Explanation:
-Filter help — delete some big unreacted, undesirable species (norit is probably from what you are sorting through, its only carbon which cleans up things)
— extract with DCM because you are probably in an aqueous phase, and some butanoate is in it
- Anhydrous sodium absorbs excess of water (dries the material)
-evaporation in the hood to clear the DCM and crystallize the product.
The four sets of lines in the hydrogen emission spectrum are known as Balmer, Brackett, Paschen, and Lyman. For each series, assign the energy (infrared, ultraviolet, or visible), the nr value, and all possible n, values up to 7 Series Balmer Brackett Paschen Lyman Energy nf ni visible) infrared ultraviolet 1 2 3 6 7
Answer:
Hydrogen spectrum
Explanation:
Balmer series - Observed in the visible region
Brackett series - Observed in the infrared region
Paschen series - Observed in the infrared region
Lyman series - Observe in the Ultraviolet region.
There are four sets of lines seen in the hydrogen emission spectra are Lyman series in the UV region, Balmer in the visible region and bracket and paschen in the IR region.
What is emission spectra?Emission spectrum is set of radiations emitted by an atom in the order of increasing frequency or decreasing wavelengths. Electrons in atoms transit between various energy levels.
Electrons excites from the lower energy level to the higher energy level by the absorption of light and return back to the ground state by emitting a complementary light.
The lyman series in the hydrogen emission spectrum is starting from the energy level n= 1. It is seen in the UV-region. Balmer series is found in visible region and their n value starts from n = 2.
Similarly Bracket starts from n= 3 and paschen series starts from n= 4. Both are overlapping lines together and they are seen in IR region.
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The ionization constant of water at a temperature above 25°C is 1.7 x 10-14. What is the pH of pure water at this temperature?
2H2O(l)--> H3O+(aq) + OH-(aq)
A) 7.88
B) 5.68
C) 6.88
D) 13.85
E) 7.00
Answer:
The pH of pure water at this temperature is 6.88.
Explanation:
Ionization constant of water = [tex]K_w=1.7 x 10^{-14}[/tex]
[tex]2H_2O(l)\rightleftharpoons H_3O^+(aq) + OH^-(aq)[/tex]
Water will dissociate into equal amount of hydronium and hydroxide ions
[tex][H_3O^+] = [OH^-]= x[/tex]
[tex]K_w=[H_3O^+]\times [OH^-][/tex]
[tex]K_w=1.7 x 10^{-14}=[H_3O^+]\times [OH^-][/tex]
[tex]1.7 x 10^{-14}=x^2[/tex]
[tex]x=1.304\times 10^{-7}[/tex]
The pH of the solution is a negative logarithm of hydronium ions concentration.
The pH of pure water at this temperature:
[tex]pH=-\log [H_3O^+][/tex]
[tex]pH=-\log (1.304\times 10^{-7})=6.88[/tex]
The pH of pure water is:
C) 6.88.
Calculation for pH:Ionization constant of water = [tex]1.7 * 10^{-14}[/tex]
Chemical reaction for ionization of water:
[tex]2H_2O(l)---- > H_3O^+(aq) + OH^-(aq)[/tex]
Water dissociates into equal amount of hydronium and hydroxide ions
[tex][H_3O^+]=[OH^-]=x\\\\\K_w=[H_3O^+]*[OH^-]\\\\1.7*10^{-14}=x^2\\\\x=1.307*10^{-7}[/tex]
The pH of the solution is a negative logarithm of hydronium ions concentration.
The pH of pure water at this temperature:
[tex]pH=-log[H_3O^+]\\\\pH=-log(1.304*10^{-7})\\\\pH=6.88[/tex]
Thus, correct option is C.
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Identify the group of elements that corresponds to each of the following generalized electron configurations and indicate the number of unpaired electrons for each: (a) [Noble gas] ns^2 np^5 (b) [noble gas] ns^2 (n-1)d^2 (c) [noble gas] ns^2 (n-1)d^10 np^1 (d)[noble gas] ns^2 (n-2)f^6
Answer:
(a) [Noble gas] ns² np⁵: Number of unpaired electron is 1 and belongs to group 17 i.e. halogen group of the periodic table.
(b) [noble gas] ns² (n-1)d²: Number of unpaired electron is 2 and belongs to group 4 of the periodic table.
(c) [noble gas] ns² (n-1)d¹⁰ np¹: Number of unpaired electron is 1 and belongs to group 13 of the periodic table.
(d)[noble gas] ns² (n-2)f⁶ : Number of unpaired electron is 6 and belongs to group 8 of the periodic table.
Explanation:
In the Periodic table, the chemical elements are arranged in 7 rows, called periods and 18 columns, called groups. They are organized in increasing order of atomic numbers.
(a) [Noble gas] ns² np⁵
As the total number of electrons in the p-orbital is 5. Therefore, the number of unpaired electron is 1.
This element has 2 electrons in ns orbital and 5 electrons in np orbital. So there are 7 valence electrons.
Therefore, this element belongs to the group 17 i.e. halogen group of the periodic table.
(b) [noble gas] ns² (n-1)d²
As the total number of electrons in the d-orbital is 2. Therefore, the number of unpaired electrons is 2.
This element has 2 electrons in ns orbital and 2 electrons in (n-1)d orbital. So there are 4 valence electrons.
Therefore, this element belongs to the group 4 of the periodic table.
(c) [noble gas] ns² (n-1)d¹⁰ np¹
As the total number of electrons in the p-orbital is 1. Therefore, the number of unpaired electron is 1.
This element has 2 electrons in ns orbital and 1 electron in np orbital. So there are 3 valence electrons.
Therefore, this element belongs to the group 13 of the periodic table.
(d)[noble gas] ns² (n-2)f⁶
As the total number of electrons in the f-orbital is 6. Therefore, the number of unpaired electron is 6.
This element has 2 electrons in ns orbital and 6 electrons in (n-2)f orbital. So there are 8 valence electrons.
Therefore, this element belongs to the group 8 of the periodic table.
The group of elements that corresponds to each of the following are:
(a) [Noble gas] ns² np⁵: Number of unpaired electron is 1 and belongs to group 17.
(b) [noble gas] ns² (n-1)d²: Number of unpaired electron is 2 and belongs to group 4.
(c) [noble gas] ns² (n-1)d¹⁰ np¹:Number of unpaired electron is 1 and belongs to group 13.
(d) [noble gas] ns² (n-2)f⁶ : Number of unpaired electron is 6 and belongs to group 8.
Periodic Table:Periods are horizontal rows (across) the periodic table, while groups are vertical columns (down) the table. The elements are arranged in increasing order of their atomic number.
Group 17 is known as Halogen group, it has only on unpaired electron that means it needs one electron more to complete its octet or attain noble gas configuration. Group 4 is the second group of transition metals in the periodic table. Group 13, is also known as Boron group and it lies in p block elements. Group 8, is also known as Iron family.
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Consider the following generic chemical equation. 2A+4B→3C What is the limiting reactant when each of the following amounts of A and B are allowed to react? Part A 2 molA; 5 molB 2 ; 5 B A SubmitRequest Answer Part B 1.8 molA; 4 molB 1.8 ; 4 B A SubmitRequest Answer Part C 3 molA; 4 molB 3 ; 4 B A SubmitRequest Answer Part D 22 molA; 40 molB 22 ; 40 B A
Answer:
A. Limiting reactant is A.
B. Limiting reactant is A.
C. Limiting reactant is B.
D. Limiting reactant is B.
Explanation:
It is possible to find the limiting reactant for a reaction taking the moles of a reactant that will react and using the chemical equation find the moles of the other reactant you will need.
For the reaction:
2A + 4B → 3C
A. 2 moles A requires:
2molA×[tex]\frac{4molB}{2molA}[/tex]= 4moles of B
As you have 5 moles of B, limiting reactant is A.
B. 1,8 moles A requires:
1,8 molA×[tex]\frac{4molB}{2molA}[/tex]= 3,6 moles of B
As you have 4 moles of B, limiting reactant is A.
C. 3 moles A requires:
3 molA×[tex]\frac{4molB}{2molA}[/tex]= 6 moles of B
As you have just 4 moles of B, limiting reactant is B.
D. 22 moles A requires:
22 molA×[tex]\frac{4molB}{2molA}[/tex]= 44moles of B
As you have just 40 moles of B, limiting reactant is B.
I hope it helps!
For each scenario, the limiting reactant is determined by comparing the moles used according to the stoichiometry of the reaction. In Parts A, B, and D, A is the limiting reactant; in Part C, B is limiting.
To determine the limiting reactant, compare the moles of each reactant to the stoichiometric coefficients in the balanced equation (2A + 4B → 3C).
1. Part A (2 mol A; 5 mol B):
- Moles of A used = 2 (coefficient in front of A)
- Moles of B used = [tex]\frac{5}{2}[/tex] (coefficient in front of B divided by 2)
- Since Moles of B used > Moles of A used, A is the limiting reactant.
2. Part B (1.8 mol A; 4 mol B):
- Moles of A used = 1.8 (coefficient in front of A)
- Moles of B used = [tex]\frac{4}{4}[/tex] (coefficient in front of B divided by 4)
- Since Moles of A used < Moles of B used, A is the limiting reactant.
3. Part C (3 mol A; 4 mol B):
- Moles of A used = 3 (coefficient in front of A)
- Moles of B used = [tex]\frac{4}{2}[/tex] (coefficient in front of B divided by 2)
- Since Moles of B used < Moles of A used, B is the limiting reactant.
4. Part D (22 mol A; 40 mol B):
- [tex]\( \text{Moles of A used} = \frac{22}{2} \)[/tex] (coefficient in front of A divided by 2)
- Moles of B used = 40 (coefficient in front of B)
- Since Moles of A used < Moles of B used, A is the limiting reactant.
In summary:
- Part A: Limiting reactant is A.
- Part B: Limiting reactant is A.
- Part C: Limiting reactant is B.
- Part D: Limiting reactant is A.
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The work function of indium is 6.56 x 10^-19 J. What is the minimum frequency of a photon that will eject an electron from the indium metal surface?
1. 9.90 x 10^14 Hz
2. 2.97 x 10^23 Hz
3. 3.42 x 10^14 Hz
4. 4.57 x 10^26 Hz
5. 2.19 x 10^-27 Hz
Answer:
1. 9.90 x 10^14 Hz
Explanation:
The work function is the minimum energy required to eject electron's from the surface of the metal. The frequency of photons that are capable of doing this is called the threshold frequency. Given the work function, the threshold frequency can be calculated in the following way
Work function = Planck's constant * frequency
6.56 x 10^-19 J = 6.62607004 × 10-34 m²-kg/s * f
f = 9.90 x 10^14 Hz
Use the standard half-cell potentials listed below to calculate the standard cell potential for the following reaction occurring in an electrochemical cell at 25�C. (The equation is balanced.)
3 Cl2(g) + 2 Fe(s) --> 6 Cl-(aq) + 2 Fe3+(aq)
Cl2(g) + 2 e- --> 2 Cl-(aq); E� = +1.36 V
Fe3+(aq) + 3 e- -->Fe(s); E� = -0.04 V
+1.32 V
-1.32 V
-1.40 V
+1.40 V
+4.16 V
Answer: + 1.40 V
Explanation:
The balanced chemical equation is:
[tex]3Cl_2(g)+2Fe(s)\rightarrow 6Cl^-(aq)+2Fe^{3+}(aq)[/tex]
Here Fe undergoes oxidation by loss of electrons, thus act as anode. Chlorine undergoes reduction by gain of electrons and thus act as cathode.
[tex]E^0=E^0_{cathode}- E^0_{anode}[/tex]
Where both [tex]E^0[/tex] are standard reduction potentials.
[tex]E^0_{[Fe^{3+}/Fe]}=-0.04V[/tex]
[tex]E^0_{[Cl_2/Cl^-]}=+1.36V[/tex]
[tex]E^0=E^0_{[Cl_2/Cl^-]}- E^0_{[Fe^{3+}/Fe]}[/tex]
[tex]E^0=+1.36-(-0.04V)=+1.40V[/tex]
The standard cell potential for the reaction is +1.40 V
Write a balanced chemical equation describing the oxidation of Cl2 gas by Cu3+ to form chlorate ion (ClO3-) and Cu2+ in an acidic aqueous solution. Use the smallest whole-number coefficients possible and indicate states of matter in your balanced reaction.
The detailed answer provides the balanced chemical equation for the oxidation of Cl2 by Cu3+ in an acidic solution, with states of matter indicated.
The balanced chemical equation describing the oxidation of Cl2 gas by Cu3+ to form chlorate ion (ClO3-) and Cu2+ in an acidic aqueous solution is:
3Cu^3+(aq) + 6Cl¯(aq) + 18H^+(aq) + Cl2(g) → 3Cu^2+(aq) + ClO3^-(aq) + 9H2O(l)
States of matter: (s) solid, (l) liquid, (g) gas, (aq) aqueous solution.
The oxidation of [tex]Cl_2[/tex] gas by [tex]Cu^{3+}[/tex] to form chlorate ion and [tex]Cu^{2+}[/tex] in an acidic aqueous solution is represented by the balanced chemical equation provided.
The given question concerns the concept of balancing chemical equations in chemistry and deriving correct chemical reactions for the mentioned reactants and products. Also, oxidation is mentioned which refers to addition of oxygen to an entity or removal of hydrogen from entity. The balanced chemical equation for the oxidation of [tex]Cl_2[/tex] gas by [tex]Cu^{3+}[/tex] in an acidic aqueous solution is:
[tex]2Cl_2(g) + 2Cu^{3+} (aq) + 4H^+ (aq) \rightarrow 2ClO^{3-} (aq) + 2Cu^{2+} (aq) + 2H_2O (l)[/tex]
This reaction results in the formation of chlorate ion ([tex]ClO^{3-}[/tex]) and [tex]Cu^{2+}[/tex] in the solution. The states of matter as gas, aqueous, and liquid are indicated in the equation.
Calculate the percentage by mass of water in magnesium sulfate heptahydrate, MgSO4•7H2O
Enter your answer with 3 significant figures
Answer:
The percentage by mass of water in magnesium sulfate heptahydrate is 51.2 %
Explanation:
Step 1: Data given
Molar mass of MgSO4*7H2O = 246.5 g/mol
Molar mass of H2O = 18.02 g/mol
Molar massof MgSO4 = 120.37 g/mol
Step 2: Calculate % water in magnesium sulfate heptahydrate
Since we have 7 molecules of water in the heptahydrate, we will divide the molar mass of 7 molecules water by the molar mass of the heptahydrate.
m%(H2O) = (7*18.02)/ 246.5
m%(H2O) = (126.14 /246.5)*100%
m%(H2O) = 51.2 %
To controle this we will calculate the mass % of MgSO4
m%(MgSO4) = (120.37/ 246.50)*100%
m%(MgSO4) = 48.8%
51.2 + 48.8 = 100%
The percentage by mass of water in magnesium sulfate heptahydrate is 51.2 %
Answer:from reffered dfn of %by mass
%by M=mass of component particular/total mass
But mass directly proportional to Molar mass
% by M=molar mass of H2O cpn/total M
%by M=7((1×2)+16)/(7((1×2)+16)+(24+(16×4)+32))
=0.573
What is the value for the reaction: N2(g) + 2 O2(g) --> N2O4(g) in terms of K values from the reactions:
½ N2(g) + ½ O2(g) ---> NO(g)
K1
2 NO(g) + O2(g) ---> N2O4(g)
K2
K1^2 + K2
2 K1 x K2
K1^2 x K2
K1 + K2
Answer : The correct expression will be:
[tex]K=(K_1)^2\times K_2[/tex]
Explanation :
The chemical reactions are :
(1) [tex]\frac{1}{2}N_2(g)+\frac{1}{2}O_2(g)\rightleftharpoons NO(g)[/tex] [tex]K_1[/tex]
(2) [tex]2NO(g)+O_2(g)\rightleftharpoons N_2O_4(g)[/tex] [tex]K_2[/tex]
The final chemical reaction is :
[tex]N_2(g)+2O_2(g)\rightleftharpoons N_2O_4[/tex] [tex]K=?[/tex]
Now we have to calculate the value of [tex]K[/tex] for the final reaction.
Now equation 1 is multiply by 2 and then add both the reaction we get the value of 'K'.
If the equation is multiplied by a factor of '2', the equilibrium constant will be the square of the equilibrium constant of initial reaction.
If the two equations are added then equilibrium constant will be multiplied.
Thus, the value of 'K' will be:
[tex]K=(K_1)^2\times K_2[/tex]
The electron configuration of a neutral atom is 1s22s22p63s2. Write a complete set of quantum numbers for each of the electrons. Name the element.
Answer:
The four quantum number for each electron will be:
[tex]1s^{2}[/tex]
[tex]n=1;l=0;m=0;s=+\frac{1}{2}/-\frac{1}{2}[/tex]
[tex]2s^{2}[/tex]
[tex]n=2;l=0;m=0;s=+\frac{1}{2}/-\frac{1}{2}[/tex]
[tex]2p^{6}[/tex]
[tex]n=1;l=1;\\m=+1,0.-1 \\s=+\frac{1}{2}/-\frac{1}{2}[/tex]
[tex]3s^{2}[/tex]
[tex]n=3;l=0;m=0;s=+\frac{1}{2}/-\frac{1}{2}[/tex]
Explanation:
As the element is neutral, the number of protons will be equal to number of electrons which will be the atomic number of the element.
Number of electrons =12
Atomic number = 12
Element : Magnesium
The principal shell is represented by "n"
i) For "s" subshell the value of l =0 (azimuthal quantum number) thus m (magnetic quantum number)= 0
The two electrons in s subshell will have either plus half or minus half spin quantum number
ii) for "p" subshell the value for l =1
thus m = 0 or +1 or -1
The two electrons in each orbital will have either plus half or minus half spin quantum number
Final answer:
The complete set of quantum numbers for each electron in the given electron configuration, which corresponds to the element magnesium (Mg), is derived from the configuration 1s²2s²2p⁶3s², by stating the values for the principal, angular momentum, magnetic, and spin quantum numbers for each electron.
Explanation:
The electron configuration of a neutral atom is 1s22s22p63s2. To write a complete set of quantum numbers for each of the electrons, we can consider the principles of quantum mechanics. Each electron in an atom is described by four quantum numbers: the principal quantum number (n), angular momentum quantum number (l), magnetic quantum number (ml), and spin quantum number (ms).
For the 1s subshell (2 electrons), the quantum numbers for each electron are (1, 0, 0, +1/2) and (1, 0, 0, -1/2). For the 2s subshell (2 electrons), the quantum numbers are similar, with the principal quantum number being 2: (2, 0, 0, +1/2) and (2, 0, 0, -1/2). For the 2p subshell (6 electrons), there are three magnetic quantum numbers (-1, 0, +1), and two spin states. Thus, the six sets of quantum numbers are (2, 1, -1, +1/2), (2, 1, -1, -1/2), (2, 1, 0, +1/2), (2, 1, 0, -1/2), (2, 1, +1, +1/2), and (2, 1, +1, -1/2). For the 3s subshell (2 electrons), the quantum numbers are (3, 0, 0, +1/2) and (3, 0, 0, -1/2).
Based on this configuration, the neutral atom would be magnesium (Mg), which has an atomic number of 12, corresponding to the 12 electrons described.
Consider the perchlorateanion.
(a) What is the central atom?
(b) How many lone pairs are around the central atom?
(c) What is the ideal angle between the chlorine-oxygen bonds?
(d) Compared to the ideal angle, you would expect the actual angle between the chlorine-oxygen bonds to be ___
Answer:
(a) Chlorine atom
(b) 0
(c) 109.5°
(d) about the same
Explanation:
Perchlorate is a negatively charged molecule that is composed of one carbon atom and four oxygen atoms. In this molecule, the central chlorine atom that is present in +7 oxidation state, is covalently bonded to four oxygen atoms.
Thus the number of bond pairs is 4 and the number of lone pairs on the central chlorine atom is 0.
So according to the VSEPR theory, perchlorate ion has a tetrahedral geometry. Therefore, the ideal bond angle for the perchlorate ion is 109.5°.
The actual bond angle of perchlorate is about the same as ideal bond angle. This is because the perchlorate is resonance stabilized and thus all the chlorine-oxygen bonds are equivalent.
In the perchlorate anion, chlorine is the central atom with zero lone pairs. The ideal angle between the chlorine-oxygen bonds is 109.5 degrees, though the actual angle is likely slightly less due to repulsion between bond pairs.
Explanation:Considering the perchlorate anion (ClO4−):
(a) The central atom in the perchlorate anion is chlorine.(b) There are zero lone pairs around the central chlorine atom in the perchlorate anion.(c) The ideal angle between the chlorine-oxygen bonds in the perchlorate anion is 109.5 degrees, which is characteristic of a tetrahedral electron geometry.(d) However, due to the repulsion between bond pairs, the actual angle between the chlorine-oxygen bonds is expected to be slightly less than 109.5 degrees.Learn more about Perchlorate Anion here:https://brainly.com/question/28296182
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The equilibrium: 2 NO2(g) \Longleftrightarrow⇔ N2O4(g) has Kc = 4.7 at 100ºC. What is true about the rates of the forward (ratefor) and reverse (raterev) reactions initially and at equilibrium if an empty container is filled with just NO2?
Initial: forward rate < reverse rate Equilibrium: forward rate > reverse rate
Initial: forward rate > reverse rate Equilibrium: forward rate > reverse rate
Initial: forward rate > reverse rate Equilibrium: forward rate = reverse rate
Initial: forward rate = reverse rate Equilibrium: forward rate = reverse rate
Answer:
Initial: forward rate > reverse rate Equilibrium: forward rate = reverse rateExplanation:
2NO₂(g) → N₂O₄(g) Kc=4.7
The definition of equilibrium is when the forward rate and the reverse rate are equal.
Because in the initial state there's only NO₂, there's no possibility for the reverse reaction (from N₂O₄ to NO₂). Thus the forward rate will be larger than the reverse rate.
Which statements are true?
a. The evaporation of water is an exothermic process.
b. A combustion reaction is exothermic.
c. When energy is transferred as heat from the surroundings to the system, Deta H is negative.
d. When energy is transferred as heat from the system to the surroundings, Deta H is negative.
e. for an exothermic reaction Deta H is positive.
f. for an endothermic reaction Deta H is positive.
Explanation:
Evaporation is defined as a process in which liquid state of water is changing into vapor state.So, we need to break the bonds of liquid substance in order to convert it into vapor state. And, energy is absorbed for breaking of bonds which means that evaporation is an endothermic process.
Hence, the statement evaporation of water is an exothermic process is false.
When a hydrocarbon reacts with oxygen and leads to the formation of carbon dioxide and water then this type of reaction is known as combustion reaction.A combustion reaction will always release heat energy. Hence, combustion reaction is exothermic in nature.
When energy is transferred as heat from the surroundings to the system then it means energy is being absorbed by the system. And, absorption of heat is an endothermic process for which [tex]\Delta H[/tex] is positive.Whereas when energy is transferred from system to the surrounding then it means energy is released by the system which is an exothermic process.Hence, for an exothermic process value of [tex]\Delta H[/tex] is negative.
Thus, we can conclude that statements which are true are as follows.
A combustion reaction is exothermic.When energy is transferred as heat from the system to the surroundings, [tex]\Delta H[/tex] is negative.For an endothermic reaction Deta H is positive.The most common method for the synthesis of unsymmetrical ethers is the Williamson synthesis, a reaction (SN2) of an alkoxide ion with an alkyl halide. Two pathways are possible, but often one is preferred. Construct the preferred pathway for the synthesis of 2-propoxypropane from propene, with propene-derived alkyl halide and alkoxide intermediates, by dragging the appropriate intermediates and reagents into their bins. Not every given reagent or intermediate will be used.
Answer:
2- propanol treated with metal base NaH followed by hydrogen bromide to form 2- propoxypropane.
Explanation:
Formation of secondary halide:
Firstly propene reacts with sulfuric acid or water to form 2- Propanol.
Formation of primary halide:
Also, according to the Anti markovinikov rule propene undergoes addition to the hydrogen bromide in the presence of peroxide to form 1- bromopropane.
The secondary halide i.e, 2- propanol treated with metal base NaH to form sodium isopropoxide. Further treated with hydrogen bromide to form 2- Propoxypropane. This reaction follows [tex]S_{N}2[/tex] mechanism.
The overall reaction is as follows.
The preferred pathway for the synthesis of 2-propoxypropane from propene in the Williamson synthesis involves the reaction of propene with an alkyl halide and an alkoxide ion.
Explanation:The preferred pathway for the synthesis of 2-propoxypropane from propene in the Williamson synthesis involves the reaction of propene with an alkyl halide and an alkoxide ion. The alkoxide ion is formed by treating an alcohol with a strong base, such as sodium hydride (NaH). The alkyl halide is then added to the alkoxide ion, resulting in the formation of the desired product, 2-propoxypropane.
Alcohol + Strong Base (NaH) → Alkoxide Ion
Alkoxide Ion + Alkyl Halide → 2-Propoxypropane
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In an experiment, 34.8243g of copper (II) nitrate hydrate, Cu(NO3)2•zH2O was heated to a constant mass of 27.0351g. Calculate the percent water in this hydrate?
What was the mass of water lost? What is the value of z (the number of waters in the formula)?
Answer:
1) The mass of water lost = 7.7892 grams
2) Z = 3: Cu(NO3)2*3H2O
Explanation:
Step 1: Data given
Mass of copper (II) nitrate hydrate, Cu(NO3)2•zH2O = 34.8243 grams
Mass of substance after heating = 27.0351 grams
Molar mass of Cu(NO3)2 = 187.56 g/mol
Molar mass of H2O = 18.02 g/mol
Step 2: Calculate mass of water
The mass of water is the mass lost after heating.
Mass water = 34.8243 - 27.0351 = 7.7892 grams of water
Step 3: Calculate moles of Cu(NO3)2
Moles Cu(NO3)2 = Mass Cu(NO3)2 / Molar mass Cu(NO3)2
Moles Cu(NO3)2 = 27.0351 grams / 187.56 g/mol
Moles Cu(NO3)2 = 0.144 moles
Step 4: Calculate moles of H2O
Moles H2O = 7.7892 grams / 18.02 g/mol
Moles H2O = 0.432 moles
Step 5: Calculate Z
z = moles H2O / moles Cu(NO3)2
Z = 0.432/0.144
Z = 3
This means we have 3 water molecules in the formula. This makes the formula ofthe hydrate: Cu(NO3)2*3H2O
Final answer:
To find the percent water in the hydrate, subtract the final mass after heating from the initial mass of the hydrate, then divide by the initial mass and multiply by 100. The mass of water lost is 7.7892g, and the percent water is 22.36%. To determine the value of z, calculate the moles of water and anhydrous Cu(NO3)2 to find z equals 3, making the formula Cu(NO3)2•3H2O.
Explanation:
The question involves determining the percent water in a hydrate, the mass of water lost during heating, and the value of z in the formula Cu(NO3)2•zH2O. Firstly, to calculate the mass of water lost, subtract the final mass after heating from the initial mass of the copper (II) nitrate hydrate. Then, to find the percent water, the mass of water lost is divided by the initial mass of the hydrate, multiplied by 100. Lastly, to determine the value of z, the moles of water lost are divided by the moles of anhydrous Cu(NO3)2.
To calculate the mass of water lost: 34.8243g (initial mass) - 27.0351g (final mass) = 7.7892g (mass of water lost).
To calculate the percent water in the hydrate: (7.7892g / 34.8243g) x 100 = 22.36%.
To find z, the number of waters, you need the molar mass of Cu(NO3)2 and H2O. Assume the molar mass of Cu(NO3)2 is approximately 187.55 g/mol. The mass of Cu(NO3)2 after heating is 27.0351g, which corresponds to 27.0351g / 187.55g/mol = 0.1441 mol. The moles of water lost is 7.7892g / 18.015g/mol = 0.4323 mol. Hence, z = 0.4323 mol / 0.1441 mol = 3, indicating the formula is Cu(NO3)2•3H2O.
The change in entropy, Δ S ∘ rxn , is related to the the change in the number of moles of gas molecules, Δ n gas . Determine the change in the moles of gas for each of the reactions and decide if the entropy increases, decreases, or has little or no change. A. 2 H 2 ( g ) + O 2 ( g ) ⟶ 2 H 2 O ( l )
Answer:
The entropy decreases.
Explanation:
The change in the standard entropy of a reaction (ΔS°rxn) is related to the change in the number of gaseous moles (Δngas), where
Δngas = n(gaseous products) - n(gaseous reactants)
If Δngas > 0, the entropy increasesIf Δngas < 0, the entropy decreases.If Δngas = 0, there is little or no change in the entropy.Let's consider the following reaction.
2 H₂(g) + O₂(g) ⟶ 2 H₂O(l)
Δngas = 0 - 3 = -3, so the entropy decreases.
A coal-burning power plant produces 500 megawatts of electricity while burning 225,000 kg of coal per hour (almost 250 tons each hour!). If the coal has an average energy content of 20 million Joules per kg, determine the efficiency of the power plant.
Answer:
40%
Explanation:
The efficiency of a plant is determined by
[tex]eff=\frac{E_{produced}}{E_{in}}[/tex]
Our produced energy is
[tex]500 \ MW[/tex]
While the input energy can be calculated
[tex]E_{in}=number\ of \ coal \ in \ * energy\ per\ coal\ unit\\ =\frac{225000}{3600}coal/ s\ *20\ MJ/coal\\ = 1250\ MW[/tex]
So, the efficiency of plant is
[tex]eff=\frac{500}{1250}}\\ 0.4[/tex]
or 40% efficient
Final answer:
The efficiency of the coal-burning power plant in question is 40%, calculated by comparing the electrical energy output to the energy input from burning coal.
Explanation:
To determine the efficiency of the coal-burning power plant, we compare the energy output in the form of electricity to the energy input from burning coal. First, calculate the total energy input per hour by multiplying the amount of coal burned by the average energy content of the coal. Then, convert the electrical energy output from megawatts to joules per hour. Finally, calculate efficiency using the formula: Efficiency (%) = (Energy output / Energy input) × 100.
Here are the calculations:
Energy input per hour = 225,000 kg/hour × 20× 106 J/kg = 4.5× 1012 J/hourEnergy output per hour = 500 MW × 3.6× 106 J/MWs = 1.8× 1012 J/hourEfficiency = (1.8× 1012 J/hour / 4.5× 1012 J/hour) × 100 = 40%Thus, the efficiency of the power plant is 40%.
Determine the molarity and mole fraction of a 1.15 m solution of acetone (CH3COCH3) dissolved in ethanol (C2H5OH). (Density of acetone = 0.788 g/cm3; density of ethanol = 0.789 g/cm3.) Assume that the final volume equals the sum of the volumes of acetone and ethanol.
Answer:
The molarity is 0.85M
The mole fraction = 0.05
Explanation:
Step 1: Data given
Molarity of acetone = 1.15 m
Density of acetone = 0.788 g/cm³
Density of ethanol = 0.789 g/cm³
Molar mass of acetone = 58.08 g/mol
Step 2: Calculate number of moles acetone
Molality = moles solute / kg solvent = moles acetone / kg ethanol. (To make it easy, we will suppose we have 1 kg ethanol)
1.15 m = 1.15 moles acetone / 1 kg ethanol
Step 3: Calculate mass of acetone
1.15 moles acetone * (58.08 g/mol) = 66.792 g acetone
Step 4: Calculate volume of acetone
66.792 g acetone / 0.788 g/mL acetone = 84.76 mL acetone
Step 5: Calculate volume of ethanol
1000 g ethanol / 0.789 g/mL ethanol = 1267.43 mL ethanol
Step 6: Calculate total volume solution
Total solution volume = 84.76 + 1267.43 = 1352.2 mL = 1.3522L
Step 7: Calculate molarity
Molarity of acetone = moles acetone /volume solution = 1.15moles / 1.3522L
Molarity = 0.85 M
Step 8: Calculate moles ethanol
moles ethanol= mass/ molar mass = 1000g/ 46.0g/mol = 21.74 moles
Step 9: Calculate mole fraction
mole fraction acetone = (moles acetone / total moles) = (1.15 / (1.15 + 21.74)) = 0.05
Two nonpolar organic liquids, hexane (C6H14) and heptane (C7H16), are mixed. Do you expect ΔHsoln to be a large positive number, a large negative number, or close to zero? Fill in the following sentence: ΔHsoln is determined by the relative magnitudes of the "old" solute-solute (ΔHsolute) and solvent-solvent (ΔHsolvent) interactions and the "new" solute-solvent (ΔHmix) interactions. In this process, the energy of mixing hexane and heptane (ΔHmix) is the energy of separating them individually (ΔHsolute + ΔHsolvent), so ΔHsoln is expected to be .
Answer:
ΔH of solution is expected to be close to zero.
Explanation:
When we mix two non polar organic liquids like hexane and heptane,the resulting mixture formed is an ideal solution.An ideal solution is formed when the force of attraction between the molecules of the two liquids is equal to the force of attraction between the molecules of the same type.
For instance if liquids A and B are mixed,
[tex]F_{A-A}[/tex] = [tex]F_{B-B}[/tex] = [tex]F_{A-B}[/tex]
Hence the condition before and after mixing remains unchanged.
Since enthalpy change is associated with inter molecular force of attraction the enthalpy change for ideal solution is zero.
More examples of ideal solutions are:
1. Ethanol and Methanol
2. Benzene and Toluene
3. Ethyl bromide and Ethyl iodide
If a radioactive isotope has a half-life of 26.5 days, how many days does it take for a sample of the isotope to decrease by 35.0% of its original value? View Available Hint(s) If a radioactive isotope has a half-life of 26.5 days, how many days does it take for a sample of the isotope to decrease by 35.0% of its original value? 40.1 days 16.5 days 7.15 days 0.0163 days
Answer:
16.5 days
Explanation:
Given that:
Half life = 26.5 days
[tex]t_{1/2}=\frac {ln\ 2}{k}[/tex]
Where, k is rate constant
So,
[tex]k=\frac {ln\ 2}{t_{1/2}}[/tex]
[tex]k=\frac {ln\ 2}{26.5}\ days^{-1}[/tex]
The rate constant, k = 0.02616 days⁻¹
Using integrated rate law for first order kinetics as:
[tex][A_t]=[A_0]e^{-kt}[/tex]
Where,
[tex][A_t][/tex] is the concentration at time t
[tex][A_0][/tex] is the initial concentration
Given:
35.0 % is decomposed which means that 0.35 of [tex][A_0][/tex] is decomposed. So,
[tex]\frac {[A_t]}{[A_0]}[/tex] = 1 - 0.35 = 0.65
t = 7.8 min
[tex]\frac {[A_t]}{[A_0]}=e^{-k\times t}[/tex]
[tex]0.65=e^{-0.02616\times t}[/tex]
t = 16.5 days.
A piston system absorbs 50.5 J of energy in the form of heat from the surroundings. The piston is working against a pressure of 0.491 atm. The final volume of the system is 56.2 L. What was the initial volume of the system if the internal energy of the system decreased by 106.0 J? (To convert between L∙atm and J, use 1 L∙atm = 101.3 J.)
Answer:
The initial volume of the system: V₁ = 53.06 L
Explanation:
Given: heat absorbed by system: q = 50.5 J, Pressure: P = 0.491 atm, Final volume: V₂ = 56.2 L, The change in the internal energy: ΔE = -106.0 J
Initial volume: V₁ = ? L
According to the First Law of Thermodynamics:
ΔE = q - PΔV
⇒ PΔV = q - ΔE = 50.5 J - (-106.0 J) = 156.5 J
As, 1 L∙atm = 101.3 J ⇒ 1 J = (1 ÷ 101.3) L∙atm
⇒ PΔV = 156.5 J = (156.5 ÷ 101.3) L∙atm = 1.54 L∙atm
So,
ΔV = 1.54 L∙atm ÷ P = 1.54 L∙atm ÷ 0.491 atm = 3.14 L
∵ ΔV = V₂ - V₁ = 3.14 L
⇒ V₁ = V₂ - 3.14 L = 56.2 L - 3.14 L = 53.06 L
Therefore, the initial volume of the system: V₁ = 53.06 L
What is the entropy change for the freezing process of 1 mole of liquid methanol at its freezing temperature (–97.6˚C) and 1 atm? Please report your answer two points past the decimal with the unit J/molK. ∆H˚fus = 3.17 kJ/mol.
Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K
Explanation :
Formula used :
[tex]\Delta S=\frac{\Delta H_{freezing}}{T_f}[/tex]
where,
[tex]\Delta S[/tex] = change in entropy
[tex]\Delta H_{fus}[/tex] = change in enthalpy of fusion = 3.17 kJ/mol
As we know that:
[tex]\Delta H_{fus}=-\Delta H_{freezing}=-3.17kJ/mol=-3170J/mol[/tex]
[tex]T_f[/tex] = freezing point temperature = [tex]-97.6^oC=273+(-97.6)=175.4K[/tex]
Now put all the given values in the above formula, we get:
[tex]\Delta S=\frac{\Delta H_{freezing}}{T_m}[/tex]
[tex]\Delta S=\frac{-3170J/mol}{175.4K}[/tex]
[tex]\Delta S=-18.07J/mol.K[/tex]
Therefore, the value of change in entropy for freezing process is, -18.07 J/mol.K
Answer:
Entropy is defined as the measure of possible arrangement of the atoms or molecules. Entropy change for the the freezing process mole of 1 liquid methanol at -97.6 degree celcius and 1 atm will be -18.07 J/mol.K
Explanation:
The entropy change can be calculated by the given formula:
[tex]\begin{aligned} \Delta \text S &= \frac{\Delta \text H_{\text {freezing}}}{\text T_{f}} \end{aligned}[/tex]
in which,
[tex]\Delta \text S[/tex] = change in entropy
[tex]\Delta \text H_{\text {fusion}} &=[/tex] change in enthalpy of fusion that is -3.17 kJ/mol (as given).
By the formula of enthalpy change,
[tex]\begin{aligned} \Delta \text H_{\text {fusion}} &= - \Delta \text H_{\text {freezing}} &= -3.17 {\text {kJ/mol}} &= -3170 \text {J/mol}\end{aligned}[/tex]
Now,
[tex]{\text T_{f}}[/tex] = Freezing Point Temperature = [tex]-97.6^{o} \text C[/tex] = 273 + (-97.6) = -175.4 K
Substituting the values in above equation of entropy change, we get,
[tex]\begin{aligned} \Delta \text S &= \frac{\Delta \text H_{\text {freezing}}}{\text T_{f}} \end{aligned}[/tex]
[tex]\begin{aligned} \Delta \text S &= \frac{-3170 \text {J/mol}}{175.4 \text {K}} \\\\Delta \text S& = -18.07 \text {J/mol.K}\end{aligned}[/tex]
Thus, the value of change in entropy for freezing process of 1 mole of liquid methanol is -18.07 J/mol.k
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how many grams are in 0.5 moles of C10H16?
Please show your work for marked brainliest!
Answer:
68 g
Explanation:
Molar mass (C10H16) = 10*12.0 g/mol + 16*1.0 g/mol = (120+16)g/mol =
= 136 g/mol
m (C10H16) = n(C10H16)*M(C10H16) = 0.5 mol*136 g/mol = 68 g
n(C10H16) - number of moles of C10H16
M(C10H16) - molar mass of C10H16
Part A Describe the electrodes in this nickel-copper galvanic cell. Drag the appropriate items to their respective bins. View Available Hint(s) ResetHelp Nickel Copper Standard reduction potentials for nickel(II) and copper(II) The standard reduction potential for a substance indicates how readily that substance gains electrons relative to other substances at standard conditions. The more positive the reduction potential, the more easily the substance gains electrons. Consider the following: Ni2+(aq)+2e−→Ni(s),Cu2+(aq)+2e−→Cu(s), E∘red=−0.230 V E∘red=+0.337 V Part B What is the standard potential, E∘cell, for this galvanic cell? Use the given standard reduction potentials in your calculation as appropriate.
Answer:
Part-A:
Anode is Nickel and Cathode is copper.
Part -B:
[tex]E^{0}_{cell}[/tex] of the reaction is 0.567V.
Explanation:
Nickel-Copper cell electrodes are "Ni" and "Cu" rod.
Nickel electrode is dipped in [tex]Ni^{+2}[/tex] solution.
Copper electrode dipped in [tex]Cu^{+2}[/tex] solution.
Part -A:
Anode:
At anode oxidation takes place
[tex]Ni(s) \rightarrow Ni^{+2}(aq)+2e^{-}[/tex]
Hence, anode is Nickel.
Cathode:
At cathode reduction takes place.
[tex]Cu^{+2}+2e^{-} \rightarrow Cu(s)[/tex]
Hence, Cathode is copper.
Part-B:
[tex]E^{0}_{cell}=E^{0}_{cathode}-E^{0}_{anode}[/tex]
[tex]=0.337-(-0.230)=0.567V[/tex]
Hence, [tex]E^{0}_{cell}[/tex] of the reaction is 0.567V.
Anodes and cathodes are the electrodes where oxidation and reduction take place.
What are the electrodes in galvanic cell?Electrochemical cells have two electrodes which is called anode and cathode. The anode is the electrode where oxidation occurs while the other hand, cathode is the electrode where reduction takes place so we can conclude that anodes and cathodes are the electrodes where oxidation and reduction take place.
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Write balanced half-reactions for the following redox reaction: 5Cl2(g)+2Mn+2(aq)+8H2O(l)→ 10Cl−(aq)+2MnO−4(aq)+16H+(aq)a. reduction: __.b. oxidation: ___.
Answer:
a. 5Cl₂ + 10e⁻ → 10Cl⁻
b. 4H₂O + Mn²⁺ → MnO₄⁻ + 5e⁻ + 8H⁺
Explanation:
5Cl2(g)+2Mn+2(aq)+8H2O(l)→ 10Cl−(aq)+2MnO4-(aq)+16H+(aq)
First of all, think in all the oxidation numbers for each compound. That's the way, you can notice the half reaction.
When the oxidation number, decrease, you have reduction. Compound is wining electrons.
When the oxidation number increase, you have oxidation. Compound is losing electrons.
Cl2(g) - Atoms in ground state, has 0 as oxidation number. In products side, you have the anion chloride which act with -1, so chlorine has been reduced.
Mn2+ - The manganese ion is already telling you with 2, that is its oxidation number. On the side products, the element was transformed into the permanganate anion; how oxygen acts with -2 and there are 4 atoms, it has -8 as oxidation state but since the general charge is -1 the Mn is acting with +7. From + 2 it went to +7, which means that it increased, so it has oxidized.
5Cl₂ + 10e⁻ → 10Cl⁻ - REDUCTION
The chlorine had to gain 1 electron to go from 0 to -1, but being a diatomic molecule were 2 electrons, but finally so that the charges are balanced and because there are 5 moles, it ends up gaining 10 electrons.
Mn²⁺ → MnO₄⁻ - OXIDATION
Look that in main reaction we have H⁺, that is the clue to notice us, that we are in acidic medium. So if we have 4 O, in MnO₄⁻, we have to complete with 4H₂O in the other side of O. And to ballance the protons, we have to add 8H⁺ in product side
4H₂O + Mn²⁺ → MnO₄⁻ + 5e⁻ + 8H⁺ OXIDATION
How do u finish this?. You have to multipply .2, the oxidation one to balance the e⁻ so, they can be cancelled.
5Cl₂ + 10e⁻ → 10Cl⁻
(4H₂O + Mn²⁺ → MnO₄⁻ + 5e⁻ + 8H⁺ ).2
8H₂O + 2Mn²⁺ → 2MnO₄⁻ + 10e⁻ + 16H⁺
5Cl₂ + 10e⁻ + 8H₂O + 2Mn²⁺ → 10Cl⁻ + 2MnO₄⁻ + 10e⁻ + 16H⁺
5Cl₂ + 8H₂O + 2Mn²⁺ → 10Cl⁻ + 2MnO₄⁻ + 16H⁺
The half-reactions for the given redox reaction includes the reduction of Mn2+(aq) to MnO4-(aq) with the addition of 5 electrons and 8 hydrogen ions and the oxidation of Cl2(g) to 10 Cl-(aq), losing 10 electrons in the process.
Explanation:The redox reaction is related to the process of reduction and oxidation which often takes place in electrochemical cells. This process includes half-reactions, which separately represent the oxidation and reduction in the redox reaction. For the given equation, the oxidation and reduction half-reactions are as follows:
a. Reduction: Mn2+(aq) + 5e- + 8H+(aq) → MnO4-(aq) + 4H2O(l).
b. Oxidation: Cl2(g) → 10e- + 10Cl-(aq).
These equations represent the transformation of Mn2+ and Cl2 in the given redox reaction where Mn2+ is being reduced and Cl2 is being oxidized.
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An unknown quantity of sugar was completely dissolved in water at 75 degrees Celsius to form a clear and colorless solution. The temperature of the solution was then lowered to 25 degrees Celsius while being mixed.
Question 1: Was the solution at 75 degrees Celsius saturated, unsaturated or supersaturated?
Question 2: Explain your reason for choosing the answer in question 1.
Answer:
67
Explanation:
Question 1: The solution at 75 degrees Celsius was likely saturated.
Question 2: The reason for choosing "saturated" is that the sugar was completely dissolved in water at 75 degrees Celsius, forming a clear and colorless solution.
What happens in such solutionsIn a saturated solution, the maximum amount of solute (sugar, in this case) that can dissolve at that temperature has already dissolved. When the solution is clear and colorless at that temperature, it suggests that the solution contains as much sugar as it can hold under those conditions.
If it were unsaturated, there would still be room for more sugar to dissolve, and if it were supersaturated, it would become unstable and potentially precipitate out excess solute when the temperature was lowered.
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A sample of N2 gas in a flask is heated from 27 Celcius to 150 Celcius. If the original gas is @ pressure of 1520 torr, what is the pressure of the final sample (in atm)?A.) 1.4atmB.) 2.8atmC.) 3.2atmD.) 4.3atm
Answer:
[tex]\large \boxed{\text{B.) 2.8 atm}}[/tex]
Explanation:
The volume and amount are constant, so we can use Gay-Lussac’s Law:
At constant volume, the pressure exerted by a gas is directly proportional to its temperature.
[tex]\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}[/tex]
Data:
p₁ = 1520 Torr; T₁ = 27 °C
p₂ = ?; T₂ = 150 °C
Calculations:
(a) Convert the temperatures to kelvins
T₁ = ( 27 + 273.15) K = 300.15 K
T₂ = (150 + 273.15) K = 423.15 K
(b) Calculate the new pressure
[tex]\begin{array}{rcl}\dfrac{1520}{300.15} & = & \dfrac{p_{2}}{423.15}\\\\5.064 & = & \dfrac{p_{2}}{423.15}\\\\5.064\times423.15&=&p_{2}\\p_{2} & = & \text{2143 Torr}\end{array}\\[/tex]
(c) Convert the pressure to atmospheres
[tex]p = \text{2143 Torr} \times \dfrac{\text{1 atm}}{\text{760 Torr}} = \textbf{2.8 atm}\\\\\text{The new pressure reading will be $\large \boxed{\textbf{2.8 atm}}$}[/tex]
Gaseous methane (CH4) reacts with gaseous oxygen gas (02) to produce gaseous carbon dioxide (CO2) and gaseous water (H20). What is the theoretical yield of carbon dioxide formed from the reaction of 1.28 g of methane and 10.1 g of oxygen gas? Be sure your answer has the correct number of significant digits in it. 02
Answer:
Theoretical yield = 3.51 g
Explanation:
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
For [tex]CH_4[/tex] :-
Mass of [tex]CH_4[/tex] = 1.28 g
Molar mass of [tex]CH_4[/tex] = 16.04 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{1.28\ g}{16.04\ g/mol}[/tex]
[tex]Moles_{CH_4}= 0.0798\ mol[/tex]
For [tex]O_2[/tex] :-
Mass of [tex]O_2[/tex] = 10.1 g
Molar mass of [tex]O_2[/tex] = 31.998 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{10.1\ g}{31.998\ g/mol}[/tex]
[tex]Moles_{O_2}= 0.3156\ mol[/tex]
According to the given reaction:
[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]
1 mole of methane gas reacts with 2 moles of oxygen gas
0.0798 mole of methane gas reacts with 2*0.0798 moles of oxygen gas
Moles of oxygen gas = 0.1596 moles
Available moles of oxygen gas = 0.3156 moles
Limiting reagent is the one which is present in small amount. Thus, [tex]CH_4[/tex] is limiting reagent.
The formation of the product is governed by the limiting reagent. So,
1 mole of methane gas on reaction produces 1 mole of carbon dioxide.
0.0798 mole of methane gas on reaction produces 0.0798 mole of carbon dioxide.
Mole of carbon dioxide = 0.0798 mole
Molar mass of carbon dioxide = 44.01 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]0.0798\ moles= \frac{Mass}{44.01\ g/mol}[/tex]
Mass of [tex]CO_2[/tex] = 3.51 g
Theoretical yield = 3.51 g
The theoretical yield of carbon dioxide formed from the reaction of 1.28 g of methane and 10.1 g of oxygen gas is 3.51 g.
The balanced equation for the reaction between methane (CH4) and oxygen gas (O2) to produce carbon dioxide (CO2) and water (H2O) is:
CH4 + 2O2 → CO2 + 2H2O
To find the theoretical yield of carbon dioxide, we need to calculate the moles of methane and oxygen and use the stoichiometry of the balanced equation.
Given: Mass of methane (CH4) = 1.28 g
Mass of oxygen gas (O2) = 10.1 g
Step 1: Calculate the moles of methane and oxygen using their molar masses:
Moles of CH4 = mass / molar mass = 1.28 g / 16.04 g/mol = 0.0798 mol
Moles of O2 = mass / molar mass = 10.1 g / 32.00 g/mol = 0.3156 mol
Step 2: Determine the limiting reactant (the reactant that is completely consumed first) by comparing the mole ratios of CH4 and O2 in the balanced equation. From the equation, for every 1 mol of CH4 there are 2 moles of O2 needed. Therefore, the mole ratio of CH4 to O2 is 1:2.
Since the mole ratio of CH4 to O2 is 1:2, and there are fewer moles of CH4 (0.0798 mol) compared to the moles of O2 (0.3156 mol), CH4 is the limiting reactant.
Step 3: Calculate the moles of CO2 produced using the mole ratio from the balanced equation:
Moles of CO2 = moles of CH4 x (1 mol of CO2 / 1 mol of CH4) = 0.0798 mol x (1 mol / 1 mol) = 0.0798 mol
Step 4: Convert the moles of CO2 to grams using the molar mass of CO2:
Mass of CO2 = moles x molar mass = 0.0798 mol x 44.01 g/mol = 3.51 g
Therefore, the theoretical yield of carbon dioxide formed from the reaction of 1.28 g of methane and 10.1 g of oxygen gas is 3.51 g (to the correct number of significant digits).
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Elemental phosphorus reacts with chlorine gas according to the equation: P4(s)+6Cl2(g)❝4PCl3(l) A reaction mixture initially contains 45.55g P4 and 131.9g Cl2.
Once the reaction has occurred as completely as possible, what mass (in g) of the excess reactant is left?
Answer:
So after reaction completion. 7.11 g of P₄ will be left over.
Explanation:
So for number of moles of P₄ = Given mass/ molar mass = 45.55/ 123.88
= 0.367 moles
Number of moles of Cl₂ = Given mass/ molar mass = 131.9/70 = 1.884 moles
We will divide the values by dividing excess of reactants with their coefficient for finding excess of reactants.
P₄ = 0.367/ coefficient 1 = 0.367
Cl₂ = 1.884/ 6= 0.314
Phosphorus is thus the excess reactant.
We need to use the mole value of the limiting reactant in order to find the mass of P₄ used. Using dimensional analysis,
1.884 mol Cl₂ ( 1mole P₄/6 mol Cl₂ ) ( 123.88 g P₄/ 1 mol P₄) = 38.44 g P₄
This is the amount which is used up. So by subtracting this amount from the Initial amount
45.55 – 38.44 = 7.11 g of P₄
So after reaction completion. 7.11 g of P₄ will be left over.
Final answer:
To find the mass of the excess reactant, we need to determine the limiting reactant first. By comparing the moles of each reactant, we can find that the excess reactant is chlorine (Cl2) and the mass left is 127.7 g.
Explanation:
To determine the mass of the excess reactant, we need to find the limiting reactant first. We can do this by comparing the moles of each reactant to their stoichiometric coefficients. Firstly, we need to convert the given masses of P4 and Cl2 to moles using their molar masses: 123.9 g/mol for P4 and 70.9 g/mol for Cl2. This gives us 0.368 mol of P4 and 1.86 mol of Cl2. From the balanced equation, we see that 1 mole of P4 reacts with 6 moles of Cl2, so we need 1/6th of the moles of Cl2 to react with all of the P4. This means that we need 0.368/6 = 0.061 mol of Cl2. The excess moles of Cl2 is then 1.86 - 0.061 = 1.799 mol. Finally, we convert the excess moles to grams using the molar mass of Cl2: 1.799 mol * 70.9 g/mol = 127.7 g. Therefore, the mass of the excess reactant (Cl2) is 127.7 g.
Chemistry: An oxygen gas container has a volume of 20.0 L. How many grams of oxygen are in the container, if the gas has a pressure of the 845mmHg at 22degrees Celsius?
Answer:
There are 29.4 grams of oxygen in the container
Explanation:
Step 1: Data given
Volume = 20.0 L
Pressure = 845 mmHg
Temperature = 22.0 °C
Molar mass of O2 = 32 g/mol
Step 2: Ideal gas law
p*V = n*R*T
⇒ p = the pressure of the gas = 845 mmHg = 1.11184
⇒ V = the volume of the gas = 20.0 L
⇒ n = the number of moles = TO BE DETERMINED
⇒R = the gasconstant = 0.08206 L*atm/K*mol
⇒ T = the temperature = 22°C + 273 = 295 Kelvin
n = (p*V)/(R*T)
n = (1.11184*20.0)/(0.08206*295)
n = 0.9186 moles
Step 3: Calculate mass of NO2
Mass of O2 = Moles O2 * Molar mass O2
Mass of O2 = 0.9186 moles * 32 g/mol
Mass of O2 = 29.4 grams
There are 29.4 grams of oxygen in the container