Answer:
Option B. 2096.1 K
Explanation:
Data obtained from the question include the following:
Enthalpy (H) = +1287 kJmol¯¹ = +1287000 Jmol¯¹
Entropy (S) = +614 JK¯¹mol¯¹
Temperature (T) =.?
Entropy is related to enthalphy and temperature by the following equation:
Change in entropy (ΔS) = change in enthalphy (ΔH) / Temperature (T)
ΔS = ΔH / T
With the above formula, we can obtain the temperature at which the reaction will be feasible as follow:
ΔS = ΔH / T
614 = 1287000/ T
Cross multiply
614 x T = 1287000
Divide both side by 614
T = 1287000/614
T = 2096.1 K
Therefore, the temperature at which the reaction will be feasible is 2096.1 K
How many valence electrons are in the electron dot structures for the elements in group 3A(13)?
Answer:
here, as we have known the elements of group 3A(13) such as aluminium , boron has three valance electron and in perodic table the elements are kept with similar proterties in same place so, their valance electron is 3.
hope it helps...
The number of valence electrons are in the electron dot structures for the elements in group 3A(13) is three.
What are Groups in the Periodic Table?The periodic table is organized into groups (vertical columns), periods (horizontal rows), and families (groups of elements that are similar). Elements in the same group have the same number of valence electrons.
Groups are the columns of the periodic table, and periods are the rows. There are 18 groups, and there are 7 periods plus the lanthanides and actinides.
There are two different numbering systems that are commonly used to designate groups, and you should be familiar with both.
The traditional system used in the United States involves the use of the letters A and B. The first two groups are 1A and 2A, while the last six groups are 3A through 8A. The middle groups use B in their titles.
Therefore, The number of valence electrons are in the electron dot structures for the elements in group 3A(13) is three.
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Chemistry question. Image attached.
Answer:
The answer to your question is given below
Explanation:
The balanced equation for the reaction is given below:
CaO(s) + CH4(g) + 2H2O(g) <=> CaCO3(s) + 4H2(g)
1. Writing an expression for the equilibrium constant, K.
The equilibrium constant, K 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.
Thus, we can write the equilibrium constant, K for the reaction as follow:
CaO(s) + CH4(g) + 2H2O(g) <=> CaCO3(s) + 4H2(g)
K = [CaCO3] [H2]⁴ / [CaO] [CH4] [H2O]²
2. Based on the value of K, more products will be in the equilibrium mixture since the value of K is a positive large number.
HELPPP.
Which of the following is a property of matter?
O A. It takes up space.
OB. It is everywhere.
O C. It is constantly changing.
O D. It cannot be divided.
Answer:
a
Explanation:
Matter can be volume or density. So, this concludes that it is when it takes up space.
Answer: A.
Explanation:
it takes up space
The thermochemical equation is for the reaction of hydrogen bromide gas to form hydrogen gas and bromine liquid. 2HBr(g) = H 2 (g)+ Br 2 (l) 72.6 kJ How many grams of HBr (g) would be made to react if 11.4 energy were provided?
Answer:
the mass of HBr that would react is 25.41 g of HBr
Explanation:
attached is the calculations.
A sample of gas is observed to effuse through a pourous barrier in 4.98 minutes. Under the same conditions, the same number of moles of an unknown gas requires 6.34 minutes to effuse through the same barrier. The molar mass of the unknown gas is:________.
g/mol.
Answer:
The molar mass of the unknown gas is [tex]\mathbf{ 51.865 \ g/mol}[/tex]
Explanation:
Let assume that the gas is O2 gas
O2 gas is to effuse through a porous barrier in time t₁ = 4.98 minutes.
Under the same conditions;
the same number of moles of an unknown gas requires time t₂ = 6.34 minutes to effuse through the same barrier.
From Graham's Law of Diffusion;
Graham's Law of Diffusion states that, at a constant temperature and pressure; the rate of diffusion of a gas is inversely proportional to the square root of its density.
i.e
[tex]R \ \alpha \ \dfrac{1}{\sqrt{d}}[/tex]
[tex]R = \dfrac{k}{d}[/tex] where K = constant
If we compare the rate o diffusion of two gases;
[tex]\dfrac{R_1}{R_2}= {\sqrt{\dfrac{d_2}{d_1}}[/tex]
Since the density of a gas d is proportional to its relative molecular mass M. Then;
[tex]\dfrac{R_1}{R_2}= {\sqrt{\dfrac{M_2}{M_1}}[/tex]
Rate is the reciprocal of time ; i.e
[tex]R = \dfrac{1}{t}[/tex]
Thus; replacing the value of R into the above previous equation;we have:
[tex]\dfrac{R_1}{R_2}={\dfrac{t_2}{t_1}}[/tex]
We can equally say:
[tex]{\dfrac{t_2}{t_1}}= {\sqrt{\dfrac{M_2}{M_1}}[/tex]
[tex]{\dfrac{6.34}{4.98}}= {\sqrt{\dfrac{M_2}{32}}[/tex]
[tex]M_2 = 32 \times ( \dfrac{6.34}{4.98})^2[/tex]
[tex]M_2 = 32 \times ( 1.273092369)^2[/tex]
[tex]M_2 = 32 \times 1.62076418[/tex]
[tex]\mathbf{M_2 = 51.865 \ g/mol}[/tex]
What is the core charge of helium and why?
Answer:
Formula for effective nuclear charge is as follows. So, for He atom value of S = 0.30 because the electrons are present in 1s orbital. Therefore, calculate the effective nuclear charge for helium as follows. Thus, we can conclude that the effective nuclear charge for helium is 1.7
Explanation:
The effective nuclear charge experienced by a 1s electron in helium is +1.70.
Consider the 1H NMR spectrum for the following compound:
CH3CH2CH3
Predict the first-order splitting patterns of the indicated proton. This question uses specific splitting patterns instead of the often used generic term "multiplet."
a. doublet of quartets
b. triplet of triplets
c. septet
d. quartet of quartets
e. quintet
Answer:
See explanation
Explanation:
In this case, we have to check the neighbors of each carbon in the molecule. In propane, we have two types of carbons (see figure 1) (blue and red ones). The red carbons are equivalent. (Both have the same neighbors). Now, we can analyze each carbon:
Blue carbon
In the blue carbon, we have 6 hydrogens neighbors (three on each methyl). If we follow the n+1 rule, we will have:
6+1= 7
For the blue carbon, we will have a Septet.
Red carbons
In the red carbon, we have only 2 neighbors (the carbon in the middle only have 2 hydrogens). If we follow the n+1 rule, we will have:
2+1=3
For the red carbon, we will have a triplet.
See figure 2
I hope it helps!
The rate at which two methyl radicals couple to form ethane is significantly faster than the rate at which two tert-butyl radicals couple. Offer two explanations for this observation.
Answer:
1. stability factor
2. steric hindrance factor
Explanation:
stability of ethane is lesser to that of two tert-butyl, so ethane will be more reactive and faster.
ethane is less hindered and more reactive, while two tert-butyl is more hindered and less reactive
The molecular weight of table salt, NaCl, is 58.5 g/mol. A tablespoon of salt weighs 6.37 grams. Calculate the number of moles of salt in one tablespoon.
Finally, solve (remember significant figures):
Answer:
0.109 mol/tablespoon
Explanation:
6.37 g/ 58.5 mol = 0.10888888 mol (0.109 significantly)
Answer:
A: 0.109
Explanation:
Edge 2020
If 3.10 moles of P4010 reacted with excess water, how many grams of H3PO4
would be produced?
P4010 +6H20 + 4H3PO4
You Answered
126 g
0 0.007918
Correct Answer
O 1220 g
0.1278
75.98
Answer:
1.22 × 10³ g
Explanation:
Step 1: Write the balanced equation
P₄O₁₀ + 6 H₂O ⇒ 4 H₃PO₄
Step 2: Calculate the moles of H₃PO₄ produced by 3.10 moles of P₄O₁₀
The molar ratio of P₄O₁₀ to H₃PO₄ is 1:4. The moles of H₃PO₄ produced are 4/1 × 3.10 mol = 12.4 mol
Step 3: Calculate the mass corresponding to 12.4 moles of H₃PO₄
The molar mass of H₃PO₄ is 97.99 g/mol.
[tex]12.4 mol \times \frac{97.99g}{mol} = 1.22 \times 10^{3} g[/tex]
List the following compounds in order from strongest acid to weakest acid. Rank the acids from strongest to weakest.
CH2CHCH2COOH CH2CH2CH2COOH CH3CHCH2COOH CH3CH2CH2COOH
Strongest Weakest
Answer:
CH3CH2CH2COOH<CH2(F)CH2CH2COOH<CH3CH(F)CH2COOH<CH2(F)CH(F)CH2COOH
Explanation:
We know that the presence of highly electronegative elements in carboxylic acid molecules lead to -I inductive effect. This implies that electrons are withdrawn along the chain towards the electronegative element. As electrons are withdrawn towards the electronegative element, the electron cloud of the carbonyl- hydrogen bond in the acid weakens and the hydrogen can now be easily lost as a proton, that is , the molecule becomes more acidic.
The -I inductive effect increases with increase in the number of electronegative elements present in the molecule and the proximity of the electronegative element to the carbonyl group. The closer the electronegative element is to the carbonyl group, the greater the acidity of the molecule since the -I inductive effect dies out with increasing distance from the carbonyl group. Also, the more the number of electronegative elements in the molecule, the greater the - I inductive effect and the greater the acidity of the molecule, hence the answer.
Which of the following is NOT one of the types of bonds? A. Ionic B. Metallic C. Covalent D. Valence
Considering the definition of bond and the different type of bonds, valence is not one of the types of bonds.
What is a chemical bondA chemical bond is defined as the force by which the atoms of a compound are held together. These are electromagnetic forces that give rise to different types of chemical bonds.
In other words, a chemical bond is the force that joins atoms to form chemical compounds and confers stability to the resulting compound.
Covalent bondThe covalent bond is the chemical bond between atoms where electrons are shared, forming a molecule. Covalent bonds are established between non-metallic elements, such as hydrogen H, oxygen O and chlorine Cl. These elements have many electrons in their outermost level (valence electrons) and have a tendency to gain electrons to acquire the stability of the electronic structure of noble gas. The shared electron pair is common to the two atoms and holds them together.
Ionic bondAn ionic bond is produced between metallic and non-metallic atoms, where electrons are completely transferred from one atom to another. During this process, one atom loses electrons and another one gains them, forming ions.
Metallic bondMetallic bonds are a type of chemical bond that occurs only between atoms of the same metallic element. In this way, metals achieve extremely compact, solid and resistant molecular structures, since the atoms that share their valence electrons.
SummaryIn summary, valence is not one of the types of bonds. The types of bonds are covalent, ionic and metallic.
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A 1.00 liter solution contains 0.31 M sodium acetate and 0.40 M acetic acid. If 0.100 moles of barium hydroxide are added to this system, indicate whether the following statements are TRUE or FALSE . (Assume that the volume does not change upon the addition of barium hydroxide.)
a. The number of moles of CH3COOH will remain the same.
b. The number of moles of CH3COO- will increase.
c. The equilibrium concentration of H3O+ will decrease.
d. The pH will decrease.
e. The ratio of [CH3COOH] / [CH3COO-] will remain the same.
Answer and Explanation:
The buffer solution is composed by sodium acetate (CH₃COONa) and acetic acid (CH₃COOH). Thus, CH₃COOH is the weak acid and CH₃COO⁻ is the conjugate base, derived from the salt CH₃COONa.
If we add a strong base, such as barium hydroxide, Ba(OH)₂, the base will dissociate completely to give OH⁻ ions, as follows:
Ba(OH)₂ ⇒ Ba²⁺ + 2 OH⁻
The OH⁻ ions will react with the acid (CH₃COOH) to form the conjugate base CH₃COO⁻.
Initial number of moles of CH₃COOH = 0.40 mol/L x 1 L = 0.40 mol
Initial number of moles of CHCOO⁻= 0.31 mol/L x 1 L = 0.31 mol
moles of OH- added: 2 OH-/mol x 0.100 mol/L x 1 L = 0.200 OH-
According to this, the following are the answers to the sentences:
a. The number of moles of CH₃COOH will remain the same ⇒ FALSE
The number of moles of CH₃COOH will decrease, because they will react with OH⁻ ions
b. The number of moles of CH₃COO⁻ will increase ⇒ TRUE
Moles of CH₃COO⁻ will be formed from the reaction of the acid (CH₃COOH) with the base (OH⁻ ions)
c. The equilibrium concentration of H₃O⁺ will decrease ⇒ FALSE
The equilibrium concentration of OH⁻ is increased
d. The pH will decrease⇒ FALSE
pKa for acetic acid is 4.75. We add the moles of base to the acid concentration and we remove the same number of moles from the conjugate base in the Henderson-Hasselbach equation to calculate pH:
[tex]pH= pKa + log \frac{[conjugate base + base]}{[acid - base]}[/tex]
pH = 4.75 + log (0.31 mol + 0.20 mol)/(0.40 mol - 0.20 mol) = 5.15
Thus, the pH will increase.
3. A student carries out the clay-catalyzed dehydration of cyclohexanol starting with 10 moles of cyclohexanol and obtains 500 mL of pure cyclohexene product. Calculate the student’s percent yield for this reaction. Show your work.
Answer:
[tex]49.45~%[/tex]
Explanation:
In this case, we have to start with the chemical reaction:
[tex]C_6H_1_2O~->~C_6H_1_0~+~H_2O[/tex]
So, if we start with 10 mol of cyclohexanol ([tex]C_6H_1_2O[/tex]) we will obtain 10 mol of cyclohexanol ([tex]C_6H_1_0[/tex]). So, we can calculate the grams of cyclohexanol if we calculate the molar mass:
[tex](6*12)+(10*1)=82~g/mol[/tex]
With this value we can calculate the grams:
[tex]10~mol~C_6H_1_0\frac{82~g~C_6H_1_0}{1~mol~C_6H_1_0}=820~g~C_6H_1_0[/tex]
Now, we have as a product 500 mL of [tex]C_6H_1_0[/tex]. If we use the density value (0.811 g/mL). We can calculate the grams of product:
[tex]500~mL\frac{0.811~g}{1~mL}=405.5~g[/tex]
Finally, with these values we can calculate the yield:
[tex]%~=~\frac{405.5}{820}x100~=~49.45%[/tex]%= (405.5/820)*100 = 49.45 %
See figure 1
I hope it helps!
Based on the data given, the percentage yield of the student's work is 49.45 %.
What is the equation of the reaction?The equation of the clay-catalyzed dehydration of cyclohexanol is given below:
C₆H₁₂O ----> C₆H₁₀ + H₂OFrom the equation of the reaction, 1 mole of cyclohexanol yields 1 mole of cyclohexene.
Therefore 10 moles of cyclohexanol should yield 10 moles of cyclohexene.
What is the moles of cyclohexene obtained?First we determine the mass of cyclohexene obtained.
Mass = density * volume
volume of cyclohexene = 500 mL
density = 0.811 g/mL
mass of cyclohexene = 500 * 0.811
mass of cyclohexene = 405.5 g
Number of moles of cyclohexene = mass/molar mass
molar mass of cyclohexene = 82 g
moles of cyclohexene = 405.5 g/82
moles of cyclohexene = 4.945 moles
What is the percentage yield?Percentage yield = actual yield /expected yield * 100%Percentage yield = 4.945/10 * 100%
Percentage yield = 49.45%
Therefore, the percentage yield of the student's work is 49.45 %.
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what is the electron configuration of the iodide ion?
A. 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²
B. 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶
C. 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶
D. 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰
Answer:
the answer to this question is C
The electron configuration of the iodide ion is 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶. The correct option is option C.
What is electron configuration ?The arrangement of an atom's or molecule's electrons in their respective atomic or molecular orbitals is known as its electron configuration; for instance, the electron configuration of the neon atom is 1s2 2s2 2p6.
According to electronic configurations, electrons move individually within each orbital while interacting with the average field produced by all other orbitals. The corrosion potential and reactivity of an atom may be calculated from its electron configuration. The electron configuration of the iodide ion is 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶.
Therefore, the correct option is option C.
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How fast are the atoms moving if the temperature of a gas is cold?
A. very, very slowly
B. they are stagnant
C. very, very quickly
Answer:
i think option a is correct answer because when there is low temperature then the kinetic enegry will be very less and the atoms moves very slowly.
Answer:
A. very, very slowly
Explanation:
A is the answer because atoms will move faster in hot gas than in cold gas.
What is the absolute magnitude of the rate of change for [NH3] if the
rate of change for [Hz] is 9.00 M/s in the reaction 2 NH3(g) → N2(g) +
3 H2(g)?
Answer:
[tex]r_{NH_3,abs} =6.00\frac{M}{s}[/tex]
Explanation:
Hello,
In this case, we can write the law of mass action for the undergoing chemical reaction, based on the rates and the stoichiometric coefficients:
[tex]\frac{1}{-2}r_{NH_3} =\frac{1}{1} r_{N_2}=\frac{1}{3}r_{H_2}[/tex]
In such a way, knowing the rate of formation hydrogen (H₂), we can know the rate of change of ammonia, that must be negative for consumption:
[tex]r_{NH_3} =\frac{-2}{3}r_{H_2}=\frac{-2}{3}*9.00\frac{M}{s} \\\\r_{NH_3} =-6.00\frac{M}{s}[/tex]
Nevertheless, the absolute magnitude will be positive:
[tex]r_{NH_3,abs} =6.00\frac{M}{s}[/tex]
Best regards.
A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.01962 mol of CO2 and 0.01961 mol of H2O. The empirical formula of the compound was found to be C3H6O2. Show how this was calculated.
What does the empirical formula tell you about the compound?
The molar mass of the actual compound was found to be 222.27g/mol. Find the molecular formula of this compound. What does the molecular formula tell you about the compound?
Can you see what type of functional group this compound could have?
Answer:
See explanation.
Explanation:
Hello,
In this case, we can show how the empirical formula is found by following the shown below procedure:
1. Compute the moles of carbon in carbon dioxide as the only source of carbon at the products:
[tex]n_C=0.01962molCO_2*\frac{1molC}{1molCO_2} =0.01962molC[/tex]
2. Compute the moles of hydrogen in water as the only source of hydrogen at the products:
[tex]n_H=0.01961molH_2O*\frac{2molH}{1molH_2O}=0.03922molH[/tex]
3. Compute the mass of oxygen by subtracting the mass of both carbon and hydrogen from the 0.4647-g sample:
[tex]m_O=0.4647g-0.01962molC*\frac{12gC}{1molC}-0.03922molH*\frac{1gH}{1molH} =0.1900gO[/tex]
4. Compute the moles of oxygen by using its molar mass:
[tex]n_O=0.1900gO*\frac{1molO}{16gO}=0.01188molO[/tex]
5. Divide the moles of carbon, hydrogen and oxygen by the moles of oxygen (smallest one) to find the subscripts in the empirical formula:
[tex]C=\frac{0.01962}{0.01188}=1.65\\ \\H=\frac{0.03922}{0.01188} =3.3\\\\O=\frac{0.01188}{0.01188} =1[/tex]
6. Search for the closest whole number (in this case multiply by 2):
[tex]C_3H_6O_2[/tex]
Moreover, the empirical formula suggests this compound could be carboxylic acid since it has two oxygen atoms, nevertheless, this is not true since the molar mass is 222.27 g/mol, therefore, we should compute the molar mass of the empirical formula, that is:
[tex]M=12*3+1*6+16*2=74g/mol[/tex]
Which is about three times in the molecular formula, for that reason, the actual formula is:
[tex]C_9H_{18}O_6[/tex]
It suggest that the compound has a highly oxidizing character due to the presence of oxygen, therefore, we cannot predict the distribution of the functional groups as it could contain, carboxyl, carbonyl, hydroxyl or even peroxi.
Best regards.
How many minutes would be required to electroplate 25.0 grams of chromium by passing a constant current of 4.80 amperes through a solution containing CrCl3
Answer:
483.27 minutes
Explanation:
using second faradays law of electrolysis
A student followed the procedure of this experiment to determine the percent NaCl in a commercial bleaching solution that was found in the basement of an abandoned house. The student diluted 50.00 mL of the commercial bleaching solution to 250 mL in a volumetric flask, and titrated a 20-ml aliquot of the diluted bleaching solution. The titration required 35.46 mL of 0.1052M Na S,O, solution. A faded price label on the gallon bottle read $0.79. The density of the bleaching solution was 1.10 g ml.
(1) Calculate the mass of commercial bleaching solution titrated. 1.1 450=559
(2) Determine the percent NaOCl in the commercial bleaching solution.
(3) Calculate the mass of one gallon of the commercial bleaching solution.
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What volume of CH4(g), measured at 25oC and 745 Torr, must be burned in excess oxygen to release 1.00 x 106 kJ of heat to the surroundings
Answer:
[tex]V=27992L=28.00m^3[/tex]
Explanation:
Hello,
In this case, the combustion of methane is shown below:
[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]
And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:
[tex]n_{CH_4}=\frac{-1.00x10^6kJ}{-890.8kJ/mol}= 1122.6molCH_4[/tex]
Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:
[tex]PV=nRT\\\\V=\frac{nRT}{P}=\frac{1122.6mol*0.082\frac{atm*L}{mol*K}*298K}{0.98atm}\\\\V=27992L=28.00m^3[/tex]
Best regards.
The element nitrogen forms a(n) _______ with the charge . The symbol for this ion is , and the name is . The number of electrons in this ion is .
Answer:
The element nitrogen forms an anion with the charge -3. The symbol for this ion is N³⁻, and the name is nitride. The number of electrons in this ion is 10.
Explanation:
The element nitrogen is in the Group 15 in the Periodic Table, so it tends to gain 3 electrons (3 negative charges) to fill its valance shell with 8 electrons.
The element nitrogen forms an anion with the charge -3. The symbol for this ion is N³⁻, and the name is nitride. The number of electrons in this ion is 10 (the original 7 plus the 3 gained). It is isoelectronic with the gas Neon, which accounts for its stability.
If the vinegar were measured volumetrically (e.g., a pipet), what additional piece of data would be needed to complete the calculations for the experiment?
Answer:
the density if vinegar will also be needed
Explanation:
Because this is an experiment of volumetric analysis
which conditions make snow
Answer:
It depends on the weather.
Explanation:
Like rain and hail, snow comes from the water vapor that forms the clouds, but obviously its formation is different: snow forms when the temperature is low in the atmosphere. In these conditions the water vapor drops freeze and when they collide form tiny crystals that join together to form snowflakes, which fall to Earth when they are heavy enough.
neeeeed helpppppppppp
Answer:
Option C. Will always.
Explanation:
A spontaneous reaction is a reaction that occurs without an external supply of heat.
This implies that spontaneous reaction will always occur as no external supply of heat is needed.
Identify each of the following half-reactions as either an oxidation half-reaction or a reduction half-reaction.
H2(g) → 2H+(aq) + 2e- oxidation
Cu2+(aq) → + 2e-Cu(s) reduction
Write a balanced equation for the overall redox reaction. Use smallest possible integer coefficients.
Answer:
H₂(g) + Cu²⁺(aq) → 2H⁺(aq) + Cu(s)
Explanation:
In a redox reaction, one half-reaction is the oxidation (where the atom loss electrons) whereas the other reaction is the reduction (Where the atom is gaining electrons.
In the reactions:
H₂(g) → 2H⁺(aq) + 2e⁻ oxidation
Here, the reaction is written as the oxidation because the hydrogen H₂ is in oxidation state 0 and H⁺ in +1. That means each atom is loosing one electron.
Cu²⁺(aq) + 2e⁻ → Cu(s) reduction
And here, the Cu²⁺ is in +2 oxidation state and after the reaction is in Cu(s) 0 state. Thus, each atom is gaining 2 electrons.
The sum of both reactions is:
H₂(g) + Cu²⁺(aq) + 2e⁻ → 2H⁺(aq) + 2e⁻ + Cu(s)
Subtracting the electrons in both sides of the reaction:
H₂(g) + Cu²⁺(aq) → 2H⁺(aq) + Cu(s)Homolysis, or homolytic bond dissociation, produces a very specific type of product under certain reaction conditions. In Part 1, select all the products (in formulae and general chemical terms) that could result from homolysis. In Part 2, select the reaction conditions that are most likely to promote homolysis.
Part 1. Choose all that may occur as possible products of a homolysis reaction.
Choose one or more:_______.
a. hydride ion
b. R3CO
c. Br2
d. H
e. a carbocation
f. H3C
g. H3CO-
h. hydrogen ion
i. a carbon free radical
Part 2. Choose the conditions under which homolysis is likely to occur.
Choose one or more:_______.
a. strong base
b. ultraviolet irradiation
c. high temperature
d. strong acid
e. infrared irradiation
f. low temperature
Answer:
1) R₃CO , H, H₃C, a carbon free radical
2) high temperature, ultraviolet irradiation
Explanation:
1) Homolysis leads to the formation of free radicals (species having a free electron). Thus, answer is :
R₃CO
H
H₃C
a carbon free radical
2) Homolysis require high temperature, ultraviolet irradiation.
When 75.5 grams of phosphorus pentachloride react with an excess of water, as shown in the unbalanced chemical equation below, how many moles of hydrochloric acid will be produced? Please show all your work for the calculations for full credit. PCl5 + H2O --> H3PO4 + HC
Answer:
Explanation: M(PCL5)= 31 + 5(35.5)
=208.5g/mol
M(H20)= 18g/mol
n(PCL5) = 75.5÷208.5
= 0.362mol
n(HCl)/n(PCL5)= 5/1
n(HCl)= 5×0.362
=1.81mol of HCl
Erbium metal (Er) can be prepared by reacting erbium(III) fluoride with magnesium; the other product is magnesium fluoride. Write and balance the equation.
Answer:
2ErF3 + 3Mg → 2Er + 3MgF2
Explanation:
Erbium metal is a member of the lanthaniod series. It reacts with halogens directly to yield erbium III halides such as erbium III chloride, Erbium III fluoride etc.
Erbium metal (Er) can be prepared by reacting erbium(III) fluoride with magnesium; the products are erbium metal and magnesium fluoride. This is a normal redox process in which the Erbium metal is reduced while the magnesium is oxidized. The balanced reaction equation of this process is; 2ErF3 + 3Mg → 2Er + 3MgF2
You use 10.0 mL of solution A, 10.0 mL of solution B, and 70.0 mL of water for your first mixture. What is the initial concentration of KIO3
Complete Question
The complete question is shown on the first uploaded image
Answer:
The initial concentration is [tex]C_f = 0.0022 \ M[/tex]
Explanation:
From the question we are told that
The volume of solution A is [tex]V_i = 10.0 mL[/tex]
The concentration of A is [tex]C_i = 0.0200 \ M[/tex]
The volume of solution B is [tex]V_B = 10.0mL[/tex]
The volume of water is [tex]V_{w } = 70.0 mL[/tex]
Generally the law of dilution is mathematically represented as
[tex]C_i * V_i = C_f * V_f[/tex]
Where [tex]C_f[/tex] is the concentration of the mixture
[tex]V_f[/tex] is the volume of the mixture which is mathematically evaluated as
[tex]V_f = 10 + 10 + 70[/tex]
[tex]V_f = 90mL[/tex]
So
[tex]C_f = \frac{C_i * V_i}{ V_f}[/tex]
substituting values
[tex]C_f = \frac{0.0200 * 10 }{90}[/tex]
[tex]C_f = 0.0022 \ M[/tex]
Note the mixture obtained is [tex]KIO_3[/tex]