Answer:
Sn²⁺ + Tl³⁺ → Sn⁴⁺ + Tl⁺
Explanation:
The Sn²⁺ is oxidized to Sn⁴⁺. Whereas Tl³⁺ is reduced to Tl⁺. The half-reactions are:
Sn²⁺ → 2e⁻ + Sn⁴⁺ (Oxidation, loosing electrons)
Tl³⁺ + 2e⁻ →Tl⁺ (Reduction, gaining electrons)
The sum of the reactions gives:
Sn²⁺ + Tl³⁺ + 2e⁻ → 2e⁻ + Sn⁴⁺ + Tl⁺
Subtracting the electrons in both sides of the reaction:
Sn²⁺ + Tl³⁺ → Sn⁴⁺ + Tl⁺The following reactions all have K < 1. 1) HCOO- (aq) + C6H5COOH (aq) HCOOH (aq) + C6H5COO- (aq) 2) C9H7O4- (aq) + C6H5COOH (aq) C6H5COO- (aq) + HC9H7O4 (aq) 3) HCOOH (aq) + C9H7O4- (aq) HC9H7O4 (aq) + HCOO- (aq) Arrange the substances based on their relative acid strength.
Answer:
Explanation:
C₉H₇O₄⁻ = weakest base
C₆H₅COO⁻ = strongest base
HCOO⁻ = intermediate base
HCOOH = not a Bronsted-Lowry base
HC₉H₇O₄ = not a Bronsted-Lowry base
C₆H₅COOH = not a Bronsted-Lowry base
g The atomic mass of an element is equal to ________. The atomic mass of an element is equal to ________. its mass number one-twelfth of the mass of a carbon-12 atom a weighted average mass of all of the naturally occurring isotopes of the element its atomic number the average mass of all of the naturally occurring isotopes of the element
Answer:
Total numbe of protons and neutrons in a single atom of that element
Explanation:
Hello,
I'll answer the question by filling in the blank spaces
"The atomic mass of an element is equal to the total number of proton and neutron in a particular atom of the element. The atomic mass of an element is equal to the atomic weight. Its mass number one-twelfth of the mass of carbon-12 atom a weighted mass of all naturally occurring isotopes of the elements. Its atomic mass is the average mass of all the naturally occurring isotopes of the element."
The atomic mass of an element is the total number of protons and neutrons in a single atom of that element.
The atomic mass of an element is equal to a weighted average mass of all of the naturally occurring isotopes of the element. The correct answer is option 2.
Isotopes are elements with the same number of protons (atomic number) but differing numbers of neutrons (mass number).
Most elements exist in nature as a mixture of isotopes, each with a different mass number and abundance. The atomic mass of an element is computed by adding the masses of all isotopes, multiplying by their relative abundance, and dividing by the total abundance of all isotopes.
This gives a weighted average mass that corresponds to the normal mass of an element's atom in nature.
Therefore, the correct answer is option 2. to a weighted average mass of all of the naturally occurring isotopes of the element.
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The atomic mass of gallium is 69.72 . The density of iron is 7.87 . The atomic mass of iron is 55.847 . Calculate the number of gallium atoms in one ton (2000 pounds) of gallium. (Enter your answer to three significant figures.)
Answer:
the atomic mass of any elemet contains avogardo numberof atoms
In case of Gallium,
69.72 gram is atomic mass and it cotnains around 6.023*10^23 atoms of Gallium
but, 2000 punds = 907184.7 grams
907184.7 gram of gallium contains= 6.023*10^23* 907184/69.72
= 79 *10^26 atoms
Explanation:
A certain mass of carbon reacts with 9.53 g of oxygen to form carbon monoxide. ________ grams of oxygen would react with that same mass of carbon to form carbon dioxide, according to the law of multiple proportions.
Answer: 9.53 *2= 19.06
Explanation:
The law of multiple proportions states that if two elements combines to form more than one compound the ratio of masses of the second element which combines to the fixed mass of the first element will always be the ratios of the small whole numbers.
in case of carbon monoxide, mass of carbon will be the same of mass of oxygen.
But in case of carbon dioxide, if carbon is 9.53 units then oxygen will be twice as that of carbon.
CO2, so 9.53*2= 19.06 grams of oxygen will combine with 9.53 grams of carbon to form carbon dioxide.
Give the characteristic of a zero order reaction having only one reactant. a. The rate of the reaction is not proportional to the concentration of the reactant. b. The rate of the reaction is proportional to the square of the concentration of the reactant. c. The rate of the reaction is proportional to the square root of the concentration of the reactant. d. The rate of the reaction is proportional to the natural logarithm of t
Answer:
a. The rate of the reaction is not proportional to the concentration of the reactant.
Explanation:
The rate expression for a zero order reaction is given as;
A → Product
Rate = k[A]⁰
[A]⁰ = 1
Rate = K
GGoing through the options;
a) This is correct because in the final form of the rate expression, the rate is independent of the concentration.
b) This option is wrong
c) This option is also wrong
d) Like options b and c this is also wrong becaus ethere is no relationship between either the concentration or t.
How many moles of CO2 can be produced by the complete reaction of 1.0 g of lithium carbonate with excess hydrochloric acid (balanced chemical reaction is given below)? Li2CO3(s) + 2HCl(aq) --> 2LiCl(aq) + H2O(l) + CO2(g) Question 1 options: 1.7 g 1.1 g 0.60 040 g
Answer:Mass of CO2 = 0.60g
Explanation:
Given the chemical rection
Li2CO3(s) + 2HCl(aq) --> 2LiCl(aq) + H2O(l) + CO2(g
No of moles = mass / molar mass
molar mass Li2CO3 = Molecular mass calculation: 6.941 x 2 + 12.0107 + 15.9994 x 3 =
= 73.8909 g/mol
therefore Number of moles Li2CO3 = 1.0g / 73.89 g/mol
= 0.0135 moles Li2CO3
From our given Balanced equation, shows that
Li2CO3(s) + 2HCl(aq) --> 2LiCl(aq) + H2O(l) + CO2(g
1 mole Li2CO3 produces 1 mole CO2
therefore 0.0135 mol Li2CO3 will produce 0.0135 moles of CO2
Also
No of moles = mass / molar mass
Mass = No of moles x molar mass
molar mass of CO2=12.0107 + 15.9994 x 2=44.0095 g/mol
Mass of CO2= 0.0135 X 44.0095 g/mol =0.594≈0.60g
On a hot summer day, the density of air at atmospheric pressure at 35.5°C is 1.1970 kg/m3. (a) What is the number of moles contained in 1.00 m3 of an ideal gas at this temperature and
Complete question:
On a hot summer day, the density of air at atmospheric pressure at 35.5°C is 1.1970 kg/m3. (a) What is the number of moles contained in 1.00 m3 of an ideal gas at this temperature and pressure.
Answer:
The number of moles contained by an ideal gas at this temperature and pressure is 41.32 moles.
Explanation:
Given;
density of dry air, ρ = 1.1970 kg/m³
temperature of the air, T = 35.5°C = 273 + 35.5 = 308.5 K
air volume, V = 1 m³
Apply ideal gas law for dry to calculate the air pressure;
[tex]P = \rho R_dT[/tex]
where;
P is the air pressure
ρ is the air density
Rd is gas constant for dry air = 287 J/kg/K
P = 1.197 x 287 x 308.5 = 105,981.78 Pa
(a) Now, determine the number of moles contained by an ideal gas at this temperature and pressure, by applying ideal gas law;
PV = nRT
where;
P is the pressure of the gas (Pa)
V is the volume of the gas (m³)
n is number of gas moles
R is gas constant = 8.314 m³.Pa / mol.K
T is temperature (K)
n = (PV) / (RT)
n = (105,981.78 x 1) / (8.314 x 308.5)
n = 41.32 moles
Therefore, the number of moles contained by an ideal gas at this temperature and pressure is 41.32 moles.
The number of moles of an ideal gas at this temperature and pressure is 41.5 moles.
Given that;
Density of dry air = 1.1970 kg/m3
Pressure of dry air = ?
Temperature of dry air = 35.5°C + 273 = 308.5 K
Hence;
P = Density × gas constant of dry air × Temperature
P = 1.1970 kg/m3 × 287.1 J/Kg/K × 308.5 K
P = 106019 Pa or 1.05 atm
Using the ideal gas equation;
PV = nRT
n = PV/RT
n = 1.05 atm × 1000 L/0.082 atmL/K.mol × 308.5 K
n = 41.5 moles
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Question 1
1 pts
2B+6HCI --
| --> 2BCl3 + 3H2
How many moles of boron chloride will be produced if you start with 8.752 moles of HCI
(hydrochloric acid)? (Round to 3 sig figs. Enter the number only do not include units.)
Answer:
2.92 mol
Explanation:
Step 1: Write the balanced equation
2 B(s) + 6 HCI(aq) ⇒ 2 BCl₃(aq) + 3 H₂(g)
Step 2: Establish the appropriate molar ratio
The molar ratio of hydrochloric acid to boron chloride is 6:2.
Step 3: Calculate the moles of boron chloride produced from 8.752 moles of hydrochloric acid
[tex]8.752molHCl \times \frac{2molBCl_3}{6molHCl} = 2.92molBCl_3[/tex]
A solution that is 0.135 M is diluted to make 500.0 mL of a 0.0851 M solution. How many milliliters of the original solution were required? View Available Hint(s) A solution that is 0.135 M is diluted to make 500.0 mL of a 0.0851 M solution. How many milliliters of the original solution were required? 5.74 mL 0.315 mL 793 mL 315 mL
Answer:
315mL
Explanation:
Data obtained from the question include the following:
Molarity of stock solution (M1) = 0.135 M
Volume of stock solution needed (V1) =?
Molarity of diluted solution (M2) = 0.0851 M
Volume of diluted solution (V2) = 500mL
The volume of the stock solution needed can be obtain as follow:
M1V1 = M2V2
0.135 x V1 = 0.0851 x 500
Divide both side by 0.135
V1 = (0.0851 x 500) / 0.135
V1 = 315mL
Therefore, the volume of the stock solution needed is 315mL
Enter an equation for the formation of C2H5OH(l) from its elements in their standard states. Enter any reference to carbon as C(s). Express your answer as a chemical equation. Identify all of the phases in your answer.
Answer:
C(s) + 3 H₂(g) + 1/2 O₂(g) ⇒C₂H₅OH(l)
Explanation:
Ethanol (C₂H₅OH) is an alcohol and it is formed by carbon (C), H (hydrogen) and O (oxygen) atoms. These elements in their standard states are:
C: C(s), it is solid, could be graphite, diamond, among others.
H: H₂(g), it is a diatomic gas.
O: O₂(g), it is a diatomic gas.
So, we can write the equation for the formation of C₂H₅OH from C(s), H₂ and O₂ as follows:
C(s) + H₂(g) + O₂(g) ⇒C₂H₅OH(l)
Finally, we have to balance the equation by adding the estequiometrical coefficients:
C(s) + 3 H₂(g) + 1/2 O₂(g) ⇒C₂H₅OH(l)
2C(s)+3[tex]H_{2} [/tex](g)+[tex]\frac{1}{2} [/tex][tex]O_{2} [/tex](g)→[tex]C_{2} [/tex][tex]H_{5} [/tex]OH(l)
Explanation:
Standard state of carbon: C(s)
Standard state of oxygen: [tex]O_{2} [/tex](g)
Standard State of Hydrogen: [tex]H_{2} [/tex](g)
Then balance the equation C2H5OH(l) to get 2C(s)+3[tex]H_{2} [/tex](g)+[tex]\frac{1}{2} [/tex][tex]O_{2} [/tex](g)→[tex]C_{2} [/tex][tex]H_{5} [/tex]OH(l).
what’s the SI unit of time ?
Answer:
The answer is A
Explanation:
When comparing the two chair conformations for a monosubstituted cyclohexane ring, which type of substituent shows the greatest preference for occupying an equatorial position rather than an axial position
Answer:
See the explanation
Explanation:
In this case, we have to keep in mind that in the monosubstituted product we only have to replace 1 hydrogen with another group. In this case, we are going to use the methyl group [tex]CH_3[/tex].
In the axial position, we have a more steric hindrance because we have two hydrogens near to the [tex]CH_3[/tex] group. If we have more steric hindrance the molecule would be more unstable. In the equatorial positions, we don't any interactions because the [tex]CH_3[/tex] group is pointing out. If we don't have any steric hindrance the molecule will be more stable, that's why the molecule will the equatorial position.
See figure 1
I hope it helps!
17. Write the molecular balanced equation for the recovering of copper metal. 18. Write the complete ionic balanced equation for the recovering of copper metal. 19. Write the net ionic balanced equation for the recovering of copper metal. 20. What type of reaction is this
Answer:
Explanation:
17. it goes from solid copper to aqueous copper:
Cu(s) --> Cu₂(aq) + 2e⁻
18. complete ionic:
Cu(s) --> Cu₂(aq) + 2e⁻
19. net ionic, must include only reacting species, so
Cu(s) --> Cu₂(aq) + 2e⁻
20. this type of reaction is dissolution reaction(redox reaction)
copper reduced from Cu²⁺ to Cu.
Carbon dioxide and water vapor are variable gases because _____.
Answer: their amounts vary throughout the atmosphere
Explanation:
There is very little that travels over the atmosphere
Vary=very little
Hope that helps
A certain element consists of two stable isotopes. The first has a mass of 62.9 amu and a percent natural abundance of 69.1 %. The second has a mass of 64.9 amu and a percent natural abundance of 30.9 %. What is the atomic weight of the element?
Answer:
63.518
Explanation:
The following data were obtained from the question:
Mass of Isotope A = 62.9 amu
Abundance of isotope A (A%) = 69.1%
Mass of isotope B = 64.9 amu
Abundance of isotope B (B%) = 30.9%
Atomic weight of the element =..?
The atomic weight of the element can be obtained as follow:
Atomic weight = [(Mass of A x A%)/100] + [(Mass of B x B%) /100]
Atomic weight = [(62.9 x 69.1)/100] + [(64.9 x 30.9)/100]
Atomic weight = 43.4639 + 20.0541
Atomic weight = 63.518
Therefore, the atomic weight of the element is 63.518.
The diagram below shows that the periodic table is divided into different blocks.
A periodic table is shown. The main table consists of seven rows; two additional rows are shown below. In each block, the first column is labeled and the remaining columns are empty. The s-block is shaded in yellow and comprises the first two columns, plus one cell at the far side of the table. The first column has seven rows with entries 1 s, 2 s, 3 s, 4 s, 5 s, 6 s, and 7 s. A lone cell labeled 1 s appears at the top far right corner, aligned with the 1 s cell in the first column. The d-block is shaded in blue and contains 10 columns and 3 or 4 rows. The first column is directly to the right of the s-block. The first entry in the first d-block column aligns with the 4 s block, and is labeled 3d; further entries in that column are 4 d, 5 d, and 6 d. The first three columns in the block are four entries long; the remaining columns are three entries long, losing the bottom entry. The p-block is shaded in orange, and has 6 columns and 5 rows. The top row aligns with the 2 s block; entrie
Elements that have complete valence electron shells are mostly found in the
s block.
d block.
p block.
Answer:
p block.
Explanation:
jus took the test
Answer:
c p block
Explanation:
What is Key for the reaction 2503(9) = 2802(9) + O2(g)?
Answer:
Option C. Keq = [SO2]² [O2] /[SO3]²
Explanation:
The equilibrium constant keq for a reaction is simply the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient.
Now, let us determine the equilibrium constant for the reaction given in the question.
This is illustrated below:
2SO3(g) <==> 2SO2(g) + O2(g)
Reactant => SO3
Product => SO2, O2
Keq = concentration of products /concentration of reactants
Keq = [SO2]² [O2] /[SO3]²
What is the Lewis structure for *OPCl3 and AlCl6^3-? What are their electron/molecular geometry and Ideal Bond Angle ?
Answer:
Here's what I get
Explanation:
1. POCl₃
(a) Lewis structure
Set P as the central atom, with O and Cl atoms directly attached to it.
Electrons available = P + O + 3Cl = 5 + 6 + 3×7 = 11 + 21 = 32
Arrange these electrons to give every atom an octet. Put a double bond between P and O.
You get the structure shown below.
(b) Geometry
There are four bond pairs and no lone pairs about the P atom.
Electron pair geometry — tetrahedral
Molecular geometry — tetrahedral
(c) Ideal bond angles
Tetrahedral bond angle = 109.5°
2. AlCl₆³⁻
(a) Lewis structure
Set Al as the central atom, with the Cl atoms directly attached to it.
Electrons available = Al + 6Cl + 3(-) = 3 + 6×6 +3 = 6 + 36 = 42
Arrange these electrons to give every atom an octet. Assign formal charges.
You get the structure shown below.
(b) Geometry
There are six bond pairs and no lone pairs about the Al.
Electron pair geometry — octahedral
Molecular geometry — octahedral
(c) Ideal bond angles
Axial-equatorial = 90°
Equatorial-equatorial = 120°
Axial-axial = 180°
At a particular temperature, an equilibrium mixture the reaction below was found to contain 0.171 atm of I2, 0.166 atm of Cl2 and 9.81 atm of ICl. Calculate the value of the equilibrium constant, Kp at this temperature.I2(g) + Cl2(g) <=> 2 ICl(g)
Answer: 3390
Explanation:
Since this problem already gives is the equilibrium values, all we have to do is to plug them into the formula for [tex]K_{p}[/tex].
[tex]K_{p} =\frac{[ICl]^2}{[I_{2}][Cl_{2}] }[/tex]
[tex]K_{p} =\frac{(9.81)^2}{(0.171)(0.166)} =3390[/tex]
Suppose that you add 27.6 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the added solute, you find that there is a freezing point depression of 3.69 oC compared to pure benzene. What is the molar mass of the unknown compound
Answer:
The molar mass of the unknown compound is 153.3 g/mol
Explanation:
Step 1: Data given
Mass of an unknown molecular compound = 27.6 grams
Mass of benzene = 0.250 kg
Kf of benzene = 5.12 °C/m
freezing point depression of 3.69 °C
Step 2: Calculate molality
ΔT = i*Kf*m
⇒with ΔT = reezing point depression of 3.69 °C
⇒with i = the van't Hoff factor of Benzene = 1
⇒with Kf = 5.12 °C/m
⇒ with m = molality = moles unknown compound / mass of benzene
3.69 = 1 * 5.12 * m
m = 0.72 molal
Step 3: Calculate moles of the unknown compound
molality = moles / mass benzene
0.72 molal = moles / 0.250 kg
Moles = 0.72 m * 0.250 kg
Moles = 0.18 moles
Step 4: Calculate molar mass of the unknown compound
molar mass = mass / moles
Molar mass = 27.6 grams / 0.18 moles
Molar mass = 153.3 g/mol
The molar mass of the unknown compound is 153.3 g/mol
Molar mass is the mass of the one mole of substance. The molar mass of the given unknown compound is 153.3 g/mol.
Molality of the compound can be calculated using
ΔT = i Kf m
Where,
ΔT = freezing point depression = 3.69 °C
i = Van't Hoff factor of Benzene = 1
Kf = constant of freezing = 5.12 °C/m
m = molality = ?
Put the values in the equation,
3.69 = 1 x 5.12 x m
m = 0.72 molal
Number of moles of the compound,
[tex]\bold {molality =\dfrac { moles} { mass\ benzene}}\\\\\bold {0.72\ molal = \dfrac {moles }{0.250\ kg}}\\\\\bold {Moles = 0.72\ m \times 0.250\ kg}\\\\\bold {Moles = 0.18}[/tex]
So, molar mass of the unknown compound,
[tex]\bold {Molar\ mass =\dfrac { mass}{ moles}}\\\\\bold {Molar\ mass = \dfrac {27.6\ grams }{0.18\ moles}}\\\\\bold {Molar\ mass = 153.3 g/mol}[/tex]
The molar mass of the given unknown compound is 153.3 g/mol.
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need helpp asapp please
Answer:
B. None of these
Explanation:
Sulfur has less ionization energy than phosphorus because sulfur has a pair of electron in its 3p subshell that increases electron repulsion in sulfur and sulfur electrons can easily remove from its sub-level.
While, there are no electron pairs in 3p subshell of phosphorus, therefore it requires more energy to remove an electron from 3p subshell.
Hence, the reason is electron repulsion and the correct answer is B.
Suppose 1.87g of nickel(II) bromide is dissolved in 200.mL of a 52.0mM aqueous solution of potassium carbonate. Calculate the final molarity of nickel(II) cation in the solution. You can assume the volume of the solution doesn't change when the nickel(II) bromide is dissolved in it.
Answer:
Molarity = 0.0428 M = 42.8 mM
Explanation:
Step 1: Data given
Mass of nickel(II) bromide = 1.87 grams
Molar mass of nickel(II) bromide = 218.53 g/mol
Volume = 200 mL = 0.200 L
Step 2: Calculate moles of nickel(II) bromide
Moles nickel (II) bromide = mass / molar mass
Moles nickel (II) bromide = 1.87 grams / 218.53 g/mol
Moles nickel (II) bromide = 0.00856 moles
Step 3: Calculate moles nickel (II) cation
For 1 mol NiBr2 we have 1 mol Ni^2+
For 0.00856 moles NiBr2 we have 0.00856 moles Ni^2+
Step 4: Calculate final molarity of Ni^2+
Molarity = moles / volume
Molarity = 0.00856 moles / 0.200 L
Molarity = 0.0428 M = 42.8 mM
A solution of benzene in methanol has a transmittance of 93.0 % in a 1.00 cm cell at a wavelength of 254 nm. Only the benzene absorbs light at this wavelength, not the methanol. What will the solution's transmittance be if it is placed in a 10.00 cm long pathlength cell
Answer:
T = 48.39%
Explanation:
In this case we need to apply the Beer law which is the following:
A = CεL (1)
Where:
A: Absorbance of solution
C: Concentration of solution
ε: Molar Absortivity (Constant)
L: Length of the cell
Now according to the given data, we have transmittance of 93% or 0.93. We can calculate absorbance using the following expression:
A = -logT (2)
Applying this expression, let's calculate the Absorbance:
A = -log(0.93)
A = 0.03152
Now that we have the absorbance, let's calculate the concentration of the solution, using expression (1).
A = CεL
C = A / εL
Replacing:
C = 0.03152 / 1 *ε (3)
Now, we want to know the transmittance of the solution with a length of 10 cm. so:
A = CεL
Concentration and ε are constant, so:
A = (0.03152 / ε) * ε * 10
A = 0.3152
Now that we have the new absorbance, we can calculate the new transmittace:
T = 10^(-A)
T = 0.4839 ----> 48.39%
A solution is prepared by mixing 5.00 mL of 0.100 M HCl and 2.00 mL of 0.200 M NaCl. What is the molarity of chloride ion in this solution?
Answer:
0.129 M
Explanation:
0.100 M HCl = 0.100 mol/L solution HCl
5.00 mL = 0.00500 L solution HCl
0.100 mol/L HCl * 0.00500 L = 0.000500 mol HCl
HCl ------> H+ + Cl-
1 mol 1 mol
0.000500 mol 0.000500 mol
0.200 M NaCl = 0.200 mol/L solution NaCl
2.00 mL = 0.00200 L solution NaCl
0.200 mol/L NaCl*0.00200 L = 0.000400 mol NaCl
NaCl ------> Na+ + Cl-
1 mol 1 mol
0.000400 mol 0.000400 mol
Chloride ion altogether (0.000500 mol + 0.000400 mol) =0.000900 mol
Solution altogether (0.00500 L+0.00200 L) = 0.00700L
Molarity (Cl-)= solute/solution = 0.000900 mol/0.00700L = 0.129 mol/L=
= 0.129 M
Describe the buffer capacity of the acetic acid buffer solution in relation to the addition of both concentrated and dilute acids and bases.
Answer:
The answer is in the explanation
Explanation:
Acetic acid, CH₃COOH, is a weak acid that will produce a buffer when its conjugate base, CH₃COO⁻, acetate ion, is added to the solution.
That is because a buffer is the mixture of a weak acid and its conjugate base or vice versa.
When an acid (HX) is added to the solution, the acetate ion will react producing acetic acid, thus:
CH₃COO⁻ + HX → CH₃COOH + X⁻
For this reason, the pH doesn't change abruptly because H⁺ ions are not produced.
Now, if a base (BOH) is added to the buffer, CH₃COOH will react producing acetate ion and water, thus:
CH₃COOH + BOH → CH₃COO⁻ + H₂O + B⁺.
In the same way, there are not produced free OH⁻ and the pH doesn't change significantly.
What are 3 characteristics of chemical reactions
Answer:
Evolution of gas.
Formation of a precipitate.
Change in color.
Explanation:
Please what's the missing minor products? And kindly explain in your own words how they were formed. Thank you!
Answer:
it's a two step elimination reaction
Explanation:
it follows a carbocationic pathway. When carbocation is stable, the equation is favourable, that is, double bond is formed by expelling hydrogen atom.
With methyl, ethyl, or cyclopentyl halides as your organic starting materials and using any needed solvents or inorganic reagents, outline syntheses of each of the following. More than one step may be necessary and you need not repeat steps carried out in earlier parts of this problem. (a) CH3I (b) I (c) CH3OH (d) OH (e) CH3SH (f) SH (g) CH3CN (h) CN (i) CH3OCH3 (j) OMe
Answer:
In the attachment you can find all the possible chemical reactions.
Some reaction can not be obtained by using alkyl halides because halides are weak leaving group which can leave compound during reaction easily but hydroxyl groups is a strong nucleophile which can not leave compound easily. So we can obtain alcohol from ethyl bromide, but we can not obtain hydroxyl ion from ethyl bromide.
Explanation:
The methyl of ethyl halides as the organic starting materials are using the needed solvents or the inorganic reagents. These can be not repeated in steps that arrive out in earlier parts.
The reaction can not be taken by the use of alkyl halides as the halides are the weakest leaving group which leave the compound during reaction easily.the hydroxyl group is the strong nucleophile that cannot leave the compound easily. Thus we can get alcohol from the ethyl bromide, but we can not obtain the hydroxyl ion from the ethyl bromide.Learn more about the methyl or the cyclopentyl.
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Tubes through which water flows as it is brought from 0.8 MPa, 150C to 240C at essentially constant pressure in the boiler of a power plant. The total mass flow rate of the water is 100 kg/s. Combustion gases passing over the tubes cool from 1067 to 547C at essentially constant pressure. The combustion gases can be modeled as air as an ideal gas. There is no significant heat transfer from the boiler to its surroundings. Surrounding (dead state) temperature and pressure are given as 25C and 1 atm, respectively. Determine i) the exergetic efficiency of the boiler ii) rate of exergy destruction as kW iii) mass flow rate of the combustion gases as kg/s
Answer:
The correct answer is i) 50.2 % ii) 13440.906 kW and iii) 71.986 kg/s.
Explanation:
In order to find the mass flow rate of the combustion of gases, there is a need to use the energy balance equation:
Mass of water × specific heat of water (T2 -T1)w = mass of gas × specific heat of gas (T2-T1)g
100 × 4.18 × [(240 + 273) - (150 + 273)] = mass of gas × 1.005 × [(1067+273) - (547+273)]
Mass of gas = 71.986 kg/s
The entropy generation of water can be determined by using the formula,
(ΔS)w = mass of water × specific heat of water ln(T2/T1)w
= 100 × 4.18 ln(513/423)
= 80.6337 kW/K
Similarly the entropy generation of water will be,
(ΔS)g = mass of gas × specific heat of gas ln(T2/T1)g
= 71.986 × 1.005 ln (820/1340)
= -35.53 kW/K
The rate of energy destruction will be,
Rate of energy destruction = To (ΔS)gen
= T₀ [(ΔS)w + (ΔS)g]
= (25+273) [80.6337-53.53)
Rate of energy destruction = 13440.906 kW
The availability of water will be calculated as,
= mass of water (specific heat of water) [(T₁-T₂) -T₀ ln T₁/T₂]
= 100 × 4.8 [(513-423) - 298 ln 513/423]
= 13591.1477 kW
The availability of gas will be calculated as,
= mass of gas (specific heat of gas) [(T₁-T₂) - T₀ ln T₁/T₂]
= 71.986 × 1.005 × [(1340-820) - 298 ln 1340/820]
= 27031.7728 kW
The exergetic efficiency can be calculated as,
= Gain of availability / loss of availability
= 13591.1477/27031.7728
= 0.502
The exergetic efficiency is 50.2%.
Enter your answer in the provided box. On a cool, rainy day, the barometric pressure is 739 mmHg. Calculate the barometric pressure in centimeters of water (cmH2O) (d of Hg = 13.5 g/mL; d of H2O = 1.00 g/mL).
Answer:
997.65cmH2O
Explanation:
Barometric pressure = 739 mmHg
density of Hg = 13.5 g/ml
density of water (H2O) = 1.00 g/ml
Calculate Barometric pressure in centimetres of water ( cmH20)
equate the barometric pressure of Hg and water
739 * 13.5 * 9.8 = x * 1 * 9.81
x ( barometric pressure of water in mmH2O ) = 739 *13.5 / 1 = 9976.5mmH2O
in cmH2O = 997.65cmH2O