During respiration, the inhalation of Oxygen (O) using the energy from Glucose (C6H12O6) is the first chain reaction.
Products formed during respiration are: Carbon Dioxide (CO2) and Water (H2O)
The general molecular formula for alkanes is CₙH₂ₙ₊₂. What is the general molecular formula for cycloalkanes?
Answer:
CₙH₂ₙ
Explanation:
Answer:
[tex]\huge \boxed{\mathrm{C_nH_{2n}}}[/tex]
Explanation:
The general molecular formula for cycloalkanes is [tex]\mathrm{C_nH_{2n}}[/tex].
If you pour 9.0 g of sodium chloride into water to produce 240 mL of solution, what will the molarity be?
Explanation:
Molarity = mol/liter
Solution 240 mL = NaCl 9.0 g
Solution 1000 mL = NaCl 9.0/240 × 1000 = ..... g
mol NaCl = g/MW
MW NaCl = ?....
...........
Molarity = ......... Molar
What are two things plants have which animal cells do not
Answer:
Animal cells don't have a dividing cell wall like plant cells do
Explanation:
Plants cells use photosynthesis from the sun, which requires them to have chloroplast filled with chlorophyll to complete this function; animal cells do not have chloroplasts
The half-life of carbon-14 is 5,715 years. After 29,000 years (5 half-lives) have elapsed, how much carbon-14 remains in a sample
that originally contained 0.32 g of carbon-14
A 0.0409
B. 0.0809
C 0.109
D. 0.0109
E 0209
After 3 half-life, 12 of the 14 of the C-14 = 18 of the C-14 would remain.
A sample of sugar (C12H22O11) contains
1.505 × 1023 molecules of sugar. How many moles of sugar are present in the sample? Answer without doing any calculations.
0.25 mol
0.50 mol
1.00 mol
2.50 mol
Answer:
0.25 mol
Explanation:
Moles= No.of molecules / Avogadros number
Answer:
0.25
Explanation:
By titration, it is found that 28.5 mL of 0.183 M NaOH(aq) is needed to neutralize 25.0 mL of HCl(aq). Calculate the
concentration of the HCl solution.
Answer:
0.209M
Explanation:
M1V1=M2V2
(28.5 mL)(0.183M)=(25.0mL)(M)
M2= 0.209M
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Watching the assignment without knowing how to solve the questions is the worst thing to experience hope someone can help with this question and the privous ones too
Answer:
- 622.5kJ
Explanation:
1) 2NH3 + 3N2O ----> 4N2 + 3H2O ΔH° 1= - 1010kJ
- 3 (N2O + 3H2 ---> N2H4 + H2O ΔH° 2 = - 317 kJ)
2) 2NH3 + 3N2O ----> 4N2 + 3H2O ΔH° 1= - 1010kJ
- 3N2O - 9 H2 ----> - 3N2H4 - 3H2O - 3ΔH° 2 = 3*317 kJ
2NH3 + 3N2H4 --->4N2+9H2, ΔH° 1 - 3*ΔH° 2
2NH3 + 3N2H4 --->4N2+9H2, ΔH° 1 - 3*ΔH° 2
3) -(2NH3 +1/2O2 ---> N2H4 + H2O, ΔH°3 = -143 kJ)
-2NH3 - 1/2O2 ---> - N2H4 - H2O, - ΔH°3 = 143 kJ
2NH3 + 3N2H4 --->4N2+9H2, ΔH° 1 - 3*ΔH° 2
-2NH3 - 1/2O2 ---> - N2H4 - H2O, - ΔH°3 = 143 kJ
4N2H4 + H2O ---> 4N2 + 9H2 + 1/2O2, ΔH° 1 - 3*ΔH° 2 - ΔH°3
4)
H2 + 1/2O2 ---> H2O, ΔH° 4 = - 286 kJ
9*(H2 + 1/2O2 ---> H2O, ΔH° 4 = - 286 kJ)
9H2+ 9/2 O2 ---> 9H2O, 9*ΔH° 4 = 9*(- 286) kJ
4N2H4 + H2O ---> 4N2 + 9H2 + 1/2O2, ΔH° 1 - 3*ΔH° 2 - ΔH°3
9H2+ 9/2 O2 ---> 9H2O, 9*ΔH° 4 = 9*(- 286) kJ
4N2H4 +4O2 --->4N2+8H2O, ΔH° 1 - 3*ΔH° 2 - ΔH°3 + 9*ΔH° 4
5)
1/4*(4N2H4 +4O2 --->4N2+8H2O, ΔH° 1 - 3*ΔH° 2 - ΔH°3 + 9*ΔH° 4
N2H4 +O2 --->N2+2H2O, 1/4( ΔH° 1 - 3*ΔH° 2 - ΔH°3 + 9*ΔH° 4)
6)
N2H4 +O2 --->N2+2,
1/4( ΔH° 1 - 3*ΔH° 2 - ΔH°3 + 9*ΔH° 4)=
=1/4(-1010kJ - 3*(-317kJ) - (-143kJ) + 9*(-286kJ))= - 622.5 kJ
What is shown by the reaction below? 2{CH2O} CO2(g) + CH4(g) how is this process related to aerobic respiration
Explanation:
[tex]CH_2O[/tex] is an energy rich molecule, which when reacts with oxygen produces carbon dioxide and energy. In aerobic reaction oxygen is consumed. IN this biochemical reaction oxygen is consumed and CO2 + CH4 is produced. Thus the reaction is aerobic one.
How do rockets work? What chemicals are used and what reactions happens with those chemicals to make a rocket fly?
Answer:
Chemical rocket engines use a fuel (something to burn) and an oxidiser (something to react with the fuel). ... As the propellant reacts inside a combustion chamber, the chemical reaction produces hot gases. It is the ejection of these rapidly expanding hot gases at high speed from the rocket nozzle that
( I found this on a website called Science Learning. Full credit to them and their work. Check it out for more information)
What is the net electric charge of an ionic compound?
Ionic compounds contain both cations and anions in a ratio that results in no net electrical charge
The name carbohydrate comes from the fact that many simple sugars have chemical formulae that look like water has simply been added to carbon. (The suffix hydrate from the Greek word hydor ("water") means "compound formed by the addition of water") The actual chemical structure of carbohydrates doesn't look anything like water molecules bonded to carbon atoms. But it is nevertheless possible to chemically extract all the hydrogen and oxygen from many simple carbohydrates as water, leaving only carbon behind. If you search the Internet for "reaction of sulfuric acid and sugar" you will find some impressive videos of this. Suppose you had (200. g) of ordinary table sugar, which chemists call sucrose, and which has the chemical formula C1,H2,0. Calculate the maximum mass of water you could theoretically extract. Be sure your answer has a unit symbol, and round it to the correct number of significant digits?
Answer:
The maximum mass of water produced is [tex]m__{H_2O}} =116 \ g[/tex]
Explanation:
From the question we are told that
The mass of sucrose is [tex]m_s = 200 \ g[/tex]
The chemical formula for sucrose is [tex]C_{12} H_{22} O_{11}[/tex]
The chemical equation for the dissociation of sucrose is
[tex]C_{12} H_{22}O_{4} \to 12C + 11H_2O[/tex]
The number of moles of sucrose can be evaluated as
[tex]n = \frac{m}{Z}[/tex]
Where Z is the molar mass of sucrose which has a constant value of
[tex]Z = 342 \ g/mol[/tex]
So
[tex]n = \frac{200}{342}[/tex]
[tex]n =0.585[/tex]
From the chemical equation one mole of sucrose produces 11 moles of water so 0.585 moles of sucrose will produce x moles of water
Therefore
[tex]x = \frac{0.585 * 11}{1}[/tex]
[tex]x = 6.433 \ moles[/tex]
Now the mass of water produced is mathematically represented as
[tex]m__{H_2O}} = x * Z__{H_2O}[/tex]
Where [tex]Z__{H_2O}[/tex] is the molar mass of water with a constant values of [tex]Z__{H_2O}} = 18 \ g/mol[/tex]
So
[tex]m__{H_2O}} = 6.43* 18[/tex]
[tex]m__{H_2O}} =116 \ g[/tex]
At a certain temperature Kc = 9.0 for the equilibrium 24() ⇔ 22(). What is
Kc at the same temperature for 2() ⇔ 1/224().
Answer:
.
Explanation:
.
The U.S. Food and Drug Administration (FDA) recommends that an adult consume less than 1.8 g
of sodium per day.
What mass of sodium chloride (in grams) can you consume and still be within the FDA guidelines? Sodium chloride is 39% sodium by mass
Answer:
less than 4.62 g
Explanation:
Sodium chloride contains 39% sodium by mass. Hence, 1 g of sodium chloride will contain:
39/100 x 1 = 0.39 g of sodium.
If 1 g of sodium chloride contains 0.39 g of sodium, then how many g of sodium chloride will contain the 1.8 g required sodium per day?
1 g sodium chloride = 0.39 g sodium
x g sodium chloride = 1.8 g sodium
x = 1.8 x 1/0.39 = 4.62 g of sodium chloride
Since the adult is supposed to consume less than 1.8 g of sodium per day, the mass of sodium that should be consume per day and still be within the FDA guidelines will be less than 4.62 g of sodium chloride.
Indicate the direction of polarity of each of the following covalent bonds. Please just write a + or - not the bonds.
1. C ---- O
2. O ---- Cl3. O ----- F4. C ----- N5. Cl ----- C6. S ----- H7. S ----- Cl
Answer:
1. C+ ---- O-
2. O+ ---- Cl-
3. O+ ----- F-
4. C+ ----- N-
5. Cl- ----- C+
6. S- ----- H+
7. S+ ----- Cl -
Explanation:
Electronegativity determines the polarity . There may be two atoms in a bond with high electronegativity, in such cases the positive charge is given to atom with comparatively lower electronegativity. Electronegativity determines the easiness with which an atom attract electrons in a chemical bond. A polar bond is formed when the difference in the electronegativity of two combining atoms is between 0.4 and 1.7. The correct direction is
1. C+ ---- O-
2. O+ ---- Cl-
3. O+ ----- F-
4. C+ ----- N-
5. Cl- ----- C+
6. S- ----- H+
7. S+ ----- Cl -
For a theoretical yield of 20 g and actual yield of 12 g, calculate the percent yield for a chemical reaction. Answer in units of %.
Please and Thank you!
Answer:
Percent yield for chemical reaction = 60%
Explanation:
Given:
Theoretical yield of chemical reaction = 20 gram
Actual yield of chemical reaction = 12 gram
Find:
Percent yield for chemical reaction = ?
Computation:
[tex]Percent\ yield \ for\ chemical\ reaction = [\frac{Actual yield}{Theoretical yield} ]100\\\\Percent\ yield \ for\ chemical\ reaction = [\frac{12}{20} ]100\\\\Percent\ yield \ for\ chemical\ reaction = [0.6 ]100\\\\Percent\ yield \ for\ chemical\ reaction = 60[/tex]
Percent yield for chemical reaction = 60%
Briefly describe how a potentiometric pHmeter works. (Hint: Describe how the pH meter
measures the amount of H+ or OH- ions in a sample)
Answer:
A general instrument, which is used to determine the concentration of hydrogen ion within the aqueous solution is known as a pH meter. The meter helps in determining the alkalinity or acidity, which is articulated in the form of pH. It is also called a potentiometric pH meter as it helps in finding the variation in electrical potential between a reference electrode and a pH electrode. This electrical potential variation is associated with the pH of the solution.
The potentiometric pH meter comprises a pair of electrodes and a basic electronic amplifier, some may even comprise a combination electrode and some sort of display that demonstrates pH units. The potentiometric pH meter generally exhibits a reference electrode or a combination electrode, and a glass electrode. The probes or electrodes are administered within a solution whose pH values are needed to be determined.
If 11.5 ml of vinegar sample (d=1g/ml) is titrated with 18.5 ml of standardized Sodium hydroxide
solution. What is the percentage of acetic acid (by mass) in the vinegar
Answer:
4.83% of acetic acid in the vinegar
Explanation:
The neutralization reaction of acetic acid (CH₃COOH) with sodium hydroxide (NaOH) is:
CH₃COOH + NaOH → CH₃COO⁻Na⁺ + H₂O
Assuming the sodium hydroxide solution has a concentration of 0.500M, moles used in the neutralization reaction are:
0.0185L × (0.500mol / L) = 0.00925 moles of NaOH = 0.00925 moles CH₃COOH
Because 1 mole of acid reacts per mole of NaOH
Using molar mass of acetic acid (60g/mol):
0.00925moles CH₃COOH ₓ (60g / mol) = 0.555g of CH₃COOH
As mass of vinegar sample is 11.5g (d = 1g/mL), percentage of acetic acid by mass is:
0.555g CH₃COOH / 11.5g vinegar × 100 = 4.83% of acetic acid in the vinegar
5. Write word equations for the following reactions
a) Burning of Sulphur in oxygen
b) Burning of carbon in excess oxygen
c) Burning of Sulphur in presence of Iron fillings
Answer:
a) Sulphur + Oxygen → Sulphur dioxide
b) Carbon + Oxygen → Carbon dioxide
c) Sulphur + Iron → Iron sulphide
PLEASE HELP ME ASAP DUE TODAY!!!!!!!!!!!!
4. A student performed an experiment to determine the density of a sugar solution and obtainedthe following results: 1.24 g/mL, 1.21 g/mL, 1.23 g/mL.The actual density is 1.50 g/mL. Circle the correct value for the average, X, in the first column and for the accuracy evaluation, RE(%), in the second column.
Answer:
1.23 g/mL
-18.2%
Explanation:
We need to find the average, which is just the sum of the numbers divided by the number of numbers. Here, the sum will be 1.24 + 1.21 + 1.23 = 3.68 g/mL. There are 3 numbers, so divide 3.68 by 3: 3.68 / 3 ≈ 1.2266...
However, we need to round this and take into account significant figures. Each trial gave a number with 3 significant figures, so we round our number off to three: 1.22666... ≈ 1.23. So, circle the first number under the X column.
We now need to find the percent error, which is RE (%). To calculate this, we take the measured value (1.23 in this case) and subtract the exact value (1.50 here) from it, and then divide that by the exact value:
(1.23 - 1.50) / 1.50 ≈ -0.1822...
Again, we need to round to 3 significant figures, which would make it:
-0.1822... ≈ -0.182
Thus, circle the last number under the RE (%) column.
The quantity of antimony in a sample can be determined by an oxidation–reduction titration with an oxidizing agent.
A 5.85 g sample of stibnite, an ore of antimony, is dissolved in hot, concentrated HCl(aq) and passed over a reducing agent so that all the antimony is in the form Sb3+(aq). The Sb3+(aq) is completely oxidized by 26.6 mL of a 0.125 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is:
BrO3-(aq) + Sb3+(aq) ------> Br-(aq) + Sb5+(aq) (unbalanced)
(A) Calculate the amount of antimony in the sample and its percentage in the ore.
Answer:
The percentage is k [tex]= 20.8[/tex]%
Explanation:
From the question we are told that
The mass of the stibnite is [tex]m_s = 5.86 \ g[/tex]
The volume of KBrO3(aq) is [tex]V = 26.6 mL = 26.6 *10^{-3} \ L[/tex]
The concentration of KBrO3(aq) is [tex]C = 0.125 M[/tex]
Now the balanced ionic equation for this reaction is
[tex]BrO_3 ^{-}+ 3Sb^{3+} + 6H^{+} \to Br^{1-} + 3Sb^{5+} + 3H_2O[/tex]
The number of moles of [tex]BrO_3 ^{-}[/tex] is
[tex]n = C *V[/tex]
substituting values
[tex]n = 26.6*10^{-3} * 0.125[/tex]
[tex]n = 0.003325 \ mols[/tex]
from the reaction we see that 1 mole of [tex]BrO_3 ^{-}[/tex] reacts with 3 moles of [tex]Sb^{3+}[/tex]
so 0.003325 moles will react with x moles of [tex]Sb^{3+}[/tex]
Therefore
[tex]x = \frac{0.003325 * 3}{1}[/tex]
[tex]x = 0.009975 \ mols[/tex]
Now the molar mass of [tex]Sb^{3+}[/tex] is a constant with a values of [tex]Z = 121.76 \ g/mol[/tex]
Generally the mass of [tex]Sb^{3+}[/tex] is mathematically represented as
[tex]m = x * Z[/tex]
substituting values
[tex]m = 1.215 \ g[/tex]
The percentage of Sb(antimony) in the overall mass of the stibnite is mathematically evaluated as
k [tex]= \frac{1.215}{5.85 } * 100[/tex]
k [tex]= 20.8[/tex]%
Answer:
Explanation:
Step 1: Data given
Mass of stibnite (Sb2S3) = 5.85 grams
The Sb3+(aq) is completely oxidized by 26.6 mL of a 0.125 M aqueous solution of KBrO3(aq).
Step 2: The balanced equation
BrO3-(aq)+ 3Sb^3+(aq) + 6 H+ → Br-(aq) + 3Sb^5+(aq) + 3H2O (l)
Step 3: Calculate moles KBrO3
Moles KBrO3 = molarity * volume
Moles KBrO3 = 0.125 M *0.0266 L
Moles KBrO3 = 0.003325 moles
Step 4: Calculate moles Bro3-
in 1 mol KBrO3 we have 1 mol K+ and 1 mol BrO3-
In 0.003325 moles KBrO3 we have 0.003325 moles BrO3-
Step 5: Calculate moles Sb
For 1 mol BrO3- we need 3 mol Sb^3+ to produce 1 mol Br- and 3 mol Sb^5+
For 0.029085 moles BrO3- we need 3*0.003325 = 0.009975 moles Sb
Step 6: Calculate mass Sb
Mass Sb = moles Sb * molar mass Sb
Mass Sb = 0.009975 moles * 121.76 g/mol
Mass Sb = 1.21 grams
Step 7: Calculate the percentage of Sb in the ore
% Sb = (mass Sb / total mass) * 100%
% Sb = (1.21 grams / 5.85 grams) * 100 %
% Sb = 20.76 %
20.76 % of the ore is antimony Given the balanced equation representing a redox reaction:
2Al + 3Cu2+ → 2Al3+ + 3Cu
Which statement is true about this reaction?
The given question is incomplete, the complete question is:
Given the balanced equation representing a redox reaction:
2Al + 3Cu2+ --> 2Al3+ + 3Cu
Which statement is true about this reaction?
(1) Each Al loses 2e- and each Cu2+ gains 3e-
(2) Each Al loses 3e- and each Cu2+ gains 2e-
(3) Each Al3+ gains 2e- and each Cu loses 3e-
(4) Each Al3+ gains 3e- and each Cu loses 2e-
Answer:
The correct answer is option 2, that is, in the given reaction each of Al loses three electrons and each of the Cu^2+ ion acquires two electrons.
Explanation:
To understand the concept, let us suppose that when an atom A loses an electron, it turns into an atom A^1+, and becomes A^2+ when it loses two electrons. On the other hand, when an atom A gains an electron, it turns into A^1- and becomes A^2- when it acquires two electrons.
Therefore, in the given case, 2Al + 3Cu^2+ --> 2Al^3+ + 3Cu, each of the Al atom turns into Al^3+ when it loses three electrons, and when each of the Cu^2+ ions acquires two electrons, it turns into Cu (Cu^2+ ^2- = Cu).
amning At Home -Chemisu
Use the specific heat values to answer the
following questions.
Cp (J/g•°C)
4.18
Which of the following has the smallest heat
capacity?
2.03
2.08
0.450
RETRY
Substance
H2O(1)
H2O(s)
H2O(g)
Fe(s)
Al(s)
Cu(s)
Sn(s)
Pb(s)
Au(s)
Hg(1)
0.897
0.385
0.227
0.129
0.129
0.140
Intro
Answer:first one: 3.0g second one: 10g H2O(I)
Explanation:
Answer:
First blank: 3.0 g Pb
Second Blank: 10g [tex]H_{2} O[/tex](l)
Explanation:
Edge '21
What reagent is need to convert 2-methylpropene to (CH3)3COH
Answer:
H2O / H+
Explanation:
In alkene hydration, water, being a very weak acid, with insufficient proton concentration, does not have the capacity to initiate the electrophilic addition reaction. An acid must be added to the medium for the reaction to take place.
The mechanism proceeds with the formation of a carbocation after adding the proton to the double bond. Alkene hydration is Markovnikov, that is, the proton is added to the less substituted carbon of the alkene.
Q6. A student mixes 50.0 mL of 1.00 M Ba(OH)2 with 86.4 mL of 0.494 M H2SO4. Calculate the
mass of solid BaSO4 formed and the pH of the mixed solution.
The mass of solid BaSO4 formed - 7.64 grams. and the pH of the mixed solution - 13.13
We can calculate the mass of the Ba(OH)2 by calculating the moles each solution contains
volume of Ba(OH)2 -n(Ba(OH)2) = 1.00M x 0.05L
= 0.05 moles
The volume of H2SO4 is :0.494M x 0.0864L
= 0.04268 moles
According to balanced reaction 1 mole Ba(OH)2 react with 1 mole H2SO4 molar ratio between Ba(OH)2 to H2SO4 is 1:1 therefore to react with 0.05 mole Ba(OH)2 required H2SO4 = 0.05 mole but H2SO4 therefore H2SO4 is limiting reactant
Ba(OH)2 + H2SO4 ----> BaSO4 + 2 H2O
0.04268<--- 0.04268
The mixed solution is Ba(OH)2 with 0.05 - 0.04268= 0.0172 mole
The concentration of mixed solution is :0.0172:(0.05 + 0.0664)
= 0.1478 M
The pH of mixed solution is:Ba(OH)2 is strong base therefore dissociate completely and 1 mole of Ba(OH)2 form to mole of OH- ion therefore concentration of OH- is double than Ba(OH)2
14 - -log[0.1478]
= 14 - 0.83
= 13.13 pH
the mass of BaSO4 is0.0328 x ( 137 + 32 + 16 x 4 )
= 7.64 grams.
Thus, The mass of solid BaSO4 formed - 7.64 grams. and the pH of the mixed solution - 13.13
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What is the freezing point of a solution of 210.0 g of glycerol, formula C3H8O3, dissolved in 350. g of water? Careful. First get molar mass and use molar mass to determine molality concentration. Then use freeze pt. depression formula to determine the change in freezing pt. Then determine the new freeze point. The freezing point depression constant for water is Kf= -1.86 oCelcius/molal. Report your answer rounded to 1 decimal point and do not include units.
Answer:
- 12.1
Explanation:
M(C3H8O3) = 3*12.0 + 8*1.0 + 3*16.0 = 92 g/mol
210.0 g C3H8O3 * 1 mol C3H8O3/92 g C3H8O3 = 210/92 mol C3H8O3
350.g = 0.350 kg H2O
Molality = mol soluty/ kg solvent = (210/92 mol C3H8O3) /(0.350 kg H2O) = =6. 522 molal
ΔT =i* Kf* m
(T2 - T1) = i* Kf* m
(T2 - 0°C) = 1*(-1.86°C/molal *6.522 molal)
T2= - 12.1°C
The change in freezing point and the new freeze point are -12.1°C and -12.1°C.
Given the following data:
Mass of glycerol = 210.0 gramsMass of water = 350.0 grams to kg = 0.35 kgChemical formula of glycerol = [tex]C_3H_8O_3[/tex]Freezing point depression constant for water, Kf = 0.512 °C/mWe know that the temperature at which water freezes is 0°C.
To determine the change in freezing point and the new freeze point:
First of all, we would determine the molar mass of glycerol:
Molar mass of glycerol ([tex]C_3H_8O_3[/tex]) = [tex]12 \times (1 \times 8)\times (16 \times 3)[/tex]
Molar mass of glycerol ([tex]C_3H_8O_3[/tex]) = [tex]12 \times (8)\times (48)[/tex]
Molar mass of glycerol ([tex]C_3H_8O_3[/tex]) = 92 g/mol.
Next, we would find the number of moles of glycerol that is required:
[tex]Number\;of\;moles = \frac{mass}{molar\;mass}\\\\Number\;of\;moles = \frac{210.0}{92}[/tex]
Number of moles = 2.28 moles
To find the molality concentration:
[tex]Molality = \frac{moles\;of \;solute}{mass\;of \;solvent} \\\\Molality = \frac{2.28}{0.35}[/tex]
Molality = 6.51 molal.
Mathematically, the freezing point elevation of a liquid is given by the formula:
[tex]\Delta T = K_f m[/tex]
Where:
[tex]\Delta T[/tex] is the change in temperature. Kf is the molal freezing point constant. m is the molality of solution.
Substituting the parameters into the formula, we have;
[tex]\Delta T = -1.86 \times 6.51[/tex]
Change in temperature = -12.1°C.
Now, we can determine the new freeze point:
[tex]\Delta T = T_n - T_f\\\\T_n = \Delta T + T_f\\\\T_n = -12.1 + 0[/tex]
New freeze point = -12.1°C.
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How many atoms are in 13.0 g Sr
Answer: 7.505
×
10
22
atoms. (rounded to third place of decimal)
Explanation:
Consider the following half reactions and their standard reduction potential values to answer the following questions.
Cu2+ (aq) + e- Cu+ (aq)E° = 0.15 V
Br2 (l) + 2e- 2Br- (aq) E° = 1.08 V
i) State which reaction occurs at the anode and which at the cathode. [2 marks]
ii) Write the overall cell reaction. [2 marks]
iii) Calculate the value of E°cell. [2 marks]
iv) Calculate the value ΔG° [2 marks]
v) What is the value of Kc at 25°C?
Answer:
Cu2+ (aq) + e- -------->Cu+ (aq)E° = 0.15 V anode
Br2 (l) + 2e- -------------> 2Br- (aq) E° = 1.08 V cathode
Explanation:
The half equation having a more positive reduction potential indicates the reduction half equation while the half equation having the less positive reduction potential indicates the oxidation half equation.
The overall redox reaction equation is;
2Cu^+(aq) + Br2(g) ----> 2Cu^2+(aq) + 2Br^-(aq)
E°cell= E°cathode -E°anode
E°cathode= 1.08 V
E°anode= 0.15 V
E°cell= 1.08V-0.15V
E°cell= 0.93V
From;
∆G=-nFE°cell
n= 2
F=96500C
E°cell= 0.93V
∆G= -(2×96500×0.93)
∆G= - 179.49 J
From ∆G= -RTlnK
∆G= - 179.49 J
R=8.314Jmol-K-1
T= 25° +273= 298K
K= ???
lnK= ∆G/-RT
lnK=-( - 179.49/(8.314×298))
lnK= 0.0724
K= e^0.0724
Kc= 1.075
Consider the following half-reactions and their standard reduction potential values to answer the following questions.
Cu2+ (aq) + e- Cu+ (aq) E° = 0.15 V
Br2 (l) + 2e- 2Br- (aq) E° = 1.08 V
i) State which reaction occurs at the anode and which at the cathode. [2 marks]
ii) Write the overall cell reaction. [2 marks]
iii) Calculate the value of E°cell. [2 marks]
iv) Calculate the value ΔG° [2 marks]
v) What is the value of Kc at 25°C? [2 marks]
Answer:
Br2 (l) + 2e- ---------> 2Br- (aq) E° = 1.08 V cathode
Cu2+ (aq) + e- --------->Cu+ (aq) E° = 0.15 V anode
Explanation:
We have to first state the fact that the reaction having the most positive reduction potential occurs at the cathode in any spontaneous electrochemical cell. The half reaction with the less positive electrode potential usually occurs at the anode.
The overall reaction equation is;
2Cu2+ (aq) + Br2 (l) ----->2Cu+ (aq) + 2Br- (aq)
E°cell= E°cathode - E°anode
E°cathode= 1.08 V
E°anode= 0.15V
E°cell = 1.08-0.15 = 0.93 V
But
∆G°= -nFE°cell
n= 2, F=96500C, E°cell= 0.93V
∆G° = -(2× 96500× 0.93)
∆G= -179490 J
But;
∆G = -RTlnK
R=8.314 JK-1
T= 25+273= 298K
Kc= the unknown
∆G° = -179490 J
Substituting values and making lnK the subject of the formula
lnK= ∆G/-RT
lnK= -( -179490/8.314 × 298)
lnK= 72.45
K= e^72.45
K= 2.91×10^31
Successful rifting can create new oceans and split apart continents. True or False
Answer:
True if by rifting you mean continental rift. If not idk.
please help me. especially with figuring out c
Answer:
A. 4Ni(s) + 3O2(g) —> 2Ni2O3(s)
B. Synthesis reaction.
C. 18.31g of O2
Explanation:
A. The equation for the reaction.
Ni(s) + O2(g) —> Ni2O3(s)
The above equation can be balance as follow:
There are 2 atoms of O on the left side and 3 atoms on the right side. It can be balance by putting 2 in front Ni2O3 and 3 in front of O2 as shown below:
Ni(s) + 3O2(g) —> 2Ni2O3(s)
There are 4 atoms of Ni on the right side and 1 atom on the left it can be balance by putting 4 in front of Ni as shown below:
4Ni(s) + 3O2(g) —> 2Ni2O3(s)
Now the equation is balanced
B. From the equation, we can see that two reactants combined to form a single product. Therefore, the reaction is termed synthesis reaction.
C. We'll begin by calculating the mass of Ni and O2 that reacted from the balanced equation. This is illustrated below:
4Ni(s) + 3O2(g) —> 2Ni2O3(s)
Molar mass of Ni = 59g/mol
Mass of Ni from the balanced equation = 4 x 59 = 236g
Molar mass of O2 = 16x2 = 32g/mol
Mass of O2 from the balanced equation = 3 x 32 = 96g
Summary:
From the balanced equation above,
236g of Ni reacted with 96g of O2.
Now, we can obtain the mass of O2 needed to react with 45g of Ni as follow:
From the balanced equation above,
236g of Ni reacted with 96g of O2.
Therefore, 45g of Ni will react with = (45 x 96)/236 = 18.31g of O2.
Therefore, 18.31g of O2 is needed to react with 45g of Ni