When performing stoichiometric calculations, the first thing we need is the balanced chemical reaction.
In this case we will have:
Al(NO₂)₃ + 3 NH₄Cl ⇒ AlCl₃ + 3 N₂ + 6 H₂O
( nitrite ion is NO₂⁻ )
Now that we have the balanced reaction, we need to calculate the number of moles, n, of Al(NO₂)₃ and NH₄Cl , and perform the calculations necessary to determine the excess reagent and its amount.
The number of moles is :
n = mass / MW where MW is the molecular weight and m the mass.
MW Al(NO₂)₃ = 40.99 g/mol
MW NH₄Cl = 53.49 g/mol
n Al(NO₂)₃ = 102.1 g / 40.99 g/mol = 2.49 mol Al(NO₂)₃
n NH₄Cl = 174.3 g / 53.49 g/mol = 3.26 mol NH₄Cl
Now lets calculate how many moles of NH₄Cl will react with 2.49 mol Al(NO₂)₃ :
( 3 mol NH₄Cl / 1 mol Al(NO₂)₃ ) x 2.49 mol Al(NO₂)₃ = 7.5 mol NH₄Cl
We only have 3.26 mol NH₄Cl . Therefore our limiting reagent is NH₄Cl , and the excess reagent is Al(NO₂)₃
Now lets calculate the number of moles Al(NO₂)₃ used to react with 3.26 mol NH₄Cl :
( 1 mol Al(NO₂)₃ / 3 mol NH₄Cl ) x 3.26 mol NH₄Cl =1.09 mol Al(NO₂)₃
The excess number of moles is:
= 2.49 mol - 1.09 mol = 1.40 mol Al(NO₂)₃
grams of Al(NO₂)₃ in excess
1.40 mol Al(NO₂)₃ x 40.99 g/mol = 57.39 g
13.3 g of benzene (C6H6) is dissolved in 282 g of carbon tetrachloride. What is the molal concentration of benzene in this solution?
Answer:
0.605 molal
Explanation:
molality is the amount of solute in a particular mass of solvent.
lets calculate the amount of benzene solute.
mass of benzene= 13.3g
molar mass of C6H6= 12*6 +1*6 =72+7=78g/mol
amount of benzene= mass/molar mass
=13.3/78
=0.1705mol
molality= amount of solute/mass of solvent in kg
mass of solvent=282g=0.282kg
molality = 0.1705/0.282
=0.605 molal
To calculate the molal concentration of benzene in the solution, divide the number of moles of benzene (0.1703 moles) by the mass of carbon tetrachloride in kilograms (0.282 kg), resulting in a concentration of 0.604 molal.
To calculate the molal concentration of benzene in the solution, first, we need to determine the number of moles of benzene. The molar mass of benzene (C6H6) is approximately 78.11 g/mol. Using the mass of benzene given (13.3 g), we can calculate the moles of benzene as follows:
Number of moles of benzene = mass of benzene / molar mass of benzene
Number of moles of benzene = 13.3 g / 78.11 g/mol = 0.1703 moles
The molal concentration is then calculated by the number of moles of solute per kilogram of solvent. Since the mass of the solvent (carbon tetrachloride) is given in grams, we convert it to kilograms:
Mass of solvent in kg = 282 g / 1000 = 0.282 kg
Now, the molal concentration (m) can be calculated as:
Molal concentration (m) = moles of solute / mass of solvent in kg
Molal concentration (m) = 0.1703 moles / 0.282 kg = 0.604 Molal
If the rate of formation of ammonia is 0.345 M/s, what is the rate of disappearance of N2?
Answer: The rate of disappearance of [tex]N_2[/tex] is 0.172 M/s
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
[tex]N_2(g)+3H_2(g)\rightarrow 2NH_3(g)[/tex]
The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.
Rate in terms of disappearance of [tex]N_2[/tex] = [tex]-\frac{1d[N_2]}{dt}[/tex]
Rate in terms of disappearance of [tex]H_2[/tex] = [tex]-\frac{1d[H_2]}{3dt}[/tex]
Rate in terms of appearance of [tex]NH_3[/tex] = [tex]\frac{1d[NH_3]}{2dt}[/tex]
The rate of formation of ammonia is = 0.345 M/s
[tex]\frac{1d[NH_3]}{dt}=0.345 M/s[/tex]
The rate of disappearance of [tex]N_2[/tex] is one by three times the rate of appearance of [tex]NH_3[/tex]
[tex]-\frac{1d[N_2]}{dt}=\frac{d[NH_3]}{2dt}[/tex]
[tex]-\frac{1d[N_2]}{dt}=\frac{0.345M/s}{2}=0.172M/s[/tex]
Thus the rate of disappearance of [tex]N_2[/tex] is 0.172 M/s
If the rate of formation of ammonia is 0.345 M/s, the rate of disappearance of nitrogen is 0.173 M/s.
Let's consider the balanced equation for the formation of ammonia.
N₂(g) + 3 H₂(g) ⇒ 2 NH₃(g)
The molar ratio of N₂ to NH₃ is 1:2. If the rate of formation of ammonia is 0.345 M/s, the rate of disappearance of nitrogen is:
[tex]\frac{0.345molNH_3}{L.s} \times \frac{1molN_2}{2molNH_3} = \frac{0.173molN_2}{L.s} = 0.173 M/s[/tex]
If the rate of formation of ammonia is 0.345 M/s, the rate of disappearance of nitrogen is 0.173 M/s.
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Identify the power of ten that defines each of these prefixes. Input your answers as 10 x where x is the power of ten. nano- _______________. kilo- _______________. centi- _______________. micro- _______________. milli- _______________. mega- _______________.
Answer: The quantity of every prefix is written below as a power of ten.
Explanation:
In the metric system of measurement, the name of multiples and subdivision of any unit is done by combining the name of the unit with the prefixes.
For Example: deka, hecto and kilo means 10, 100 and 1000 respectively. Deci, centi and milli means one-tenth, one-hundredth, and one-thousandth respectively.
The quantity of these prefixes are written as the power of 10.
For the given prefixes:
Nano: The quantity will be [tex]10^{-9}[/tex]
Kilo: The quantity will be [tex]10^3[/tex]
Centi: The quantity will be [tex]10^{-2}[/tex]
Micro: The quantity will be [tex]10^{-6}[/tex]
Milli: The quantity will be [tex]10^{-3}[/tex]
Mega: The quantity will be [tex]10^6[/tex]
Hence, the quantity of every prefix is written above as a power of ten.
Which one of the following statements is notapplicable to molecularsolids?
a. The units that occupy the lattice points are molecules.
b. The binding forces in molecular solids are dispersion forces or dispersion forces and dipole-dipole interactions.
c. Molecular solids have relatively low melting points.
d. Molecular solids are usually excellent conductors of electric current. e. Molecular solids are soft compared to covalent solids.
Answer: The answer is D. Molecular solids are usually excellent conductors of electric current
Explanation: Molecular solids are solids that have separate individual molecules held together by intermolecular forces (Van der Waal's force) rather than bonds.
Van der Waal's forces are the weak forces that contribute to intermolecular bonding between molecules.
Examples of a molecular solid are water ice, solid carbon dioxide and white phosphorus.
Molecular solids cannot conduct electricity because they have localized electrons (election localized within the bonds in each molecule).
Final answer:
The statement that is not applicable to molecular solids is 'd. Molecular solids are usually excellent conductors of electric current' because molecular solids are poor conductors due to the lack of ions or free electrons.
Explanation:
To answer which statement is not applicable to molecular solids, let us examine each option given:
a. The units that occupy the lattice points are molecules - This is true for molecular solids.
b. The binding forces in molecular solids are dispersion forces or dispersion forces and dipole-dipole interactions - This is also true. Dispersion forces are present in nonpolar molecular solids, while polar molecular solids exhibit dipole-dipole interactions and sometimes hydrogen bonds as well.
c. Molecular solids have relatively low melting points - This statement is correct since the intermolecular forces in molecular solids are weaker than ionic or covalent bonds, resulting in lower melting points.
d. Molecular solids are usually excellent conductors of electric current - This statement is not true. Molecular solids lack ions or free electrons, which makes them poor conductors of electricity.
e. Molecular solids are soft compared to covalent solids - This is generally true because the intermolecular bonds in molecular solids are weaker than the covalent bonds in covalent solids.
Therefore, the statement that is not applicable to molecular solids is d. Molecular solids are usually excellent conductors of electric current.