Answer:
D) 2.01 x 10⁻⁴ .
Explanation:
pH = 2.11
[ H⁺ ] = [tex]10^{-2.11}[/tex]
Let the acid be HA
It will ionise as follows .
HA ⇄ H⁺ + A⁻
in equilibrium .30 [tex]10^{-2.11}[/tex] [tex]10^{-2.11}[/tex]
Acid ionisation constant Ka = [tex]\frac{(10^{-2.11})^2}{0.3}[/tex]
= 2 x 10⁻⁴
Answer:
D) 2.01 x 10⁻⁴ is correct!
Explanation:
I got it in class!
Hope this Helps!! :))
You have a saturated solution of BaSO4, a slightly soluble ionic compound. What happens if you add Ba(OH)2, NaNO3, and CuSO4 to this solution
Answer:
- Addition of Ba(OH)2: favors the formation of a precipitate.
- Undergo a chemical reaction forming soluble species.
- Addition of CuSO4 : favors the formation of a precipitate.
Explanation:
Hello,
In this case, since the dissociation reaction of barium sulfate is:
[tex]BaSO_4(s)\rightleftharpoons Ba^{2+}(aq)+SO_4^{2-}(aq)[/tex]
We must analyze the effect of the common ion:
- By adding barium hydroxide, more barium ions will be added to the equilibrium system so the formation of solid barium sulfate will be favored (reaction shifts leftwards towards reactants).
- By adding sodium nitrate, the following reaction will undergo:
[tex]BaSO_4(s)+NaNO_3(aq)\rightarrow Ba(NO_3)_2(aq)+Na_2SO_4(aq)[/tex]
So the precipitate will turn into other soluble species.
- By adding copper (II) sulfate, more sulfate ions will be added to the equilibrium system so the formation of solid barium sulfate will be favored (reaction shifts leftwards towards reactants).
All of this is supported by the Le Chatelier's principle.
Best regards.
Calculate LaTeX: \DeltaΔGº for a voltaic cell with Eºcell = +0.24 V if the overall reaction involves a 3 electron reduction.
Answer:
-69 kJ
Explanation:
Step 1: Given data
Standard cell potential (E°cell): +0.24 V
Electrons involved (n): 3 mol
Step 2: Calculate the standard Gibbs free energy change (ΔG°) for the voltaic cell
We will use the following expression.
ΔG° = -n × F × E°cell
where,
F is Faraday's constant (96,485 C/mol e⁻)
ΔG° = -n × F × E°cell
ΔG° = -3 mol e⁻ × 96,485 C/mol e⁻ × 0.24 V
ΔG° = -69 kJ
Select the true statement concerning voltaic and electrolytic cells. Select one: a. Voltaic cells involve oxidation-reduction reactions while electrolytic cells involve decomposition reactions. b. Voltaic cells require applied electrical current while electrolytic cells do not. . c. all electrochemical cells, voltaic and electrolytic, must have spontaneous reactions. d. Electrical current drives nonspontaneous reactions in electrolytic cells.
Answer:
Electrical current drives nonspontaneous reactions in electrolytic cells.
Explanation:
Electrochemical cells are cells that produce electrical energy from chemical energy.
There are two types of electrochemical cells; voltaic cells and electrolytic cells.
A voltaic cell is an electrochemical cell in which electrical energy is produced from spontaneous chemical process while an electrolytic cell is an electrochemical cell where electrical energy is produced from nonspontaneous chemical processes. Current is needed to drive these nonspontaneous chemical processes in an electrolytic cell.
Answer:
electrolytic cells generate electricity through a non-spontaneous reaction while voltaic cells absorb electricity to drive a spontaneous reaction.
Explanation:
Answer via Educere/ Founder's Education
Janet observes that bubbles rise inside water when water is heated. Which of the following best names and explains the change that causes bubbles to rise?
Answer:
Boiling
Explanation:
When a liquid is heated, the vapor pressure rises steadily. When water attains a temperature of 100°C or 212°F its vapor pressure is now equal to the atmospheric pressure at sea level, this is what we mean by boiling.
When this occurs, water continues to evaporate untill the vapor pressure inside the bubbles becomes high enough to stop water bubbles from collapsing again from the pressure of the water around it so the bubbles rise and break the surface.
A mixture of 50ml of 0.1M HCOOH and 50ml of 0.05M NaOH is equivalent to
Answer:
d) a solution that is 0.025M in HCOOH and 0.025M in HCOONa
Explanation:
The reaction of a weak acid (HOOH) with NaOH is as follows:
HCOOH + NaOH → HCOONa + H₂O
Based on the reaction, 1 mole of the acid reacts with 1 mole of the base (Ratio 1:1).
The initial moles of both species are:
HCOOH: 0.050L × (0.1mol / L) = 0.0050 moles of HCOOH
NaOH: 0.050L × (0.05 mol / L) = 0.0025 moles NaOH
After the reaction, all NaOH reacts with HCOOH producing HCOONa (Because moles of NaOH < moles HCOOH).
Final moles:
HCOOH: 0.0050 moles - 0.0025 moles (After reaction) = 0.0025 moles
HCOONa: Moles HCOONa = Initial Moles NaOH: 0.0025 moles
As volume of the mixture is 100mL (50 from the acid + 50 from NaOH), molarity of both HCOOH and HCOONa is:
0.0025 moles / 0.100L = 0.025M of both HCOOH and HCOONa
Thus, the initial mixture is equivalent to:
d) a solution that is 0.025M in HCOOH and 0.025M in HCOONaA piece of solid Fe metal is put into an aqueous solution of Cu(NO3)2. Write the net ionic equation for any single-replacement redox reaction that may be predicted. Assume that the oxidation state of in the resulted solution is 2 . (Use the lowest possible coefficients for the reaction. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank. If no reaction occurs, leave all boxes blank and click on Submit.)
Answer:
Fe(s) + Cu^2+(aq) ---> Fe^2+(aq) + Cu(s)
Explanation:
The ionic equation shows the actual reaction that took place. It excludes the spectator ions. Spectator ions are ions that do not really participate in the reaction even though they are present in the system.
For the reaction between iron and copper II nitrate, the molecular reaction equation is;
Fe(s) + Cu(NO3)2(aq)----> Fe(NO3)2(aq) +Cu(s)
Ionically;
Fe(s) + Cu^2+(aq) ---> Fe^2+(aq) + Cu(s)
What is the pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid with 15.00 mL of 0.10 M KOH? Assume that the volumes of the solutions are additive. K a = 1.8 ×× 10-5 for CH3CO2H.
Answer:
pH = 8.72
Explanation:
This is like a titration of a weak acid and a strong base, in this case, we are at the equivalence point plus we have the same mmoles of acid and base. We have completely neutralized the acid.
CH₃COOH + OH⁻ ⇄ CH₃COO⁻ + H₂O
0.1M . 15 mL 0.1M . 15 mL
We only have (0.1M . 15 mL) mmoles of acetate ion. → 1.5 mmoles
As this compound acts like a base, we propose this equilibrium:
CH₃COO⁻ + H₂O ⇄ CH₃COOH + OH⁻ Kb
We need to work with Kb and we know, that Kw = Ka. Kb so, Kb = Kw/Ka
Kb = 1×10⁻¹⁴ /1×10 ⁻⁵ = 5.55×10⁻¹⁰
Concentration of CH₃COO⁻ → 1.5 mmol / 30mL (volumes of the solutions are additive) = 0.05M
So: [CH₃COOH] . [OH⁻] / [CH₃COO⁻] = Kb
x²/ 0.05-x = 5.55×10⁻¹⁰
We can avoid the quadractic equation because Kb is so small
[OH⁻] = √(5.55×10⁻¹⁰ . 0.05) = 5.27×10⁻⁶
pOH = - log [OH⁻] → 5.28
pH = 14 - pOH = 8.72
The pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid should be 8.72.
Calculation of the pH of the solution:Since the following equation should be used.
CH₃COOH + OH⁻ ⇄ CH₃COO⁻ + H₂O
0.1M . 15 mL 0.1M . 15 mL
Now
(0.1M . 15 mL) mmoles of acetate ion. → 1.5 mmoles
So,
CH₃COO⁻ + H₂O ⇄ CH₃COOH + OH⁻ Kb
Now
Kw = Ka. Kb
Kb = Kw/Ka
And,
Kb = 1×10⁻¹⁴ /1×10 ⁻⁵
= 5.55×10⁻¹⁰
Now
[CH₃COOH] . [OH⁻] / [CH₃COO⁻] = Kb
x²/ 0.05-x = 5.55×10⁻¹⁰
Now
[OH⁻] = √(5.55×10⁻¹⁰ . 0.05) = 5.27×10⁻⁶
pOH = - log [OH⁻] → 5.28
pH = 14 - pOH
= 8.72
Hence, The pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid should be 8.72.
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At what temperature in K will 0.750 moles of oxygen gas occupy 10.0 L and exert 2.50 atm of pressure
Answer:
406 K.
Explanation:
The following data were obtained from the question:
Number of mole (n) = 0.750 mole
Volume (V) = 10.0 L
Pressure (P) = 2.50 atm
Temperature (T) =.?
Note: Gas constant (R) = 0.0821 atm.L/Kmol
The temperature, T can be obtained by using the ideal gas equation as follow:
PV = nRT
2.5 x 10 = 0.75 x 0.0821 x T
Divide both side by 0.75 x 0.0821
T = (2.5 x 10) /(0.75 x 0.0821 )
T = 406 K.
Therefore, the temperature is 406 K.
Answer: 406 K
Explanation:
We can rewrite the ideal gas law to solve for T:
PV = nRT
T=PV / nR
We are given the following from the problem:
n=0.750 mol P=2.50 atm V=10.0 L
Plugging in our values and using R=0.08206 L⋅atm / K⋅mol we get:
T=(2.50 atm)(10.0 L) / (0.750 mole)(0.08206L ⋅ atm ⋅ mole K) = 406 K
One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titrating a sample of silver nitrate solution. Any chloride anions in solution will combine with the silver cations to produce bright white silver chloride precipitate. Suppose an EPA chemist tests a sample of groundwater known to be contaminated with nickel(II) chloride, which would react with silver nitrate solution like this:
Answer:
6.5 mg/L.
Explanation:
Step one: write out and Balance the chemical reaction in the Question above:
NiCl2 + 2AgNO3 =====> 2AgCl + Ni(NO3)2.
Step two: Calculate or determine the number of moles of AgCl.
So, we are given that the mass of AgCl = 3.6 mg = 3.6 × 10^-3 g. Therefore, the number of moles of AgCl can be calculated as below:
Number of moles AgCl = mass/molar mass = 3.6 × 10^-3 g / 143.32. = 2.5118 × 10^-5 moles.
Step three: Calculate or determine the number of moles of NiCl2.
Thus, the number of moles of NiCl2 = 2.5118 × 10^-5/ 2 = 1.2559 × 10^-5 moles.
Step four: detemine the mass of NiCl2.
Therefore, the mass of NiCl2 = number of moles × molar mass = 1.2559 × 10^-5 moles × 129.6 = 1.6 × 10^-3 g.
Step five: finally, determine the concentration of NiCl2.
1000/ 250 × 1.6 × 10^-3 g. = 6.5 mg/L.
Aqueous potassium nitrate (KNO3) and solid silver bromide are formed by the reaction of aqueous potassium bromide and aqueous silver nitrate (AgNO3). Write a balanced chemical equation for this reaction
Answer:
For the mentioned reaction, the balanced chemical equation is:
KBr (aq) + AgNO3 (s) ⇒ KNO3 (aq) + AgBr (s)
The number written in front of the ion, atoms, and molecules in a chemical reaction so that each of the elements on both the sides of reactants and products of the equation gets balanced is known as the stoichiometric coefficient.
From the mentioned balanced equation, the stoichiometric coefficient before KBr is 1, AgNO3 is 1, KNO3 is 1, as well as before AgBr is also 1. Thus, it is clear that 1 mole of potassium bromide reacts with 1 mole of silver nitrate to produce 1 mole of potassium nitrate and 1 mole of silver bromide.
Calculate the energy required to heat 1.30kg of water from 22.4°C to 34.2°C . Assume the specific heat capacity of water under these conditions is 4.18·J·g−1K−1 . Round your answer to 3 significant digits.
Answer:
The energy required to heat 1.30 kg of water from 22.4°C to 34.2°C is 64,121.2 J
Explanation:
Calorimetry is the measurement of the amount of heat that a body gives up or absorbs in the course of a physical or chemical process.
The sensible heat of a body is the amount of heat received or transferred by a body when undergoing a temperature variation (Δt) without there being a change in physical state. That is, when a system absorbs (or gives up) a certain amount of heat, it may happen that it experiences a change in its temperature, involving sensible heat. Then, the equation for calculating heat exchanges is:
Q = c * m * ΔT
Where Q is the heat or quantity of energy exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature (ΔT=Tfinal - Tinitial).
In this case:
[tex]c=4.18 \frac{J}{g*K}[/tex]m= 1.30 kg= 1,300 g (1 kg=1,000 g)ΔT= 34.2 °C - 22.4 °C= 11.8 °C= 11.8 °K Being a temperature difference, it is independent if they are degrees Celsius or degrees Kelvin. That is, the temperature difference is the same in degrees Celsius or degrees Kelvin.Replacing:
[tex]Q=4.18 \frac{J}{g*K}*1,300 g*11.8 K[/tex]
Q= 64,121.2 J
The energy required to heat 1.30 kg of water from 22.4°C to 34.2°C is 64,121.2 J
Why do prices increase when demand for a product is high? Companies know they can make more money by selling fewer products at higher prices. Companies know that people will be willing to spend more to get an in-demand product. Companies take advantage of the demand to make people spend more money on excess products. Companies know they can stop production and still make money on sales.
Answer:
Companies know that people will be willing to spend more to get an in-demand product.
Explanation:
When a product is really in demand, many customers are willing to part with more money order to purchase the product, as a result of that, many companies may take advantage of the increasing demand for the product to hike it's price.
Hence, the increase in price may not really have a negative impact on the quantity demanded because the demand for the product is high and customers are willing to spend more money in order to purchase an in-demand product, hence the answer above.
Prices increase when demand is high because companies know that people will be willing to spend more to get in-demand products.
Prices generally increase with higher demand for goods because the higher demand creates pressure for the supply to meet up.
Manufacturing companies can either increase their production to meet up with demand at the same price or capitalize on the situation to make more money by increasing the price without increasing the supply.
Since there is a buying pressure on the product in the market already, people would still be open to buying even at higher prices.
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The number of moles of H2O which contains 4g of oxygen?
Answer:
16G = 1 mole ; then 4G = how many moles? 4/16 = 0.25 mole; That means 4 grams of oxygen is 0.25 moles.
Explanation:
A mole of water molecules contains 2 moles of hydrogen atoms and 1 mole of oxygen atoms.
Because of movements at the Mid-Atlantic Ridge, the Atlantic Ocean widens by about 2.5 centimeters each year. Explain which type of plate boundary causes this motion.
Answer:
A divergent plate boundary
Explanation:
At a divergent boundary, the plates pull away from each other and generate new crust.
Answer:
Because the ocean becomes larger, this is a divergent plate boundary. Divergent plates cause the ocean floor to expand, making the ocean larger.
Explanation:
PLATO ANSWER
When 1604 J of heat energy is added to 48.9 g of hexane, C6H14, the temperature increases by 14.5 ∘C. Calculate the molar heat capacity of C6H14.
Answer:
THE MOLAR HEAT CAPACITY OF HEXANE IS 290.027 J/ C
Explanation:
1604 J of heat is added to 48.9 g of hexane
To calculate the molar heat capacity of hexane, it is important to note that the molar heat capacity of a substance is the measure of the amount of heat needed to raise 1 mole of a substance by 1 K.
Since 1604 J of heat = 48.9 g of hexane
Molar mass of hexane = 86 g/mol = 1 mole
then;
1604 J = 48.9 g
x = 86 g
x = 1604 * 86 / 48.9
x = 4205.4 J
Hence, 4205.4 J of heat will be added to 1 mole or 86 g of hexane to raise the temperature by 14.5 C.
In other words,
heat = molar heat capacity * temperature change
molar heat capacity = heat/ temperature change
Molar heat capacity = 4205.4 J / 14.5 C
Molar heat capacity = 290.027 J/C
The molar heat capacity of hexane is 290.027 J/ C
While balancing a chemical equation, we change the _____ to balance the number of atoms on each side of the equation.
Answer:
While balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation
Explanation:
While balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation.
What is chemical equation?To summarize in chemistry terms, a chemical equation depicts the initial chemicals, or reactants, on the left-hand side and the final compounds, or products, just on right-hand side, divided by an arrow. In the chemical equation, the number of atoms in each element as well as the total charge are the same on opposite of the equation's sides.
Chemical equations are used in chemistry to depict chemical processes by writing the reactants and products in terms of their corresponding chemical formulas. While balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation.
Therefore, while balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation.
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Zinc is used as a coating for steel to protect the steel from environmental corrosion. If a piece of steel is submerged in an electrolysis bath for 24 minutes with a current of 6.5 Amps, how many grams of zinc will be plated out? The molecular weight of Zn is 65.38, and Zn+2 + 2e– → Zn. Question 7 options: A) 3.17 g of Zn B) 1.09 g of Zn C) 6.34 g of Zn D) 12.68 g of Zn
Answer:
A) 3.17 g of Zn
Explanation:
Let's consider the reduction of Zn(II) that occurs in an electrolysis bath.
Zn⁺²(aq) + 2e⁻ → Zn(s)
We can establish the following relations:
1 min = 60 s1 A = 1 C/sThe charge of 1 mole of electrons is 96,468 C (Faraday's constant).When 2 moles of electrons circulate, 1 mole of Zn is deposited.The molar mass of Zn is 65.38 g/molThe mass of Zn deposited under these conditions is:
[tex]24min \times \frac{60s}{1min} \times \frac{6.5C}{s} \times \frac{1mol\ e^{-} }{96,468C} \times \frac{1molZn}{2mol\ e^{-}} \times \frac{65.38g}{1molZn} = 3.17 g[/tex]
Answer:
A.) 3.17
Explanation:
I got it right in class!
Hope this Helps!! :))
To calculate changes in concentration for a system not at equilibrium, the first step is to determine the direction the reaction will proceed. To do so, we calculate Q and compare it to the equilibrium concentration, K. We can then determine that a reaction will shift to the right if:__________
Answer:
We can then determine that a reaction will shift to the right if Q<K
Explanation:
Comparing Q with K allows to find out the status and evolution of the system:
If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium. If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium and will evolve spontaneously, decreasing the value of Qc until it equals the equilibrium constant. In this way, the concentrations of the products will decrease and the concentrations of the reagents will increase. In other words, the reverse reaction is favored to achieve equilibrium. Then the system will evolve to the left (ie products will be consumed and more reagents will be formed).If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium and will evolve spontaneously increasing the value of Qc until it equals the equilibrium constant. This implies that the concentrations of the products will increase and those of the reagents will decrease. In other words, to achieve balance, direct reaction is favored. Then the reaction will shift to the right, that is, reagents will be consumed and more products will be formed.In this case, we can then determine that a reaction will shift to the right if Q<K
Given the following equivalents, make the following conversion: 1.00 knop = ? knips
4 clips = 5 blips
1 knop = 6 bippy
3 blip = 18 pringle
1 clip = 10 knip
10 bippy = 8 pringle
Answer:
[tex]6.4knips[/tex]
Explanation:
Hello,
In this case, given the stated equivalences, we can use the following proportional factor in order to compute the required knips:
[tex]knips=1.00knop*\frac{6bippy}{1knop} *\frac{8pringle}{10bippy}* \frac{3blip}{18pringle} *\frac{4clips}{5blips} *\frac{10knip}{1clip} \\\\=6.4knips[/tex]
Regards.
1 knop=6.4 knips
First convert knop to bippy:-
[tex]1\ knop\times\frac{6\ bippy}{1\ knop} =6\ bippy[/tex]
Now, Convert 6 bippy to pringle:-
[tex]6\ bippy\times\frac{8\ pringle}{10\ bippy} =4.8\ pringle[/tex]
Now, convert 4.8 pringle to blip:-
[tex]4.8\ pringle\times\frac{3\ blip}{18\ priangle} =0.8\ blip[/tex]
Now, convert 0.8 blip to clips as follows:-
[tex]0.8\ blip\times\frac{4\ clips}{5\ blip} =0.64\ clip[/tex]
Now, convert 0.64 clip to knips:-
[tex]0.64\ clip\times\frac{10\ knip}{1\ clip} =6.4\ knip[/tex]
Hence, the following conversion is as follows:-
1.00 knop=6.4 knips
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4NH3(g) 5O2(g)4NO(g) 6H2O(g) Using standard thermodynamic data at 298K, calculate the free energy change when 1.81 moles of NH3(g) react at standard conditions.
Answer:
-434.14 kJ
Explanation:
Step 1: Write the balanced equation
4 NH₃(g) + 5 O₂(g) ⇒ 4 NO(g) + 6 H₂O(g)
Step 2: Calculate the standard free energy change (ΔG°r) for the reaction
We will use the following expression.
ΔG°r = 4 mol × ΔG°f(NO(g)) + 6 mol × ΔG°f(H₂O(g)) - 4 mol × ΔG°f(NH₃(g)) - 5 mol × ΔG°f(O₂(g))
ΔG°r = 4 mol × (86.55 kJ/mol) + 6 mol × (-228.57 kJ/mol) - 4 mol × (-16.45 kJ/mol) - 5 mol × (0 kJ/mol)
ΔG°r = -959.42 kJ
Step 3: Calculate the standard free energy change for 1.81 moles of NH₃
959.42 kJ are released per 4 moles of NH₃.
[tex]\frac{-959.42 kJ}{4mol} \times 1.81mol = -434.14 kJ[/tex]
Using the standard reduction potentials Ni2+(aq) + 2 e‑Ni(s) ‑0.25 volt Fe3+(aq) + e‑Fe2+(aq) +0.77 volt Calculate the value of E°cell for the cell with the following reaction. Ni2+(aq) + 2 Fe2+(aq) →Ni(s) + 2 Fe3+(aq)
Answer:
The correct answer is - 1.02 V
Explanation:
From the reduction-oxidation reaction:
Ni²⁺(aq) + 2 Fe²⁺(aq) → Ni(s) + 2 Fe³⁺(aq)
Ni²⁺ is reduced to Ni(s) while Fe²⁺ is oxidized to Fe³⁺. Thus, the half reactions are:
Reduction (cathode) : Ni²⁺(aq) + 2 e‑ → Ni(s) Eº= ‑0.25 V
Oxidation (anode) : 2 x (Fe²⁺ → Fe³⁺ + e-)(aq) Eº= -0.77 V
-------------------------------------
Ni²⁺(aq) + 2 Fe²⁺(aq) → Ni(s) + 2 Fe³⁺(aq)
In order to calculate the Eºcell, we have to add the reduction potential of the reaction in cathode (reduction) to the oxidation potential of the anode (oxidation):
Eºcell= Eºr + Eºo= (-0.25 V) + (-0.77 V) = - 1.02 V
Which of the following is a salt that could be generated by combining a weak acid and a weak base? Select the correct answer below:
a) NaCl
b) Na2SO4
c) NH4NO3
d) NH4F
Answer:
d) NH4F
Explanation:
Hello,
In this case, the base resulting from mixing a weak acid and a weak base is d) NH4F since ammonium hydroxide is a wear base and hydrofluoric acid is a weak acid.
Ammonium hydroxide is a weak base since it is not completely ionized in ammonium and hydroxyl ions:
[tex]NH_4OH\rightarrow NH_4^++OH^-[/tex]
Moreover, hydrofluoric acid is a weak acid since it is not completely ionized in hydrogen and fluoride ions:
[tex]HF\rightleftharpoons H^++F^-[/tex]
For the both of the substances, the limit is established by the basic and the acid dissociation constant respectively.
Regards.
Gaseous methane (CH4) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO) and gaseous water (H2O) . Suppose 0.963 g of methane is mixed with 7.5 g of oxygen. Calculate the minimum mass of methane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.
Answer:
0 g.
Explanation:
Hello,
In this case, since the reaction between methane and oxygen is:
[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]
If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:
[tex]n_{CH_4}=0.963gCH_4*\frac{1molCH_4}{16gCH_4}=0.0602molCH_4\\ \\n_{CH_4}^{consumed}=7.5gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2} =0.117molCH_4[/tex]
Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.
[tex]left\ over=0g[/tex]
Regards.
From the following balanced equation, CH4(g)+2O2(g)⟶CO2(g)+2H2O(g) how many grams of H2O can be formed when 1.25g CH4 are combined with 1.25×10^23 molecules O2? Use 6.022×10^23 mol−1 for Avogadro's number.
Answer:
2.81 g of H2O.
Explanation:
We'll begin by calculating mass of O2 that contains 1.25×10²³ molecules O2.
This can be obtained as follow:
From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.022×10²³ molecules. This implies that 1 mole of O2 also contains 6.022×10²³ molecules.
1 mole of O2 = 16x2 = 32 g.
Thus 6.022×10²³ molecules is present in 32 g of O2,
Therefore, 1.25×10²³ molecules will be present in =
(1.25×10²³ × 32) / 6.022×10²³ = 6.64 g of O2.
Therefore, 1.25×10²³ molecules present in 6.64 g of O2.
Next, the balanced equation for the reaction. This is given below:
CH4(g) + 2O2(g) —> CO2(g) + 2H2O(g)
Next, we shall determine the masses of CH4 and O2 that reacted and the mass of H2O produced from the balanced equation.
This can be obtained as follow:
Molar mass of CH4 = 12 + (4x1) = 16 g/mol.
Mass of CH4 from the balanced equation = 1 x 16 = 16 g
Molar mass of O2 = 16x2 = 32 g/mol.
Mass of O2 from the balanced equation = 2 x 32 = 64 g
Molar mass of H2O = (2x1) + 16 = 18 g/mol.
Mass of H2O from the balanced equation = 2 x 18 = 36 g
From the balanced equation above,
16 g of CH4 reacted with 64 g of O2 to produce 36 g if H2O.
Next, we shall determine the limiting reactant.
This can be obtained as follow:
From the balanced equation above,
16 g of CH4 reacted with 64 g of O2.
Therefore, 1.25 g of CH4 will react with = (1.25 x 64)/16 = 5 g of O2.
From the above calculations, we can see that only 5 g out of 6.64 g of O2 is needed to react completely with 1.25 g of CH4.
Therefore, CH4 is the limiting reactant.
Finally, we shall determine the mass of H2O produced from the reaction.
In this case, the limiting reactant will be used because it will give the maximum yield of H2O.
The limiting reactant is CH4 and the mass of H2O produced from the reaction can be obtained as follow:
From the balanced equation above,
16 g of CH4 reacted to produce produce 36 g if H2O.
Therefore, 1.25 g of CH4 will react to produce = (1.25 x 36)/16 = 2.81 g of H2O.
Therefore, 2.81 g of H2O were obtained from the reaction.
The mass in grams of H₂O which can be formed when 1.25g CH₄ are combined with 1.25×10²³ molecules O₂ is 2.8 grams.
What is stoichiometry?Stoichiometry of any reaction tells about the amount of species present before and after the completion of the reaction.
Given chemical reaction is:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)
Moles of CH₄ will b calculate as:
n = W/M, where
W = given mass = 1.25g
M = molar mass = 16g/mol
n = 1.25/16 = 0.078 moles
Molecues of CH₄ in 0.078 moles = 0.078×6.022×10²³ = 0.46×10²³
Given molecules of O₂ = 1.25×10²³
Required molecules of CH₄ is less as compared to the molecules of O₂, so here CH₄ is the limiting reagent and formation of water is depends on it only.
From the stoichiometry of the reaction it is clear that:
1 mole of CH₄ = will produce 2 moles of H₂O
0.078 moles of CH₄ = will produce 2×0.078=0.156 moles of H₂O
Mass of H₂O will be calculated by using its moles as:
W = (0.156)(18) = 2.8g
Hence required mass of H₂O is 2.8g.
To know more about limiting reagent, visit the below link:
https://brainly.com/question/1163339
Calculate the concentration of H3O+ in a solution that contains 5.5 × 10-5 M OH- at 25°C. Identify the solution as acidic, basic, or neutral.
Explanation:
To calculate [H3O+] in the solution we must first find the pH from the [ OH-]
That's
pH + pOH = 14
pH = 14 - pOH
To calculate the pOH we use the formula
pOH = - log [OH-]
And [OH-] = 5.5 × 10^-5 M
So we have
pOH = - log 5.5 × 10^ - 5
pOH = 4.26
Since we've found the pOH we can now find the pH
That's
pH = 14 - 4.26
pH = 9.74
Now we can find the concentration of H3O+ in the solution using the formula
pH = - log H3O+
9.74 = - log H3O+
Find the antilog of both sides
H3O+ = 1.8 × 10^ - 10 MThe solution is basic since it's pH lies in the basic region.
Hope this helps you
The substance formed on addition of water to an aldehyde or ketone is called a hydrate or a/an:_______
A) vicinal diol
B) geminal diol
C) acetal
D) ketal
Answer:
B) geminal diol
Explanation:
Hello,
In this case, considering the attached picture, you can see that the substance resulting from the hydrolysis of an aldehyde or a ketone is a geminal diol since the two hydroxyl groups are in the same carbon. Such hydrolysis could be carried out in either acidic or basic conditions depending upon the equilibrium constant.
Regards.
The absorption spectrum of argon has a line at 515 nm. What is the energy of
this line? (The speed of light in a vacuum is 3.00 x 108 m/s, and Planck's
constant is 6.626 x 10-34 Jos.)
O A. 2.59 x 1027j
O B. 3.86 x 10-28 J
O C. 3.86 x 10-19 J
O D. 2.59 x 1018 J
Answer:
OPTION C is correct
3.86 x 10-19 J
Explanation:
Energy of the line can be calculated using below formula
E= h ν.................(1)
Where E= energy
h= plank constant= 6.626 10-34 J s
c=speed of light=3 x 108 m/s
But we know that Velocity V= = c / λ
Then substitute into equation (1) we have
E = h c / λ.............(2)
We can calculate our( hc ) in nm for unit consistency
h c =( 6.626 ×10^-34)x(3×108)
h c = (1.986 x 10-16 )
hc = 1.986 x 10-16 J nm then since our (hc) and λ are in the same unit , were good to go then substitute into equation(2)
E = h c / λ = (1.986 x 10-16) / 515
E = 3.86 x 10-19 J
Therefore, the Energy is 3.86 x 10-19 J
Match the words below to the appropriate blanks in the sentences. Make certain each sentence is complete before submitting your answerβ-1,4- and α-1,6-glycosidicβ-1,4-glycosidicgalactosean unbranchedglucosea branchedfructoseα-1,6-glycosidicAmylose is ......... polymer of ....... units joined by ........ bonds. Amylopectin is ....... polymer of .......units joined by ........ bonds.
The words given are not clear, so the clear question is as follows:
Match the words below to the appropriate blanks in the sentences. Make certain each sentence is complete before submitting your answer:
A. β-1,4- and α-1,6-glycosidic
B. α-1,4-glycosidic
C. α-1,4-galactose
D. an unbranched glucose
E. a branched fructose
F. α-1,6-glycosidic
Amylose is ......... polymer of ....... units joined by ........ bonds.
Amylopectin is ....... polymer of .......units joined by ........ bonds.
Answer:
D. an unbranched glucose
C. α-1,4-galactose
B. α-1,4-glycosidic
E. a branched fructose
A. β-1,4- and α-1,6-glycosidic
F. α-1,6-glycosidic
Explanation:
Amylose and amylopectin are two types of polysaccharides that can be found in starch granules.
Amylose is linear or unbranched glucose polymer of α-1,4-galactose units that are joined by α-1,4-glycosidic.
Amylopectin is a branched fructose polymer of β-1,4- and α-1,6-glycosidic units joined by α-1,6-glycosidic bonds.
Hence, the correct answers in the sequential order are:
Amylose:
D. an unbranched glucose
C. α-1,4-galactose
B. α-1,4-glycosidic
Amylopectin:
E. a branched fructose
A. β-1,4- and α-1,6-glycosidic
F. α-1,6-glycosidic
A chemist prepares a solution of sodium chloride by measuring out 25.4 grams of sodium chloride into a 100. mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist's sodium chloride solution. Be sure your answer has the correct number of significant digits.
Answer:
The concentration in mol/L = 4.342 mol/L
Explanation:
Given that :
mass of sodium chloride = 25.4 grams
Volume of the volumetric flask = 100 mL
We all know that the molar mass of sodium chloride NaCl = 58.5 g/mol
and number of moles = mass/molar mass
The number of moles of sodium chloride = 25.4 g/58.5 g/mol
The number of moles of sodium chloride = 0.434188 mol
The concentration in mol/L = number of mol/ volume of the solution
The concentration in mol/L = 0.434188 mol/ 100 × 10⁻³ L
The concentration in mol/L = 4.34188 mol/L
The concentration in mol/L = 4.342 mol/L
How many valence electrons must two atoms share to form a single covalent bond? answers A.2 B.4 C.3 D.1
Answer:
2
Explanation:
A single covalent bond is formed when two electrons are shared between the same two atoms, one electron from each atom.
Answer:
the answer is 2
Explanation:
