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:
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.
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]
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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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.
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
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%
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
*Text me at 561-400-5105 for private tutoring if interested: I can do homework, labs, and other assignments :)
n the space below, briefly explain why the average times for the two CO2 molecules you calculated should have been similar (or at least not significantly different). If a CO2 molecule starts out surrounded by other CO2 molecules, does this influence how quickly it will reach the other side of the leaf? You might find it helpful to reset the simulation and to re-examine the initial configurations of molecules. Collisions influence how molecules move, but do molecules only collide with other molecules of the same substance?
Answer:
See explaination
Explanation:
No, the neurons don't use diffusion to transfer messages rather they employ the use of electrical impulses which are very fast for example, if we drop something accidentally on our foot we immediately move back to prevent ourselves from injury this quick reaction is possible because the signals are transferred in the form of electrical impulses.If diffusion was there to transfer the message then it would have taken very long time and our foot would have been bleeding too as diffusion process is relatively much slower.
This is due to collision of CO2 molecule with other molecules of CO2 and vander vals interaction between them which results in slow down of CO2 molecule and the CO2 molecules also collide with walls of leaf as well which also decreases the speed of CO2 molecule resulting it in taking more time.
Iron (Fe) undergoes an allotropic transformation at 912°C: upon heating from a BCC (α phase) to an FCC (γ phase). Accompanying this transformation is a change in the atomic radius of Fe—from RBCC = 0.12584 nm to RFCC = 0.12894 nm. The highest density planes in BCC structure is (110) and for FCC structure is (111). i. Compare the planar density of the two. (110) in BCC and (111) in FCC iron. EA = − 1.436 r ER = 5.8 × 10−6 r 9 ii. Do you think a (111) plane in FCC structure is more amenable to dislocation motion or (110) plane in BCC structure? What is an implication of that on the mechanical properties of materials.
Answer:
The description including its given problem is outlined in the following section on the clarification.
Explanation:
The given values are:
RBCC = 0.12584 nm
RFCC = 0.12894 nm
The unit cell edge length (ABCC) as well as the atomic radius (RBcc) respectively connected as measures for BCC (α-phase) structure:
√3 ABCC = 4RBCC
⇒ ABCC = [tex]\frac{4RBCC}{\sqrt{3} }[/tex]
⇒ = [tex]\frac{4\times 0.12584}{\sqrt{3}}[/tex]
⇒ = [tex]0.29062 \ nm[/tex]
Likewise AFCC as well as RFCC are interconnected by
√2AFCC = 4RFCC
⇒ AFCC = [tex]\frac{4RFCC}{\sqrt{2}}[/tex]
⇒ = [tex]\frac{4\times 0.12894}{\sqrt{2} }[/tex]
⇒ = [tex]0.36470 \ nm[/tex]
Now,
The Change in Percent Volume,
= [tex]\frac{V \ final-V \ initial}{V \ initial}\times 100 \ percent[/tex]
= [tex]\frac{(VFCC)unit \ cell-(VBCC)unit \ cell}{(VBCC)unit \ cell}\times 100 \ percent[/tex]
= [tex]\frac{(aFCC)^3-(aBCC)^3}{(aBCC)^3}\times 100 \ percent[/tex]
= [tex]\frac{(0.36470)^3-(0.29062)^3}{(0.29062)^3}\times 100 \ percent[/tex]
= [tex]97.62 \ percent (approximately)[/tex]
Note: percent = %
How many atoms are in 13.0 g Sr
Answer: 7.505
×
10
22
atoms. (rounded to third place of decimal)
Explanation:
nswer the following questions relating to HCl, CH3Cl, and CH3Br.HCl(g), can be prepared by the reaction of concentrated H2SO4(ag), with NaCl(s), asa.represented by the following equation.H2SO4(ag) + 2 NaCl(s) → 2 HCl(g) Na2SO4(ag)A student claims that the reaction is a redox reaction. Is the student correct?i.Justify your answer.Calculate the mass, in grams, of NaCl(s), needed to react with excess H2SO4(ag)ii.to produce 3.0 grams of HCl(g). Assume that the reaction goes to completion
Answer:
It is an example of double displacement reaction.
4.8 g of NaCl is needed to react.
Explanation:
Balanced reaction: [tex]H_{2}SO_{4}(aq.)+2NaCl(s)\rightarrow 2HCl(g)+Na_{2}SO_{4}(aq.)[/tex]
Here, oxidation states of H, S, O, Na and Cl do not change during reaction. Hence it is not a redox reaction.
In this reaction, cations and anions of the reactants interchange their partners during reaction. Hence, it is an example of double displacement reaction.
As [tex]H_{2}SO_{4}(aq.)[/tex] remain in excess amount therefore NaCl (s) is the limiting reagent. Hence production of HCl entirely depends on amount of NaCl used.
Molar mass of HCl = 36.46 g/mol
So, 3.0 g of HCl = [tex]\frac{3.0}{36.46}[/tex] mol of HCl = 0.082 mol of HCl
According to balanced equation-
2 moles of HCl are produced from 2 moles of NaCl
So, 0.082 moles of HCl are produced from 0.082 moles of NaCl
Molar mass of NaCl = 58.44 g/mol
So, mass of 0.082 moles of NaCl = [tex](0.082\times 58.44)[/tex] g = 4.8 g
Hence 4.8 g of NaCl is needed to react.
mention the raw material use in respiration and products formed during respiration.
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)
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).
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:
.
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
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 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 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].
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 -
Find the pH of a solution whose hydronium concentration is 1.0 x 10-11
Answer:
11
Explanation:
pH=-log(H+)
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.
Sketch the simple Lewis dot structures and the predicted VSEPR
shapes for each of the following species. Be sure to clearly indicate
lone pair electrons, multiple bonds (double or triple), and any three-
dimensionality (using dashes and wedges). For any ions, don’t forget
to account for the charge when calculating valence electrons! Only
expand the octet of the central atom when absolutely necessary.
A. CO2
B. CS2
C. SeCL
D. SO2
E. SO32-
F. CO32-
G. N3-
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
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 antimonyWatching 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
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.
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
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
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 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.