4) In water, Vanillin, C8H8O3, has a solubility of 0.070 moles of vanillin per liter of solution at 25C. What will be produced if 5.00 g of vanillin are added to 1 L of water at 25 C
Answer:
An unsaturated solution.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to firstly realize we need to calculate the grams of vanillin in 0.070 moles by using its molar mass of 152.15 g/mol:
[tex]m=0.070mol*\frac{152.15 g}{1mol} =10.65g[/tex]
Thus, since the solubility is 10.65 g per 1 L of solution, we can notice 5.00 g will complete dissolve and produce an unsaturated solution.
Best regards!
When CH3NO2 burns in excess oxygen, it forms carbon dioxide, nitrogen dioxide and water. How many moles of oxygen are required to burn 17.10 mole(s) of CH3NO2
Explanation:
The given reaction is the combustion of CH3NO2.
The balanced chemical equation of the reaction is:
[tex]4CH_3NO_2+ 7O_2 ->4 CO_2+4NO_2+6H_2O[/tex]
So, from the balanced chemical equation, it is clear that:
4 moles of CH3NO2 --- 7 moles of oxygen gas is required.
then,
for 17.10 moles of CH3NO2 the following number of moles of oxygen is required.
[tex]The number of moles of O_2 required=17.10 mol. x \frac{7 mol}{4 mol} \\=29.925 mol[/tex]
Answer is :
29.9 mol of oxygen gas is required.
The decomposition of ethyl amine, C2H5NH2, occurs according to the reaction: C2H5NH2(g)⟶C2H4(g)+NH3(g) At 85∘C, the rate constant for the reaction is 2.5 x 10-1 s-1. What is the half-life (in sec) of this reaction?
Answer:
2.772 seconds
Explanation:
Given that;
t1/2 = 0.693/k
Where;
t1/2 = half life of the reaction
k= rate constant
Note that decomposition is a first order reaction since the rate of reaction depends on the concentration of one reactant
t1/2 = 0.693/2.5 x 10-1 s-1
t1/2= 2.772 seconds
Please explain how to do it as well!
Write a complete, balanced equation for the following reactions:
a) The combustion of C₆H₁₂O (teachers note: You figure out products).
b) Aqueous ferric iron (III) sulfate plus barium hydroxide (teachers note: You figure out the products).
Answer:
a) C₆H₁₂O + 8.5 O₂ ⇒ 6 CO₂ + 6 H₂O
b) Fe₂(SO₄)₃ + 3 Ba(OH)₂ ⇒ 3 BaSO₄ + 2 Fe(OH)₃
Explanation:
a) A combustion is a reaction of a compound with oxygen to produce carbon dioxide and water. The corresponding equation is:
C₆H₁₂O + O₂ ⇒ CO₂ + H₂O
We will start balancing C atoms by multiplying CO₂ by 6 and H atoms by multiplying H₂O by 6.
C₆H₁₂O + O₂ ⇒ 6 CO₂ + 6 H₂O
Then, we get the balanced equation by multiplying O₂ by 8.5.
C₆H₁₂O + 8.5 O₂ ⇒ 6 CO₂ + 6 H₂O
b) This is a double displacement reaction of the general structure:
Salt 1 + Base 1 = Salt 2 + Base 2
The corresponding equation is:
Fe₂(SO₄)₃ + Ba(OH)₂ ⇒ BaSO₄ + Fe(OH)₃
First, we will balance Fe atoms by multiplying Fe(OH)₃ by 2 and S atoms by multiplying BaSO₄ by 3.
Fe₂(SO₄)₃ + Ba(OH)₂ ⇒ 3 BaSO₄ + 2 Fe(OH)₃
Then, we will get the balanced equation by multiplying Ba(OH)₂ by 3.
Fe₂(SO₄)₃ + 3 Ba(OH)₂ ⇒ 3 BaSO₄ + 2 Fe(OH)₃
Answer:
a.[tex]\bold{C_{6}H_{12}O_{6}\rightarrow 6CO_{2}+6H_{2}O}[/tex]
△H=−72 kcal
The energy required for production of 1.6 g of glucose is [molecular mass of glucose is 180 gm]
b.
[tex]\bold{Fe_{2}(SO_{4})_{3}+3Ba(OH)_{2}\rightarrow 3BaSO_{4}+2Fe(OH)_{3}}[/tex]
The iron(III) ions and chloride ions remain aqueous and are spectator ions in a reaction that produces solid barium sulfate.
Water put into a freezer compartment in the same refrigerator goes into a state of less molecular disorder when it freezes. Is this an exception to the entropy principle
Answer:
No it is not an exception to this principle
Explanation:
Work was carried out by this compressor to reduce the entropy of ice. What this means is that the ice gave out heat which is as a result of the work that the compressor was putting in. there are violations of this principle
the entropy principle has that the entropy of the universe is always going to be more than 0 (system + surrounding). in this question, the that of the system is negative while that of the surrounding is positive. As the refrigerator was cooling the water, the air outside was getting heated. Outside this refrigerator, the gain in entropy is more than the entropy that was lost in the water.
the entropy of the universe once again is more than 0.
What is the pH of a solution prepared by mixing 100.00 mL of 0.020 M Ca(OH)2 with 50.00 mL of 0.100 M NaOH? Assume that the volumes are additive?a. 12.78.b. 13.25.c. 12.67.d. 12.95.
Answer:
The pH of the solution is 12.78.
Explanation:
The pOH (or potential OH) is a measure of the basicity or alkalinity of a solution. The pOH is defined as the negative logarithm of the activity of the hydroxide ions. That is, the concentration of OH- ions:
pOH= - log [OH-]
On the other side, Molarity or Molar Concentration is the number of moles of solute that are dissolved in a certain volume. Molarity is calculated as:
[tex]Molarity= \frac{number of moles}{volume}[/tex]
Molarity is expressed in units: [tex]\frac{moles}{liter}[/tex]
In this case, the solution is prepared by mixing 100 ml (equal to 0.1 L, where 1000 mL = 1 L) of Ca(OH)₂ 0.020 M with 50 ml (equal to 0.05 L) of 0.100 M NaOH. Then, Ca(OH)₂ and NaOH are strong bases, so they dissociate completely. In the case of the first hydroxide, for each mole of Ca(OH)₂,
form two moles of OH-. In the case of sodium hydroxide, for each mole of hydroxide, one mole of OH- is formed. So, taking into account the definition of molarity, the number of moles of OH- that each hydroxide contributes to the solution is calculated as:
From Ca(OH)₂: 0.1 L* 0.02 M*2 = 0.004 moles
From NaOH: 0.05 L* 0.1 M= 0.005 moles
So, the amount of total moles of OH- is the sum that each hydroxide contributes to the solution: 0.004 moles + 0.005 moles= 0.009 moles
On the other hand, volumes are additive. Then: 0.1 L + 0.05 L= 0.15 L
Replacing in the definition of molarity the number of moles and the volume:
[tex][OH-]=\frac{0.009 moles}{0.15 L}[/tex]
Solving:
[OH-]= 0.06 [tex]\frac{moles}{liter}[/tex]
Replacing in the definition of pOH:
pOH= - log 0.06
pOH= 1.22
The following relationship can be established between pH and pOH:
pH + pOH= 14
Being pOH= 1.22 and replacing:
pH + 1.22= 14
pH= 14 - 1.22
pH= 12.78
The pH of the solution is 12.78.
Write the structure of methanamine
Answer: CH3NH2
Explanation:
A sample of pure tin metal is dissolved in nitric acid to produce 15.00 mL of solution containing Sn2+. When this tin solution is titrated, a total of 42.1 mL of 0.145 mol/L KMnO4 is required to reach the equivalence point. a. What is the concentration of the Sn2+ solution?b. Find the concentration of the Sn2+(aq) in mol/L: (give your answer to 3 decimal places)
Answer:
1.00 M
Explanation:
Sn^2+ reacts with KMNO4 as follows;
5Sn^2+(aq) + 2MnO4^-(aq) + 16H^+(aq) ----> 5Sn^4+(aq) + 2Mn^+(aq) + 8H2O(l)
The number of moles of MnO4^- reacted = 42.1/1000 L × 0.145 mol/L
= 0.0061 moles
If 5 moles of Sn^2+ reacts with 2 moles of MnO4^-
x moles of Sn^2+ reacts with 0.0061 moles of MnO4^-
x= 5 × 0.0061/2
x= 0.015 moles
Since the volume of the Sn^2+ solution is 15.00mL or 0.015 L
number of moles = concentration × volume
Concentration = number of moles/volume
Concentration= 0.015 moles/0.015 L
Concentration = 1 M
How many moles in 3.30g of iron
The answer below is correct but to give you the process, here it is:
Molar mass of iron, Fe = 55.85 g/mol
3.30g/(55.85 g/mol) = 0.0591 mol
42 Organic compound may have names ending in -ane, -ene, -ol or -oic acid. How many of these endings indicate the compounds contain double bonds in their molecules? * (1 Point)
Answer: Organic compounds ending with the name (-ene) indicate that the compounds contain double bonds in their molecules.
Explanation:
Organic compounds are those molecules that contains carbon atoms (as their main element), hydrogen and oxygen which are usually present. The presence of numerous organic compounds is due to the following properties of carbon:
--> the exceptional ability of carbon atoms to catenate, that is, to combine with one another to form straight chains, branched chains or ring compounds containing many carbon atoms.
--> The ease with which carbon combines with hydrogen, oxygen, Nitrogen and halogens
--> The ability of carbon atoms to form single, DOUBLE or triple bonds.
The organic compound that has the name ending with -ene are known as the alkenes. The members of the alkene series are formed from the alkanes by the removal of two hydrogen atoms and the introduction of a DOUBLE BOND in the carbon chain. They are named after the corresponding alkanes by changing the -ane ending to -ene.
Note: the systematic name of a compound is formed from the root hydrocarbon by adding a suffix and prefixes to denote the substitution of the hydrogen atoms.
Precipitation of an ionic compound will occur upon mixing of desired reagents if the initial ion product is:_______
A) greater than the Ksp
B) equal to the pksp
C) equal to the Ksp
D) less than the Ksp
Answer:
A) greater than the Ksp
Explanation:
Given a solid ionic compound AB, it dissociates in water into its ions, as follows:
AB(s) → A⁺(aq) + B⁻(aq)
At equilibrium, the product of the concentrations of the ions is constant, and it is called Ksp:
AB(s) ⇄ A⁺(aq) + B⁻(aq)
Ksp = [A⁺][B⁻] ⇒ (concentrations at equilibrium)
Upon mixing the reagents for the formation of AB, the compound will precipitate if the initial ion product (Q) is greater than the Ksp. If Q is equal to Ksp, the ions are at equilibrium with the solid compound AB, and if is it less than the Ksp, the ions are soluble and no solid AB is formed yet.
Q = [A⁺][B⁻] ⇒ (initial concentrations)
Q = Ksp ⇒ saturated solution (at equilibrium)
Q< Ksp ⇒ unsaturated solution (ions are soluble)
Q> Ksp ⇒ precipitation of solid compound.
Therefore, the correct option is A) greater than the Ksp
The cation of the salt is sodium ion, and the anion is aurothiosulfate ion. Based on the chemical formula of the salt, what must the charge be of the aurothiosulfate ion [Au(S2O3)2 n- ]
Answer:
3-
Explanation:
Sodium aurothiosulfate is a salt with the formula Na₃Au(S₂O₃)₂. The cation of the salt is sodium ion, and the anion is aurothiosulfate ion. We can determine the charge of the aurothiosulfate ion, considering that the sum of the positive and negative charges must be equal to the charge of the compound, which is zero.
3 × Na⁺ + 1 × Au(S₂O₃)₂ⁿ⁻ = 0
3 × +1 + 1 × Au(S₂O₃)₂ⁿ⁻ = 0
Au(S₂O₃)₂ⁿ⁻ = 3-
A chemical reaction in a bomb calorimeter evolves 3.86 kJ of energy in the form of heat. If the temperature of the bomb calorimeter increases by 4.17 K, what is the heat capacity of the calorimeter?
Answer:
925.66 J/K
Explanation:
Applying,
Q = CΔt............. Equation 1
Where Q = amount of heat, C = heat capacity of the calorimeter, Δt = rise in temperature.
make C the subject of the equation
C = Q/Δt.............. Equation 2
From the question,
Given: Q = 3.86 kJ = 3860 J, Δt = 4.17K
Substitute into equation 2
C = 3860/4.17
C = 925.66 J/K
Compute the equilibrium constant for the spontaneous reaction between Cd2 (aq) and Zn(s).
Answer:
Kc = [Zn²⁺] / [Cd²⁺]
Explanation:
Let's consider the spontaneous redox reaction between Cd²⁺ and Zn.
Cd²⁺(aq) + Zn(s) ⇄ Cd(s) + Zn²⁺(aq)
The equilibrium constant, Kc, is the ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients. It only includes gases and aqueous species.
Kc = [Zn²⁺] / [Cd²⁺]
A compound, C20H28O, produces a 1H NMR spectrum with 11 distinct signals. The steps made by the integral trace measure 52, 17, 17, 26, 17, 25, 26, 9, 9, 35, and 8 mm. Complete the following table.
Integral # Products
52 mm
17 mm
17 mm
26 mm
17 mm
25 mm
26 mm
35 mm
8 mm
Solution :
The smallest integer value represents the smaller number of protons.
In this case, in the given values, the smallest numbers are 8 mm and 9 mm, so both contains 1H each. Then next highest value is 17 mm, which contains 1H more. Thus 17 mm contains 2H each. Then the next highest is 25 mm and 26 mm which contains 3M each and so on.
Thus the tables is :
Integral Protons
52 mm [tex]6[/tex]
17 mm [tex]2[/tex]
17 mm 2
26 mm 3
17 mm 2
25 mm 3
26 mm 3
9 mm 1
9 mm 1
35 mm 4
8 mm 1
The homework question reads:
"A sample of gas in a cylinder of volume 3.42 L at 298 K
and 2.57 atm expands to 7.39 L by two different pathways.
Path A is an isothermal, reversible expansion. Path B has two
steps. In the fi rst step, the gas is cooled at constant volume to
1.19 atm. In the second step, the gas is heated and allowed to
expand against a constant external pressure of 1.19 atm until
the final volume is 7.39 L. Calculate the work for each path.
Answer:
Explanation:
this guy on brainly already did it:
Alleei
Virtuoso
4.8K answers
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Answer : The work done for path A and path B is -685.3 J and -478.1 J respectively.
Explanation :
To calculate the work done for path A :
First we have to calculate the moles of the gas.
where,
= initial pressure of gas = 2.57 atm
= initial volume of gas = 3.42 L
n = moles of gas = ?
R = gas constant = 0.0821 atm.L/mol.K
T = temperature of gas = 298 K
Now put all the given values in the above formula, we get:
According to the question, this is the case of isothermal reversible expansion of gas.
As per first law of thermodynamic,
where,
= internal energy
q = heat
w = work done
As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.
So, at constant temperature the internal energy is equal to zero.
The expression used for work done will be,
where,
w = work done on the system = ?
n = number of moles of gas = 0.359 mole
R = gas constant = 8.314 J/mole K
T = temperature of gas = 298 K
= initial volume of gas = 3.42 L
= final volume of gas = 7.39 L
Now put all the given values in the above formula, we get :
Thus, the work done of path A is, -685.3 J
To calculate the work done for path B :
The formula used for isothermally irreversible expansion is :
where,
w = work done
= external pressure = 1.19 atm
= initial volume of gas = 3.42 L
= final volume of gas = 7.39 L
Now put all the given values in the above formula, we get :
Thus, the work done of path B is, -478.1 J
Na2CO3 + CaCl2•2H2O -> CaCO3 + 2NaCl + 2H2O
Calculate how many moles of CaCl2•2H2O are present in 1.50 g of CaCl2•2H2O and then calculate how many moles of pure CaCl2 are present in 1.50 g of CaCl2•2H2O. Then determine how many grams of Na2CO3 are necessary to reach stoichiometric quantities.
For CaCl2 I got 0.0135 mol but I have seen some put 0.0102 mol. Which is it?
For the initial mol of Na2CO3 I got 0.0102 mol but again I’m not sure if I’m right.
For the grams of Na2CO3 I got 1.08 g
Can someone help me figure out if I have this correct?
Answer:
See explanation
Explanation:
Number of moles = reacting mass/molar mass
Number of moles of CaCl2•2H2O = 1.50 g/147.02 = 0.0102 moles
From the equation;
Na2CO3 + CaCl2•2H2O -> CaCO3 + 2NaCl + 2H2O
We can see is 1:1
1 mole of Na2CO3 reacts with 1 mole of CaCl2•2H2O
x moles of Na2CO3 reacts with 0.0102 moles of CaCl2•2H2O
x = 1 × 0.0102 moles/1
x = 0.0102 moles of Na2CO3
Mass of Na2CO3 = 0.0102 moles of Na2CO3 × 106g/mol = 1.08 g of Na2CO3
Approximately how much energy (in kJ) would be released during the formation of the bonds in 2.00 mol of acetone molecules
Answer:
7822 kJ
Explanation:
The formula for acetone is: CH3COCH3
From the standard bond energy(enthalpy):
C - H bond = 412
C - C bond = 348
C = O bond = 743
From the structure of an acetone
C is bonded to H in six places;
so, for C- H bond = 6 × 412 = 2472
C is only bonded to two other carbon atoms
For C - C bond = 2 × 348 = 696
Carbon is only doubly bonded to an oxygen atom
For C = O bond = 1 × 743 = 743
∴
The total net energy bond in a mole = (2472 + 696 + 743) kJ/mol
= 3911 kJ/mol
Finally, in 2 moles of acetone, the required amount of energy will be:
= 3911 kJ/moles × 2 moles
= 7822 kJ
Classify each of the following chemical reaction as a synthesis, decomposition, single-displacement, or double-displacement reaction. Drag the appropriate items to their respective bins.
CH3Br → CH3(g) + Br(g)
Zn(s) + CoCl2(aq) → ZnCl2(aq)
First Reaction is a Decomposition reaction as a single reactant hets decompoesed to form two products.Second reaction is a Synthesis reaction as two Reactant reacts together to form one product.
What is Decomposition Reaction ?Decomposition reactions are processes in which chemical species break up into simpler parts.
Usually, decomposition reactions require energy input.
Hence, First Reaction is a Decomposition reaction as a single reactant hets decompoesed to form two products.Second reaction is a Synthesis reaction as two Reactant reacts together to form one product.
Learn more about Reaction here ;
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How many grams of P4O10 (292.88 g/mol) form when phelpsphorous (P4, 125.52 g/mol) reacts with 16.2 L of O2 (33.472 g/mol) ) at standard temperature and pressure
Answer:
40.5 g of P₄O₁₀ are produced
Explanation:
We state the reaction:
P₄ + 5O₂ → P₄O₁₀
We do not have data from P₄ so we assume, it's the excess reactant.
We need to determine mass of oxygen and we only have volumne so we need to apply density.
Density = mass / volume, so Mass = density . volume
Denstiy of oxygen at STP is: 1.429 g/L
1.429 g/L . 16.2L = 23.15 g
We determine the moles: 23.15 g . 1mol / 33.472g = 0.692 moles
5 moles of O₂ can produce 1 mol of P₄O₁₀
Our 0.692 moles may produce (0.692 . 1)/ 5 = 0.138 moles
We determine the mass of product:
0.138 mol . 292.88 g/mol = 40.5 g
what are the angles a and b in the actual molecule of which this is a lewis structure note for advanced students give the ideal angles and don t worry about small differences from the ideal that might be caused by the fact that different electron groups may have slightly different sizes
Answer:
The answer is "120 C and 109.5 C".
Explanation:
The carbon atom is hybridized by sp2. This angle of connection thus is 120 degrees. Alkene, specifically both carbons which are in the C=C, are an instance of carbon with sp2 hybridized atom's nucleus. Those three hybridized orbits were linked to certain other atoms forming sigma connections. Its remaining 2p orbital makes a pi link with 2p orbit by the side-overlap of all the other carbon. O is hybridized inside the [-OH] Group. The optimal bond angle therefore is [tex]109.5^{\circ}[/tex].
[tex]a= 120 \ C\\\\b= 109.5 \ C[/tex]
Elements that have the same number of electron rings are ?
Answer:
are in the same orbital
Explanation:
Answer:
are in the same orbit
Explanation:
A covalent bond is formed by the following process
Answer:
Covalent bonding occurs when pairs of electrons are shared by atoms.
Explanation:
Atoms will covalently bond with other atoms in order to gain more stability, which is gained by forming a full electron shell. By sharing their outer most (valence) electrons, atoms can fill up their outer electron shell and gain stability.
5.60g of glyceraldehydes was dissolved in 10ml of a solvent and placed in a 50mm cell if the rotation is 1.74 calculate the specific rotation?
Answer:
6.214 degrees-mL/gdm
Explanation:
The specific rotation α' = α/LC where α = observed rotation, L = length of tube and C = concentration of solution.
Given that α = 1.74, L = length of cell = 50 mm = 0.50 dm and C = m/V where m = mass of glyceraldehyde = 5.60 g and V = volume = 10 ml
So, C = m/V = 5.60 g/10 ml = 0.560 g/ml
Since α' = α/LC
substituting the values of the variables into the equation, we have
α' = α/LC
α' = 1.74/(0.50 dm × 0.560 g/ml)
α' = 1.74/(0.28 gdm/l)
α' = 0.006214 °mL/gdm
α' = 6.214 °mL/gdm
α' = 6.214 degrees-mL/gdm
For the following list of acids, rank the acids in strength from weakest acid to strongest acid.
a. FCH2OH
b. F2CHOH
c. CH3OH
d. F3COH
Answer:
CH3OH < FCH2OH < F2CHOH < F3COH
Explanation:
Let us recall that, for a carboxylic acid, the dissociation of the acid yields;
RCOOH ⇄RCOO^- + H^+
The ease of dissociation and release of the hydrogen ion depends on the nature of the group designated R.
When R is is a highly electronegative element, the -I inductive effect causes the hydrogen to become less tightly held by the C-Cl bond.
As the number of electron withdrawing substituents increaseses, the acid ionizes much more and becomes stronger.CH3OH < FCH2OH < F2CHOH < F3COH
Hence, the order of decreasing acid strength is;
It is found that, when a dilute gas expands quasistatically from 0.40 to 5.0 L, it does 210 J of work. Assuming that the gas temperature remains constant at 300 K, how many moles of gas are present
Answer:
[tex]n=0.033mole[/tex]
Explanation:
From the question we are told that:
Initial volume [tex]V_1=0.40L[/tex]
Final Volume[tex]V_2=5.0L[/tex]
Work [tex]W=210J[/tex]
Temperature [tex]T=300k[/tex]
Generally the equation for Ideal gas is mathematically given by
[tex]W=nRTIn\frac{V_2}{V_1}[/tex]
[tex]n=\frac{W}{RTIn\frac{V_2}{V_1}}[/tex]
[tex]n=\frac{210}{8.32*300In\frac{5.0}{0.4}}[/tex]
[tex]n=0.033mole[/tex]
To determine the concentration of an EDTA solution, 4.11 g of Zn metal was used. The volume of EDTA solution needed to reach the endpoint was 28.26 mL. What was the concentration (in molarity) of the EDTA solution?
Answer:
2.23M
Explanation:
Molarity of a solution is calculated thus
Molarity = number of moles (n) ÷ volume (V)
According to this question, 4.11g of Zn metal was used in order to reach a volume of EDTA solution of 28.26 mL.
28.26mL = 28.26/1000
= 0.02826L
Using mole = mass/molar mass to calculate no. of moles of Zn
Mole = 4.11/65.4
mole = 0.0628mol
Molarity = 0.0628 ÷ 0.02826
Molarity = 2.23M
The concentration of the EDTA solution used is 2.23M
Solid aluminum (AI) and oxygen (0) gas react to form solid aluminum oxide (AIO). Suppose you have 7.0 mol of Al and 13.0 mol of o, in a reactor. Suppose as much as possible of the Al reacts. How much will be left? Round your answer to the nearest 0.1 mol mol 0.
Answer:
[tex]n_{O_2}^{leftover}=7.7mol[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out firstly necessary for us to set up the corresponding chemical equation:
[tex]4Al+3O_2\rightarrow 2Al_2O_3[/tex]
In such a way, we calculate the moles of aluminum consumed by 13.0 moles of oxygen in the reaction, by applying the 4:3 mole ratio between them:
[tex]n_{Al}=13.0molO_2*\frac{4molAl}{3molO_2} =17.3molAl[/tex]
This means that Al is actually the limiting reactant and oxygen is in excess, for that reason we calculate the moles of oxygen consumed by 7.0 moles of aluminum:
[tex]n_{O_2}=7.0molAl*\frac{3molO_2}{4molAl} =5.3molO_2[/tex]
Thus, the leftover of oxygen is:
[tex]n_{O_2}^{leftover}=13.0mol-5.3mol\\\\n_{O_2}^{leftover}=7.7mol[/tex]
Whereas all the aluminum is assumed to be consumed.
Regards!
How many molecules are in
6.0 moles of methane (CH4)?
Answer:
[tex]{ \tt{1 \: mole = 6.02 \times {10}^{23} \: molecules }} \\ { \tt{6.0 \: moles = (6 \times 6.02 \times {10}^{23}) \: molecules }} \\ = { \bf{3.612 \times {10}^{24} \: molecules}}[/tex]
Each set of quantum numbers to the correct sub shell description
