Answer:
I would say c) a closed circuit.
Hope I was right.
What is pH of a buffer made by combining 45.0mL of 0.150M nitrous acid and 20.0mL of 0.175M sodium nitrate
Answer:
3.11
Explanation:
Any buffer system can be described with the reaction:
[tex]HA~->~H^+~+~A^-[/tex]
Where is the acid and is the base. Additionally, the calculation of the pH of any buffer system can be made with the Henderson-Hasselbach equation:
[tex]pH~=~pKa~+~Log(\frac{ [A^-]}{[HA]})[/tex]
With all this in mind, we can write the reaction for our buffer system:
-) Nitrous acid: [tex]HNO_2[/tex]
-) Sodium nitrate: [tex]NaNO_2[/tex]
[tex]HNO_2~->~H^+~+~NO_2^-[/tex]
In this case, the acid is [tex]HNO_2[/tex] with a concentration of 0.150 M and a volume of 45.0 mL (0.045 L). The base is [tex]NO_2^-[/tex] with a concentration of 0.175 M and a volume of 20.0 mL (0.020 L).
We can calculate the moles of each compound is we take into account the molarity equation ([tex]M=\frac{mol}{L}[/tex]). So:
-) moles of [tex]HNO_2[/tex]:
[tex]mol=0.150~M*0.045~L=0.00657[/tex]
-) moles of [tex]NO_2^-[/tex]:
[tex]mol=0.175~M*0.020~L=0.0035[/tex]
The total volume would be:
0.020 L + 0.045 L = 0.065 L
With this in mind, we can calculate the molarity of each compound:
-) Concentration of [tex]HNO_2[/tex]
[tex]M=\frac{0.00657~mol}{0.065~L}=0.101~M[/tex]
-) Concentration of [tex]NO_2^-[/tex]
[tex]M=\frac{0.0035~mol}{0.065~L}=0.0538~M[/tex]
The pKa reported is 3.39, therefore we can plug the values into the Henderson-Hasselbach equation:
[tex]pH~=~3.39~+~Log(\frac{[0.0538~M]}{[0.101~M]})~=~3.11[/tex]
The final pH value would be 3.11
I hope it helps!
The pH of a buffer made by combining 45.0 mL of 0.150M nitrous acid and 20.0mL of 0.175M sodium nitrate is 2.87.
We have a buffer made by combining 45.0mL of 0.150 M nitrous acid and 20.0mL of 0.175M sodium nitrate.
Nitrous acid is a weak acid and nitrate ion is its conjugate base.
What is a buffer?It is a solution used to resist abrupt changes in pH when acids or bases are added.
Step 1: Calculate the moles of each species.We do so by multiplying the molar concentration by the volume in liters.
HNO₂: 0.150 mol/L × 0.0450 L = 6.75 × 10⁻³ mol
NaNO₂: 0.175 mol/L × 0.0200 L = 3.50 × 10⁻³ mol
Step 2: Calculate the total volume of the mixture.The total volume will be the sum of the volumes of each solution.
V = 45.0 mL + 20.0 mL = 65.0 mL = 0.0650 L
Step 3: Calculate the molar concentration of each species in the mixture.HNO₂: 6.75 × 10⁻³ mol/0.0650 L = 0.104 M
NaNO₂: 3.50 × 10⁻³ mol/0.0650 L = 0.0538 M
Step 4: Calculate the pH of the buffer.We can calculate the pH of a buffer system using Henderson-Hasselbach's equation.
pH = pKa + log [NaNO₂]/[HNO₂]
pH = 3.16 + log 0.0538/0.104 = 2.87
The pH of a buffer made by combining 45.0 mL of 0.150M nitrous acid and 20.0mL of 0.175M sodium nitrate is 2.87.
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Calculate the molar solubility of CaF2 at 25°C in a solution that is 0.010 M in Ca(NO3)2. The Ksp for CaF2 is 3.9 x 10-11.
Answer:
[tex]Molar \ solubility=3.12x10^{-5}M[/tex]
Explanation:
Hello,
In this case, for the dissociation of calcium fluoride:
[tex]CaF_2(s)\rightleftharpoons Ca^{2+}+2F^-[/tex]
The equilibrium expression is:
[tex]Ksp=[Ca^{2+}][F^-]^2[/tex]
In such a way, via the ICE procedure, including an initial concentration of calcium of 0.01 M (due to the calcium nitrate solution), the reaction extent [tex]x[/tex] is computed as follows:
[tex]3.9x10^{-11}=(0.01+x)(2*x)^2\\\\x=0.0000312M[/tex]
Thus, the molar solubility equals the reaction extent [tex]x[/tex], therefore:
[tex]Molar \ solubility=3.12x10^{-5}M[/tex]
Regards.
The molar solubility of Calcium fluoride has been calculated as [tex]3.12\;\times\;10^-^5\;\rm M[/tex].
The dissociation of calcium fluoride has been given by:
[tex]\rm CaF_2\;\rightarrow\;Ca^2^+\;+\;2\;F^-[/tex]
The solubility constant, ksp has been given as:
[tex]ksp=\rm[Mg^2^+]\;[F^-]^2[/tex]
From the dissociation of Calcium nitrate, the concentration of Ca ion in the solution has been 0.01 M.
The dissociation of Calcium fluoride x M has been resulted in x M Ca and 2x M F ions.
The concentration of Ca in the solution has been resulted as x + 0.01 M.
The solubility product can be given as:
[tex]3.9\;\times\;10^-^1^1=[x+0.01]\;[2x]^2\\3.9\;\times\;10^-^1^1=[x+0.01]\;4x^2\\x=3.12\;\times\;10^-^5[/tex]
The molar solubility of Calcium fluoride has been calculated as [tex]3.12\;\times\;10^-^5\;\rm M[/tex].
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Name the following compound from the concise formula:______.
CH3CH(CH3)CHCHCH(CH3)CH2CH3
A. 2,4-dimethyl-3-heptene
B. 2,5-dimethyl-3-heptene
C. 3,5-dimethyl-3-heptene
D. 2,5-dimethyl-4-heptene
Answer:
B. 2,5-dimethyl-3-heptene
Explanation:
Answer:
B. 2,5-dimethyl-3-heptene
Explanation:
Give the formula of each coordination compound. Include square brackets around the coordination complex. Do not include the oxidation state on the metal. Use parentheses only around polyatomic ligands.
For ethylenediamine, use (en) in the formula.
a) sodium hexachloroplatinate(IV)
b) dibromobis(ethylenediamine)cobalt(III) bromide
c) pentaamminechlorochromium(III) chloride
Answer:
sodium hexachloroplatinate(IV)- Na2[PtCl6]
dibromobis(ethylenediamine)cobalt(III) bromide- [Co(en)2Br2]Br
pentaamminechlorochromium(III) chloride-[Cr(NH3)5Cl]Cl2
Explanation:
The formulas of the various coordination compounds can be written from their names taking cognisance of the metal oxidation state as shown above. The oxidation state of the metal will determine the number of counter ions present in the coordination compound.
The number ligands are shown by subscripts attached to the ligand symbols. Remember that bidentate ligands such as ethylenediamine bonds to the central metal ion via two donors.
Identify the term that matches each electrochemistry definition. The electrode where oxidation occurs Cathode The electrode where reduction occurs Choose... An electrochemical cell powered by a spontaneous redox reaction Choose... An electrochemical cell that takes in energy to carry out a nonspontaneous redox reaction Choose... A chemical equation showing either oxidation or reduction Choose...
Answer:
An electrochemical cell that takes in energy to carry out a nonspontaneous redox reaction
Both chlorine and fluorine are represented by a green modeling piece that has 4 holes. Is using the same piece for two different atoms acceptable? Why or why not
Answer:
Yes, same piece can be used.
Explanation:
The same piece can be used for two different atoms are acceptable because both atoms has 7 electrons in their outermost valance shell. Both atoms belong to same group i. e. halogens so same piece can be used for both atoms. If the atoms belong to different groups and they have different number of electrons in their outermost shell so using same piece will be a problem so it is recommended to use different pieces for different atoms.
The use of the same modeling piece for chlorine and fluorine has been accepted as it has consisted of the same properties and belongs to the same group.
Chlorine and fluorine have been the elements of group 17. The elements are halogens with the presence of 7 valence electrons.
The elements have been belonging to the same group and have the same number of valence electrons thus resembling each other in the chemical properties.
Since both the elements are similar to each other, the use of the same piece for two different atoms has been acceptable.
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What are the products in the following chemical reaction Pb(NO3)+KCI
Answer:
The products are KNO3 + PbCl2.....
Espero que te sirva.
The concentration of glucose, C6H12O6, in normal spinal fluid is 75 mg/100g. What is the molality of the solution
Answer:
4.16x10⁻³m
Explanation:
Molality is defined as the ratio between moles of a solute, in this case glucose, and kg of solvent.
As there are 100g of solvent, the kg are 0.1. Thus, we only need to calculate from the mass of glucose its moles to solve the molality of the solution.
Moles glucose:
There are 75mg = 0.075g of glucose. To conver mass to moles it is necessary molar mass.
Molar mass glucose:
6C = 12.01g/mol*6 = 72.06g/mol
12H = 12*1.008g/mol = 12.10g/mol
6O = 6*16g/mol = 96g/mol
72.06 + 12.10 + 96 = 180.16g/mol
Moles of 0.075g of glucose:
0.075g * (1 mol / 180.16g) =
4.16x10⁻⁴ moles of glucose
Molality of the solution:
4.16x10⁻⁴ moles of glucose / 0.1kg of solvent =
4.16x10⁻³mThe molarity of the solution is 4.16x10⁻³m
Calculation of the molarity:We know that the molarity refers to the ratio that arise between the moles of a solute.
Since there are 100 g of solvent so here the kg should be 0.1.
Likewise there is 75 mg so it should be 0.075g
Now the Molar mass glucose should be
6C = 12.01g/mol*6 = 72.06g/mol
12H = 12*1.008g/mol = 12.10g/mol
6O = 6*16g/mol = 96g/mol
So,
= 72.06 + 12.10 + 96
= 180.16g/mol
Now
Moles of 0.075g of glucose:
0.075g * (1 mol / 180.16g) =
4.16x10⁻⁴ moles of glucose
Now finally
Molality of the solution:
= 4.16x10⁻⁴ moles of glucose / 0.1kg of solvent
=4.16x10⁻³m
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If an individual proton has mass 1.007825 amu, and an individual neutron has mass 1.008665 amu, what's the calculated mass of a neptunium-236 nucleus? options: A) 237.92482 amu B) 236.99873 amu C) 237.96682 amu D) 237.04817 amu
Answer:
C) 237.96682 amu
Explanation:
The symbol for neptunium-236 is given as;
²³⁶₉₃Np
This element has 93 protons and (236 - 93 = 143) neutrons.
Mass Number =Total mass of Protons + Total mass of neutrons
Total Mass pf protons = 93 * 1.007825 amu, = 93.727725 amu
Total mass of Neutrons = 143 * 1.008665 amu = 144.239095 amu
Mass = 144.239095 + 93.727725 = 237.96682 amu
Correct option is option C.
Calculate the molarity of a solution containing 29g of glucose (C 6 H 12 O 6 ) dissolved in 24.0g of water. Assume the density of water is 1.00g/mL.
Answer:
whats the ph ofpoh=9.78
Explanation:
After combining hydrogen ions and hydroxide ions combine to form water molecules, the next step in balancing a redox reaction under basic conditions is to: ______
Explanation:
After combining hydrogen ions and hydroxide ions combine to form water molecules, the next step is the last step when balancing a redox reaction under basic condition.
The last step is to cancel all common terms that arises as a result of the formation of the water molecule. Usually, there's need to balance the water molecules in the reactant and product side of the reaction.
A sample of argon gas (molar mass 40 g) is at four times the absolute temperature of a sample of hydrogen gas (molar mass 2 g). Find the ratio of the rms speed of the argon molecules to that of the hydrogen. Assume hydrogen molecule has only translational degree of freedom.
Answer:
Ratio of Vrms of argon to Vrms of hydrogen = 0.316 : 1
Explanation:
The root-mean-square speed measures the average speed of particles in a gas, and is given by the following formula:
Vrms = [tex]\sqrt{3RT/M}[/tex]
where R is molar gas constant = 8.3145 J/K.mol, T is temperature in kelvin, M is molar mass of gas in Kg/mol
For argon, M = 40/1000 Kg/mol = 0.04 Kg/mol, T = 4T , R = R
Vrms = √(3 * R *4T)/0.04 = √300RT
For hydrogen; M = 1/1000 Kg/mol = 0.001 Kg/mol, T = T, R = R
Vrms = √(3 * R *T)/0.001 = √3000RT
Ratio of Vrms of argon to that of hydrogen = √300RT / √3000RT = 0.316
Ratio of Vrms of argon to that of hydrogen = 0.316 : 1
A voltaic cell is set up as follows: Anode: Zn electrode in a solution of 0.050 M Zn(NO 3 ) 2 Cathode: Pt electrode with 0.500 atm H 2 (g) in 0.010 M HNO 3 a) Write the overall balanced cell reaction.
Answer:
[tex]Zn(s) + 2H+(aq)\Rightarrow Zn^{2+}(aq) + H_{2}(g)[/tex]
Explanation:
Given that,
Anode : Zn electrode in a solution of 0.050 M Zn(NO₃)₂
Cathode : Pt electrode with 0.500 atm H₂(g) in 0.010 M HNO₃
Anode :
[tex]Zn(s)\Rightarrow Zn^{2+}(aq) + 2e^{-}[/tex]
Cathode :
[tex]2H+(aq)+2e^{-}\Rightarrow H_{2}(g)[/tex]
We need to write the overall balanced cell reaction
Using anode and cathode
[tex]Zn(s) + 2H+(aq)\Rightarrow Zn^{2+}(aq) + H_{2}(g)[/tex]
Hence, This is required answer.
g Which ONE of the following molecules and ions has trigonal planar molecular geometry? (NOTE: You must first determine what the Lewis structure of each substance is.) A) PCl3 B) HCN C) CO3 2– D) H3O+ E) NF3
Answer: C) [tex]CO^{2-}_{3}[/tex]
Explanation: The Valence Shell Electron Pair Repulsion Model (VSEPR Model) shows bonding and nonbonding electron pairs present in the valence, outermost, shell of an atom connecting to other atoms. It also gives the molecular geometric shape of a molecule.
To determine molecular geometry:
1) Draw Lewis Structure, i.e., a simplified representation of the valence shell electrons;
2) Count the number of electron pairs (count multiple bonds as 1 pair);
3) Arrange electron pairs to minimise repulsion;
4) Position the atoms to minimise the lone pair;
5) Name the molecular geometry from the atom position;
Trigonal planar molecular geometry is a model which molecule's shape is triangular and in one plane. Such molecule has three regions of electron density extending out from the central atom and the repulsion will be at minimum when angle between any two is 120°.
The Lewis structure of each molecule is shown in the attachment.
Analysing each one, it can be concluded that molecule with trigonal planar geometry is [tex]CO^{2-}_{3}[/tex]
Using the standard reduction potentials Ni2+(aq) + 2 e‑Ni(s) ‑0.25 volt Fe3+(aq) + e‑Fe2+(aq) +0.77 volt Calculate the value of E°cell for the cell with the following reaction. Ni2+(aq) + 2 Fe2+(aq) →Ni(s) + 2 Fe3+(aq)
Answer:
The correct answer is - 1.02 V
Explanation:
From the reduction-oxidation reaction:
Ni²⁺(aq) + 2 Fe²⁺(aq) → Ni(s) + 2 Fe³⁺(aq)
Ni²⁺ is reduced to Ni(s) while Fe²⁺ is oxidized to Fe³⁺. Thus, the half reactions are:
Reduction (cathode) : Ni²⁺(aq) + 2 e‑ → Ni(s) Eº= ‑0.25 V
Oxidation (anode) : 2 x (Fe²⁺ → Fe³⁺ + e-)(aq) Eº= -0.77 V
-------------------------------------
Ni²⁺(aq) + 2 Fe²⁺(aq) → Ni(s) + 2 Fe³⁺(aq)
In order to calculate the Eºcell, we have to add the reduction potential of the reaction in cathode (reduction) to the oxidation potential of the anode (oxidation):
Eºcell= Eºr + Eºo= (-0.25 V) + (-0.77 V) = - 1.02 V
a sample of gas occupies a volume of 2.62 liters at 25 C and 1.00 atm. what will be the volume at 50 C and 2 atm
Answer:2.62 L
Explanation:
A sample of gas occupies a volume of 2.62 liters at 25° C and 1.00 atm. and the volume at 50° C and 2 atm then volume is 2.62 liters.
What is ideal gas law ?The equation of state for a fictitious perfect gas is known as the ideal gas law, sometimes known as the generic gas equation. Although it has significant drawbacks, it is a decent approximation of the behavior of many gases under various situations.
An ideal gas is one in which there are no intermolecular attraction forces and all collisions between atoms or molecules are entirely elastic. It may be seen as a group of perfectly hard spheres that collide but do not else interact with one another.
By using ideal gas equation,
P₁ V ₁ ÷ T = P₂V₂ ÷ T
1 × 2.62 ÷ 25 = 2 × V₂ ÷ 50
V₂ = 1 × 2.62 × 50 ÷ 25 × 2
V₂ = 2.62 liters.
Thus, a sample of gas occupies a volume of 2.62 liters at 25° C and 1.00 atm. and the volume at 50° C and 2 atm then volume is 2.62 liters.
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What's the concentration of hydronium ions if a water-base solution has a temperature of 25°C (Kw = 1.0×10–14), with a concentration of hydroxide ions of 2.21×10–6 M? answer options: A) 2.8×10–8 M B) 4.52 ×10–9 M C) 1.6×10–9 M D) 3.1×10–6 M
Answer:
ITS NOT D. ITS B. 4.52x10^-9 M
Explanation:
Answer:
4.52 ×10–9 M
Explanation:
When titrating a strong acid with a strong base, after the equivalence point is reached, the pH will be determined exclusively by: Select the correct answer below:
A) hydronium concentration
B) hydroxide concentration
C) conjugate base concentration
D) conjugate acid concentration
Answer:
B) hydroxide concentration
Explanation:
Hello,
In this case, since we are talking about strong both base and acid, since the base is the titrant and the acid the analyte, once the equivalence point has been reached, some additional base could be added before the experimenter realizes about it, therefore, since the titrant is a strong base, it completely dissociates in hydroxide ions and metallic ions which allows us to compute the pOH of the solution by known the hydroxide ions concentration.
After that, due to the fact that the pH is related with the pOH as shown below:
pH=14-pOH
We can directly compute the pH.
Best regards.
The condition that a reaction takes place without outside help Choose... Solution in which no more solute can be dissolved in the solvent Choose... Difference of the enthalpy (of a system) minus the product of the entropy and absolute temperature Choose... The extent of randomness in a system Choose... Sum of the internal energy plus the product of the pressure and volume for a reaction
Answer:
Difference of the enthalpy (of a system) minus the product of the entropy and absolute temperature
Explanation:
The basis of spontaneity in a chemical reaction is that ∆G must be negative. ¡∆G is known as the change in free energy of a system. If ∆G is negative, then the reaction will occur without any external help (the reaction is spontaneous at room temperature).
∆G is given by;
∆G= ∆H -T∆S
Where;
∆H= change in enthalpy of the system
T= absolute temperature of the system
∆S= change in entropy
Hence; when ∆H -T∆S gives a negative result, the reaction proceeds without any external help.
A mixture of 50ml of 0.1M HCOOH and 50ml of 0.05M NaOH is equivalent to
Answer:
d) a solution that is 0.025M in HCOOH and 0.025M in HCOONa
Explanation:
The reaction of a weak acid (HOOH) with NaOH is as follows:
HCOOH + NaOH → HCOONa + H₂O
Based on the reaction, 1 mole of the acid reacts with 1 mole of the base (Ratio 1:1).
The initial moles of both species are:
HCOOH: 0.050L × (0.1mol / L) = 0.0050 moles of HCOOH
NaOH: 0.050L × (0.05 mol / L) = 0.0025 moles NaOH
After the reaction, all NaOH reacts with HCOOH producing HCOONa (Because moles of NaOH < moles HCOOH).
Final moles:
HCOOH: 0.0050 moles - 0.0025 moles (After reaction) = 0.0025 moles
HCOONa: Moles HCOONa = Initial Moles NaOH: 0.0025 moles
As volume of the mixture is 100mL (50 from the acid + 50 from NaOH), molarity of both HCOOH and HCOONa is:
0.0025 moles / 0.100L = 0.025M of both HCOOH and HCOONa
Thus, the initial mixture is equivalent to:
d) a solution that is 0.025M in HCOOH and 0.025M in HCOONaConsider the bond dissociation energies listed below in kcal/mol. CH3-Br 70 CH3CH2-Br 68 (CH3)2CH-Br 68 (CH3)3C-Br 65 These data show that the carbon-bromine bond is weakest when bromine is bound to a __________.
Answer:
The answer is "Tertiary carbon".
Explanation:
Accent to the results, the carbon-bromine bond is weak, whenever, the bromine is connected to tertiary carbon so, bonding energy is separation for methyl-carbon, which is connected to the bromine = 70 kcal/mol and for the primary energy to the secondary energy is= 68 kcal/mol, and for tertiary CO2 = 65 kcal/mol.
The stronger the energy dissociating connection and the weaker, its power dissociation connection and its weaker bond becomes connecting with a tertiary carbon, that's why "Tertiary carbon" is the correct answer.
Candle wax melts low temperature, it is not conductive to electricity, it is insoluble in water and partially soluble in solvents nonpolar, like gasoline. Than type of links are present in the candle wax?
A. Electrostatics.
B. Apolar.
C. lónicos.
D. Hydrogen bridges.
electrostatic and ionic are definitely not the answer because they have high melting point
hydrogen bonds are too weak and not permanent.
so the answer is apolar as it is soluble in polar solvents (water)
Answer:
B. Nonpolar
Explanation:
The low melting point tells you the compound is not ionic, metallic, or a network solid.
It is almost certainly a molecular solid.
It does not conduct electricity, so it is not metallic (which we have already ruled out).
It is insoluble in polar solvents (water) and soluble in nonpolar solvents (gasoline).
Since like dissolves like, the molecule is nonpolar.
The type of links must be nonpolar.
For the following set of volume/temperature data, calculate the missing quantity after the change is made. Assume that the pressure and the amount of gas remain constant.
V=2.91 L at 23.0 °C
V= 4.20 L at ? °C
Answer:
155 °C
Explanation:
Step 1: Given data
Initial volume (V₁): 2.91 LInitial temperature (T₁): 23.0°CFinal volume (V₂): 4.20 LFinal temperature (T₂): ?Step 2: Convert the initial temperature to Kelvin
We will use the following expression.
K = °C + 273.15 = 23.0°C + 273.15 = 296.2 K
Step 3: Calculate the final temperature
Assuming an ideal gas behavior, we can calculate the final temperature using Charles' law.
V₁/T₁ = V₂/T₂
T₂ = V₂ × T₁/V₁
T₂ = 4.20 L × 296.2 K/2.91 L
T₂ = 428 K
Step 4: Convert the final temperature to Celsius
We will use the following expression.
°C = K - 273.15 = 428 - 273.15 = 155 °C
If a substance has a half-life of 55.6 s, and if 230.0 g of the substance are present initially, how many grams will remain after 10.0 minutes?
Answer:
[tex]m=0.127g[/tex]
Explanation:
Hello,
In this case, for a first-order reaction, we can firstly compute the rate constant from the given half-life:
[tex]k=\frac{ln(2)}{t_{1/2}} =\frac{ln(2)}{55.6s}=0.0125s^{-1}[/tex]
In such a way, the integrated first-order law, allows us to compute the final mass of the substance once 10.0 minutes (600 seconds) have passed:
[tex]m=m_0*exp(-kt)=230.0g*exp(-0.0125s^{-1}*600s)\\\\m=0.127g[/tex]
Best regards.
Based on relative bond strengths, classify these reactions as endothermic (energy absorbed) or exothermic (energy released).
Strongest Bond
A-B
A-A
B-B
C-C
B-C
A-C
1. A2 + C2 rightarrow 2AC
2. B2 + C2 rightarrow 2BC
3. A + BC rightarrow AB + C
4. A2 + B2 rightaarrow 2AB
5. AB + C rightarrow AC + B
a. endothermic
b. exothermic
Answer:
1. Exothermic 2. Exothermic 3. Endothermic 4. Endothermic 5. Exothermic.
Explanation:
1. An A-A and a C-C bond results in 2 A-C bonds which are lower than the A-A and C-C bonds so this reaction is exothermic.
2. A B-B bond and a C-C bond results in 2 B-C bonds which are lower than the first 2 bonds so this reaction is also exothermic.
3. There is no bond for single A, a single B-C bond results in a A-B bond and a C molecule. A-B bond is stronger than the B-C bond so the reaction absorbed energy along the way. This shows that it is endothermic.
4. An A-A bond and a B-B bond results in 2 A-B bonds which are stronger than the first two bonds so this reaction is also endothermic.
5. An A-B bond and a C molecule result in an A-C bond and a B molecule. A-C bond is weaker than the A-B bond so there is energy released. This reaction is exothermic.
I hope this answer helps.
Predict the most likely bond type for the following.
a. Cu (Copper)
b. KCl (Potassium Chloride)
c. Si (Silicon)
d. CdTe (Cadmium Telluride)
e. ZnTe (Zinc Telluride)
Answer:
The three major types of bond are ionic, polar covalent, and covalent bonds. Ionic occurs majorly between metals and non-metals, which allows sharing of electrons to form an ionic compound. Whereas covalent bonding calls for complete transfer of electrons between atoms. Polar covalent bonds have unequaly shared electron-pair between two atoms.
Explanation:
a. Cu (Copper)- ionic bonding
b. KCl (Potassium Chloride) - ionic bonding
c. Si (Silicon) - covalent bonding
d. CdTe (Cadmium Telluride) - polar covalent bonding
e. ZnTe (Zinc Telluride)- polar covalent bonding
What will be formed when 2,2,3-trimethylcyclohexanone reacts with hydroxylamine?
Answer:
Following are the solution to this equation:
Explanation:
In the given-question, an attachment file of the choices was missing, which can be attached in the question and its solution can be defined as follows:
In the given question "Option (iii)" is correct, which is defined in the attachment file.
When 2,2,3-trimethylcyclohexanone reacts with hydroxylamine it will produce the 2,2,3-trimethylcyclohexanoxime.
A 5-column table with 2 rows. Column 1 is labeled number of protons, with entries 20 and 9; column 2 is number of neutrons, with entries 20 and D; Column 3 is atomic number, with entries A and E; Column 4 is Mass Number, with entries B and 19, and Column 5 is Element (symbol) with entries C and F. Using the periodic table, complete the table to describe each atom. Type in your answers
Answer:
A ⇒ 20
B ⇒ 40
C ⇒ Ca
D ⇒ 10
E ⇒ 9
F ⇒ F
Explanation:
edge 2021
Answer:
the person above is correct
Explanation:
How many grams of H2O will be formed when 32.0 g H2 is mixed with 73.0 g of O2 and allowed to react to form water
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g What is the molarity of hydrochloric acid if 40.95 mL of HCl is required to neutralize 0.550 g of sodium oxalate, Na2C2O4
Answer:
0.0002 M
Explanation:
The molarity of the HCl required would be 0.0002 M.
First, let us consider the balanced equation of the reaction:
[tex]Na_2C_2O_4 + 2HCl = 2NaCl + H_2 + 2CO_2[/tex]
Stoichiometrically, 1 mole of [tex]Na_2C_2O_4[/tex] reacts with 2 moles of [tex]HCl[/tex] for a complete neutralization reaction.
Recall that: mole = [tex]\frac{mass}{molar mass}[/tex]
Mole of 0.550 g sodium oxalate = 0.550/134 = 0.0041 mole
If 1 mole [tex]Na_2C_2O_4[/tex] requires 2 moles HCl, then 0.0041 mole will require:
0.0041 x 2 = 0.0082 mole HCl
Volume of the HCl = 40.95 L
Molarity = mole/volume
Hence, molarity of the HCl = 0.0082/40.95 = 0.0002 M
