Explanation:
Lanthanide series= E4
Boron=Si
Chalogen=O
Alkaline Earth metal =M9
using the balanced equation below how many grams of lead(||) sulfate would be produced from the complete reaction of 23.6 g lead (|V) oxide
Answer:
59.8 g of PbSO₄.
Explanation:
The balanced equation for the reaction is given below:
Pb + PbO₂ + 2H₂SO₄ —> 2PbSO₄ + 2H₂O
Next, we shall determine the mass of PbO₂ that reacted and the mass of PbSO₄ produced from the balanced equation. This can be obtained as follow:
Molar mass of PbO₂ = 207 + (16×2)
= 207 + 32
= 239 g/mol
Mass of PbO₂ from the balanced equation = 1 × 239 = 239 g
Molar mass of PbSO₄ = 207 + 32 + (16×4)
= 207 + 32 + 64
= 303 g/mol
Mass of PbSO₄ from the balanced equation = 2 × 303 = 606 g
SUMMARY:
From the balanced equation above,
239 g of PbO₂ reacted to produce 606 g of PbSO₄.
Finally, we shall determine the mass of PbSO₄ that will be produced by the reaction of 23.6 g of PbO₂. This can be obtained as follow:
From the balanced equation above,
239 g of PbO₂ reacted to produce 606 g of PbSO₄.
Therefore, 23.6 g of PbO₂ will react to produce = (23.6 × 606) / 239 = 59.8 g of PbSO₄.
Thus, 59.8 g of PbSO₄ were obtained from the reaction.
Which of the following reasons correctly explains the color changes that take place when ethylenediamine (C2N2H8) is added to the solution of cobalt(II) chloride?
a. Addition of the liquid ethylenediamine dilutes the concentration of cobalt(II) chloride in the solution resulting in a color change.
b. The ethylenediamine is oxidized and the resulting solution is deeply colored.
c. The water ligands surrounding the cobalt metal center are being replaced by ethylenediamine and chloride ligands which results in a different crystal field splitting. Thus, the energy associated with electron transitions between the do-orbitals will differ for the two compounds showing a color change.
Answer:
The water ligands surrounding the cobalt metal center are being replaced by ethylenediamine and chloride ligands which results in a different crystal field splitting. Thus, the energy associated with electron transitions between the do-orbitals will differ for the two compounds showing a color change.
Explanation:
The five d-orbitals are usually degenerate. Upon approach of a ligand, the d-orbitals split into two sets of orbitals depending in the nature of the crystal field.
The magnitude of crystal field splitting is affected by the nature of the ligand. Ligands having filled p-π orbitals such as ethylenediamine lead to greater crystal field splitting.
The change in the colour that takes place when ethylenediamine is added to the solution of cobalt(II) chloride occurs due to a different crystal field splitting pattern. Thus, the energy associated with electron transitions between the d-orbitals now differ for the two compounds showing a color change.
81.5 g of metal was heated from 11 degrees Celsius to 69 degrees Celsius. If 6739 joules of heat energy were used, what is the specific heat capacity of the metal?
Answer:
the metal become red hot
Analyze the data and determine the actual concentration of calcium chloride in the solution. Show all calculations and report in % wt/v concentration.
Known; Mass of CaCl2 present in original solution, based on actual yield= 1.77g moles
CaCl2 present in original solution, based on actual yield= 1.77g/molar mass of CaCl2=1.77g/110.98g/mol=0.016 moles
Total Volume of solution =V, which is 80ml
Answer:
2.21% wt/v
Explanation:
The mass/volume percentage, %wt/v, is an unit of concentration used in chemistry defined as 100 times the ratio of the mass of solute in g (In this case, CaCl2 = 1.77g) and the volume of solution in mL = 80mL
The %wt/v of this solution is:
%wt /v = 1.77g / 80mL * 100
%wt/v = 2.21% wt/v
What is the hydrogen atoms in 39.6g of ammonium sulphate,NH4 2SO4
Propane gas reacts with oxygen according to this balanced equation: C subscript 3 H subscript 8 space (g )space plus space 5 space O subscript 2 space (g )space rightwards arrow 3 space C O subscript 2 space (g )space plus space 4 space H subscript 2 O space (g )How many liters of carbon dioxide are produced at STP when 44 g of C3H8 completely reacts with oxygen
Explanation:
The balanced chemical equation of the reaction is:
[tex]C_3H_8(g)+ 5O_2 (g)->3CO_2(g)+4H_2O(g)[/tex]
From the balanced chemical equation,
1 mole of propane forms ------ 3 mol. of [tex]CO_2[/tex] gas.
The molar mass of propane is 44.1 g/mol.
One mole of any gas at STP occupies --- 22.4 L.
Hence, 44 g of propane forms (3x22.4 L=) 67.2 L of CO2 gas at STP.
Answer:
Thus, 67.2 L of CO2 is formed at STP.
PLEASE HELP ASAP
A total of 132.33g C3H8 is burned in 384.00 g O2. Use the following questions to determine the amounts of products formed.
• How many grams of CO2 and H2O will be produced? (2 points)
b. If the furnace is not properly adjusted, the products of combustion can include other gases, such as CO and unburned hydrocarbons. If only 269.34 g of CO2 were formed in the above reaction, what would the percent yield be? (2 points)
What is ethane?
A. A polymer
B. An alkyne
C. An alkane
D. An alkene
Answer:
D. An alkene
Explanation:
because Ethane is C2H4
Answer:
It's a alkANE. C.
Explanation:
The easiest way to memorize this is to look at the endings. Substances that end in -ANE are alkANEs. Substances that end in -ENE are alkENEs. Substances that end in -YNE are alkYNEs.
Select the number of valence electrons for hydrogen.
Answer:
Vanlency of hydrogen - 11
Electrons of hydrogen - 1
Answer:
The answer is: 1
Hope this helps :) <3
Explanation:
What is the balanced form of the following equation?
Br2 + S2O32- + H2O → Br1- + SO42- + H+
Answer:
5 Br₂ + S₂O₃²⁻ + 5 H₂O ⇒ 10 Br⁻ + 2 SO₄²⁻ + 10 H⁺
Explanation:
We will balance the redox reaction through the ion-electron method.
Step 1: Identify both half-reactions
Reduction: Br₂ ⇒ Br⁻
Oxidation: S₂O₃²⁻ ⇒ SO₄²⁻
Step 2: Perform the mass balance, adding H⁺ and H₂O where appropriate
Br₂ ⇒ 2 Br⁻
5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺
Step 3: Perform the charge balance, adding electrons where appropriate
2 e⁻ + Br₂ ⇒ 2 Br⁻
5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺ + 10 e⁻
Step 4: Make the number of electrons gained and lost equal
5 × (2 e⁻ + Br₂ ⇒ 2 Br⁻)
1 × (5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺ + 10 e⁻)
Step 5: Add both half-reactions
5 Br₂ + S₂O₃²⁻ + 5 H₂O ⇒ 10 Br⁻ + 2 SO₄²⁻ + 10 H⁺
How does the number of molecules in one mole of carbon dioxide compare with the number of molecules in one mole of water?
ОА.
There are four times as many molecules in one mole of carbon dioxide as there are in one mole of water.
ОВ.
There are twice as many molecules in one mole of carbon dioxide as there are in one mole of water.
OC
There are three times as many molecules in one mole of carbon dioxide as there are in one mole of water.
OD
There are the same number of molecules in one mole of carbon dioxide as there are in one mole of water.
Answer:
d
Explanation:
What is the concentration of Htions at a pH = 11?
mol/L
What is the concentration of Htions at a pH = 6?
mol/L
How many fewer Htions are there in a solution at a
pH = 11 than in a solution at a pH = 6?
9. Consider a magnesium atom with charge +2. How many overall electrons are on this particle?
Hint: Magnesium's atomic number is 12.
10
12
14
A student was asked to determine the percent mass of sodium nitrate in a mixture of sodium nitrate (NaNO3) and calcium carbonate (CaCO3). The mass of the mixture used was 3.2 g. The student extracted NaNO3 from the mixture with water and separated the insoluble CaCO3 from the solution by filtration. After evaporating the filtrate, the student recovered and dried the NaNO3, and found that it weighted 0.45 g. The student dried the insoluble residue of CaCO3 and found that it weighted 2.23 g. Calculate the percent mass of NaNO3 in the mixture and round your final answer to the correct number of significant figures.
Answer:
自分の仕事をする translate to english
Explanation:
Urea, CH4N2O (s), is manufactured from NH3 (g) and CO2 (g). H2O (l) is another product of this reaction. An experiment is started with 2.6 grams of NH3 (g) added into a reaction vessel with CO2 (g).
Write the balanced equation for this reaction, being sure to include physical states. Based on the balanced equation above, calculate the following:
a. the theoretical yield of urea in grams that can be made from the NH3
b. the actual amount of urea made if the percent yield for this reaction is 34%.
Answer:
a. 4.41 g of Urea
b. 1.5 g of Urea
Explanation:
To start the problem, we define the reaction:
2NH₃ (g) + CO₂ (g) → CH₄N₂O (s) + H₂O(l)
We only have mass of ammonia, so we assume the carbon dioxide is in excess and ammonia is the limiting reactant:
2.6 g . 1mol / 17g = 0.153 moles of ammonia
Ratio is 2:1. 2 moles of ammonia can produce 1 mol of urea
0.153 moles ammonia may produce, the half of moles
0153 /2 = 0.076 moles of urea
To state the theoretical yield we convert moles to mass:
0.076 mol . 58 g/mol = 4.41 g
That's the 100 % yield reaction
If the percent yield, was 34%:
4.41 g . 0.34 = 1.50 g of urea were produced.
Formula is (Yield produced / Theoretical yield) . 100 → Percent yield
A substance that donates a pair of electrons to form coordinate covalent bond is called
Lewis base: any species that can donate a pair of electrons and form a coordinate covalent bond. ligand: molecule or ion that surrounds a transition metal and forms a complex ion; ligands act as Lewis bases
What is the oxidation number of the metal ion in the coordinate complex [Fe(CN)6]3–?
A. NCO-
B. -OH
C. -CN
D. -SCN
Answer:
The options are incorrect.................
Explanation:
The Oxidation no. is +3
1) Recall the two written definitions of an oxidation-reduction reaction provided in our lessons. Which of these definitions is
most inclusive of redox reactions? Explain your answer:
A redox reaction is where the oxidation and reduction reaction takes place at the same time, the oxidation half
reaction involves losing electrons and in the reduction half reaction involves gaining electrons. So in a redox
reaction an electron is lost by the reducing agent.
Explain how the reaction below meets these definitions. Which substance is being oxidized and which is
being reduced?
4Ag(s) + 2H2S(g) + O2(g)
2Ag2S(s) + 2H20(9)
Answer:
A redox reaction in which a change of oxidation number of the reacting species occurs either by oxidation or by reduction which occur simultaneously and to the same extent during the reaction.
Silver atoms are oxidized while oxygen atoms are reduced by a loss of electrons and a gain of electrons respectively.
Explanation:
A redox reaction in which a change of oxidation number of the reacting species occurs either by oxidation or by reduction which occur simultaneously and to the same extent during the reaction.
In a redox reaction,the two reactions occurring simultaneously can be divided into two half reactions; an oxidation half-reaction and a reduction half-reaction.
The oxidation half-reaction involves losing electrons and thus an increase in oxidation number of the species being oxidized. Whereas, the reduction half reaction involves gaining electrons and thus, a reduction innthe oxidation number of the species being reduced.
The species which oxidizes another species is known as an oxidizing agent and isnitself reduced due to its accepting electrons from the species being oxidized. Th reducing agent reduces another species and is itself oxidized as it loses electrons to the oxidized agent.
In the given reaction as shown below:
4 Ag (s) + 2 H₂S (g) + O₂ (g) ---> 2 Ag₂S (s) + 2 H₂0 (g)
The reaction is a redox reaction as a change innthe oxidation number of the reacting species; both oxidation and reduction occurs simultaneously and to the same extent.
The metallic silver atoms, have an oxidation number of zero initially. However, each of the four moles of atoms give up one mole of a electrons each to become oxidized to silver (i) ions, Ag+.
On the other hand, molecular oxygen gas also having oxidation number of zero becomes reduced to oxygen ion, O²-. Each of the two moles of atom in the oxygen gas molecule accept two electrons each donated by the metallic silver atoms to become reduced to oxygen ion, O²-.
The oxidation numbers of hydrogen ion and sulfide ion do not change.
(in the image)
What Is the noble gas configuration for strontium (atomic number = 38)? Given: He (atomic number = 2) Ne (atomic number = 10) Ar (atomic number = 18) Kr (atomic number = 36) B В A [Kr 45Kr 5
Answer:
[Kr] 5s²
Explanation:
From the question given above, the following data were obtained:
Atomic number of strontium (Sr) = 38
Electronic configuration =?
Next, we shall determine the electronic configuration of the noble gas element before strontium (Sr).
The noble gas element before strontium (Sr) is krypton (Kr). Thus, the electronic configuration of krypton (Kr) is given below:
Kr (36) => 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s²4p⁶
Finally, we shall determine the electronic configuration of strontium (Sr). This can be obtained as follow:
Kr (36) => 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s²4p⁶
Sr (38) =>?
Sr (38) => 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s²4p⁶ 5s²
But
Kr (36) => 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s²4p⁶
Therefore,
Sr (38) => [Kr] 5s²
For which of the following reactions is the enthalpy change equal to the second ionization energy of nitrogen?
Answer:
"[tex]N^+(g) \rightarrow N^{2+}(g) + e^-[/tex]" is the appropriate answer.
Explanation:
Whenever one electron or particle must be removed from some kind of gas atom or molecule, it requires that the very first amount of energy necessary.Two electrons must be removed from such a mono-positive exhaust gases structure or position of ion before they may become a dipositive gaseous ion.Thus the above is the correct answer.
HELP ASAP PLS
Reactions, products and leftovers
Answer:
See the answer below
Explanation:
From the original equation in the image, the mole ratio of C:CO2:CO is 1:1:2. This means that for every 1 mole of C and CO2, 2 moles of CO would be produced.
Now, looking at the simulation below the equation of the reaction, 3 moles of C and 8 moles of CO2 were supplied as input. Applying this to the original equation of reaction, C seems to be a limiting reagent for the reaction because the ratio of C to CO2 should 1:1.
Hence, taking all the 3 moles of C available means that only 3 moles out of the available 8 for CO2 would be needed. 3 moles c and 3 moles CO2 means that 6 moles CO would be produced (remember that the ratio remains 1:1:3 for C, CO2, and CO). This means that 5 moles CO2 would be leftover.
In other words, all the 3 moles C would be consumed, 3 out of 8 moles CO2 would be consumed, and 6 moles CO would be produced while 5 moles CO2 would be leftover.
Consider the following chemical reaction:
2SO2 (g) + O2 (g) -----------> 2SO3 (g)
1.50 L. of sulfur trioxide at the pressure of 1.20 atm. and temperature of 25 oC is mixed with excess of oxygen.
Calclate volume of the product in L. at STP.
A. 11.2 L.
B. 1.65 L.
C. 16.5 L.
D. 0.129 L.
Answer:
B. 1.65 L
Explanation:
Step 1: Write the balanced equation
2 SO₂(g) + O₂(g) ⇒ 2 SO₃(g)
Step 2: Calculate the moles of SO₂
The pressure of the gas is 1.20 atm and the temperature 25 °C (298 K). We can calculate the moles using the ideal gas equation.
P × V = n × R × T
n = P × V / R × T
n = 1.20 atm × 1.50 L / (0.0821 atm.L/mol.K) × 298 K = 0.0736 mol
Step 3: Calculate the moles of SO₃ produced
0.0736 mol SO₂ × 2 mol SO₃/2 mol SO₂ = 0.0736 mol SO₃
Step 4: Calculate the volume occupied by 0.0736 moles of SO₃ at STP
At STP, 1 mole of an ideal gas occupies 22.4 L.
0.0736 mol × 22.4 L/1 mol = 1.65 L
Standard hydrogen electrode acts as both anode and cathode.Explain.
Answer:
A Standard Hydrogen Electrode is an electrode that scientists use for reference on all half-cell potential reactions. The value of the standard electrode potential is zero, which forms the basis one needs to calculate cell potentials using different electrodes or different concentrations. It is important to have this common reference electrode, just as it is important for the International Bureau of Weights and Measures to keep a sealed piece of metal that is used to reference the S.I. Kilogram.
Answer:
The role of an electrode as cathode or anode depends on the nature and electrode potential of the other electrode with which it forms the complete electrochemical cell.
When a cell is to be made with zinc electrode and hydrogen electrode, the hydrogen electrode will behave as a cathode and the zinc electrode will behave as anode because zinc is present above hydrogen in the activity series. That is zinc is more electropositive than hydrogen.
If the cell is made with a copper electrode and hydrogen electrode, the hydrogen electrode will behave as anode and the copper electrode as a cathode. This is due to the fact that Cooper is present below hydrogen in the activity series. Copper is less electropositive than hydrogen.
Explanation:
What is Bose Einstein state of matter
Write an essay on sensipar (cinacalcet)
Explanation:
The menstrual cycle is a reproductive cycle that takes place in the females of the group of primates. The menstrual cycle is divided into four phases:
(i) Menstrual phase: It extends from 1
st
to 4
th
day of the cycle. It occurs in the absence of fertilisation. During this phase, bleeding occurs as the endometrium of the uterus is sloughed off. The menstrual flow consists of secretion of endometrial glands, cell debris, unfertilized ovum. After 4
th
day, once again the FSH secretion from the pituitary is resumed and the new follicle starts developing.
(ii) Follicular phase: When the ovary is in this phase, the uterus enters in the proliferative phase. This takes place from 5
th
to 13
th
day of the cycle. During this phase new primordial follicle in the ovary develops due to the action of FSH from the pituitary. It gradually changes into the Graafian follicle and the production of estrogen starts. Only one follicle develops in one cycle. Corresponding to the changes in the ovary, the uterus also undergoes proliferation. Endometrial glands, stimulated by estrogen do repair process of the uterus.
(iii) Ovulatory phase: During this phase, ovulation takes place. It usually occurs on 14
th
day. Mature Graafian follicle ruptures due to LH secreted by the pituitary. Graafian follicle bursts and releases the ovum. This ovum along with the follicular fluid is picked up by the fimbriae of the infundibulum of the fallopian tube. It passes through the fallopian tube, where, if it happens to meet a sperm, it is fertilised. If not fertilised, the ovum degenerates.
(iv) Luteal phase: It corresponds with the secretory phase in the uterus. It takes place between 15
th
to 28
th
day of the cycle.
Ovarian changes: In the ovary, corpus luteum is formed from an empty Graafian follicle. Progesterone is secreted now. If the ovum is fertilised, corpus luteum is retained. LH and LTH from pituitary help in the maintenance of corpus luteum. If the ovum is not fertilised, corpus luteum degenerates and forms corpus albicans.
Uterine changes: Under the influence of progesterone, there is an increase in the thickness of the endometrium. Endometrial glands grow and become secretory. Progesterone is responsible for the maintenance of pregnancy. When fertilised ovum reaches the uterus, it is implanted and the placenta is formed. Till placenta becomes functional corpus luteum keeps on producing progesterone. But when progesterone source is cut off, endometrium sloughs off and menstruation begins
Identify the correct descriptions of beta particles.
a. A beta particle has a negative charge.
b. A beta particle contains neutrons.
c. A beta particle is less massive than a gamma ray.
d. A beta particle is a high-energy electron.
Answer:
a. A beta particle has a negative charge. d. A beta particle is a high-energy electron.
Explanation:
Identify the correct descriptions of beta particles.
a. A beta particle has a negative charge. YES. A beta particle is originated in the following nuclear reaction: ¹₀n ⇒ ¹₁H + ⁰₋₁e (beta particle.)
b. A beta particle contains neutrons. NO. It is a electron originated in the nucleus.
c. A beta particle is less massive than a gamma ray. NO. Gamma rays don't have mass while a beta particle has a mass which is half of one thousandth of the mass of a proton.
d. A beta particle is a high-energy electron. YES. Beta particles are nuclear originated hig-energy electrons.
Consider the Fischer ester synthesis of methyl benzoate from benzoic acid and methanol in the presence of sulfuric acid as a catalyst. A reaction was performed in which 3.3 g of benzoic acid was reacted with excess methanol to make 1.7 g of methyl benzoate. Calculate the theoretical yield and percent yield for this reaction.
Answer:
46.2%
Explanation:
Number of moles benzoic acid reacts = 3.3g/122.12 g/mol = 0.027 moles
Since the reaction is 1:1, 0.027 moles of methyl benzoate is formed.
Hence;
Theoretical yield of methyl benzoate = 0.027 moles × 136.15 g/mol = 3.68 g
% yield = actual yield/theoretical yield × 100
% yield = 1.7 g/3.68 g × 100
% yield = 46.2%
) The C O bond dissociation energy in CO2 is 799 kJ/mol. The maximum wavelength of electromagnetic radiation required to rupture this bond is ________.
Answer:
λ = 150 nm
Explanation:
For C-O bond rupture:
The required energy to rupture C-O bond = bond energy of C-O bond
= 799 kJ/mol
[tex]\mathsf{= 799 \ kJ/mol \times ( \dfrac{1 \ mol }{6.023 \times 10^{23} \ C-O \ bonds })}[/tex]
[tex]\mathsf{= 1.3265 \times 10^{-21} \ kJ/ C-O \ bond}[/tex]
[tex]\mathsf{= 1.33 \times 10^{-18} \ J/C-O \ bond}[/tex]
Recall that the wavelength associated with energy and frequency is expressed as:
[tex]E = \dfrac{hc}{\lambda}[/tex]
[tex]\lambda = \dfrac{hc}{E}[/tex]
[tex]\lambda = \dfrac{(6.626 \times 10^{-34} \ J.s^{-1}) \times (3.0 \times 10^8 \ ms^{-1})}{ 1.33 \times 10^{-18} \ J/C-O \ bond}}[/tex]
[tex]\mathsf{\lambda = 1.50 \times 10^{-7} \ m}[/tex]
λ = 150 nm
The reaction A + B <-------> C + D has been studied at five widely different temperature and the equilibrium tabulated.
Equilibrium constant K (at varies temperatures)
K at T1 1 x 10^-2
K at T2 2.25
K at T3 1.0
K at T4 81
K at T5 4 x 10^1
Which temperature is the products favored?
If K is greater than 1, then products are favored
Determine the number of water molecules in 0.2830g Na.
Answer:
7.38*10^21
Explanation:
2Na+2H20=2NaOH+H2
nNa=0.0123
number of water moles: 0.012*6*10^23=7.38*10^21
