Answer:
Oxygen, O₂ is the limiting reactant
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
2H₂ + O₂ —> 2H₂O
Next, we shall determine the masses of H₂ and O₂ that reacted from the balanced equation. This can be obtained as follow:
Molar mass of H₂ = 2 × 1 = 2 g/mol
Mass of H₂ from the balanced equation = 2 × 2 = 4 g
Molar mass of O₂ = 16 × 2 = 32 g/mol
Mass of O₂O from the balanced equation = 1 × 32 = 32 g
SUMMARY:
From the balanced equation above,
4 g of H₂ reacted with 32 g of O₂.
Finally, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
4 g of H₂ reacted with 32 g of O₂.
Therefore, 10 g of H₂ will react with
= (10 × 32)/4 = 80 g of O₂.
From the calculations made above, we can see that a higher mass (i.e 80 g) of O₂ than what was given (i.e 5 g) is required to react completely with 10 g of H₂. Therefore, O₂ is the limiting reactant.
Oxygen has been the limiting reactant in the reaction.
A limiting reactant can be defined as the reactant in the reaction in which the product concentration has been dependent.
The balanced equation for the formation of water has been:
[tex]\rm 2\;H_2\;+\;O_2\;\rightarrow\;2\;H_2O[/tex]
For the formation of reaction to form 2 moles of water, 2 moles of hydrogen reacts with 1 mole of oxygen.
The moles can be calculated as:
Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]
The moles of Hydrogen in 10 g [tex]\rm H_2[/tex]:
Moles = [tex]\rm \dfrac{10}{2}[/tex]
Moles of hydrogen = 5 mol.
Moles of Oxygen in 5 grams Oxygen:
Moles = [tex]\rm \dfrac{5}{32}[/tex]
Moles of oxygen = 0.156 mol.
For the reaction with 2 moles of Hydrogen 1 mole of Oxygen has been required.
For reacting with 5 mol of Hydrogen, moles of oxygen required are:
Moles of oxygen = [tex]\rm \dfrac{1}{2}\;\times\;5[/tex]
Moles of oxygen required = 2.5 moles.
The available oxygen = 0.156 moles.
Since the moles of oxygen available is lesser than required, the formation of the product has been dependent on the concentration of the oxygen.
Thus, oxygen has been the limiting reactant in the reaction.
For more information about the limiting reactant, refer to the link:
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Observe as equações e de acordo com Brönsted-Lowry, os compostos destacados são, respectivamente
answer
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A student observes a chemical
reaction where two liquids are mixed together. After the liquids are mixed the beaker feels cold to touch. This reactions is an
example of a
reaction.
Synthesis
Combustion
Exothermic
Endothermic
Answer:
Endothermic
Explanation:
Endothermic reaction is one in which the enthalpy increases. What this implies is that, it is a closed system which absorbs heat from its surroundings and thus after reaction become cooler than prior to the reaction.
In this case, the mixture of the two liquids feel colder to touch. Thus, it is an endothermic reaction from the definition earlier given.
Mava=mbvb ma x5.0ml = 5.2ml x 0.10m
Answer:
[tex]M_{a}[/tex] = 0.104 m
Explanation:
This expression is used to determine either the mass or volume of an acid or a base used during titration process.
So that;
[tex]M_{a}[/tex][tex]V_{a}[/tex] = [tex]M_{b}[/tex][tex]V_{b}[/tex]
[tex]M_{a}[/tex] is the mass of the acid
[tex]V_{a}[/tex] is the volume of the acid
[tex]M_{b}[/tex] is the mass of the base
[tex]V_{b}[/tex] is the volume of the base
Given that:
[tex]M_{a}[/tex] x 5.0 m l = 5.2 ml x 0.10 m
[tex]M_{a}[/tex] x 5.0 = 0.52 m
[tex]M_{a}[/tex] = [tex]\frac{0.52}{5.0}[/tex]
= 0.104
[tex]M_{a}[/tex] = 0.104 m
The mass of the acid used is 0.104 m.
if a bottle of vinegar has 4.78g of acetic acid (CH3COOH) per 100.0 g of solution (mixed with water, what is the molarity of the vinegar? Density of the solution is 1.00g/mL.
Answer:
0.796 M
Explanation:
Step 1: Given data
Gravimetric concentration (Cg): 4.78 g%g
Density of the solution (ρ): 1.00 g/mL
Step 2: Calculate the volumetric concentration of the solution (Cv)
We will use the following expression.
Cv = Cg × ρ
Cv = 4.78 g%g × 1.00 g/mL = 4.78 g%mL
Step 3: Calculate the molarity of the solution (M)
The volumetric concentration is 4.78 g%mL, that is, there are 4.78 g of acetic acid per 100 mL of solution. We can calculate the molarity using the following expression.
M = mass solute / molar mass solute × liters of solution
M = 4.78 g / 60.05 g/mol × 0.1 L = 0.796 M
How many liters of hydrogen can be produced at a pressure of 2 atm and a temperature of 298 K
Answer:
1.17 L of H₂
Explanation:
We'll begin by calculating the number of mole in 2.3 g of Mg. This can be obtained as follow:
Mass of Mg = 2.3 g
Molar mass of Mg = 24 g/mol
Mole of Mg =?
Mole = mass /molar mass
Mole of Mg = 2.3 / 24
Mole of Mg = 0.096 mole
Next, we shall determine the number of mole of H₂ produced by the reaction of 2.3 g (i.e 0.096 mole) of Mg. This can be obtained as follow:
Mg + 2HCl —> MgCl₂ + H₂
From the balanced equation above,
1 mole of Mg reacted to 1 mole of H₂.
Therefore, 0.096 mole of Mg will also react to produce 0.096 mole of H₂.
Finally, we shall determine volume of H₂ produced from the reaction. This can be obtained as follow:
Number of mole (n) of H₂ = 0.096 mole
Pressure (P) = 2 atm
Temperature (T) = 298 K
Gas constant (R) = 0.0821 atm.L/Kmol
Volume (V) of H₂ =?
PV = nRT
2 × V = 0.096 × 0.0821 × 298
Divide both side by 2
V = (0.096 × 0.0821 × 298) /2
V = 1.17 L
Therefore, 1.17 L of H₂ were obtained from the reaction.
The speed of sound depends on the __?_____ and ____?____ of the medium through which it travels
Answer:
Density and rigidity
Cis-4-tertButylcyclohexyl bromide (compound 1) and Trans-4 tert Butylcyclohexylbromide (compound 2) are reacted with Potassium Tertiary butoxide in Tertiary butanol to produce 4-tertbutylcyclohexene. The following statement is completely true?A. In compound 1 the Tert butyl group occupies the equatorial position and the Bromine occupies the axial position and in compound 2 both the Tert butyl and the bromine occupy equatorial positions. Compound 1 reacts faster than compound 2.
B. In compound 1 the Tert butyl group occupies the axial position and the Bromine occupies the axial position and in compound 2 both the Tert butyl and the bromine occupy equatorial positions. Compound 1 reacts faster than compound 2.
C. In compound 1 the Tert butyl group occupies the equatorial position and the Bromine occupies the equatorial position and in compound 2 both the Tert butyl and the bromine occupy equatorial positions. Compound 1 reacts faster than compound 2.
D. In compound 1 the Tert butyl group occupies the equatorial position and the Bromine occupies the axial position and in compound 2 the Tert butyl occupies the axial and the bromine occupies equatorial positions. Compound 1 reacts faster than compound 2.
E. In compound 1 the Tert butyl group occupies the equatorial position and the Bromine occupies the axial position and in compound 2 both the Tert butyl occupies the equatorial and the bromine occupies axial position. Compound 1 reacts faster than compound 2.
F. In compound 1 the Tert butyl group occupies the equatorial position and the Bromine occupies the axial position and in compound 2 both the Tert butyl and the bromine occupy equatorial positions. Compond 2 reacts faster than compound 1.
Answer:
In compound 1 the Tert butyl group occupies the equatorial position and the Bromine occupies the axial position and in compound 2 the Tert butyl occupies the axial and the bromine occupies equatorial positions. Compound 1 reacts faster than compound 2.
Explanation:
In cyclic organic compounds, substituents may occupy the axial or equatorial positions. The axial positions are aligned parallel to the symmetry axis of the ring while the equatorial positions are around the plane of the ring.
Bulky substituents have more room in the equatorial than in the axial position. This means that compound 1 is more stable than compound 2.
This is clear on the basis of stability of the molecules because compound 1 will react faster than compound 2 since the bulky tertiary butyl group in compound 1 occupy equatorial and not axial positions.
important of organic chemistry in our daily life
Answer:
Human life has become very simple by using organic chemicals. The importance of organic chemicals in the daily life and industrial area can be explained as follows. (i) Food Vitamins, proteins, sugar, flour, fats etc. ... (ii) Agriculture is an important place for organic chemical for the growth of agricultural production.
True or false: Boron contains 2s22p1 valence electrons, so only one p orbital is needed to form molecular orbitals.
Answer:
True
Explanation:
The valence orbitals of boron are 2s2 2p1. We have to recall that all the valence orbitals whether full or empty are involved in the formation of molecular orbitals.
The number of molecular orbitals formed is equal to the number of atomic orbitals that are combined.
Since there are two valence orbitals and there is only one p orbital among the valence orbitals, it is true that only one p orbital is needed to form molecular orbitals in boron.
Calculate the moment of inertia of a CH³⁵CL₃ molecule around a rotational axis that contains the C-H bond. The C-Cl bond length is 177pm and the HCCl angle is 107⁰f
Answer:
The correct answer is "[tex]4.991\times 10^{-45} \ kg.m^2[/tex]".
Explanation:
According to the question,
[tex]R_{C-Cl} = 177 \ pm[/tex]
or,
[tex]=1.77\times 10^{-10} \ m[/tex]
[tex]\alpha = 107^{\circ}[/tex]
[tex]m_{Cl}=34.97 \ m.u[/tex]
or,
[tex]=34.97\times 1.66\times 10^{-27}[/tex]
[tex]=5.807\times 10^{-26} \ kg[/tex]
The moment of inertia around the rotational axis will be:
⇒ [tex]I=3\times m_{Cl}\times (R_{C-Cl})^2 \ Sin^2 \alpha[/tex]
By putting the values, we get
[tex]=3\times 5.807\times 10^{-26}\times (1.77\times 10^{-10})^2 \ Sin^2 (107)[/tex]
[tex]=3\times 5.807\times 10^{-26}\times (1.77\times 10^{-10})^2\times 0.91452[/tex]
[tex]=4.991\times 10^{-45} \ kg.m^2[/tex]
(4.184 J——> 1 calorie ? Take your answer from number 1 and convert to the answer into calories?
Answer:
There are 1.195 calories in 5 J.
Explanation:
The relation between calorie and Joules is as follow :
4.184 J = 1 calorie
or
1 J = (1/4.184) calorie
Let the taken number is 5 J
So,
5 J = (5/4.184) calorie
5 J = 1.195 calorie
So, there are 1.195 calories in 5 J.
In a solution, the solvent is
ANSWER:
A. always water
B. dissolved in the solute
C. present in larger amount than the solute is
D. always nonpolar
Answer:
dissolved in the solute
Explanation:
A solvent is the component that dissolves the solute and is present in larger amount. The type of solution is determined by the state of the solute and solvent. If you have NaCl, a solid, dissolved in water, a liquid, the type of solution is a solid/liquid solution.
#6 and #7. How many carbon atoms are in a mixture of 7.00 mol c2F2 and 0.400 mol carbon dioxide and also #7
Answer:
#6 8.67x10²⁴ atoms
#7
1. Atom
2. Formula unit
3. Molecule
4. Ion
Explanation:
#6 First we calculate how many carbon moles are there in 7.00 moles of C₂F₂, keeping in mind that there are 2 C moles per C₂F₂ mol:
7.00 mol C₂F₂ * 2 = 14.00 mol CAs for carbon dioxide, there are 0.400 C moles in 0.400 moles of CO₂.
We calculate the total number of C moles:
14.00 mol + 0.400 mol = 14.4 mol CFinally we calculate the number of atoms in 14.4 C moles, using Avogadro's number:
14.4 mol * 6.023x10²³ atoms/mol = 8.67x10²⁴ atoms#7
1. Radon - Atom (Ra)2. Formula unit (It is a crystalline solid, BaBr₂)3. Molecule (NH₃)4. Ion (It has a formal charge, +2)Which list contains gases found only in "clean" air?
A. carbon dioxide, nitrogen dioxide, sulfur dioxide, helium
B. helium, methane, carbon dioxide, xenon
hydrogen, oxygen, chlorine, ammonia
C. argon, neon, krypton, radon
D. fluorine, chlorine, bromine, iodine
Answer:
hydrogen, oxygen, chlorine, ammonia
Explanation:
Air is a mixture of gases. When we say "clean" air here, we are referring to air that does not contain pollutant gases.
Some components of air such as water vapour, methane, CO2, and N2O are greenhouse gases. They are known to contribute towards global warming.
Some gases such as SO2 and NO2 contribute towards acid rain. The oxides of nitrogen are particularly involved in the formation of photochemical smog.
The halogens are known to lead to the depletion of the ozone layer and radon is a radioactive gas.
Hence, hydrogen, oxygen, chlorine, ammonia have no negative environmental impact hence they are found in clean air.
A sample of 10.6 g of KNO3 was dissolved in 251.0 g of water at 25 oC in a calorimeter. The final temperature of the solution was 21.5 oC. What is the molar heat of solution of KNO3
Answer:
36.55kJ/mol
Explanation:
The heat of solution is the change in heat when the KNO3 dissolves in water:
KNO3(aq) → K+(aq) + NO3-(aq)
As the temperature decreases, the reaction is endothermic and the molar heat of solution is positive.
To solve the molar heat we need to find the moles of KNO3 dissolved and the change in heat as follows:
Moles KNO3 -Molar mass: 101.1032g/mol-
10.6g * (1mol/101.1032g) = 0.1048 moles KNO3
Change in heat:
q = m*S*ΔT
Where q is heat in J,
m is the mass of the solution: 10.6g + 251.0g = 261.6g
S is specififc heat of solution: 4.184J/g°C -Assuming is the same than pure water-
And ΔT is change in temperature: 25°C - 21.5°C = 3.5°C
q = 261.6g*4.184J/g°C*3.5°C
q = 3830.87J
Molar heat of solution:
3830.87J/0.1048 moles KNO3 =
36554J/mol =
36.55kJ/mol
he equation for the dissociation of pyridine is
C5H5N(aq) + H2O(l) ⇌ C 5H5NH+(aq) + OH-(aq) Kb = 1.9 × 10-9
Calculate the pH of a pyridine solution that has a concentration of 9.2 M. Round your answer to two decimal places.
Answer:
10.10
Explanation:
Step 1: Write the basic dissociation reaction for pyridine
C₅H₅N(aq) + H₂O(l) ⇌ C₅H₅NH⁺(aq) + OH⁻(aq) Kb = 1.9 × 10⁻⁹
Step 2: Calculate [OH⁻]
For a weak base, we will use the following expression.
[OH⁻] = √(Cb × Kb) = √(9.2 × 1.9 × 10⁻⁹) = 1.3 × 10⁻⁴ M
Step 3: Calculate pOH
We will use the definition of pOH.
pOH = -log [OH⁻] = -log 1.3 × 10⁻⁴ = 3.9
Step 4: Calculate pH
We will use the following expression.
pH = 14 - pOH = 14 - 3.9 = 10.10
in a breaker of water ,the water -water bonds can properly be called
Answer:
LICK
Explanation:
Complete and balance the equations for the given single displacement reactions. Write the reaction in molecular form. Phases are optional. If you need to clear your work and reset the equation, click the CLR button.Li(s) + H2O---------- Ca(s) + H2O ---------------
Answer:
Li(s) + H2O(l) -----> LiOH(aq) + H2(g)
Ca(s) + 2H2O(l) ---------------> Ca(OH)2(aq) + H2(g)
Explanation:
Metals react with water to yield the metal hydroxide and hydrogen gas as shown in the answer above.
The reaction equations were balanced, the number of atoms of each element on both side of the reaction equation is exactly the same.
This is the way to write a balanced reaction equation for the species shown in the question.
When the pH of a solution is 12.83, what is [H +]?
A 7.4 × 10 -12 M
B 9.7 10 -11 M
C 1.5 X 10 -13 M
D 12.8 x 10 -2 M
Answer:
B.9.710-11M
Explanation:
plss tell me if im wrongUsing the following reaction:
H2SO4 (aq) + 2NaOH (aq) → Na2SO4 (aq) + 2H2O (l)
Calculate the molarity of the H2SO4 solution if 14.92 mL of NaOH was necessary to reach the endpoint of a titration. The molarity of the NaOH solution was 0.83 M and 25.18 mL of H2SO4 was added to the Erlenmeyer flask.
Answer:
The molar concentration of the H₂SO₄ solution is 0.28 M
Explanation:
Molar concentration = number of moles / volume in litres
Number of moles = molar concentration × volume
From the equation of reaction, molar ratio of acid to base = 1 : 2
Using the formula; Na/Nb = CaVa/CbVb
Where Na is the number of moles of acid; Nb = number of moles of base; Ca = concentration of acid; Va = volume of acid; Cb = concentration of base; Vb = volume of base; Na/Nb = mole ratio of acid to base
Substituting the given values in the equation:
1/2 = Ca × 25.18 / 0.93 × 14.92
Ca = 0.93 × 14.92/ 25.18 × 2
Ca = 0.28M
Therefore, the molar concentration of the H₂SO₄ solution is 0.28 M
Typhoon signals rise due to what? wind speed or wind strength or both?
Which would most likely be reduced when combined with Zn(s)?
A. Fe²⁺
B. K⁺
C. Ni
D. Al³⁺
The reduction of the species defines the gain of electrons. The iron is most likely to be reduced when reacts with zinc. Thus option A is correct.
What is oxidizing agent?Oxidizing agents are the species that gain electrons and get reduced, their oxidation number gets reduced when the metal reacts.
In the reactivity, series zinc is placed before iron and hence is a reducing agent that gets oxidized. Down the series, the reducing ability decreases while the oxidizing increases.
Therefore, option A. iron will be reduced when reacts with zinc.
Learn more about the reactivity series here:
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For each of the following circumstances, indicate whether the calculated molarity of NaOH would be lower, higher or unaffected. Explain your answer in each case. a. The inside of the pipet used to transfer the standard HCl solution was wet with water.b. you added 40 mL of water to the titration flask rather than 25ml. c. The buret, wet with water, was not rinsed with NaOH solution before filling the buret with NaOH solution. d. Five (5) drops of phenolphthalein were added to the solution to be titrated rather than three (3) drops.
Answer:
a)calculated molarity of NaOH would be lower
b) calculated molarity of NaOH would be lower
c) calculated molarity of NaOH would be lower
d) calculated molarity of NaOH would be unaffected
Explanation:
Let us recall that the reaction of NaOH and HCl is as follows;
NaOH(aq) + HCl(aq) ----> NaCl(aq) + H2O(l)
Since the reaction is 1:1, when the number of moles of HCl reacting with NaOH is low due to dilution, the calculated molarity of NaOH also becomes less than it's accurate value.
When 40mL of water is added to the titration flask rather than 25ml of water, the acid is more dilute hence less number of moles of acid than necessary reacts with the base thereby yielding a less than accurate value of the molarity of NaOH.
If the burette wet with water is not rinsed with NaOH solution, the concentration of the NaOH in the burette decreases due to dilution with water and a less than accuracy value is calculated for the molarity of NaOH.
If five drops of phenolphthalein is used instead of one or two drops, there is no qualms since enough phenolphthalein may be added to ensure that a sharp end point is obtained.
howtocalculatethevolumeofcarbondioxideproducedwhen400gofmarblewereats.t.p
Answer:
so 0.15 moles X 22.4 dm3/mole=3.36 dm3. Next we find the moles of hexane combusted, and then the moles of CO2. Finally, we find the volume of CO2 using the fact that at STP, 1 mole of gas = 22.4 dm3.
3. Does entropy increase or decrease in the following processes?
A. Complex carbohydrates are metabolized by the body, converted into simple sugars.
Answer: Increase
es-lesund
B. Steam condenses on a glass surface.
Answer:
decreare
-->
MgCl2(s)
C. Mg(s) + Cl2(g)
correct
Answer:
Answer:
hope it helps much as you can
A 250 mL sample of gas at 1.00 atm and has the temperature increased to and the volume increased to 500 mL. What is the new pressure
Answer:
0.53atm = P2
Explanation:
Gas at 1.00atm and 20°C. Temperature increased to 40°C...
We can solve this question using combined gas law:
P1*V1 / T1 = P2*V2 / T2
Where P is pressure, V volume and T absolute temperature of 1, initial state and 2, final state.
Compunting the values of the problem:
P1 = 1.00atm
V1 = 250mL
T1 = 20°C + 273.15 = 293.15K
P2 = ?
V2 = 500mL
T2 = 40°C + 273.15 = 313.15K
1.00atm*250mL / 293.15K = P2*500mL / 313.15K
0.53atm = P2
Rank the following chemical species from lowest absolute entropy (So) (1) to highest absolute entropy (5) at 298 K?
a. Al (s)
b. H2O (l)
c. HCN (g)
d. CH3COOH (l)
e. C2H6 (g)
Answer:
Rank the following chemical species from lowest absolute entropy (So) (1) to highest absolute entropy (5) at 298 K?
a. Al (s)
b. H2O (l)
c. HCN (g)
d. CH3COOH (l)
e. C2H6 (g)
Explanation:
Entropy is the measure of the degree of disorderness.
In solids, the entropy is very less compared to liquids and gases.
The entropy order is:
solids<liquids<gases
Among the given substances, water in liquid form has a strong intermolecular H-bond.
So, it has also less entropy.
Next acetic acid.
Between the gases, HCN, and ethane, ethane has more entropy due to very weak intermolecular interactions.
HCN has slight H-bonding in IT.
Hence, the entropy order is:
Al(s) < CH3COOH (l) <H2O(l) < HCN(g) < C2H6(g)
1) Write a balanced equation to show the reaction of gaseous ethane with gaseous oxygen to form carbon monoxide gas and water vapor.
Answer:
C₂H₆(g) + 2.5 O₂(g) ⇒ 2 CO(g) + 3 H₂O(g)
Explanation:
Let's consider the unbalanced equation in which gaseous ethane reacts with gaseous oxygen to form carbon monoxide gas and water vapor. This is an incomplete combustion reaction.
C₂H₆(g) + O₂(g) ⇒ CO(g) + H₂O(g)
We will balance it using the trial and error method.
First, we will balance C atoms by multiplying CO by 2 and H atoms by multiplying H₂O by 3.
C₂H₆(g) + O₂(g) ⇒ 2 CO(g) + 3 H₂O(g)
Then, we get the balanced equation by multiplying O₂ by 2.5.
C₂H₆(g) + 2.5 O₂(g) ⇒ 2 CO(g) + 3 H₂O(g)
An unknown weak acid with a concentration of 0.530 M has a pH of 5.600. What is the Ka of the weak acid
Answer:
Ka = 3.45x10⁻⁶
Explanation:
First we calculate [H⁺], using the given pH:
pH = -log[H⁺][H⁺] = [tex]10^{-pH}=10^{-5.6}[/tex] [H⁺] = 2.51x10⁻⁶ MTo solve this problem we can use the following formula describing a monoprotic weak acid:
[H⁺] = [tex]\sqrt{C*Ka}[/tex]We input the data that we already know:
2.51x10⁻⁶ = [tex]\sqrt{0.530*Ka}[/tex]And solve for Ka:
Ka = 3.45x10⁻⁶Pls pls help me me pls
Answer:
Danger
Explanation: