Answer:
13g of CO₂ is the maximum amount that could be produced
Explanation:
The reaction of hexane with oxygen is:
C₆H₁₄ + 19/2O₂ → 6CO₂ + 7H₂O
Where 19/2 moles of oxygen react per mole of C₆H₁₄
To solve this question we need to find theoretical yield finding limiting reactant :
Moles C₆H₁₄:
4.3g C₆H₁₄ * (1mol / 86.18g) = 0.0499 moles
Moles O₂:
25.8g * (1mol / 32g) = 0.806 moles
For a complete reaction of 0.0499 moles of C₆H₁₄ are needed:
0.0499 moles of C₆H₁₄ * (19/2 mol O₂ / 1mol C₆H₁₄) = 0.474 moles of O₂.
As there are 0.806 moles, O₂ is in excess and C₆H₁₄ is limiting reactant
In theoretical yield, the moles of hexane added = 6Moles of CO₂ produced. Moles of CO₂ are:
0.0499 moles C₆H₁₄ * (6mol CO₂ / 1mol C₆H₁₄) = 0.299 moles CO₂
In grams:
0.299 moles CO₂ * (44.01g / mol) = 13g CO₂
13g of CO₂ is the maximum amount that could be produced
Write the equilibrium expression of each chemical equation.
2H2S(g) 2H2(g) + S2(g)
Answer:
[H2]2[S2][H2S]2Kc=[H2]2[S2][H2S]2
Explanation:
2H2S(g)⇋2H2(g)+S2(g)2H2S(g)⇋2H2(g)+S2(g)
The equilibrium constant expression in terms of concentrations is:
Kc=[H2]2[S2][H2S]2Kc=[H2]2[S2][H2S]2.
The equilibrium expression for the given reaction can be written in terms of equilibrium constant which is the ratio of power of molar concentration of the product to the product of power of molar concentration of the reactants.
What is equilibrium?Equilibrium is a state for a reversible reaction where, the rate of forward reaction is equal to the rate of backward reaction. The rate of a reaction is the rate of decrease in the concentration of reactants or the rate of increase in the concentration of the products.
The given reaction at equilibrium state is written as:
[tex]\rm 2H_{2}S (g)\leftrightharpoons 2H_{2} (g)+ S_{2}(g)[/tex]
The equilibrium constant Kb is ratio of power of molar concentration of the product to the product of power of molar concentration of the reactants.
[tex]Kb = \rm \frac{[H_{2}S]^{2}}{[H_{2}]^{2} [S_{2}]}[/tex]
The rate of the reaction will be r = Kb [H₂]² [S₂].
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For each of the following substituents, indicate whether it withdraws electrons inductively, donates electrons by hyperconjugation, withdraws electrons by resonance, or donates electrons by resonance. (Effects should be compared with that of a hydrogen; remember that many substituents can be characterized in more than one way).
A. Br
1. Withdraws electrons inductively.
2. Donates electrons by hyperconjugation.
3. Withdraws electrons by resonance.
4. Donates electrons by resonance.
B. CH2CH3
1. Withdraws electrons inductively.
2. Donates electrons by hyperconjugation.
3. Withdraws electrons by resonance.
4. Donates electrons by resonance.
C. NHCH31. Withdraws electrons inductively.2. Donates electrons by hyperconjugation.3. Withdraws electrons by resonance.4. Donates electrons by resonance.D. OCH31. Withdraws electrons inductively.2. Donates electrons by hyperconjugation.3. Withdraws electrons by resonance.4. Donates electrons by resonance.E. +N(CH3)31. Withdraws electrons inductively.2. Donates electrons by hyperconjugation.3. Withdraws electrons by resonance.4. Donates electrons by resonance.
Answer:
Br- Withdraws electrons inductively
Donates electrons by resonance
CH2CH3 - Donates electrons by hyperconjugation
NHCH3- Withdraws electrons inductively
Donates electrons by resonance
OCH3 - Withdraws electrons inductively
Donates electrons by resonance
+N(CH3)3 - Withdraws electrons inductively
Explanation:
A chemical moiety may withdraw or donate electrons by resonance or inductive effect.
Halogens are electronegative elements hence they withdraw electrons by inductive effect. However, they also contain lone pairs so the can donate electrons by resonance.
Alkyl groups donate electrons by hyperconjugation involving hydrogen atoms.
-NHCH3 and contain species that have lone pair of electrons which can be donated by resonance. Also, the nitrogen and oxygen atoms are very electron withdrawing making the carbon atom to have a -I inductive effect.
+N(CH3)3 have no lone pair and is strongly electron withdrawing by inductive effects.
If given 12 moles of Cl2, how many moles of HCl can form?
CH4+ 4Cl2——CCI4 + 4HCI
A. 4 moles HCl
B. 12 moles HCI
C. 1 mole HCI
D. 3 moles HCI
Answer:
B
Explanation:
4moles of Cl2 produces 4moles of HCl therefore 12moles of Cl2 will produce 12moles of HCl
Volume is the independent or dependent variable
Answer:
Independent
Explanation:
Independent Variable is the volume of the object. Dependent Variable is the mass of the object. So it
SOMEONE PLEASE HELP MEEE!!!!
What is the mass in grams of 6.25 mol of copper (II) nitrate ,Cu(NO3)2?
Answer:
lol
Explanation:
lol
can energy transfer even if the objects are in the same temprature
Answer:
Yes
Explanation:
When the temperature of a reaction is raised from 278 K to 300 K, the reaction rate increases by a factor of 17. Determine the activation energy, Ea, of the reaction. R = 8.314 J/mol.K
Answer: The activation energy, Ea, of the reaction is 89195 Joules
Explanation:
The effect of temperature on rate constant is given by Arrhenius equation:
[tex]ln \frac{k_{2}}{k_{1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}][/tex]
where
[tex]k_2[/tex] = rate constant at temperature [tex]T_2[/tex]
[tex]k_1[/tex] = rate constant at temperature [tex]T_1[/tex]
[tex]E_a[/tex]= activation energy
R= gas constant
[tex]T[/tex] = temperature
[tex]ln \frac{17k_1}{k_{1}} = \frac{-E_{a}}{8.314}[\frac{1}{300} - \frac{1}{278}][/tex]
[tex]2.83=\frac{-E_{a}}{8.314}[\frac{1}{300} - \frac{1}{278}][/tex]
[tex]E_a=89195J[/tex]
The activation energy, Ea, of the reaction is 89195 Joules
A geochemist in the field takes a 25.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X. He notes the temperature of the pool, 26.° C, and caps the sample carefully. Back in the lab, the geochemist first dilutes the sample with distilled water to 350. mL. Then he filters it and evaporates all the water under vacuum. Crystals of X are left behind. The researcher washes, dries and weighs the crystals. They weigh 3.00 g.1) Using only the information above, can you calculate the solubility of X in water at 26 degrees Celsius.? Yes or No2) If yes than calculate the solubility of X. Round your answer to 3 significant digits
Answer:
Yes we can calculate the solubility
The solubility of the mineral is 120 g/L
Explanation:
We have the information that the sample taken originally from the rock pool is 25.0 mL
So,
25.0 mL of the sample contained 3.000 g of the sample
10000mL of the sample now contains 1000 * 3.000/25 = 120 g
This means that 120 g of the sample dissolves in 1000mL or 1L of solution.
Therefore, the solubility of the mineral is 120 g/L
In order to use a pipet, place a at the top of the pipet. Use this object to fill the pipet such that the of the liquid is even with the volume line. Release the liquid, touching the tip of the pipet to the side of the container if necessary to release the last drop the pipet tip.
Answer:
bulb or pump, meniscus, outside
Explanation:
In order to use a pipet, place a BULB OR PUMP at the top of the pipet. Use this object to fill the pipet such that the MENISCUS of the liquid is even with the volume line. Release the liquid, touching the tip of the pipet to the side of the container if necessary to release the last drop OUTSIDE the pipet tip.
Find odd one out.
H2SO4 , HCL, NaOH, HNO3
Answer:NaOH
Explanation:all are acids except NaOH
Answer:
it is NaOH
Explanation:
Because NaOH is base, while the others are acid
How many mL of 0.774 M HBr are needed to dissolve 6.73 g of CaCO3?
2HBr(aq) + CaCO3(s) -> CaBr2(aq) + H2O(1) + CO2(g)
____mL
Answer:
173.9 mL of HBr
Explanation:
We'll begin by calculating the number of mole in 6.73 g of CaCO₃. This can be obtained as follow:
Mass of CaCO₃ = 6.73 g
Molar mass of CaCO₃ = 40 + 12 + (16×3)
= 40 + 12 + 48
= 100 g/mol
Mole of CaCO₃ =?
Mole = mass / Molar mass
Mole of CaCO₃ = 6.73 / 100
Mole of CaCO₃ = 0.0673 mole
Next, we shall determine the number of mole of HBr that will react with 6.73 g (i.e 0.0673 mole) of CaCO₃. This can be obtained as follow:
2HBr + CaCO₃ —> CaBr₂ + H₂O + CO₂
From the balanced equation above,
2 moles of HBr reacted with 1 mole of CaCO₃.
Therefore, Xmol of HBr will react with 0.0673 mole of CaCO₃ i.e
Xmol of HBr = 2 × 0.0673
Xmol of HBr = 0.1346 mole
Thus, 0.1346 mole of HBr reacted.
Next, we shall determine the volume of HBr needed for the reaction. This can be obtained as follow:
Mole of HBr = 0.1346 mole
Molarity of HBr = 0.774 M
Volume =?
Molarity = mole / Volume
0.774 = 0.1346 / volume
Cross multiply
0.774 × volume = 0.1346
Divide both side by 0.774
Volume = 0.1346 / 0.774
Volume = 0.1739 L
Finally, we shall convert 0.1739 L to mL. This can be obtained as follow:
1 L = 1000 mL
Therefore,
0.1739 L = 0.1739 L × 1000 mL / 1 L
0.1739 L = 173.9 mL
Thus, 173.9 mL of HBr is needed for the reaction.
The Lewis dot notation for two atoms is shown.
Mg is written with two dots on its right. O is written on the right of Mg. There are six dots around O. Two arrows point from the dots near Mg to O.
What is represented by this notation?
Mg gains two protons from O.
Mg donates two protons to O.
Mg gains two electrons from O.
Mg donates two electrons to O.
Answer:
Mg donates two electrons to O
Explanation:
Lewis dot notation uses dots and crosses to represent valence electrons on atoms.
Magnesium is a metal and would donate or lose electrons during bonding.
Oxygen is a non metal and would gain electrons during bonding.
The correct option is;
Mg donates two electrons to O
Which chemical equation models the law of conservation of mass?
A. 3KOH + H3PO4 - K3PO4 + 3H20
B. 2KOH + H3PO4 - K3PO4 + 2H20
C. 3KOH + 2H3PO4 - 2K3PO4 + 3H20
D. KOH + H3PO4 - K3PO4.
The Law of conservation of mass states the creation and destruction of mass to be impossible. The reaction between 3 moles of Potassium hydroxide and Phosphoric acid is conserved.
What is conservational mass?The law of mass of conservation states that the mass of the reactants and the products of the reaction cannot be added or removed.
In the first reaction, the mass of the atoms in the reaction is balanced, while in the second reaction the mass of potassium and hydrogen is not balanced.
In the third and fourth reactions, the mass of the elements is not balanced and not conserved hence not following the law.
Therefore, option A. [tex]\rm 3KOH + H_{3}PO_{4} \rightarrow K_{3}PO_{4} + 3H_{2}O[/tex] follows the law of conservation of mass.
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If one DNA strand reads CCGTAATGCAT, what will be the sequence of the complementary
strand?
Si una hebra de ADN lee CCGTAATGCAT, ¿cuál será la secuencia de la hebra
complementaria?
Write your answer here:
Escriba su respuesta aquí:
a) CCGTAATGCAT
b) GGCATTACGTA
c) CCGATTAGCTA
d) GGCTAATCGAT
If one DNA strand reads CCGTAATGCAT, the sequence of the complementary strand will be - b) GGCATTACGTA
Complementary base pairingThe nucleic acid sequence of bases can form a double-stranded structure by matching base pairs.Adenine matches with thymine (uracil in RNA) and vice versaGuanine matches with cytosine and vice versaFor example, the complementary sequence to C-A-T-G (where each letter stands for one of the bases in DNA) is G-T-A-C.So, DNA strand - C-C-G-T-A-A-T-G-C-A-T
Complementary DNA strand - G-G-C-A-T-T-A-C-G-T-A
Thus, If one DNA strand reads CCGTAATGCAT, the sequence of the complementary strand will be - b) GGCATTACGTA
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Predict the missing product of this equation
1 MgF2 + 1 Li2CO3 -> 1 ______ +2LiF
Answer:
MgCO₃
Explanation:
From the question given above, we obtained:
MgF₂ + Li₂CO₃ —> __ + 2LiF
The missing part of the equation can be obtained by writing the ionic equation for the reaction between MgF₂ and Li₂CO₃. This is illustrated below:
MgF₂ (aq) —> Mg²⁺ + 2F¯
Li₂CO₃ (aq) —> 2Li⁺ + CO₃²¯
MgF₂ + Li₂CO₃ —>
Mg²⁺ + 2F¯ + 2Li⁺ + CO₃²¯ —> Mg²⁺CO₃²¯ + 2Li⁺F¯
MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF
Now, we share compare the above equation with the one given in the question above to obtain the missing part. This is illustrated below:
MgF₂ + Li₂CO₃ —> __ + 2LiF
MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF
Therefore, the missing part of the equation is MgCO₃
A gas produced as a by-product from the carbonization of coal has the following composition, mole %: carbon dioxide 4, carbon monoxide 15, hydrogen 50, methane 12, ethane 2, ethylene 4, benzene 2, balance nitrogen. Using the data given in Appendix C (available online at booksite .Elsevier/Towler), calculate the gross and net calorific values of the gas. Give your answer in MJ/m3, at standard temperature and pressure.
Answer:
6059.63 kcal/kg.
Explanation:
CH4 consist of Hydrogen and carbon. Therefore, Hydrogen in CH4 = 12 × 4 = 48 kg, carbon in CH4 = 12 × 12 = 144kg.
For ethane, the amount of hydrogen present = 2 × 6 = 12kg and that of carbon in ethane = 2 × 24 = 48kg.
The weight of carbon in CO = 15 × 12 = 18kg and the weight of Hydrogen in CO = 15 × 16 = 240kg.
For hydrogen, its weight in H2 = 50 × 2 = 100kg.
For CO2, carbon has = 4 × 12 = 48 kg and oxygen has = 4 × 32 = 128kg.
For C6H6, carbon has 2 × 72 = 144kg and hydrogen has 2 × 6 = 12kg.
For N2, the amount of nitrogen= 11 × 28 = 308 kg.
For CH2= CH2, carbon has 4 × 24 = 96kg and hydrogen = 4 × 4 = 16kg.
The gross calofiric value = 1/100 [ 8080 C + 34500 + ( H - O/8) + 22405].
Where the total weight = 128 + 180+ 48 + 240 + 100 + 144 + 48 + 12 + 48 + 16 + 96 + 308 + 12 + 144 = 1524 kg.
The percentage by weight of carbon = total weight of carbon/total weight × 100.
The total Weight of carbon= 48 + 180 + 144 + 48 +144 + 96 = 660kg.
The percentage weight of carbon = 660/1524 × 100 = 43.3 %.
The percentage weight of hydrogen = total weight of hydrogen/total weight × 100.
The total weight of hydrogen = 100 + 12 + 48 + 16 +12 = 188.
The percentage weight of Hydrogen = 188/ 1524 × 100 = 12.33%.
Percentage weight of oxygen = total weight of oxygen/total weight × 100.
Percentage weight of oxygen = ( 128 + 240) / 1524 × 100 = 24.15%.
The gross calorific value = 1/100 [ 8080 × 43.3 ,+ 34500 ( 13.33 - 24.15/8) ] = 6711.02 kcal/kg.
Net calorific value = 6711.02 - 0.09 × 12.3 × 587 = 6059.63kcal/kg.
Nayla grew skin in a lab by adding cells to a synthetic material. The skin functioned normally for 12 days. Then Nayla separated some
of the skin cells into cell membranes, cytoplasm, and vacuoles to studythem. WhichofthefollowingwerealiveduringNayla’s experiment?
a. The skin and the cytoplasm
b. The skin and the skin cells
c. The cell membranes and the skin cells
d. The cell membranes and the cytoplasm
Answer:B) The Skin and The Skin cells
Explanation:
Hope this helped
All of the following are characteristics of matter except
A.matter can disappear and reappear
B.matter has mass
C.matter occupies space
D.all things are composed of matter
how much heat is required to convert 29g of ice at -4.0c to water vapor at 105
Answer:
45 degrees at 29 so 105 is lower than 45 and 105 wants to be 32 so 45 can be higher. so then the water vapor will be 35 because 105 is too much of water gallon
Carbon dioxide is an example of a molecular compound.
O True
O False
Aqueous hydrochloric acid HCl will react with solid sodium hydroxide NaOH to produce aqueous sodium chloride NaCl and liquid water H2O. Suppose 0.365 g of hydrochloric acid is mixed with 0.18 g of sodium hydroxide. Calculate the maximum mass of sodium chloride that could be produced by the chemical reaction. Round your answer to 2 significant digits.
Answer:
0.26g of NaCl is the maximum mass that could be produced
Explanation:
Based on the reaction:
HCl + NaOH → NaCl + H₂O
Where 1 mol of HCl reacts per mol of NaOH to produce 1 mol of NaCl
To solve this question we need to find limiting reactant. The moles of limiting reactant = Moles of NaCl produced:
Moles HCl -Molar mass: 36.46g/mol-:
0.365g HCl * (1mol / 36.46g) = 0.010 moles HCl
Moles NaOH -Molar mass: 40g/mol-:
0.18g NaOH * (1mol / 40g) = 0.0045 moles NaOH
As the reaction is 1:1 and moles NaOH < moles HCl, limiting reactant is NaOH and maximum moles produced of NaCl are 0.0045 moles.
The mass of NaCl is:
Mass NaCl -Molar mass: 58.44g/mol-:
0.0045 moles * (58.44g/mol) =
0.26g of NaCl is the maximum mass that could be producedWhat kind of thermal energy transfer is illustrated in the diagram?
Answer:
I think radiaction
Explanation:
For 100.0 mL of a solution that is 0.040M CH3COOH and 0.010 M CH3COO, what would be the pH after adding 10.0 mL 50.0 mM HCl?
Answer:
The pH of the buffer is 3.90
Explanation:
The mixture of a weak acid CH3COOH and its conjugate base CH3COO produce a buffer that follows the equation:
pH = pKa + log [A-] / [HA]
Where pH is the pH of the buffer, pKa is the pKa of acetic acid (4.75), and [A-] could be taken as the moles of the conjugate base and [HA] the moles of thw weak acid.
To solve this question we need to find the moles of the CH3COOH and CH3COO- after the reaction with HCl:
CH3COO- + HCl → CH3COOH + Cl-
The moles of CH3COO- are its initial moles - the moles of HCl added
And moles of CH3COOH are its initial moles + moles HCl added
Moles CH3COO-:
Initial moles = 0.100L * (0.010mol / L) = 0.00100moles
Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles
Moles CH3COO- = 0.000500 moles
Moles CH3COOH:
Initial moles = 0.100L * (0.040mol / L) = 0.00400moles
Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles
Moles CH3COO- = 0.003500 moles
pH is:
pH = 4.75 + log [0.000500] / [0.00350]
pH = 3.90
The pH of the buffer is 3.90
The pH of the buffer after addition of 10.0 mL of 50.0mM HCl is 3.90
What is the Henderson-Hasselbach equation?The Henderson-Hasselbach equation is given as:
pH = pKa + log [A-] / [HA]where
pKa of acetic acid (4.75), [A-] is moles of the conjugate base[HA] the moles of thw weak acid.How to determine the moles of CH3COO- and CH3COOHThe formula for calculating number of moles is:
Moles = concentration × volumeThe equation of the reaction is given below:
CH3COO- + HCl → CH3COOH + Cl-
moles of CH3COO- = initial moles - moles of HCl added
moles of CH3COOH = initial moles + moles HCl added
Moles CH3COO-
Molarity = 0.010 M
volume = 100 mL = 0.100 L
Initial moles = 0.100 L * 0.010 M = 0.001 moles
Moles HCl = 0.010L * 0.050 M= 0.0005 moles
Moles CH3COO- = 0.001 - 0.0005 moles
Moles of CH3COO- = 0.000500 moles
Moles CH3COOH:
Molarity = 0.040 M
volume = 100 mL = 0.100 L
Initial moles = 0.100L * (0.040mol / L) = 0.00400moles
Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles
Moles CH3COO- = 0.004 - 0.0005 moles
Moles CH3COO- = 0.003500 moles
Substituting the calculated values:
pH = 4.75 + log [0.000500] / [0.00350]
pH = 3.90
Therefore, the pH of the buffer after addition of the 10.0 mL 50.0mmHg HCL is 3.90
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Which of the following represent chemical processes? Which represent physical processes?
A). Calcium chloride dihydrate (CaCl2 * 2H2O) slowly heated in a crucible to become calcium chloride (anhydrous).
B). A hydrocarbon such as propane (C3H8) undergoes combustion to power a grill
C). A rock climber’s rope becomes frayed and turns the color of the rocks
D). A dog urinates on an air conditioner coil and the coil becomes corroded
Answer:
A). Calcium chloride dihydrate (CaCl2 * 2H2O) slowly heated in a crucible to become calcium chloride (anhydrous).
Dehyration is a physical process
B). A hydrocarbon such as propane (C3H8) undergoes combustion to power a grill
Combustion is a chemical process.
C). A rock climber’s rope becomes frayed and turns the color of the rocks
This is physical process
D). A dog urinates on an air conditioner coil and the coil becomes corroded
Corrsion is a chemical process.
Explanation:
The chemical process is combustion and corrosion i.e. B and D. The physical change has been the heating of calcium chloride, and fraying of rock i.e. A and C.
A chemical process has been given as the reaction in which the composition of the sample changes. It has been an irreversible process.
A physical process has been described as a change in the physical properties of substances. It is a reversible process.
The following reaction can be given as:
A. The reaction between calcium chloride and water has been mediated by heating. It is a reversible process and does not change the chemical composition. It is a physical process.
B. The combustion results in the change in the chemical constituents of propane. It is a chemical process.
C. There has been no change in the chemical composition of materials. It has been a physical process.
D. The corrosion has resulted from the chemical change in the iron. It has been a chemical process
Thus, the chemical process is combustion and corrosion. The physical change has been the heating of calcium chloride, and fraying of rock.
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WILL GIVE BRAINLIEST how do you make a potion of fire resistance
A. Blaze powder, water bottle and nether wart
B. water bottle, nether wart, and magma cream
C. wither skull, water bottle and netherrack
D. water bottle, golden carrot and god apple
Answer:
B is the answer
Please give brainliest
Explanation:
I play Minecraft and Dream made it once
Limiting factors for an Antelope include:
O A. Lions
O B. Water
O C. Grass
O D. All of the above
Answer:
D all the above
Explanation:
Answer:
Answer:- Lions..
Explanation:
Hope you have got your answer..
Happy Learning!..
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How many moles are there in 6.33x10^23 molecules of NH3?
Glycerol (C3H8O3), also called glycerine, is widely used in the food and pharmaceutical industries. Glycerol is polar and dissolves readily in water and polar organic solvents like ethanol. Calculate the mole fraction of the solvent in a solution that contains 2.51 g glycerol dissolved in 21.10 mL ethanol (CH3CH2OH; density
Answer: The mole fraction of the solvent in a solution that contains 2.51 g glycerol dissolved in 21.10 mL ethanol is 0.93
Explanation:
Given : Volume of ethanol (solvent) = 21.10 ml
density of ethanol (solvent)= 0.789 g/ml
Mass of ethanol (solvent) = [tex]0.789g/ml\times 21.10ml=16.6g[/tex]
Mass of glycerol (solute) = 2.51 g
Mole fraction of a component is the ratio of moles of that component to the total moles present.
moles of ethanol =[tex]\frac{\text {given mass}}{\text {molar mass}}=\frac{16.6g}{46g/mol}=0.36mol[/tex]
moles of glycerol =[tex]\frac{\text {given mass}}{\text {molar mass}}=\frac{2.51g}{92g/mol}=0.027mol[/tex]
mole fraction of ethanol (solvent) = [tex]\frac{\text {moles of ethanol}}{\text {moles of ethanol + moles of glycerol}}=\frac{0.36}{0.36+0.027}=0.93[/tex]
The mole fraction of the solvent in a solution that contains 2.51 g glycerol dissolved in 21.10 mL ethanol is 0.93
What do you call the new material that are created in chemical
The standard heat of combustion is shown in the following chemical equation CgH 20 (g) + 140 2(g) 9CO 2(g) + 10H 2 o (1) delta
The given question is incomplete. The complete question is:
The standard heat of combustion is shown in the following chemical equation [tex]C_9H_{20}(g)+14O_2(g)\rightarrow 9CO_2(g)+10H_2O[/tex][tex]\Delta H_{rxn}=-6125.21kJ/mol[/tex]. If 130 g of nonane combusts , how much heat is released?
Answer: 6211.21 kJ
Explanation:
Heat of combustion is the amount of heat released on complete combustion of 1 mole of substance.
Given :
Amount of heat released on combustion of 1 mole of nonane = 6125.21 kJ
According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP, weighs equal to the molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
1 mole of nonane [tex](C_9H_{20})[/tex] weighs = 128.2 g
Thus we can say:
128.2 g of nonane on combustion releases = 6125.21 kJ
Thus 130 g of [tex]C_4H_{10}[/tex] on combustion releases =[tex]\frac{6125.21}{128.2}\times 130=6211.21kJ[/tex]
Thus the heat of combustion of 130 g of nonane is 6211.21 kJ