Answer:
185.49 grams of Zinc would react with 454g (1lb) of copper sulfate
Explanation:
Yo know the following balanced reaction:
CuSO₄(aq)+ Zn(s) →Cu(s) + ZnSO₄(aq)
You can see that by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reagents and products are part of the reaction:
CuSO₄: 1 moleZn: 1 moleCu: 1 moleZnSO₄: 1 moleBeing:
Cu: 63.54 g/moleS: 32 g/moleO: 16 g/moleZn: 65.37 g/molethe molar mass of the compounds participating in the reaction is:
CuSO₄:63.54 g/mole + 32 g/mole + 4*16 g/mole= 159.54 g/mole ≅ 160 g/moleZn: 65.37 g/moleCu: 63.54 g/moleZnSO₄: 65.37 g/mole + 32 g/mole + 4*16 g/mole= 161.37 g/moleThen, by stoichiometry of the reaction, the following amounts of mass of reagent and product participate in the reaction:
CuSO₄: 1 moles* 160 g/mole= 160 gZn: 1 mole* 65.37 g/mole= 65.37 gCu: 1 mole* 63.54 g/mole= 63.54 gZnSO₄: 1 mole* 161.37 g/mole= 161.37 gNow you can apply the following rule of three: if 160 grams of CuSO₄ react with 65.37 grams of Zn by this reaction stoichiometry, 454 grams of CuSO₄ with how much mass of Zn will it react?
[tex]mass of Zn=\frac{454 grams of CuSO_{4} *65.37 grams of Zn}{160 grams of CuSO_{4}}[/tex]
mass of Zn= 185.49 grams
185.49 grams of Zinc would react with 454g (1lb) of copper sulfate
A piece of wood near a fire is at 23°C. It gains 1,160 joules of heat from the fire and reaches a temperature of 42°C. The specific heat capacity of
wood is 1.716 joules/gram degree Celsius. What is the mass of the piece of wood?
ОА. 16 g
OB. 29 g
ОC. 36 g
OD. 61 g
Answer:
35.578g or 36g if you round
Explanation:
Q=mc ∆∅ where ∅ is temperature difference
1160= m x 1.716 x (42-23)
m = 1160/ 1.716 x19
m=35.578g
m = 36g to nearest whole number
Answer: C. 36 g
Explanation: I got this right on Edmentum.
what are the differences between strong and weak acids?
Strong acids are completely ionised and weak acids are partly ionised
Answer:
Como forman los iones en soluciión
Explanation:
Los ácidos fuertes y las bases fuertes se refieren a especies que se disocian completamente para formar los iones en solución. Por el contrario, los ácidos y bases débiles se ionizan solo parcialmente y la reacción de ionización es reversible.
Vanadium (V) and oxygen (O) form a series of compounds with the following compositions: Mass % V 76.10 67.98 61.42 56.02 Mass % O 23.90 32.02 38.58 43.98 Compound Mass % N 1 33.28 2 39.94 Mass % Si 66.72 60.06 10. What are the relative numbers of atoms of oxygen in the compounds for a given mass of vanadium
Answer:
For every given mass of Vanadium, the relative number of oxygen atoms present or the mole ratio of Oxygen to Vanadium is:
A. 1:1
B. 3:2
C. 2:1
D. 5:2
Note: The question is stated more clearly below:
Vanadium (V) and oxygen (O) form a series of compounds with the following compositions: Mass % V 76.10 67.98 61.42 56.02 Mass % O 23.90 32.02 38.58 43.98 Compound Mass % N 1 33.28 2 39.94.
What are the relative numbers of atoms of oxygen in the compounds for a given mass of vanadium?
Explanation:
Number of moles in 100 g mass = % mass / molar mass
Molar mass of Vanadium, V = 51 g/mol
Molar mass of oxygen atom, O = 16 g/mol
1. Percentage mass of V and O is 76.10% and 23.90% respectively.
Number of moles of each atom;
V = 76.10/51.0 = 1.5 moles
O = 23.9/16 = 1.5 moles
Mole ratio of oxygen to vanadium = 1.5/1.5 = 1 : 1
2. Percentage mass of V and O is 67.98% and 32.02% respectively
Number of moles of each atom:
V = 67.98/51 = 1.33
O = 32.02/16 = 2
Mole ratio of oxygen to vanadium = 2/1.33 = 1.5 : 1 = 3 : 2
3. Percentage mass of V and O is 61.42% and 38.58% respectively
Number of moles of each atom:
V = 61.42/51 = 1.2
O = 38.58/16 = 2.4
Mole ratio of oxygen to vanadium = 2.4/1.2 = 2 : 1
4. Percentage mass of V and O is 56.02% and 43.98% respectively
Number of moles of each atom:
V = 56.02/51 = 1.10
O = 43.98/16 = 2.75
Mole ratio of oxygen to vanadium = 2.75/1.10 = 2.5 : 1 = 5 : 2
Mass of the Vanadium, number of O2 atoms present, or the mole ratio of 1:1 , 3:2 , 2:1 , 5:2 . As Vanadium (V) and oxygen (O) form a series of compounds is given with masses of 76.10 67.98, 23.90 32.02, 33.28 2 39.94, etc.
As per No of moles in 100 g mass = % mass / molar mass Mass of Vanadium, V = 51 g/ mol e, Mass of oxygen atom, O = 16 g/mole O = 23.9/16 = 1.5 moles for oxygen to vanadium = 1.5/1.5 = 1 : 1 2. Percentage mass of V and O is 67.98% and 32.02%. Mole ratio of oxygen to vanadium = 2/1.33 = 1.5 : 1 = 3 : 2 3. Percentage mass of V and O is 61.42% and 38.58% Mole ratio of oxygen to vanadium = 2.4/1.2 = 2 : 1 4. Percentage mass of V and O is 56.02% and 43.98%. Mole ratio of oxygen to vanadium = 2.75/1.10 = 2.5 : 1 = 5 : 2Learn more about the Vanadium (V) and oxygen (O).
brainly.com/question/2145642.
The value of ΔG°′ΔG°′ for the conversion of glucose-6-phosphate to fructose-6-phosphate (F6P) is +1.67 kJ/mol+1.67 kJ/mol . If the concentration of glucose-6-phosphate at equilibrium is 2.65 mM2.65 mM , what is the concentration of fructose-6-phosphate? Assume a temperature of 25.0°C25.0°C .
Answer:
The concentration of fructose-6-phosphate F6P ≅ 1.35 mM
Explanation:
Given that:
ΔG°′ is the conversion of glucose-6-phosphate to fructose-6-phosphate (F6P) = +1.67 kJ/mol = 1670 J/mol
concentration of glucose-6-phosphate at equilibrium = 2.65 mM
Assuming temperature = 25.0°C
=( 25 + 273)K
= 298 K
We are to find the concentration of fructose-6-phosphate
Using the relation;
ΔG' = -RT In K_c
where;
R = 8.314 J/K/mol
1670 = - (8.314 × 298 ) In K_c
1670 = -2477.572 × In K_c
1670/ 2477.572 = In K_c
0.67 = In K_c
[tex]K_c = e^{-0.67}[/tex]
[tex]K_c =[/tex] 0.511
Now using the equilibrium constant [tex]K_c[/tex]
[tex]K_c = \dfrac{[F6P]}{[G6P]}[/tex]
[tex]0.511 = \dfrac{[F6P]}{[2.65]}[/tex]
F6P = 0.511 × 2.65
F6P = 1.35415
F6P ≅ 1.35 mM
The simplest carboxylic acid is called *
O Formaldehye
O formic acid
acetic acid
O
acetone
Of Sr or Ba , the element with the higher first ionization energy is
Answer:
Sr
Explanation:
Sr has an ionization of 550 whereas Ba has an ionization of 503
Thermal decomposition of 5.00 metric tons of limestone to lime and carbon dioxide requires 9.00 × 106 kJ of heat. Convert this energy to (a) joules; (b) calories; (c) British thermal units. Give your answers in scientific notation.
Answer:
Take a look at the attachment below
Explanation:
Hope that helps!
Consider the three isomeric alkanes n-hexane,2,3-dimethylbutane, and 2-methylpentane. Which of the following correctly lists these compounds in order of increasing boiling point
a. 2,3-dimethylbutane < 2-methylpentane < n-hexane
b. 2-methylpentane
c. 2-methylpentane < 2,3-dimethylbutane
d. n-hexane < 2-methylpentane < 2,3-dimethylbutane
e. n-hexane < 2,3-dimethylbutane < 2-methylpentane
Answer:
a. 2,3-dimethylbutane < 2-methylpentane < n-hexane
Explanation:
The boiling point of alkanes is highly affected by the degree of branching in the molecule. Branched alkanes generally have a lower boiling point than unbranched alkanes.
The reason for the higher boiling point of unbranched alkanes is because they have greater vanderwaals forces acting between their molecules due to their larger surface area. Recall that branched alkanes have a lesser surface area compared to unbranched alkanes.
n-hexane is an unbranched alkane hence it will have the highest boiling point followed by 2-methyl pentane and lastly 2,3-dimethyl butane. The boiling point continues to decrease as the extent of branching increases.
A certain reaction with an activation energy of 155 kJ/mol was run at 495 K and again at 515 K . What is the ratio of f at the higher temperature to f at the lower temperature
Answer:
4.32 is the ratio of f at the higher temperature to f at the lower temperature
Explanation:
Using the sum of Arrhenius equation you can obtain:
ln (f₂/f₁) = Eₐ / R ₓ (1/T₁ - 1/T₂)
Where f represents the rate constant of the reaction at T₁ and T₂ temperatures. Eₐ is the energy activation (155kJ / mol = 155000J/mol) and R is gas constant (8.314J/molK)
Replacing:
ln (f₂/f₁) = 155000J/mol / 8.314J/molK ₓ (1/495K - 1/515)
Where 2 represents the state with the higher temperature and 1 the lower temperature.
ln (f₂/f₁) = 155000J/mol / 8.314J/molK ₓ (1/495K - 1/515)
ln (f₂/f₁) = 1.4626
f₂/f₁ = 4.32
4.32 is the ratio of f at the higher temperature to f at the lower temperature
Analyze: The metallic character of an element is determined by how readily it loses electrons. Elements that lose electrons most easily have the greatest metallic character
A. Which group has the greatest metallic character?
B. Which group has the lowest metallic character?
C. What is the relationship between metallic character and ionization energy?
Answer:
Group 1 or akali metals have the greatest metallic property.
Group 17 has the lowest metallic character.
C. As you move from right to lefton the periodic table, metallic character increases which is the ability to lose electrons. Ionization energy decrease as we move from right to left on the periodic table.
Explanation:
Akali metals in group 1 have the greatest metallic property and they are the most reactive metals. Francium metal on the group has the most metallic characteristics. It is rare and very radioactive. Group 17 has the lowest metallic character. This is because while moving across the period, the number of electrons in the outermost shell increases. This make it difficult for atoms to leave see electrons and become electropositive . Group 17 has the highest tendency of accepting electrons.
Ionization energy is the energy use to remove electron from an atom in gaseous stage. Ionization energy decrease as we move from right to left on the periodic table and metallic character increases as we move from right to left on the periodic table.
The volume of a sample of water is 2.5 mL the volume of the sample in liters is
Answer:
0.0025Litters
Explanation:
2.5ml= 2.5x10^-3l
2.5ml= 0.0025l
Answer:
AAAAAAAA
Explanation:
Write the complete balanced equation for the neutralization reaction that occurs when aqueous hydroiodic acid, HI, and sodium hydrogen carbonate, NaHCO3, are combined. Include physical states.
Answer:
[tex]HI_(_a_q_)~+~NaHCO_3_(_a_q_)~->~NaI_(_a_q_)~+~H_2O_(_l_)~+~CO_2_(_g_)[/tex]
Explanation:
In this case, we will have a neutralization reaction. We have a base ([tex]HI[/tex]) and a base ([tex]NaHCO_3[/tex]). Additionally, we have a strong acid and a strong base, therefore both will be soluble on water, so we will have an aqueous state for these compounds. If we will have a neutralization reaction, we will have as a salt as a product. With this in mind the reaction would be:
[tex]HI_(_a_q_)~+~NaHCO_3_(_a_q_)~->~NaI_(_a_q_)~+~H_2O_(_l_)~+~CO_2_(_g_)[/tex]
All the sodium salts are soluble in water, therefore we will have an aqueous state. Water is a liquid and carbon dioxide is a gas.
I hope it helps!
glucose 6‑phosphate+H2O⟶glucose+Pi glucose 6‑phosphate+H2O⟶glucose+Pi K′eq1=270 K′eq1=270 ATP+glucose⟶ADP+glucose 6‑phosphate ATP+glucose⟶ADP+glucose 6‑phosphate K′eq2=890 K′eq2=890 Using this information for equilibrium constants determined at 25∘C,25∘C, calculate the standard free energy of hydrolysis of ATP. standard free energy:
Answer:
-30.7 kj/mol
Explanation:
The standard free energy for the given reaction that is the hydrolysis of ATP is calculated using the formula: ∆Go ’= -RTln K’eq
where,
R = -8.315 J / mo
T = 298 K
For reaction,
1. K′eq1=270,
∆Go ’= -RTln K’eq
= - 8.315 x 298 x ln 270
= - 8.315 x 298 x 5.59
= - 13,851.293 J / mo
= - 13.85 kj/mol
2. K′eq2=890
∆Go ’= -RTln K’eq
= - 8.315 x 298 x ln 890
= - 8.315 x 298 x 6.79
= - 16.82 kj/mol
therefore, total standard free energy
= - 13.85 + (-16.82)
= -30.7 kj/mol
Thus, -30.7 kj/mol is the correct answer.
If sodium arsenite is Na3AsO3, the formula for calcium arsenite would be
Answer:
Ca₃(AsO₃)₂
Explanation:
Sodium arsenite, with the chemical formula Na₃AsO₃, is formed by the cation Na⁺ and the anion AsO₃³⁻. For the molecule to be neutral, 3 cations Na⁺ and 1 anion AsO₃³⁻ are required.
Calcium arsenite would be formed by the cation Ca²⁺ and the anion AsO₃³⁻. For the molecule to be neutral, we require 3 cations Ca²⁺ and 2 anions AsO₃³⁻. The resulting chemical formula is Ca₃(AsO₃)₂.
Using appropriate chemical equation distinguish between cation and anion hydrolysis
Answer:
HCO3- (aq) + H2O (I) <--> H2CO3 (aq) + OH- (aq)
Explanation:
The equation to distinguish between cation and anion hydrolysis is given below :
HCO3- (aq) + H2O (I) <--> H2CO3 (aq) + OH- (aq)
The important thing to remember is their origin. The anions can react with water and can produce hydroxide ions while hydroxide ions make a solution basic.
25.00 mL of a H2SO4 solution with an unknown concentration was titrated to a phenolphthalein endpoint with 28.11 mL of a 0.1311 M NaOH solution. What is the concentration of the H2SO4 solution
Answer:
Concentration of the H₂SO₄ solution is 0.0737 M
Explanation:
Equation of the neutralization reaction between the acid, H₂SO₄, and the base, NaOH, is given below:
H₂SO₄ + 2NaOH -----> Na₂SO₄ + 2H₂O
From the above equation, one mole of acid requires 2 moles of base for complete neutralization which occurs at phenolphthalein endpoint.
mole ratio of acid to base, nA/nB = 1:2
Concentration of the base, Cb = 0.1311 M
Volume of base, Vb, = 28.11 mL
Concentration of acid, Ca = ?
Volume of acid, Va + 25.0 mL
Using the formula, CaVa/CbVb = nA/nB
making Ca subject of the formula, Ca = Cb*Vb*nA/Va*nB
substituting the values into the equation
Ca = (0.1311 * 28.11 * 1) / 25.0 * 2 = 0.0737 M
Therefore, concentration of the H₂SO₄ solution is 0.0737 M
under the same conditions carbon (iv) oxide,propane and nitrogen (i) oxide diffuse at the same rate.Explain
Answer:
Rate of diffusion is same .
Explanation:
As we know that Rate of the diffusion is directly proportional to the [tex]\frac{1}{\sqrt{M} }[/tex] .They have same mass if there is same rate and similar condition therefore the mass of carbon (iv) oxide,propane and nitrogen (i) oxide will be similar.
The mass is directly proportional to the Rate of the diffusion.Therefore the rate of diffusion is similar in all carbon (iv) oxide,propane and nitrogen (i) oxide .Electrophilic substitution on 3-phenylpropenenitrile occurs at the meta position. Draw resonance structures to show how the ring is electron-poor at the ortho and para positions.
Answer:
See figure 1
Explanation:
In this question, we have to remember that a poor electron carbon is a carbon in which we have a positive charge, a carbocation. Therefore we have to start with the production of the carbocation. First, a double bond from the benzene is moved to the carbon in the top to produce a new double bond generating a positive charge in a carbon with ortho position (electron-poor). Then we can move another double bond inside the ring to produce a positive charge in the para carbon. Finally, we can move the last double bond to produce again another positive charge in the second ortho carbon.
See figure 1.
I hope it helps!
A student mixes baking soda and vinegar in a glass. Are there any new substances created from this mixture?
Answer:
Explanation:
1. A student mixes baking soda and vinegar in a glass. The results are shown at left. ... Yes I do belive that new substances are being formed because there is a chemical reaction between the baking soda and vinegar turning it into a bubbly substances instead of a powder and liquid.
Yes, there are new substances created from this mixture.
how to write the lewis dot structure for H2CCl2
Answer:
H
° * . .
H ° * C * · Cl :
* · . .
: Cl :
. .
Explanation:
Carbon has 4 valent electrons
*
*C*
*
Hydrogen has 1 electron
H°
Cl has 7 electrons on the last level.
. .
: Cl·
. .
H
° * . .
H ° * C * · Cl :
* · . .
: Cl :
. .
The Lewis dot structure for [tex]H_2CCl_2[/tex] is explained in the explanation part below.
A Lewis structure is a symbolic depiction of a molecule or ion that depicts the arrangement of atoms and valence electrons.
It is also known as a Lewis dot structure or electron dot structure. Gilbert N. Lewis, an American chemist, invented it.
The total number of valence electrons in the molecule must be determined before writing the Lewis dot structure for H2CCl2 (dichloroethylene).
H has one valence electron, while C has four.
Cl has seven valence electrons.
Thus, the Lewis dot structure for the given compound is attached below as image.
For more details regarding Lewis dot structure, visit:
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Cathodic protection of iron involves using another more reactivemetal as a sacrificial anode. Classify each of thefollowing metals by whether they would or would not act as asacrificial anode to iron.
a. Sn
b. Cu
c. Zn
d. Au
e. Pb
f. Ag
g. Mg
An old iron beam was coated with an unknown metal. There is a crackon the coating and it is observed that the iron is rusting at thefracture. The beam is in a structure that experiences high stress,resulting in frequent fractures to the coating.
What was the old metal coating likely made of and what metal youwould use to repair the fractures to avoid further corrosion?
Choices: tin, aluminum, gold
1. The old coating was made of __________________.
2. __________________would be a good choice for repairing thefracture.
Answer:
1.) zinc and aluminum
2.)
a.) The old coating was made of tin.
b.) Aluminum would be a good choice for repairing the fracture.
a) What substances are present in an aqueous buffer composed of HC2H3O2 and C2H3O2 - ?b) What happens when LiOH is added to a buffer composed of HC2H3O2 and C2H3O2 - ? Write a chemical equation for that reaction.c) What happens when HBr is added to this buffer? Write a chemical equation for that reaction.
Answer:
a) HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻
b) HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂
c) C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻
Explanation:
a) In a HC₂H₃O₂/C₂H₃O₂⁻ buffer system, the following reactions take place:
HC₂H₃O₂ + H₂O ⇄ C₂H₃O₂⁻ + H₃O⁺
C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻
Thus, the species present are: HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻.
b) When LiOH is added to the buffer system, it is partially neutralized according to the following equation.
HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂
c) When HBr is added to the buffer system, it is partially neutralized according to the following equation.
C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻
Naturally occurring sulfur consists of four isotopes: 32S (31.97207 u, 95.0%); 33S (32.97146 u, 0.76%); 34S (33.96786 u, 4.22%); and 36S (35.96709 u, 0.014%). Calculate the average atomic mass of sulfur in atomic mass units.
Answer:
32.062
Explanation:
The following data were obtained from the question:
Mass of isotope A (32S) = 31.97207 u
Abundance of isotope A (A%) = 95.0%
Mass of isotope B (33S) = 32.97146 u Abundance of isotope B (B%) = 0.76%
Mass of isotope C (34S) = 33.96786 u
Abundance of isotope C (C%) = 4.22%
Mass of isotope D (36S) = 35.96709 u Abundance of isotope D (D%) = 0.014%
Average atomic mass of S =..?
The average atomic mass of sulphur, S can be obtained as follow:
Average atomic mass = [(Mass of A x A%)/100] + [(Mass of B x B%)/100] + [(Mass of C x C%)/100] + [(Mass of D x D%)/100]
Average atomic mass of sulphur =
[(31.97207 x 95)/100] + [(32.97146 x 0.76)/100] + [(33.96786 x 4.22)/100] + [(35.96709 x 0.014)/100]
= 30.373 + 0.251 + 1.433 + 0.005
= 32.062
Therefore, the average atomic mass of sulphur is 32.062
An experiment calls for 10.0 mL of bromine (d = 3.12 g/mL). Since an accurate balance is available, it is decided to measure the bromine by mass. How many grams should be measured out? Multiple Choice 3.21 32.1 3.12 31.2 0.312
Answer:
31.2g
Explanation:
The following data were obtained from the question:
Volume of bromine = 10mL
Density of bromine = 3.12 g/mL
Mass of bromine =...?
The Density of the substance is related to it's mass and volume by the following equation:
Density = Mass /volume
With the above equation, we can calculate the mass of bromine as follow:
Density = Mass /volume
Volume of bromine = 10mL
Density of bromine = 3.12 g/mL
Mass of bromine =...?
Density = Mass /volume
3.12 = Mass /10
Cross multiply
Mass of bromine = 3.12 x 10
Mass of bromine = 31.2g
Therefore, the mass of bromine is 31.2g
Classify the substances as atomic elements, molecular elements, molecular compounds, or ionic compounds. a. Agb. Cdc. MgCl2d. F2f. HIg. NO2h. NaCli. Cl2
Answer:
Explanation:
Hello,
We'll be doing some classification of some chemical substances based on molecules, elemental state or ionic or electrovalent properties.
A) Ag = atomic element : silver (Ag) in its elemental state is an atomic element.
B) Cd = atomic element : Cadmium (Cd) is an element of the periodic table and belongs to transition metal.
C) MgCl = ionic compounds: this is a compound formed between magnesium (Mg) and chlorine (Cl) to give MgCl. This compound has ionic or electrovalent properties since electron transfer occurred between the cation (Mg) and anion (Cl).
D) F₂ = moleculer element : Fluorine F₂ is moleculer element since two elements of fluorine combine together to form a molecule.
E) HI = molecular compound : this is a compound formed from the reaction between hydrogen and iodine. It's a molecular compound because they are two different elements combining together to form a compound.
F) NO₂ = molecular compound
G) NaCl = ionic compound
H) Cl₂ = molecular element
Which of the following would be more reactive than magnesium (Mg)?
A. Calcium (Ca)
B. Potassium (K)
C. Argon (Ar)
D. Beryllium (Be)
Answer:potassium is more reactive than Mg because both lie in the same group and the element potassium has more electropositivity than magnesium
Explanation:
I hope it will help you
Answer: B. Potassium(K)
Explanation:
PdPd has an anomalous electron configuration. Write the observed electron configuration of PdPd. Express your answer in complete form in order of orbital filling. For example, 1s22s21s22s2 should be entered as 1s^22s^2. View Available Hint(s)
Answer:
1s²,2s²,2p⁶,3s²,3p⁶,4s²,3d¹⁰,4p⁶,5s⁰,4d¹⁰.
Explanation:
Palladium is a chemical element with the symbol Pd and atomic number 46.
The electronic configuration is;
[Kr] 4d¹⁰
The full electronic configuration observed for palladium is given as;
1s²,2s²,2p⁶,3s²,3p⁶,4s²,3d¹⁰,4p⁶,5s⁰,4d¹⁰.
The reason for for the anomlaous electron configuration is beacuse;
1. Full d orbitals are more stable than partially filled ones.
2. At higher energy levels, the levels are said to be degenerated which means that they have very close energies and then electrons can jump from one orbital to another easily.
find the mass of h2 produced Binary compounds of alkali metals and hydrogen react with water to produce H2(g). The H2H2 from the reaction of a sample of NaH with an excess of water fills a volume of 0.505 L above the water. The temperature of the gas is 35 ∘C∘C and the total pressure is 755 mmHg
Answer: Mass of hydrogen produced is 0.0376 g.
Explanation:
The reaction equation will be as follows.
[tex]NaH(aq) + H_{2}O(l) \rightarrow H_{2}(g) + NaOH(aq)[/tex]
Now, formula for total pressure will be as follows.
[tex]P_{total} = P_{H_{2}} + P_{H_{2}O}[/tex]
Hence, [tex]P_{H_{2}} = P_{total} - P_{H_{2}O}[/tex]
= 755 mm Hg - 42.23 mm Hg
= 712.77 mm Hg
[tex]P_{H_{2}} = \frac{712.77 \times 1 atm}{760 mm Hg}[/tex]
= 0.937 atm
Now, we will calculate the moles of [tex]H_{2}[/tex] as follows.
[tex]P_{H_{2}}V = nRT[/tex]
[tex]0.937 atm \times 0.505 L = n \times 0.0821 \times 308.15 K[/tex]
n = [tex]\frac{0.473}{25.29}[/tex] mol
= 0.0187 mol
Therefore, mass of [tex]H_{2}[/tex] will be calculated as follows.
[tex]m_{H_{2}} = \frac{0.0187 mol \times 2.0158 g}{1 mol}[/tex]
= 0.0376 g
Thus, we can conclude that mass of hydrogen produced is 0.0376 g.
chemical equation for potassium sulfate and lead(II) acetate
Answer:
K₂SO₄ + Pb(C₂H₃O₂)₂ →PbSO₄ + 2KC₂H₃O₂
A chemical equation is a symbolic representation of a chemical reaction. The chemical equation for the reaction between potassium sulfate ([tex]K_2SO_4[/tex]) and lead(II) acetate ([tex]Pb(CH_3COO)_2[/tex]) can be written as follows:
[tex]K_2SO_4 + Pb(CH_3COO)_2 = PbSO_4 + 2CH_3COOK[/tex]
A basic chemical equation consists of two main parts: the reactant side (left side) and the product side (right side), separated by an arrow indicating the direction of the reaction. Reactants are substances that undergo a chemical change, while products are substances formed as a result of the reaction.
In this reaction, potassium sulfate reacts with lead(II) acetate to form lead(II) sulfate and potassium acetate. It is important to note that the equation is balanced with stoichiometric coefficients, ensuring that the number of atoms of each element is the same on both sides of the equation.
Therefore, the chemical equation for the reaction between potassium sulfate ([tex]K_2SO_4[/tex]) and lead(II) acetate ([tex]Pb(CH_3COO)_2[/tex]) can be written as follows:
[tex]K_2SO_4 + Pb(CH_3COO)_2 = PbSO_4 + 2CH_3COOK[/tex]
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4. Which of the following statements explains the cause of lanthanide contraction?
A. All lanthanides and actinides are radioactive
B. Protons exhibit a stronger pull on outer f orbitals
C. The d orbitals in lanthanides have unpair electrons
D. The d orbitals in actinides have paired electrons
Answer:
B. PROTONS EXHIBIT STRONGER PULL ON OUTER f ORBITALS
Explanation:
Lanthanide contraction is the greater than normal decrease in the ionic radius of the lanthanide series from atomic number 57 to atomic number 71. This decrease is rather not expected of the ionic radii of these elements and they result in the greater decrease in the subsequent series of the lanthanides from the atomic number 72. The cause of which is as a result of the poor shielding effects of the nuclear charge around the electrons of the f orbitals. So therefore, protons are strongly pulled out of the 4f orbital and as a result of the poor shielding effect which causes the electrons of the 6s orbitals to be drawn more closer to the nucleus and hence resulting in a smaller atomic radii. It is worthy to note that the shielding effects of the inner electrons decreasing from s orbital to the f orbital; that is s > p > d > f. So from the decrease in the shielding effects from s to the f orbitals, lanthanide contraction results from the inability of the orbitals far away from s like the 4f orbiatls to shield the outermost shells of the lanthanide elements. So the cause of lanthanide contraction is the action of the protons which strongly pull the electrons of the f orbitals because of the poor shielding effects due to the distance of this orbital from the nucleus.
Answer:
B) Protons exhibit a stronger pull on outer f orbitals than on d orbitals.
Explanation: