Answer:
42.64 M
Explanation:
Step 1: Given data
Moles of HCl (solute): 8.1 molVolume of the solution: 189.98 mLStep 2: Convert the volume of the solution to liters
We will use the relationship 1 L = 1,000 mL.
[tex]189.98 mL \times \frac{1L}{1,000mL} = 0.18998L[/tex]
Step 3: Calculate the molarity of the solution
The molarity is equal to the moles of solute divided by the liters of solution.
[tex]M= \frac{8.1mol}{0.18998L} = 42.64 M[/tex]
What is the main side reaction that competes with elimination when a primary alkyl halide is treated with alcoholic potassium hydroxide, and why does this reaction compete with elimination of a primary alkyl halide but not a tertiary alkyl halide
Answer:
The main competing reaction when a primary alkyl halide is treated with alcoholic potassium hydroxide is SN2 substitution.
Explanation:
The relative percentage of products of the reaction between an alkyl halide and alcoholic potassium hydroxide generally depends on the structure of the primary alkylhalide. The attacking nucleophile/base in this reaction is the alkoxide ion. Substitution by SN2 mechanism is a major competing reaction in the elimination reaction intended.
A more branched alkyl halide will yield an alkene product due to steric hindrance, similarly, a good nucleophile such as the alkoxide ion may favour SN2 substitution over the intended elimination (E2) reaction.
Both SN2 and E2 are concerted reaction mechanisms. They do not depend on the formation of a carbocation intermediate. Primary alkyl halides generally experience less steric hindrance in the transition state and do not form stable carbocations hence they cannot undergo E1 or SN1 reactions.
SN2 substitution cannot occur in a tertiary alkyl halides because the stability of tertiary carbocations favours the formation of a carbocation intermediate. The formation of this carbocation intermediate will lead to an SN1 or E1 mechanism. SN2 reactions is never observed for a tertiary alkyl halide due to steric crowding of the transition state. Also, with strong bases such as the alkoxide ion, elimination becomes the main reaction of tertiary alkyl halides.
A(n) _____ reaction occurs when an acid and a base are present in the same solution.
Answer:
The answer is Neutralization reaction
It occurs when an acid and a base are present in the same solution and react to form salt and water only
Hope this helps you
When solutions of hydrochloric acid and sodium hydroxide are mixed, a chemical reaction occurs forming aqueous sodium chloride and water. What would you expect to observe if you ran the reaction in the laboratory
Answer:
a change in temperature would be observed(ΔH is -ve)
Explanation:
Hydrochloric acid react with sodium hydroxide to give salt(sodium chloride) and water
HCl(aq) + NaOH(aq) =====> NaCl(aq) + H2O(l)
There would be no notable change since sodium chloride dissolved in water but there would be a change in temperature.
Since neutralization is exothermic(heat is evolved), therefore ΔH is negative
Do you think you could go a week without causing any chemical reactions?
yes yes yes yes
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plllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllls help me
Answer:
Explanation:
equilibrium constant
Kc = [ C ]² / [ A ] [ B ]
= .5² / .2 x 3
= .4167
Let moles of A to be added be n
concentration of A unreacted becomes .2 + n M
increase of product C by .2 M will require use of A and B be .1 M
So unreacted A = .2 + n - .1 = n + .1
Kc = [ C ]² / [ A ] [ B ]
.4167 = .7² / ( n + .1 ) ( 3 - .1 )
n + .1 = .4
n = . 3 moles .
So .3 moles of A to be added .
Write the name for the following molecular compounds. Remember to use the correct prefix for each compound.
a. CS2
b. PBr3
c. NO
d. CF4
e. P2O5
Answer:
Hey there!
CS2) Carbon Disulfide.
PBr3) Phosphorus Tribromide
NO) Nitric Oxide
CF4) Carbon Tetrafluoride
P2O5) Phosphorus Pentoxide
Let me know if this helps :)
Enter an equation for the formation of CaCO3(s) from its elements in their standard states. Enter any reference to carbon as C(s). Express your answer as a chemical equation. Identify all of the phases in your answer.
Answer:
CaF2 + CO3- ----> CaCO3 + 2 F-
Explanation:
The chemical compounds found on the left side of the date are the reagents and those found on the right are the products, where calcium carbonate appears.
Calcium carbonate is a quaternary salt
The constant pressure molar heat capacity of argon, C_{p,m}C
p,m
, is
20.79\text{ J K}^{-1}\text{ mol}^{-1}20.79 J K
−1
mol
−1
at 298\text{ K}298 K. What
will be the value of the constant volume molar heat capacity of argon,
C_{V,m}C
V,m
, at this temperature?
Answer:
Constant-volume molar heat capacity of argon is 12.47 J K ⁻¹mol⁻¹
Explanation:
Argon is a monoatomic gas that behaves as an ideal gas at 298K.
Using the first law of thermodinamics you can obtain:
Work, Q, for constant pressure molar heat capacity,CP:
CP = (5/2)R
For constant-volume molar heat capacity,CV:
CV = (3/2)R
That means:
2CP/5 = 2CV/3
3/5 = CV / CP
As CP of Argon is 20.79 J K ⁻¹mol⁻¹, CV will be:
3/5 = CV / CP
3/5 = CV / 20.79 J K ⁻¹mol⁻¹
12.47 J K ⁻¹mol⁻¹ = CV
Constant-volume molar heat capacity of argon is 12.47 J K ⁻¹mol⁻¹a piece of copper weighing 850 grams is placed in a cup with 450 ml of water at 21 C and the Cp of the cup is 47 J/K, how many grams of gasoline would it take to heat the entire system to 110 C?
Answer:
4.2g of gasoline
Explanation:
In the problem, you need to give energy to the cup from the combustion of gasoline. The energy you need to give is:
Qcup + QWater + QCopper
As you need to increase (110ºC - 21ºC = 89º = Increase 89K) 89K, the Qcup is:
Qcup = 89K × (47J/K) = 4183J.
You can find Qwater using its specific heat, C (4.18Jg⁻¹K⁻¹), its mass (450mL = 450g) and the change of temperature, 89K:
QWater = CₓmₓΔT
QWater = 4.184Jg⁻¹K⁻¹ ₓ 450g×89K
QWater = 167569J
And Q of Copper, QCu, could be obtained in the same way (Specific heat Cu: 0.387 J/g⁻¹K⁻¹:
QCu = CₓmₓΔT
QCu = 0.387 J/g⁻¹K⁻¹ₓ850gₓ89K
QCu = 29277J
Thus, total heat you need is:
Q = Qcup + QWater + QCopper
Q = 4183J + 167569J + 29277J
Q = 201029J = 201kJ
The combustion of gasoline (Octane) produce 47.8kJ/g (Its heat of combustion). that means to produce 201kJ of energy you require:
201kJ × (1g / 47.8kJ) =
4.2g of octane = Gasoline you requireIdentify the correct structure of 5-bromo-4-isopropylheptanoic acid.
Answer:
See attached picture.
Explanation:
Hello,
In this case, given the IUPAC name, we can infer we have a seven-carbon carboxylic acid that has a bromine at the fifth carbon, an isopropyl at the fourth carbon and the carboxyl functional group (COOH) at the first carbon, thus, on the attached document, you will find the correct structure.
Best regards.
How many grams of the excess reactant remain after the limiting reactant is completely consumed? Express your answer in grams to three significant figures.
The question is incomplete, the complete question is;
One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of NH3 to NO How many grams of NO and of H20 form? Enter your answers numerically separated by a comma. 4NH3(g) +502(g)------->4NO(g)+6H2O(g)
In a certain experiment, 1.10 g of NH3 reacts with 2.02 g of O2. How many grams of the excess reactant remain after the limiting reactant is completely consumed? Express your answer in grams to three significant figures.
Answer:
Mass of excess ammonia 0.034 g of ammonia
Mass of water formed= 1.37g
Mass of NO formed = 1.50g
Explanation:
The limiting reactant is the reactant that yields the least number of moles of product.
For NH3, molar mass of ammonia = 17g mol-1
Number of moles of ammonia reacted= 1.10g/17 gmol-1 = 0.065 moles of ammonia
According to the reaction equation;
4 moles of ammonia yields 4 moles of NO
Hence 0.065 moles of ammonia will yield 0.065 ×4/4 = 0.065 moles of NO
For oxygen, molar mass of oxygen gas = 32gmol-1
Number of moles of oxygen gas= 2.02g/32gmol-1 = 0.063 moles of oxygen
From the reaction equation;
5 moles of oxygen gas yields 4 moles of NO
0.063 moles of oxygen will yield 0.063 ×4 /5 = 0.050 moles of NO
Hence oxygen is the limiting reactant and ammonia is the excess reactant.
Amount of excess ammonia = Amount of ammonia - amount of oxygen
Amount of excess ammonia= 0.065-0.063= 2×10^-3 moles
Mass of excess ammonia = 2×10^-3 moles × 17 gmol-1 = 0.034 g of ammonia
Mass of NO formed is obtained from the limiting reactant. Since molar mass of is 30gmol-1. Then mass of NO formed = 0.050 moles of NO × 30gmol-1 = 1.50 g of NO
For water;
5 moles of oxygen yields 6 moles of water
Hence 0.063 moles of oxygen yields 0.063 × 6/5 = 0.076 moles of water
Molar mass of water = 18gmol-1
Hence mass of water = 0.076 moles × 18gmol-1 = 1.37g of water
The activation energy of the uncatalyzed reaction is about 3.98 times that of the catalyzed reaction with activation energy of 4.6 kJ. The uncatalyzed reaction has such a large activation energy that its rate is extremely slow. What is the activation energy for the uncatalyzed reaction
Answer:
18.308 KJ
Explanation:
From the given above, we obtained the following:
Activation energy for the catalyzed reaction = 4.6 kJ.
Activation energy for the uncatalyzed reaction =..?
Now, a careful observation of the question revealed that the activation energy of the uncatalyzed reaction is about 3.98 times that of the catalyzed reaction.
With this vital information, we can thus, calculate the activation energy of the uncatalyzed reaction as follow:
Activation energy for the uncatalyzed reaction = 3.98 times that of the catalyzed reaction.
Activation energy for the uncatalyzed reaction = 3.98 x 4.6 kJ = 18.308 KJ
Therefore, the activation energy of the uncatalyzed reaction is 18.308 KJ.
2.
Name the following compounds:
a. Rb20
Answer:
Rubidium oxide
Explanation:
Draw the structure 2 butylbutane
Answer:
please look at the picture below.
Explanation:
A critical reaction in the production of energy to do work or drive chemical reactions in biological systems is the hydrolysis of adenosine triphosphate, ATP, to adenosine diphosphate, ADP, as described by the reactionATP(aq)+ H2O(l) → ADP(aq)+ HPO4^-2 (aq)for which ΔGrxn = -30.5 kj/mol at 37.0C and pH 7.0. Required:a. Calculate the value of ΔGrxn in a biological cell in which [ATP] = 5.0 mM, [ADP] = 0.30 mM, and HPO4^-2= 5.0mMb. Is the hydrolysis of ATP spontaneous under these conditions?
Answer:
Δ [tex]G_{rxn}[/tex] = −51. 4 kJ/mol
However, since Δ [tex]G_{rxn}[/tex] is negative. The hydrolysis of ATP for this reaction is said to be spontaneous
Explanation:
From the question; The equation for this reaction can be represented as :
[tex]ATP_{(aq)} + H_2O_{(l)} \to ADP_{(aq)}+ HPO_4^{2-}} _{(aq)}[/tex]
where:
[tex]\Delta G ^0 _{rxn} =[/tex]-30.5 kJ/mol
= -30.5 kJ/mol × 1000 J/ 1 kJ
= -30.5 × 10 ⁻³ J/mol
Temperature T = 37 ° C
= (37+273)
= 310 K
pH = 7.0
[ATP] = 5.0 mM
= 5.0mM × 1M/1000mM
= 0.005 M
[ADP] = 0.30 mM
= 0.30 mM × 1M/1000mM
= 0.0003 M
[tex][HPO_4^{2-}}][/tex] = 5.0 mM
= 5.0mM × 1M/1000mM
= 0.005 M
The objective is to calculate the value for Δ [tex]G_{rxn}[/tex] in the biological cell and to determine if the hydrolysis of ATP is spontaneous under these conditions.
Now;
From the equation given; the equilibrium constant [tex]K_{eq}[/tex] can be expressed as:
[tex]K_{eq} = \dfrac{[ADP][ HPO_4^{2-}]} {[ATP]}[/tex]
[tex]K_{eq} = \dfrac{(0.0003 \ M)(0.005 \ M)} {(0.005 \ M)}[/tex]
[tex]K_{eq} = 3*10^{-4}[/tex]
The Δ [tex]G_{rxn}[/tex] in the biological cell can now be calculated as:
Δ [tex]G_{rxn}[/tex] = [tex](-30.5 * 10 ^3 \ J/mol) + (8.314 \ J/mol.K)(310 K ) In ( 3*10^{-4})[/tex]
Δ [tex]G_{rxn}[/tex] = [tex](-30.5 * 10 ^3 \ J/mol) + (-20906.68126)[/tex]
Δ [tex]G_{rxn}[/tex] = −51406.68 J/mol
Δ [tex]G_{rxn}[/tex] = −51. 4 × 10³ J/mol
Δ [tex]G_{rxn}[/tex] = −51. 4 kJ/mol
Thus since Δ [tex]G_{rxn}[/tex] is negative. The hydrolysis for this reaction is said to be spontaneous
Determine the the mass of one molecule of hydrogen sulfide gas.
Answer:
the molecular mass of hydrogen sulphide, which contains two atoms of hydrogen and one atom of sulphur is = 2 — 1 + 1 — 32 = 34 a.m.u.
Identify the structure of S (molecular formula C7H14O2). Compound S the odor of rum, (1H NMR data (ppm) at 0.93 (doublet, 6 H), 1.15 (triplet, 3 H), 1.91 (multiplet, 1 H), 2.33 (quartet, 2 H), and 3.86 (doublet, 2 H) ppm.Compound S:_______.
Answer:
Following are the answer to this question:
Explanation:
The structure of the S molecular formula [tex]C_7H_{14}O_2[/tex] defined in the attachment file.
Please find the attachment file.
An aqueous solution of glucose (C6H12O6), called D5W, is used for intravenous injection. D5W contains 54.30 g of glucose per liter of solution. What is the molar concentration of glucose in D5W
Answer:
The correct answer is 0.30 M
Explanation:
The molar concentration or molarity of a solution is defined as moles of solute per liter of solution. We found the moles of solute (glucose) by dividing the mass (54.30 g) into the molecular weight (MW) of glucose (C₆H₁₂O₆):
MW(C₆H₁₂O₆)= (12 g/mol x 6) + (1 g/mol x 12) + (16 g/mol x 6) = 180 g/mol
Moles of glucose= mass/MW= 54.30 g/(180 g/mol)= 0.30 mol
There is 0.30 mol of solute per liter of solution, thus the molarity is:
M= moles solute/L solution= 0.30 mol/1 L = 0.30 M
Searches related to If 0.75 grams of iron (Fe) react according to the following reaction, how many grams of copper (Cu) will be produced? Fe + CuSO4 -> Cu + FeSO4
Answer:
0.83 g
Explanation:
Step 1: Write the balanced equation
Fe + CuSO₄ ⇒ Cu + FeSO₄
Step 2: Calculate the moles corresponding to 0.75 g of Fe
The molar mass of Fe is 55.85 g/mol.
[tex]0.75g \times \frac{1mol}{55.85g} = 0.013 mol[/tex]
Step 3: Calculate the moles of Cu produced from 0.013 moles of Fe
The molar ratio of Fe to Cu is 1:1. The moles of Cu produced are 1/1 × 0.013 mol = 0.013 mol.
Step 4: Calculate the mass corresponding to 0.013 moles of Cu
The molar mass of Cu is 63.55 g/mol.
[tex]0.013mol \times \frac{63.55g}{mol} = 0.83 g[/tex]
Answer:
If 0.75 grams of iron (Fe) react, 0.85 grams of copper (Cu) will be produced.
Explanation:
You know the following balanced reaction:
Fe + CuSO₄ ⇒ Cu + FeSO₄
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following quantities react and are produced:
Fe: 1 moleCuSO₄: 1 moleCu: 1 moleFeSO₄: 1 moleBeing:
Fe: 55.85 g/moleCu: 63.54 g/moleS: 32 g/moleO: 16 g/molethe molar mass of the compounds participating in the reaction is:
Fe: 55.85 g/moleCuSO₄: 63.54 g/mole + 32 g/mole+ 4* 16 g/mole= 159.54 g/moleCu: 63.54 g/moleFeSO₄: 55.85 g/mole + 32 g/mole+ 4* 16 g/mole= 151.85 g/moleThen, by stoichiometry of the reaction, the amounts of reagent and product that participate in the reaction are:
Fe: 1 mole*55.85 g/mole= 55.85 gCuSO₄: 1 mole* 159.54 g/mole= 159.54 gCu: 1mole* 63.54 g/mole= 63.54 gFeSO₄: 1 mole* 151.85 g/mole= 151.85 gThen you can apply a rule of three as follows: if 55.85 grams of Fe produces 63.54 grams of Cu, 0.75 grams of Fe how much mass of Cu does it produce?
[tex]mass of Cu=\frac{0.75 grams of Fe*63.54 grams of Cu}{55.85 grams of Fe}[/tex]
mass of Cu= 0.85 grams
If 0.75 grams of iron (Fe) react, 0.85 grams of copper (Cu) will be produced.
Hypochlorous acid is formed in situ by reaction of aq. sodium hypochlorite solution with acetic acid. Draw balanced chemical equations to show the formation of hypochlorous acid and protonated hypochlorous acid.
Answer:
NaClO + CH₃COOH ----> HClO + CH3CO- + Na
Explanation:
This reaction occurs between the combination of a salt and an acid, that is, an oxide-reduction reaction
Suppose that you add 26.7 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the added solute, you find that there is a freezing point depression of 2.74 oC compared to pure benzene. What is the molar mass of the unknown compound
Answer: The molar mass of the unknown compound is 200 g/mol
Explanation:
Depression in freezing point is given by:
[tex]\Delta T_f=i\times K_f\times m[/tex]
[tex]\Delta T_f=2.74^0C[/tex] = Depression in freezing point
i= vant hoff factor = 1 (for molecular compound)
[tex]K_f[/tex] = freezing point constant = [tex]5.12^0C/m[/tex]
m= molality
[tex]\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]
Weight of solvent (benzene)= 0.250 kg
Molar mass of solute = M g/mol
Mass of solute = 26.7 g
[tex]2.74^0C=1\times 5.12\times \frac{26.7g}{Mg/mol\times 0.250kg}[/tex]
[tex]M=200g/mol[/tex]
Thus the molar mass of the unknown compound is 200 g/mol
The molar mass of an unknown solute compound in the solution has been 199.626 g/mol.
With the addition of the solute to the solution, there has been a depression in the freezing point. The depression in the freezing point can be expressed as:
Depression in freezing point = Van't Hoff factor × Freezing point constant × molality
The molality can be defined as the moles of solute per kg solvent
Molaity = [tex]\rm \dfrac{Mass\;of\;solute\;(g)}{Molecular\;mass\;of\;solute}\;\times\;\dfrac{1}{Mass\;of\;solvent\;(kg)}[/tex]
The depression in freezing point can be given as:
Depression in freezing point = Van't Hoff factor × Freezing point constant × [tex]\rm \dfrac{Mass\;of\;solute\;(g)}{Molecular\;mass\;of\;solute}\;\times\;\dfrac{1}{Mass\;of\;solvent\;(kg)}[/tex] ......(i)
Given, the depression in freezing point = 2.74 [tex]\rm ^\circ C[/tex]
Van't Hoff factor = 1 (Molecular compound)
Freezing point constant (Kf) = 5.12 [tex]\rm ^\circ C[/tex]/m
Mass of solute = 26.7 g
Mass of solvent = 0.250 kg
Substituting the values in equation (i):
2.74 [tex]\rm ^\circ C[/tex] = 1 × 5.12
[tex]\rm \dfrac{2.74}{5.12}[/tex] = [tex]\rm \dfrac{1}{Molecular\;mass\;of\;solute}\;\times\;\dfrac{26.7}{0.250\;kg}[/tex]
0.535 = [tex]\rm \dfrac{1}{Molecular\;mass\;of\;solute}\;\times\;106.8[/tex]
Molecular mass of solute = [tex]\rm \dfrac{106.8}{0.535}[/tex] g/mol
Molecular mass of solute = 199.626 g/mol
The molar mass of an unknown solute compound in the solution has been 199.626 g/mol.
For more information about the molality of the compound, refer to the link:
https://brainly.com/question/7229194
What is the standard cell potential for the spontaneous voltaic cell formed from the given half-reactions
Answer:
because it is
The molar mass of an unknown gas was measured by an effusion experiment. It was found that it took 63 s for the gas to effuse, whereas nitrogen gas (N2) required 48 s. The molar mass of the unknown gas is-
Answer:
8.13 g/mol.
Explanation:
The following formula gives us the relationship between the effusion rates of two gases and their molar masses:
[tex]\sqrt{\frac{MM_{x} }{MM_{y} } } = \frac{rate_{y} }{rate_{x} }[/tex]
where x and y are respective sample gases and MM and rate are molar mass and rate of effusion respectively.
⇒[tex]\sqrt{\frac{14}{y} } = \frac{63}{48}[/tex]
[tex]\frac{14}{y} = 1.3125^{2}[/tex]
y= 14 / [tex]1.3125^{2}[/tex] = 8.13 g/mol.
2) Which type movement do pivot joints allow?
Which substance is a base? HCOOH RbOH H2CO3 NaNO3
Answer:
RbOH
Explanation:
For this question, we have to remember what is the definition of a base. A base is a compound that has the ability to produce hydroxyl ions [tex]OH^-[/tex], so:
[tex]AOH~->~A^+~+~OH^-[/tex]
With this in mind we can write the reaction for each substance:
[tex]HCOOH~->~HCOO^-~+~H^+[/tex]
[tex]RbOH~->~Rb^+~+~OH^-[/tex]
[tex]H_2CO_3~->~CO_3^-^2~+~2H^+[/tex]
[tex]NaNO_3~->~Na^+~+~NO_3^-[/tex]
The only compound that fits with the definition is [tex]RbOH[/tex], so this is our base.
I hope it helps!
RbOH or rubidium hydroxide is a base.
• RbOH or rubidium hydroxide is the inorganic compound.
• It comprises hydroxide anions and an equal number of rubidium cations.
• Rubidium hydroxide, like other strong bases is highly corrosive.
• The formation of rubidium hydroxide takes place when the metal rubidium reacts with water.
• The bases refers to the substance, which gets dissociate in an aqueous solution to produce OH- or hydroxide ions.
• Rubidium hydroxide is a base, when it is dissolved in water it give rise to an alkali, when reacts with acid it generates a rubidium salt.
Thus, RbOH is a base.
To know more about:
https://brainly.com/question/18616956
How many grams of H 2O are produced from 28.8 g of O 2? (Molar Mass of H 2O = 18.02 g) (Molar Mass of O 2=32.00 g) 4 NH 3 (g) + 7 O 2 (g) → 4 NO 2 (g) + 6 H 2O (g)
Answer: 13.9 g of [tex]H_2O[/tex] will be produced from the given mass of oxygen
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} O_2=\frac{28.8g}{32.00g/mol}=0.900moles[/tex]
The balanced chemical reaction is:
[tex]4NIO_2(g)+7O_2(g)\rightarrow 4NO_2(g)+6H_2O(g)[/tex]
According to stoichiometry :
7 moles of [tex]O_2[/tex] produce = 6 moles of [tex]H_2O[/tex]
Thus 0.900 moles of [tex]O_2[/tex] will produce =[tex]\frac{6}{7}\times 0.900=0.771moles[/tex] of [tex]H_2O[/tex]
Mass of [tex]H_2O=moles\times {\text {Molar mass}}=0.771moles\times 18.02g/mol=13.9g[/tex]
Thus 13.9 g of [tex]H_2O[/tex] will be produced from the given mass of oxygen
Select the oxidation reduction reactions??
Answer:
Explanation:
1 ) Cl₂ + ZnBr₂ = ZnCl₂ + Br₂
In this reaction , oxidation number of Cl decreases from 0 to -1 so it is reduced and oxidation number of Br increases from -1 to 0 so it is oxidised . Hence this reaction is oxidation - reduction reaction .
2 )
Pb( ClO₄)₂ + 2KI = PbI₂ + 2KClO₄
In this reaction oxidation number of none is changing so it is not an oxidation - reduction reaction.
3 )
CaCO₃ = CaO + CO₂
In this reaction also oxidation number of none is changing so it is not an oxidation - reduction reaction.
So only first reaction is oxidation - reduction reaction.
2nd option is correct.
The authors state in the general procedures that the reaction was monitored by TLC. How would this be done? What would you spot in each lane? How would you know the reaction was done?
Answer:
Thin Layer Chromatography (TLC) can be used to analyze chemical reactions. During this reaction monitoring, a typical TLC plate would have three spots: the reactant lane, the reaction mixture lane, and a "co-spot" where reaction product would be spotted directly on top of reactant.
The co-spot serves as a reference point and is vital for reactions where reactant and product have similar Rfs, and many other variations of eluent tracking.
To indicate completion of the reaction, the disappearance of a spot (usually the starting reactant) is observed.
Write the condensed electron configurations for the Ca atom. Express your answer in condensed form as a series of orbitals. For example, [He]2s22p2 should be entered as [He]2s^22p^2.
Answer:
[Ar] 4s²
Explanation:
Ca is the symbol for Calcium. It is the 20th element and it has 20 electrons.
The full electronic configuration for calcium is given as;
1s²2s²2p⁶3s²3p⁶4s²
The condensed electronic configuration is given as;
[Ar] 4s²
The activation energy for the decomposition of HI is 183 kJ/mol. At 573 K, the rate constant was measured to be 2.91 x 10^{-6} M/s. At what temperature in Kelvin does the reaction have a rate constant of 0.0760 M/s
Answer:
[tex]T_2=453.05K[/tex]
Explanation:
Hello,
In this case, the temperature-variable Arrhenius equation is written as:
[tex]\frac{k(T_2)}{k(T_1)}=exp(\frac{Ea}{R}(\frac{1}{T_2}-\frac{1}{T_1} ))[/tex]
Now, for us to solve for the temperature by which the reaction rate constant is 0.0760M/s we proceed as shown below:
[tex]ln(\frac{k(T_2)}{k(T_1)})=\frac{Ea}{R}(\frac{1}{T_2}-\frac{1}{T_1} )\\ln(\frac{0.0760M/s}{0.00000291M/s} )=\frac{183000J/mol}{8.314J/(mol*K)} *(\frac{1}{T_2} -\frac{1}{573K} )\\\frac{1}{T_2} -\frac{1}{573K} =\frac{10.17}{22011.06K^{-1}} \\\\\frac{1}{T_2}=4.62x10^{-4}K^{-1}+\frac{1}{573K}\\\\\frac{1}{T_2}=2.21x10^{-3}K^{-1}\\\\T_2=453.05K[/tex]
Regards.