Answer:
[tex]2NH_{3 (g)} + 3O_{2 (g)} +2CH_{4 (g)} --> 2HCN_{(aq)} + 6H_{2}O_{(l)}[/tex]
Explanation:
Sulfur and oxygen form both sulfur dioxide and sulfur trioxide. When samples of these are decomposed, the sulfur dioxide produces 3.52 g oxygen and 3.53 g sulfur, while the sulfur trioxide produces 9.00 g oxygen and 6.00 g sulfur.
Calculate the mass of oxygen per gram of sulfur for each sample and show that these results are consistent with the law of multiple proportions
Answer:
See explanation
Explanation:
For sulphur dioxide;
3.53 g of sulphur combines with 3.52 g of oxygen
1.00 g of sulphur combines with 1.00 × 3.52/ 3.53 = 0.997 g
For sulphur trioxide
6.00 of sulphur combines with 9.00 g of oxygen
1.00 g of sulphur combines with 1.00 × 9.00/6.00 =1.5 g
Ratio of mass of oxygen;
1 : 1.5
The law of multiple proportions states that, if two elements A and B combine to form more than one chemical compounds, then the various masses of element A which combines separately with a fixed mass of element B are in simple multiple ratio.
We can see that the mass of oxygen which combines separately with sulphur in sulphur dioxide and sulphur trioxide are in simple ratio of 1:1.5.
High-density polyethylene may be flourinated by inducing the random substitution of Flourine atoms for hydrogen. (a) Determine the concentration of Flourine (in wt%) that must be added if this substitution occurs for 12% of all the original hydrogen atoms.
Answer:
Concentration of Flourine = 24.756%
Explanation:
Given that :
High-density polyethylene may be flourinated by inducing the random substitution of Flourine atoms for hydrogen.
the objective is to determine he concentration of Flourine (in wt%) that must be added if this substitution occurs for 12% of all the original hydrogen atoms.
At standard conditions , the atomic weight of the these compounds are as follows:
Carbon = 12.01 g/mol
Chlorine = 35.45 g/mol
Fluorine = 19.00 g/mol
Hydrogen = 1.008 g/mol
Oxygen = 16.00 g/mol
The chemical formula for polyethylene = (CH₂ - CH₂)ₙ
Therefore, for two carbons, there will be 4 hydrogens;
i.e
(CH₂ - CH₂)₂
( C₂H₄ - C₂H₄ )
Suppose the number of original hydrogen = 4moles
number of moles of Flourine F = 12% of 4
= 0.12 × 4
= 0.48 mol
∴ the number of remaining moles of Hydrogen is:
= 4 - 0.48
= 3.52 moles
number of moles of Carbon = 2 moles
∴ the mass of flourine F = number of moles of F × molar mass of F
= 0.48 × 19
= 9.12
The total mass of the compound now is = (0.48 × 19 ) + (3.52 × 1) + (2× 12)
= 9.12 + 3.52 + 24
= 36.64
Concentration of Flourine = (mass of flourine/total mass) × 100
Concentration of Flourine = (9.12/36.84 ) × 100
Concentration of Flourine = 0.24756 × 100
Concentration of Flourine = 24.756%
Which is not a compound? water silicon dioxide oxygen gas carbon dioxide gas
Answer: Oxygen gas
Explanation: Gradpoint
Which of the following conditions characterizes a system in a state of chemical equilibrium?
a) Reactants are being consumed at the same rate they are being produced.
b) Concentrations of reactants and products are equal.
c) Product concentrations are greater than reactant concentrations.
d) Reactant molecules no longer react with each other.
e) Rate of forward reaction has dropped to zero.
Answer:
a) Reactants are being consumed at the same rate they are being produced.
Explanation:
In a chemical reaction, chemical equilibrium is a state in which there is no change in concentration of the reactant and product. This means that as the reactants are consumed, they are also produced. Same thing with the products; immediately they are formed, they are consumed.
The option that accurately depicts a reaction at equilibrium is;
a) Reactants are being consumed at the same rate they are being produced.
The density of mercury is 13.6 g/mL what is the density in lbs/L ( 1 lb hint =0.454 kg )
Answer:
[tex]30.0\frac{lb}{L}[/tex]
Explanation:
Hello,
In this case, since 454 g are equivalent to 1 pound and 1000 millilitres are equivalent to 1 liter, the required density is computed below by applying the corresponding conversion factor:
[tex]=13.6\frac{g}{mL} *\frac{1lb}{454g} *\frac{1000mL}{1L} \\\\=30.0\frac{lb}{L}[/tex]
Regards.
In a first-order reaction, the half-life is 139 minutes. What is the rate constant?
a. 8.31 x 10-5 s-1
b. 5770 s-1
c. 0.299 s-1
d. 4.99 x10-3s-1
e. 1.20 x 10-4s-1
Answer:
a. 8.31 x 10-5 s-1
Explanation:
The general first-order reaction is:
ln[A] = ln[A]₀ -kt
Where [A] is acutal concentration of reactant, initial concentration is [A]₀, k is rate constant and t is time pass
And the equation of the half-life, t 1/2, is:
[tex]t_{1/2} = \frac{ln 2}{K}[/tex]
Has half-life is 139min:
139min * (60s / 1min) = 8340s
[tex]t_{1/2} = \frac{ln 2}{8340s}[/tex]
Half-life is 8.31x10⁻⁵ s⁻¹
a. 8.31 x 10-5 s-1Chromium(III) oxide reacts with hydrogen sulfide (H_2S) gas to form chromium(III) sulfide and water:______.
Cr_2 O_3 (s) + H_2S (g) ? Cr_2S_3 (s) + 3H_2O (l)
To produce 346 g of Cr_2S_3,
(a) how many moles of Cr_2 O_3 are required?
......... mol
(b) how many grams of Cr_2 O_3 are required?
......... g
Answer:
a. 1.728 moles.
b. 262.7g of Cr₂O₃ are required
Explanation:
Based on the reaction:
Cr₂O₃(s) + 3H₂S(g) → Cr₂S₃(s) + 3H₂O(l)
The important thing in the reaction is that 1 mole of Cr₂O₃ produce 1 mole of Cr₂S₃
a. To produce 346g of Cr₂S₃ we must know how many moles of Cr₂S₃ must be produced, and, as 1 mole of Cr₂O₃ produce 1 mole of Cr₂S₃ we can know moles of Cr₂O₃ that are required.
Moles of 346g Cr₂S₃ (Molar mass: 200.19g/mol):
346g Cr₂S₃ * (1mol / 200.19g) = 1.728 moles of Cr₂S₃
Based on the reaction, moles of Cr₂O₃ that are required are
1.728 moles of Cr₂O₃b. Again, to conver the 1.728 moles of Cr₂O₃ to grams we must use molar mass of Cr₂O₃ (151.99g/mol):
1.728 moles Cr₂O₃ * (151.99g / mol) =
262.7g of Cr₂O₃ are required72. A piece of unknown metal with mass 30 g is heated to 1100 C and dropped into 100.0 g of water at 200 C. The final temperature of the system is 25 degree Celsius. What is the specific heat of the metal
Answer:
THE SPECIFIC HEAT OF THE METAL IS 28.69 J/g C
Explanation:
Mass of unknown metal = 30 g
Mass of water = 100 g
Initial Temperature of metal = 110 C
Initial temperature of water = 200 C
Final temperature of the system = 25 C
Specific heat of water = 4200 J/ g C
Specific heat of metal = unknown
Heat lost = Heat gained\
M C Q = M C Q
30 * C * ( 25 - 110) = 100 * 4200 * (25 -200)
C = 100 * 4.18 * -175 / 30 * - 85
(the minus sign cancels out and we have):
C = 100 * 4.18 * 175 / 30 * 85
C = 73150 / 2550
C = 28.69 J/ g C
The specific heat of the unknown metal is hence 28.69 J/ g C
At what atom of the pyridine ring will oxidation and reduction reactions take place?a. 2.b. 4.c. 3.d. 1.
Answer:
1
Explanation:
Pyridine is a six membered hetrocyclic aromatic compound. Oxidation of the pyridine ring occurs at position-1.
Just like benzene, pyridine is stable towards mild oxidizing agents, however, pyridine is oxidized by peracetic acid to yield pyridine N-oxide.
Consider a galvanic cell consisting of the following two redox couplesAlB+ 3e Al , Mga + 2e Mg, (ag. E® = 1.676 V ) E° = 2.356 VA) Write the equation for the half-reaction occurring at the cathode.B) Write the equation for the half-reaction occurring at th anodeC) Write the overall equation for the cell reation.D) what is the standard cell potential EoCell, for the Cell?
Answer:
0.68 V
Explanation:
For anode;
3Mg(s) ---->3Mg^2+(aq) + 6e
For cathode;
2Al^3+(aq) + 6e -----> 2Al(s)
Overall balanced reaction equation;
3Mg(s) + 2Al^3+(aq) ----> 3Mg^2+(aq) + 2Al(s)
Since
E°anode = -2.356 V
E°cathode = -1.676 V
E°cell=-1.676 -(-2.356)
E°cell= 0.68 V
What is the concentration of OH− and pOH in a 0.00072 M solution of Ba(OH)2 at 25 ∘C? Assume complete dissociation.
Given :
0.00072 M solution of [tex]Ba(OH)_2[/tex] at [tex]25^oC[/tex] .
To Find :
The concentration of [tex]OH^-[/tex]and pOH .
Solution :
1 mole of [tex]Ba(OH)_2[/tex] gives 2 moles of [tex]OH^-[/tex] ions .
So , 0.00072 M mole of [tex]Ba(OH)_2[/tex] gives :
[tex][OH^-]=2 \times 0.00072\ M[/tex]
[tex][OH^-]=0.00144\ M[/tex]
[tex][OH^-]=1.44\times 10^{-3}\ M[/tex]
Now , pOH is given by :
[tex]pOH=-log[OH^-]\\\\pOH=-log[1.44\times 10^{-3}]\\\\pOH=2.84[/tex]
Hence , this is the required solution .
Classify each as a physical or chemical change.
Explain each answer.
a) When limestone is heated carbon dioxide and
calcium oxide result.
b) Digesting a candy bar.
c) Copper can be drawn into thin wires.
d) Ripening of fruits.
e) Ice floats on water
Answer:
a) When limestone is heated in a kiln, the calcium carbonate breaks down into calcium oxide and carbon dioxide. This type of reaction is called thermal decomposition. The process is called calcining. ... Slaked lime is used to reduce the acidity of the soil, which helps increase its fertility.
b) Mechanical digestion occurs in the mouth, stomach, and small intestine. Food is chemically changed indigestion when new, smaller substances are formed. These chemical changes are examples of chemical digestion. Chemical digestion begins in the mouth when enzymes in saliva begin to break down carbohydrates.
c) Copper is a ductile metal. This means that it can easily be shaped into pipes and drawn into wires. Copper pipes are lightweight because they can have thin walls. They don't corrode and they can be bent to fit around corners.
d) Ripening of fruits is a chemical change. During ripening the pigment present in the skin of the fruit changes and this can not be reversed back. it is a chemical change as once a fruit ripens it cannot turn raw again.
e) Ice is a unique substance because it is solid-state — ice — is less dense than its liquid state. Because of this property, ice floats in water. Physical properties are characteristics of a substance. They do not change. Physical properties include colour, smell, freezing/melting point, and density.
If 5.0 kJ of energy is added to a 15.5-g sample of water at 10.°C, the water is Group of answer choices
Answer:
The water is completely vaporized at this stage.
Explanation:
The complete question is
If 5.0 kJ of energy is added to a 15.5-g sample of water at 10.°C, the water is
-boiling
-completely vaporized
-frozen solid
-decomposed
-still a liquid
Energy added = 50 kJ = 50000 J
mass of water = 15.5 g = 0.0155 kg
temperature of water = 10 °C
We know that the energy posses by a mass of water at a given temperature is given as
H = mcT
where H is the energy possessed by the mass of water
m is the mass of the water
c is the specific heat capacity of water = 4200 J/ kg- °C
T is the temperature of the water
substituting values, the energy of this amount of water is
H = 0.0155 x 4200 x 10 = 651 J
If 50 kJ is added to the water, the energy increases to
50000 J + 651 J = 50651 J
Temperature of this water at this stage will be gotten from
H = mcT
we solve for the new temperature
50651 = 0.0155 x 4200 x T
50651 = 65.1 x T
T = 50651/65.1 = 778.05 °C
This temperature is well over 100 °C, which is the vaporization temperature of water, but less than 3000 °C for its molecules to decompose.
A student uses paper towels to clean up a small chemical spill in the lab.
Where should the dirty paper towels be placed?
A in the lab sink
B in the wastebasket
C in the chemical waste container
D in the hazardous waste container
Answer:
C. in the chemical wastebasket
If a urine sample has a mass of 244 g and a volume of 242 mL, what is its density in g/mL?
Answer:
244/242 alright
Explanation:
yesss
Answer:
244/242 or 1.00826446 g/mL
Explanation:
Density can be calculated using the following formula.
d=m/v
where m is the mass and v is the volume.
The mass is 244 grams and the volume is 242 milliliters.
m= 244 g
v= 242 mL
Substitute the values into the formula.
d= 244 g / 242 mL
Divide 244 g by 242 mL.
d= 244/242 g/mL
d= 1.00826446 g/mL
The density is 244/242 or 1.00826446 grams per milliliter.
Will a deflated football sink in water?
Answer:
no
Explanation:
light objects float
SpongeBob & Friends and the Scientific Method
Scenario A. Krusty Krabs Breath Mints
Mr. Krabs created a secret ingredient for a breath mint that he thinks will "cure the bad breath people get from
cating crabby patties at the Krusty Krab. He asked 100 customers with a history of bad breath to try his new
breath mint. He had fifty customers (Group A) cat a breath mint after they finished eating a crabby patty. The
other fifty (Group B) also received a breath mint after they finished the sandwich, however, it was just a regular
breath mint and did not have the secret ingredient. Both groups were told that they were getting the breath mint
that would cure their bad breath. Two hours after cating the crabby patties, thirty customers in Group A and ten
customers in Group B reported having better breath than they normally had after cating crabby patties.
1. Which people are in the control group?
2. What is the independent variable?
3. What is the dependent variable?
4. What should Mr. Krabs' conclusion be?
5. Was this a quantitative or a qualitative measurement?
6. Why do you think 10 people in group B reported fresher breath?
Answer:
1. Which people are in the control group? The people who received the mint without the secret ingredient
(Group B) would be the control group.
2. What is the independent variable? Secret ingredient in the breath mint
3. What is the dependent variable? Amount of breath odor (or bad breath)
4. What should Mr. Krabs’ conclusion be? The breath mint with the secret ingredient appears to reduce the
amount of breath odor more than half the time, but it is not 100% effective.
5. Why do you think 10 people in group B reported fresher breath? This may be due to the placebo effect.
3.25 kcal is the same amount of energy as
Answer:
3.25 kcal has the same amount of energy as 13600 J
interpretacao gerais das praticas
sobre densidade de amostra solido e liquido
Answer:
O sólido tem densidade mais alta em comparação ao líquido.
Explicação:
A densidade da amostra sólida é maior do que a densidade do líquido porque há pouco espaço entre as partículas do sólido. A densidade tem relação inversa com o volume de uma substância, se uma substância ocupa mais espaço então sua densidade é menor, enquanto se a substância ocupa menos espaço então tem maior densidade. As substâncias sólidas ocupam menos espaço em comparação com as substâncias líquidas, então podemos dizer que a densidade do sólido é maior do que as substâncias líquidas.
Consider the following numbered processes: 1. A → 2B 2. B → C + D 3. E → 2D ΔH for the process A → 2C + E is
Answer:
ΔH = ΔH₁ + ΔH₂ - ΔH₃
Explanation:
Given that:
1. A → 2B
2. B → C + D
3. E → 2D
Assuming from the corresponding ΔH for process 1, 2 and 3 are ΔH₁, ΔH₂, ΔH₃ respectively.
To estimate the ΔH for the process A → 2C + E
We multiply 2 with equation 2 where (B → C + D)
2B → 2C + 2D ⇒ 2ΔH₂
Also, let's switch equation (3), such that we have,
2D → E -ΔH₃
The summation of all the equation result into :
A → 2C + E
where; ΔH = ΔH₁ + ΔH₂ - ΔH₃
Based on Hess law of constant heat summation, the ΔH for the process A → 2C + E is ΔH1 + ΔH2 - ΔH3
What is the enthalpy of the series steps of reaction steps?According to Hess' law of constant heat summation, the heat of any reaction ΔH∘f for a specific reaction is equal to the sum of the heats of reaction for any set of reactions which in sum are equivalent to the overall reaction.
Thus, ΔH for the process A → 2C + E is calculated as follows:
Reaction step 1: A → 2B ; ΔH1Reaction step 2: B → C + D ; ΔH2Reactipn step 3: E → 2D ; ΔH3The A → 2C + E can be obtained from the summation of the processes above:
A ---> 2B
2B ---> 2C + 2D
2D ---> E by reversing reaction step 3
Thus, ΔH = ΔH1 + ΔH2 - ΔH3
Therefore, the ΔH for the process A → 2C + E is ΔH1 + ΔH2 - ΔH3
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Fatty acids that have multiple double bonds are ?
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _
( 15 letter word )
. A sample of crude oil has a density of 0.87 g/mL. What volume (in liters) does a 3.6 kg sample of this oil occupy
Answer:
The volume is 4.13793 L
Explanation:
Density is a quantity that expresses the relationship between the mass and the volume of a body, so it is defined as the quotient between the mass and the volume of a body:
[tex]density=\frac{mass}{volume}[/tex]
Density is a characteristic property of every body or substance.
The most commonly used units of density are [tex]\frac{kg}{m^{3} }[/tex] or [tex]\frac{g}{cm^{3} }[/tex] for solids, and [tex]\frac{kg}{L}[/tex] or [tex]\frac{g}{mL}[/tex] for liquids and gases.
In this case, you know:
density= 0.87 [tex]\frac{g}{mL}[/tex]mass= 3.6 kg= 3,600 g (being 1 kg=1,000 g)volume= ?Replacing:
[tex]0.87\frac{g}{mL} =\frac{3,600 g}{volume}[/tex]
Solving:
[tex]volume =\frac{3,600 g}{0.87\frac{g}{mL}}[/tex]
volume= 4,137.93 mL
Being 1,000 mL=1 L, then volume= 4,137.93 mL= 4.13793 L
The volume is 4.13793 L
which variable is directly proportional to frequency?
velocity
wavelength
position
energy
Answer:
its letter D Energy I think
PLEASE HELP I WILL GIVE BRAINLIEST 40+ POINTS. PLEASE
This is the periodic table(image below)
Using information from the periodic table, which diagram shows a
correct model for an atom neon(also image below)
Answer:
A.
Explanation:
Neon 19 is shown on image A.
Diagram A shows the correct model for an atom neon.
What are the properties of periodic table ?The arrangement of elements by their periodic properties and physical and chemical characteristic make a periodic table.
The chemical and physical properties of the group of the element belongs to Alkali Metals, Less dense, One loosely bound valence electron, Highly reactive, Low ionization energy, Low electronegativity
The lanthanides and actinides are also transition metals, these are softer and nonmetallic properties, all of these elements tend to have a shiny, metallic appearance, radioisotopes of other elements, actinides are radioactive.
Learn more about periodic table, here:
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Which of the following is the most effective buffer system for a pH value of 4.45? a) H3PO4/H2PO4 (Kal for H3PO4 is 6.9 times 10-3) b) H_2CO_3/HCO_3^- (K_a1 for H_2CO_3 is 4.8 times 10^-7) c) HCO_3^-/CO_3^2- (K_a2 for H2CO_3 is 4.8 times 10^-11) d) H_2S/HS^- (K_a1 for H_2S is 8.9 times 10^-8) e) HC_2O_4^-/C_2O_4^2- (K_a2 for H_2C_2O_4 is 5.1 times 10^-5)
Answer:
Option e.
Acid oxalate / oxalate
Explanation:
To determine the most efecttive buffer, we must look the pKas of the system.
If pKa value is similiar to buffer's pH, the variation of pH by the addition of a small amount of strong acid or base is minimal. It would be a good buffer.
a. H₃PO₄ / H₂PO₄⁻ Ka 1 = 6.9×10⁻³
pKa = - log Ka → - log 6.9×10⁻³ = 2.16
b. H₂CO₃ / HCO₃⁻ Ka1 = 4.8×10⁻⁷
pKa = - log Ka → - log 4.8×10⁻⁷ = 6.32
c. HCO₃⁻ / CO₃⁻² Ka2 = 4.8×10⁻¹¹
pKa = - log Ka → - log 4.8×10⁻¹¹ = 10.32
d. H₂S / HS⁻ Ka1 = 8.9×10⁻⁸
pKa = - log Ka → - log 8.9×10⁻⁸ = 7.05
e. HC₂O₄⁻ / C₂O₄⁻² Ka2 = 5.1×10⁻⁵
pKa = - log Ka → - log 5.1×10⁻⁵ = 4.29
pKa which is so close to the ph's value is e, acid oxalate / oxalate
Which sentence describes a chemical property?
A. Cars that are missing paint can rust over time.
B. A pot of water gets hotter on a burner.
O C. Paper can be cut into smaller pieces.
O D. A pitcher can hold several glasses of water.
Answer:
A. Cars that are missing paint can rust over time shows a chemical property of rusting.
4Fe + 3O² + 6H²O → 4Fe(OH)³.
(Iron) (Air) (Water) (Rust)
Note: The powers in equation should be written below .
Hope it helps.
In the laboratory you dissolve 23.8 g of manganese(II) nitrate in a volumetric flask and add water to a total volume of 125 . mL. What is the molarity of the solution
Answer:
The correct answer is 1.06 M
Step-by-step explanation:
We have to calculate the molarity (M), which is:
M= moles solute/ 1 L solution.
The chemical formulae of manganese (II) nitrate is Mn(NO₃)₂. So, we first calculate its molecular weight (Mw) as follows:
Mw(Mn(NO₃)₂)= molar mass Mn + (2 x molar mass N) + (6 x molar mass N)= 55 g/mol + (2 x 14 g/mol) + (6 x 16 g/mol) = 179 g/mol
Then, with Mw we calculate the number of moles there is in the given mass of Mn(NO₃)₂:
moles Mn(NO₃)₂= mass/Mw= 23.8 g/(179 g/mol)= 0.133 mol
Now, we need the final volume in liters, so we convert the volume from mL to L:
125 mL x 1 L/1000 mL = 0.125 L
Finally, we divide the moles of Mn(NO₃)₂ into the volume in L, to obtain the molarity in mol/L:
M= 0.133 moles/0.125 L = 1.06 mol/L= 1.06 M
Draw the neutral organic product expected under these reaction conditions.
Answer:
The neutral organic product is Imine.
Explanation:
From the image attached below, we will see an organic compound the question intended us to work out.
The proposed reaction is between cyclohexane and a stick structure of butylamine to yield imine as the product. The stepwise process and the mechanism carried out to get the product of this reaction is also shown in the image attached below.
What concentration of monosodium phosphate would be required to make a solution of pH 7.4 with 0.2 M disodium phosphate ( pKa
Answer:
The concentration of monosodium phosphate is 0.1262M
Explanation:
The buffer of H₂PO₄⁻ / HPO₄²⁻ (Monobasic phosphate and dibasic phosphate has a pKa of 7.2
To determine the pH you must use Henderson-Hasselbalch equation:
pH = pKa + log [A⁻] / [HA]
Where [A⁻] is molarity of the conjugate base of the weak acid, [HA].
For H₂PO₄⁻ / HPO₄⁻ buffer:
pH = 7.2 + log [HPO₄⁻² ] / [H₂PO₄⁻]
As molarity of the dibasic phosphate is 0.2M and you want a pH of 7.4:
7.4 = 7.2 + log [0.2] / [H₂PO₄⁻]
0.2 = log [0.2] / [H₂PO₄⁻]
1.58489 = [0.2] / [H₂PO₄⁻]
[H₂PO₄⁻] = 0.1262M
The concentration of monosodium phosphate is 0.1262M
I know how to solve it with D=M/V and M1V1 however the answer isn’t correct. Help me please
Answer:
23.28 g of O2.
Explanation:
We'll begin by calculating the mass of hexane. This can obtain as follow:
Volume of hexane = 10 mL
Density of hexane = 0.66 g/mL
Mass of hexane =?
Density = mass /volume
0.66 = mass of hexane /10
Cross multiply
Mass of hexane = 0.66 x 10
Mass of hexane = 6.6 g
Next, we shall write the balanced equation for the reaction. This is given below:
2C6H14 + 19O2 —> 12CO2 + 14H2O
Next, we shall determine the masses of C6H14 and O2 that reacted from the balanced equation. This can be obtained as follow:
Molar mass of C6H14 = (12.01x6) + (1.008 x 14)
= 72.06 + 14.112
= 86.172 g/mol
Mass of C6H14 from the balanced equation = 2 x 86.172 = 172.344 g
Molar mass of O2 = 16x2 = 32 g/mol
Mass of O2 from the balanced equation = 19 x 32 = 608 g
From the balanced equation above,
172.344 g of C6H14 reacted with 608 g of O2.
Finally, we shall determine the mass of O2 needed to react with 10 mL (i.e 6.6 g) of hexane, C6H14. This can be obtained as follow:
From the balanced equation above,
172.344 g of C6H14 reacted with 608 g of O2.
Therefore, 6.6 g of C6H14 will react with = (6.6 x 608)/172.344 = 23.28 g of O2.
Therefore, 23.28 g of O2 is needed for the reaction.