The biochemistry that takes place inside cells depends on various elements, such as sodium, potassium, and calcium, that are dissolved in water as ions. These ions enter cells through narrow pores in the cell membrane known as ion channels. Each ion channel, which is formed from a specialized protein molecule, is selective for one type of ion. Measurements with microelectrodes have shown that a 0.30-nm-diameter potassium ion (K+) channel carries a current of 1.8 pA. How many potassium ions pass through if the ion channel opens for 1.0 ms? What is the current density in the ion channel?
Answers
Answer:
Explanation:
The biochemistry that takes place inside cells depends on various elements, such as sodium, potassium, and
calcium, that are dissolved in water as ions. These ions enter cells through narrow pores in the cell membrane
known as ion channels. Each ion channel, which is formed from a specialized protein molecule, is selective
for one type of ion. Measurements with microelectrodes have shown that a 0.30-nm-diameter potassium ion (K+) channel carries a current of 1.8 pA.
Part A. How many potassium ions pass through if the ion channel opens for 1.0 ms?
Part B. What is the current density in the ion channel?
Solution: In 1.0 ms, the charge that passes through is
Q = I ∆t =
( 1.8 × 10−12 A ) (1.0 × 10−3 s )
= 1.8 × 10−15 C
Since each ion has a +1 charge (measured in electron charges), this represents
NK+ = Q e = (1.8 × 10−15 C ) (1.60 × 10−19 C) = 11250
The current density is calculated from the current and the size of the channel.
J = I A = ( 1.8 × 10−12 A ) ( π (0.30 × 10−9 m/2)2 ) = 2.55 × 107 A/m2
A) The number of potassium ions that will pass through the ion channel; 11250
B) The current density in the ion channel = 2.55 × 10⁷ A/m²
Given data;
Measurement with microelectrodes = 0.30-nm- diameter
K⁺ = 1.8 pA
a) calculate the number of potassium ions that passes through the ion channel
given that the channel opens for 1.0 ms
first step ; determine the value of charge ( Q )
Q = I*Δt = ( 1.8 × 10⁻¹² A )*(1.0 × 10⁻³ s ) = 1.8 × 10⁻¹⁵ C
where ; I = 1.8 × 10⁻¹² A
Δt = 1.0 × 10⁻³ s
next step : determine number of K⁺ ions passing through
NK⁺ = Q*e = ( 1.8 × 10⁻¹⁵ C )*( 1.60 × 10⁻¹⁹ C) = 11250
∴ number of K⁺ ions passing through = 11250 ions.
B) determine the current density in the ion channel
current density = current * size of channel
= ( 1.8 * 10⁻¹² ) * ( π * ( 0.30 * 10⁻⁹ )²
= 2.55 × 10⁻⁷ A/m²
Hence we can conclude that the number of potassium ions that will pass through the ion channel; 11250 and The current density in the ion channel = 2.55 × 10⁷ A/m²
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Related Questions
26. The heat of neutralization of HCl(aq) by NaOH(aq) is produced. If 50.00 mL of 1.05 M NaOH is added to 25.00 mL of 1.86 M HCl, with both solutions originally at what will be the final solution temperature? (Assume that no heat is lost to the surrounding air and that the solution produced in the neutralization reaction has a density of and a specific heat capacity of) Petrucci, Ralph H.. General Chemistry (p. 291). Pearson Education. Kindle Edition.
Answers
Answer:
T₂ = 33.2⁰C
Explanation:
Needed in problem text is the temperature of the acid and base solutions before reaction and the accepted (published) molar heat of neutralization (strong acids) for HCl by NaOH. => Assuming solution temperature is 25⁰C* for both acid and base solutions before reaction and the molar heat of neutralization for HCl by NaOH is 55.7Kj/mole,** then …
* Standard Thermodynamic conditions => 25⁰C (298K) & 1.00 Atm.
**(https://chemdemos.uoregon.edu/demos/Heat-of-Neutralization-HClaq-NaOHaq)
NaOH + HCl => NaCl + H₂O
=> 50ml(1.05M NaOH) + 25ml(1.86M HCl)
=> 0.05(1.05)mole NaOH + 0.025(1.86)mole HCl
=> 0.0525mole NaOH + 0.0465mole HCl
=> (0.0525 – 0.0465)mole NaOH excess + 0.0465mole NaCl + H₂O + Heat
=> 0.0060mole NaOH in excess + 0.0465mole NaCl + H₂O + Heat
Note => NaOH neutralizes 0.0465mole HCl (Limiting Reactant) and produces 0.0465mole NaCl & H₂O + Heat of Neutralization.
----------------------------------------------------------
Heat flow (Q) = Heat received by solvent water from the NaOH + HCl reaction
=> Q = mcΔT = mc(T₂ - T₁) = specific heat produced by 0.0465mole HCl
=> Q(m) = Molar Heat of Neutralization = Q/mole = mcΔT/n
- m = mass of solvent water receiving heat = (50ml + 25ml)1g/ml = 75g
- c = specific heat of water = 4.184j/g⁰C
- T₂ - T₁ = T₂ - 25⁰C
- Q(m) = 55,700 joules/mole (published heat of neutralization)
- n = moles of HCl neutralized = 0.0465mole HCl
=>Q(m) = mcΔT/n = 55,700j/mole = (75g)(4.184j/g⁰C)(T₂ - 25⁰C) /0.0465mole
Solving for T₂ => T₂ = 33.2⁰C
Final answer:
The final temperature for the neutralization of HCl by NaOH cannot be determined without specific values for initial temperatures, heat of neutralization, specific heat capacity, and density. For a similar reaction with provided values, the final temperature can be calculated based on the produced heat and the specific heat capacity.
Explanation:
The question pertains to the heat of neutralization of HCl(aq) by NaOH(aq). To find out the final temperature of the solution, we would typically calculate the amount of heat produced during the reaction and use it together with the specific heat capacity of the solution. However, the details required for calculation (initial temperatures, heat of neutralization value, specific heat capacity, density) are not provided in the question. In an example provided in the reference, when 50.0 mL of 1.00 M HCl and 50.0 mL of 1.00 M NaOH, both at 22.0 °C in a coffee cup calorimeter, the mixture reached 28.9 °C. The exothermic reaction between HCl and NaOH produces NaCl and H₂O, and the heat produced can be calculated using the change in temperature, the mass of the solution, and the specific heat capacity.
Given the following equation: 2 C4H10 13 O2 > 8 CO2 10 H20 + How many grams of CO2 are produced if 12.4 grams of C4H10 reacts with 56.9 grams of O2?
Answers
Answer : The mass of [tex]CO_2[/tex] produced will be, 37.488 grams.
Explanation : Given,
Mass of [tex]C_4H_{10}[/tex] = 12.4 g
Mass of [tex]O_2[/tex] = 56.9 g
Molar mass of [tex]C_4H_{10}[/tex] = 58 g/mole
Molar mass of [tex]O_2[/tex] = 32 g/mole
Molar mass of [tex]CO_2[/tex] = 44 g/mole
First we have to calculate the moles of [tex]C_4H_{10}[/tex] and [tex]O_2[/tex].
[tex]\text{Moles of }C_4H_{10}=\frac{\text{Mass of }C_4H_{10}}{\text{Molar mass of }C_4H_{10}}=\frac{12.4g}{58g/mole}=0.213moles[/tex]
[tex]\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{56.9g}{32g/mole}=1.778moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O[/tex]
From the balanced reaction we conclude that
As, 2 moles of [tex]C_4H_{10}[/tex] react with 13 mole of [tex]O_2[/tex]
So, 0.213 moles of [tex]C_4H_{10}[/tex] react with [tex]\frac{13}{2}\times 0.213=1.385[/tex] moles of [tex]O_2[/tex]
From this we conclude that, [tex]O_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]C_4H_{10}[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]CO_2[/tex].
As, 2 moles of [tex]C_4H_{10}[/tex] react to give 8 moles of [tex]CO_2[/tex]
So, 0.213 moles of [tex]C_4H_{10}[/tex] react to give [tex]\frac{8}{2}\times 0.213=0.852[/tex] moles of [tex]CO_2[/tex]
Now we have to calculate the mass of [tex]CO_2[/tex].
[tex]\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass of }CO_2[/tex]
[tex]\text{Mass of }CO_2=(0.852mole)\times (44g/mole)=37.488g[/tex]
Therefore, the mass of [tex]CO_2[/tex] produced will be, 37.488 grams.
H2 (g) + Br2 (g) <=> 2 HBr (g)
the equilibrium constant is 13485. At equilibrium the H2 concentration is 0.05 M, while the Br2 concentration is 0.023 M. Calculate the HBr concentration at equilibrium, to 1 decimal. Be careful with the units.
Answers
Answer:
[HBr] = 4.7M at equilibrium
Explanation:
When CO2(g) reacts with H2(g) to form CO(g) and H2O(g) , 9.85 kcal of energy are absorbed for each mole of CO2(g) that reacts. Write a balanced equation for the reaction with an energy term in kcal as part of the equation.
Answers
Answer: The balanced chemical equation is written below.
Explanation:
A balanced chemical equation is defined as the equation in which total number of individual atoms on the reactant side must be equal to the total number of individual atoms on the product side.
For the reaction of carbon dioxide with hydrogen gas, 9.85 kcal of energy is absorbed. So, this energy term will be written on the reactant side.
Thus, the balanced chemical equation for the reaction of carbon dioxide with hydrogen gas follows:
[tex]CO_2(g)+H_2(g)+9.85kcal\rightarrow CO(g)+H_2O(g)[/tex]
Hence, the balanced chemical equation is written above.
A reaction in which A, B, and C react to form products is zero order in A, one-half order in B, and second order in C. By what factor does the reaction rate change if [A] is doubled (and the other reactant concentrations are held constant)?
Answers
Answer: There will be no change in rate.
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
[tex]A+B+C\rightarrow Products[/tex]
[tex]Rate=k[A]^x[B]^y[C]^z[/tex]
k= rate constant
x = order with respect to A = 0
y = order with respect to B = [tex]\frac{1}{2}[/tex]
z= order with respect to C = 2
Thus [tex]Rate=k[A]^0[B]^{\frac{1}{2}}[C]^2[/tex]
Given : when [A] is doubled and the other reactant concentrations are held constant.
Thus the new rate law is [tex]Rate'=k[2A]^0[B]^{\frac{1}{2}}[C]^2[/tex]
[tex]Rate'=k[2]^0[A]^0[B]^{\frac{1}{2}}[C]^2[/tex]
[tex]Rate'=k[A]^0[B]^{\frac{1}{2}}[C]^2[/tex] [tex](2^0=1)[/tex]
[tex]Rate'=Rate[/tex]
Thus the reaction rate would not change.
Doubling the concentration of A has no effect on the reaction rate because the reaction is zero order in A. The rate law for the reaction is rate = k[B]^1/2[C]², indicating independence from [A].
The question pertains to chemical kinetics, specifically how the reaction rate changes with respect to changes in reactant concentrations. Since the reaction is zero order with respect to A, doubling the concentration of A ([A]) will have no effect on the rate of the reaction. The rate law for this reaction could be represented as rate =k[B]^1/2[C]², which clearly shows that the reaction rate is independent of the concentration of A. Therefore, the reaction rate remains unchanged if the concentration of A is doubled while keeping B and C constant.
A student needs to determine the volume occupied by a gas in a 125 mL flask using the experimental setup described in the procedure. The student measures the volume of the flask to be 157 mL to the top of the flask. The student measures the volume of the flask with a stopper in it to be 140 mL. The student performs the experiment by reacting the strip of magnesium with 5 mL of HCl solution. What is the volume of the flask occupied by the hydrogen gas?
Answers
Answer:
see explanation...
Explanation:
Given 125ml standard flask
Measured volume to top = 157ml
Measured volume with stopper = 140ml
------------------------------------------------------------------------------------------
Based on the wording of the question, the volume of the flask is ...
= 157ml without stopper*, or
= 140ml with stopper.
If interest is in volume occupied by H₂(g) above 5ml of rxn solution ...
=> Volume without stopper* = 157ml - 5ml = 152ml above rxn solution
=> Volume with stopper = 140ml - 5ml = 135ml (Assuming stopper doesn't pop out of container). :-)
*Assuming gas fills available volume without escaping container.
Answer:
Volume of flask without stopper=157 mL-5mL=152 mL
Volume of flask with stopper=140-5=135 mL
A compound is 54.53% C,54.53% C, 9.15% H,9.15% H, and 36.32% O36.32% O by mass. What is its empirical formula? Insert subscripts as needed. empirical formula: CHOCHO The molecular mass of the compound is 132 amu.132 amu. What is its molecular formula? Insert subscripts as needed. molecular formula: CHO
Answers
The empirical formula and the molecular formula for the given compound will be [tex]C_2H_4O[/tex] and [tex]C_6H_{12} O_3[/tex] respectively.
The simplest whole-number ratio of atoms in a molecule is the empirical formula of a chemical compound in chemistry.
Let us consider that, we have 100g of the sample, then carbon would be 54.53 g, hydrogen would be 9.15g and oxygen would be 36.32g and the number of moles for each element will be = [tex]\frac{GivenMass}{Molar Mass}[/tex]
Number of moles of carbon = [tex]\frac{54.53}{12} = 4.544[/tex] mol
Similarly, the number of moles of Hydrogen and Oxygen would be:
Number of moles of Oxygen = [tex]\frac{36.32}{16} = 2.27\\[/tex] mol
Number of moles of Hydrogen = [tex]\frac{9.15}{1} = 9.15 mol[/tex]
Molar ratio of the elements = C:H:O = 4.54 : 9.15 : 2.27
The molar ratio = 2:4:1,
So, the empirical formula = [tex]C_2H_4O[/tex] and empirical mass = [tex]2*12 + 4*1 + 16 = 44[/tex]
The empirical formula and the molecular formula are often related in the following way: (Molecular Formula = n [tex]*[/tex] Empirical Formula). This may be used to determine the compound's empirical formula as well as its molecular formula. The term “n” in the relationship denotes the proportion between the compound's molecular mass and empirical mass.
[tex]n = \frac{Molecular Mass}{Empirical Mass} = \frac{132}{44} = 3[/tex]
So, molecular formula = [tex]3* C_2H_4O = C_6 H_{12} O_3[/tex]
Thus, the molecular formula and empirical formula for the given compound are [tex]C_2H_4O[/tex] and [tex]C_6H_{12} O_3[/tex].
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The empirical and molecular formula of a compound with 54.53% carbon, 9.15% hydrogen, and 36.32% oxygen and molar mass 132 amu, is CH₄O₂.
Explanation:The process to find the empirical and molecular formula of a compound uses percentage composition of its constituent elements and molar mass. First, we convert the percentage to grams (assuming 100g of the compound) which gives: C 54.53g, H 9.15g, and O 36.32g. Using atomic masses of C(12.01), H(1.01), and O(16.00), we find moles of C, H, and O.
Dividing the obtained values by the smallest number of moles, gives us the ratio of the elements. In this case, the empirical formula comes out to be CH₄O₂. We then calculate the molecular mass of the empirical formula and divide the molar mass of the compound by the empirical formula mass. The quotient gives the number of empirical formula units present in the compound's molecular formula. If this quotient is anything other than '1', we must multiply the subscript of each element in the empirical formula by this number to get the molecular formula. However, in this case, the empirical and molecular formulas are the same, i.e., CH₄O₂.
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Which of the following reactions will not occur as written? Which of the following reactions will not occur as written? Sn (s) + 2AgNO3 (aq) → 2Ag (s) + Sn(NO3)2 (aq) Zn (s) + Pb(NO3)2 (aq) → Pb (s) + Zn(NO3)2 (aq) Co (s) + 2HI (aq) → H2 (g) + CoI2 (aq) Mg (s) + Ca(OH)2 (aq) → Ca (s) + Mg(OH)2 (aq) Co (s) + 2AgCl (aq) → 2Ag (s) + CoCl2 (aq)
Answers
Answer: Option (4) is the correct answer.
Explanation:
According to the reactivity series, a more reactive metal has the ability to replace a less reactive metal in a chemical reaction.
It is known that calcium is more reactive than magnesium. So, in a chemical reaction magnesium can never replace calcium.
For example, [tex]Mg(s) + Ca(OH)_2(aq) \rightarrow Ca(s) + Mg(OH)_2(aq)[/tex]
Therefore, this given reaction is not possible. But rest all given reactions are possible.
Thus, we can conclude that out of the given options [tex]Mg(s) + Ca(OH)_2(aq) \rightarrow Ca(s) + Mg(OH)_2(aq)[/tex] reaction will not occur as written.
OPTION D.
The reaction Mg (s) + Ca(OH)₂ (aq) → Ca (s) + Mg(OH)₂ (aq) will not occur as written because in the activity series of metals, magnesium (Mg) is more reactive than calcium (Ca), and hence, Mg cannot displace Ca from its compound.
Explanation:In the given set of reactions, Mg (s) + Ca(OH)₂ (aq) → Ca (s) + Mg(OH)₂ (aq) will not occur as written. The explanation is based on activity series of metals. In this series, magnesium (Mg) is more reactive than calcium (Ca). So, Mg cannot displace Ca from its compound. In other words, a more reactive metal can displace a less reactive metal from its compound in a solution, but not the other way round. The other reactions would occur as represented since in each of them the free metal is more reactive than the one in the compound.
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A 298 lb person must receive Heparin, determine the number of units given every hour (Heparin 8.0 units/kg per hour). Enter your answer with 1 decimal place and no units (the understood unit in this problem is units).
Answers
Answer:
1,081.1 units of heparin should be given to 298 lb person.
Explanation:
We are given:
Weight of the person = 298 lb = 135.143 kg
Conversion factor used:
1 lb = 0.4535 kg
Number of unit of heparin to be given in an hour = 8.0 units/kg
Number of units given to the patient weighing 135.143 kg :
[tex]8.0 units/kg\times 135.143 kg=1,081.144 units\approx 1,081.1 units[/tex]
1,081.1 units of heparin should be given to 298 lb person.
Electron configurations are a shorthand form of an orbital diagram, describing which orbitals are occupied for a given element. For example, 1s22s22p1 is the electron configuration of boron. Use this tool to generate the electron configuration of arsenic (As).
Answers
Answer:
[Ar] 4s² 3d¹⁰ 4p³
Explanation:
Arsenic has an atomic number of 33, thus 33 protons and 33 electrons. The electron configuration is:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p³
The shorthand notation includes the previous noble gas to represent the inner electrons. The gas previous to arsenic is argon, with 18 electrons. The shorthand notation for arsenic is:
[Ar] 4s² 3d¹⁰ 4p³
Arsenic (Ar) is located in the 5th period (row) and 15th group (column) of the periodic table. Its atomic number is 33, which means it has 33 electrons. The electrons' configuration is: [tex]1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^3[/tex].
This is the electron configuration of arsenic (Ar). It indicates that arsenic has two electrons in the 1s orbital, two electrons in the 2s orbital, six electrons in the 2p orbital, two electrons in the 3s orbital, six electrons in the 3p orbital, two electrons in the 4s orbital, ten electrons in the 3d orbital, and three electrons in the 4p orbital.
The electronic configuration of Arsenic is [tex]1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^3[/tex].
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A KCl solution is prepared by dissolving 40.0 g KCl in 250.0 g of water at 25°C. What is the vapor pressure of the solution if the vapor pressure of water at 25°C is 23.76 mm Hg?
Answers
Answer:
22.0 mmHgExplanation:
The vapor pressure of a solution is a colligative property, which means that it is determined by the number of particles (molecules or ions) of solute present in a solution.
Raoult's law permits the calculations of the change of the vapor pressure of a solvent when a solute is added.
The equation is:
P solvent - P solution = ΔP = X solute × P solvenWhere:
P solvent = vapor pressure of the pure solvent.P solution = vapor pressure of the solutionX solute = molar fraction of the soluteIn the case of ionic solutes, you must take into account the number of ions that result from the ionization.
Calculating the molar fraction:
number of moles = mass in grams / molar massnumber of moles of KCl: 40.0 g / 74.5513 g/mol = 0.567 molmoles of ions = 2× number of moles of KCl = 1.134 molmoles of water: 250.0g / 18.015 g/mol = 13.877 moltotal moles = 1.134 mol + 13.877 mol = 15.011 molX solute = moles of ions / total moles = 1.134 mol / 15.011 mol = 0.0755Calculating the change in the vapor pressure of the solution:
ΔP = X solute × P solvent = 0.0755 × 23.76 mmHg = 1.78 mmHgVapor pressure of the solution:
P solution = P solvent + ΔP = 23.76 mmHg - 1.79 mm Hg = 21.97mmHgRounding to three significant figures (because 40.0g has three significant figures): 22.0 mmHg ← answer.
Raoult's law states that for a given solution, the partial pressure of each component is equal to the mole fraction in that solution. The vapor pressure of the solution is 22 mmHg.
The vapor pressure is defined as the force exerted by the vapors in the walls of a container. It is a colligative property, such that the amount of substance increased or decreased is dependent on the amount of solute present.
The equation can be represented as:
P[tex]_{\text{solvent}}[/tex] - P[tex]_{\text{solution}}[/tex] = [tex]\Delta[/tex] P = X [tex]_{\text{solute}}[/tex] x P[tex]_{\text{solvent}}[/tex]where,
P[tex]_{\text{solvent}}[/tex] = vapor pressure of the pure solventP[tex]_{\text{solution}}[/tex] = vapor pressure of the solutionX[tex]_{\text{solute}}[/tex] = molar fraction of the soluteNow, calculating the mole fraction, where:
Moles of ions = 2 × number of moles of KCl = 1.134 molNumber of moles KCl = [tex]\dfrac{40}{74.55}&= 0.567[/tex] molesTotal moles are: 1.134 mol + 13.877 mol = 15.011 molX[tex]_{\text{solute}}[/tex] = [tex]\dfrac{1.134}{15.01}&= 0.075[/tex]Now, the vapor pressure of the solution can be calculated as:
[tex]\Delta[/tex] P = X[tex]_{\text{solute}}[/tex] x P[tex]_{\text{solvent}}[/tex]
[tex]\Delta[/tex] P = 0.0755 × 23.76 mmHg = 1.78 mmHg
Hence, the vapor pressure of the solution:
P[tex]_{\text{solution}}[/tex] = [tex]\Delta[/tex] P + P[tex]_{\text{solvent}}[/tex]
P[tex]_{\text{solution}}[/tex] = 23.76 mmHg - 1.79 mm Hg = 21.97mmHg
Therefore, the vapor pressure of the solution is approximately 22 mmHg.
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Which one of the following is an oxidation-reduction reaction?
NaOH + HNO3 --> H2O + KNO3
NaOH + HNO3 --> H2O + KNO3
SO3 + H2O --> H2SO4
CaCl2 + Na2CO3 --> CaCO3 + 2 NaCl
CH4 + 2 O2 --> CO2 + 2 H2O
Al2(SO4)3 + 6 KOH --> 2 Al(OH)3 + 3 K2SO4
Answers
Answer:
CH4 + 2 O2 --> CO2 + 2 H2O
Explanation:
CH4 + 2 O2 --> CO2 + 2 H2O is the only reaction where an element (oxygen) undergoes a change in oxidation state. In this reaction oxygen changes disproportionately to O⁻². That is ...
O₂ → CO₂ + 4e⁻ ==> oxidation
O₂ + 4e⁻ → H₂O ==> reduction
2O₂ + 4e⁻ → CO₂ + H₂O + 4e⁻ ==> Net oxidation-reduction
=> 4e⁻ gained by one mole O₂ in formation of CO₂ = 4e⁻ lost by the other mole O₂ in forming H₂O.
Then...
Including CH₄ (whose elements do not undergo changes in oxidation states) requires doubling reaction to balance by mass thus giving ...
2CH₄ + 2O₂ + 8e⁻ → 2CO₂ + 2H₂O + 8e⁻
Cancelling 8 reduction electrons on left with 8 oxidation electrons on right gives...
2CH₄ + 2O₂ → 2CO₂ + 2H₂O
Answer:
CH₄ + 2O₂ ⟶ CO₂ + 2H₂O
Explanation:
To identify an oxidation-reduction reaction, you must determine the oxidation number of every atom involved in the reaction and see if it changes.
The only reaction where two elements change oxidation number is the oxidation of methane.
Here's the oxidation number of every atom involved:.
[tex]\stackrel{\hbox{-4}}{\hbox{C}}\stackrel{\hbox{+1}}{\hbox{H}}_{4} +\stackrel{\hbox{0}}{\hbox{O}}_{2} \, \longrightarrow \, \stackrel{\hbox{+4}}{\hbox{C}}\stackrel{\hbox{-2}}{\hbox{O}}_{2} + \stackrel{\hbox{+1}}{\hbox{H}}_{2}\stackrel{\hbox{-2}}{\hbox{O}}[/tex]
We see that some elements change oxidation numbers.
C: -4 ⟶ +4; increase in oxidation number = oxidation
O: 0 ⟶ -2; decrease in oxidation number = reduction
H: +1 ⟶ +1; no change.
The reaction is an oxidation-reduction reaction, because carbon is oxidized, and oxygen is reduced.
Which of the following statements is completely correct? a. NH3 is a weak base, and H2CO3 is a strong acid. b. H2CO3 is a strong acid, and NaOH is a strong base. c. NH3 is a weak base, and HCl is a strong acid. d. H2CO3 is a weak acid, and NaOH is a weak base. e. NH3 is a strong base, and HCl is a weak acid.
Answers
Answer: The correct answer is Option c.
Explanation:
Weak acid is defined as the acid which does not get completely dissociated into its ions when dissolved in water. For Example: [tex]CH_3COOH,H_2CO_3[/tex] etc..
Strong acid is defined as the acid which gets completely dissociated into its ions when dissolved in water. For Example: [tex]HCl,HNO_3[/tex] etc..
Weak base is defined as the base which does not get completely dissociated into its ions when dissolved in water. For Example: [tex]NH_3,NH_4OH[/tex] etc..
Strong base is defined as the base which gets completely dissociated into its ions when dissolved in water. For Example: [tex]NaOH,KOH[/tex] etc..
From the above information, it is clearly visible that the correct answer is Option c.
The correct statement is that C. NH3 is a weak base, and HCl is a strong acid.
Explanation:When a weak base, such as ammonia (NH3), reacts with a strong acid, like hydrochloric acid (HCl), a chemical reaction occurs. The acid donates protons (H+) to the base, forming water and the conjugate acid of the weak base. For example, in the reaction between NH3 and HCl, [tex]NH3 + HCl -- NH4+ + Cl-,[/tex] ammonium chloride is formed. The resulting solution is acidic due to the presence of excess H+ ions.
Thus, the correct statement among the options is c. NH3 is a weak base, and HCl is a strong acid. Ammonia (NH3) is a weak base because it donates only a partial amount of its lone pair of electrons. Hydrochloric acid (HCl) is a strong acid because it completely ionizes in water to produce a high concentration of hydrogen ions.
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An aqueous solution of hydroiodic acid is standardized by titration with a 0.186 M solution of calcium hydroxide. If 26.5 mL of base are required to neutralize 20.3 mL of the acid, what is the molarity of the hydroiodic acid solution? M hydroiodic acid
Answers
Answer: The molarity of hydroiodic acid in the titration is 0.485 M.
Explanation:
To calculate the molarity of acid, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HI[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is [tex]Ca(OH)_2[/tex]
We are given:
[tex]n_1=1\\M_1=?M\\V_1=20.3mL\\n_2=2\\M_2=0.186M\\V_2=26.5mL[/tex]
Putting values in above equation, we get:
[tex]1\times M_1\times 20.3=2\times 0.186\times 26.5\\\\M_1=0.485M[/tex]
Hence, the molarity of hydroiodic acid is 0.485M.
Final answer:
The molarity of the hydroiodic acid solution is calculated using the stoichiometry of its reaction with calcium hydroxide and the volumes and molarity of the titrant, resulting in a molarity of 0.485 M for the hydroiodic acid.
Explanation:
To find the molarity of the hydroiodic acid solution, we need to first understand the reaction occurring between hydroiodic acid (HI) and calcium hydroxide (Ca(OH)₂). The balanced chemical equation for this reaction is: 2HI(aq) + Ca(OH)₂(aq) → CaI₂(aq) + 2H₂O(l). This equation indicates that two moles of HI react with one mole of Ca(OH)₂.
Using the titration information provided: 26.5 mL of 0.186 M Ca(OH)₂ were required to neutralize 20.3 mL of HI. From this, we can calculate the moles of Ca(OH)₂ used (Moles = Molarity × Volume in L), which is (0.186 M) × (0.0265 L) = 0.004929 moles of Ca(OH)₂. Given the stoichiometry of the reaction, there are twice as many moles of HI, so 0.009858 moles of HI were neutralized.
To find the molarity of the hydroiodic acid solution, we use the formula Molarity = Moles/Volume in L. Here, it is 0.009858 moles / 0.0203 L = 0.485 M HI.
Lattice energy is __________. A. the energy required to convert a mole of ionic solid into its constituent ions in the gas phase B. the sum of ionization energies of the components in an ionic solid C. the sum of electron affinities of the components in an ionic solid D. the energy required to produce one mole of an ionic compound from its constituent elements in their standard states E. the energy given off when gaseous ions combine to form one mole of an ionic solid
Answers
Answer: Option (A) is the correct answer.
Explanation:
Lattice energy is defined as the energy needed by an ionic solid to break its constituent components into gaseous ions.
For example, [tex]NaCl(s) \rightarrow Na^{+}(g) + Cl^{-}(g)[/tex]
Two factors that help in determining lattice energy are changes on the ions and size of ions.
So, larger is the charge on ions, smaller will be the size of ions. As a result, more energy is required to break the bond between ions. Hence, lattice energy will also increase.
Whereas when smaller is the charge on ions, larger will be the size of ions. As a result, less energy is required to break the bond between ions. Hence, lattice energy will also decrease.
Lattice energy is the energy required to convert a mole of ionic solid into its constituent ions in the gas phase. It is an important factor in the ionic bond strength and varies depending on the charges of the ions and the distances between them. The lattice energy can be calculated via Coulomb's law.
Explanation:Lattice energy is the energy required to convert a mole of ionic solid into its constituent ions in the gas phase, as option A suggests. Just to give an example, for the ionic solid MX, the lattice energy is the enthalpy change of the process: AH lattice MX (s) Mn+ (g) + X(g). Lattice energy gives us an insight into ionic bond strength, which varies depending on the charges of the ions and the distances between them.
The lattice energy of an ionic crystal can be expressed by an equation derived from Coulomb's law, which governs the forces between electric charges. The lattice energy increases as the charges of the ions increase and the sizes of the ions decrease. For instance, the lattice energy of LiF is 1023 kJ/mol, whereas that of MgO is significantly higher at 3900 kJ/mol due to increased charges.
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Magnesium and nitrogen react in a combination reaction to produce magnesium nitride: 3 Mg + N2→ Mg3N2 In a particular experiment, a 5.65-g sample of N2 reacts completely. The mass of Mg consumed is ________ g.
Answers
Answer: The mass of magnesium consumed will be 14.731 g.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
Given mass of nitrogen gas = 5.65 g
Molar mass of nitrogen gas = 28 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of }N_2=\frac{5.65g}{28g/mol}=0.202mol[/tex]
For the given chemical equation:
[tex]3Mg+N_2\rightarrow Mg_3N_2[/tex]
By Stoichiometry of the reaction:
1 mole of nitrogen gas reacts with 3 moles of magnesium.
So, 0.202 moles of nitrogen gas will react with = [tex]\frac{3}{1}\times 0.202=0.606mol[/tex] of magnesium.
Now, calculating the mass of magnesium by using equation 1, we get:
Moles of magnesium = 0.606 moles
Molar mass of magnesium = 24.31 g/mol
Putting values in equation 1, we get:
[tex]0.606mol=\frac{\text{Mass of magneisum}}{24.31g/mol}\\\\\text{Mass of magnesium}=14.731g[/tex]
Hence, the mass of magnesium consumed will be 14.731 g.
(a) Write the balanced neutralization reaction that occurs between H2SO4 and KOH in aqueous solution. Phases are optional. (b) Suppose 0.750 L of 0.480 M H2SO4 is mixed with 0.700 L of 0.290 M KOH. What concentration of sulfuric acid remains after neutralization?
Answers
These are two questions and two answers
Answer:
Question 1:
H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l)Question 2:
0.201 MExplanation:
Question 1:
The neutralization reaction that occurs between H₂SO₄ and KOH is an acid-base reaction.
The products of an acid-base reaction are salt and water.
This is the sketch of such neutralization reaction:
1) Word equation:
sulfuric acid + potassium hydroxide → potassium sulfate + water↑ ↑ ↑ ↑
acid base salt water
2) Skeleton equation (unbalanced)
H₂SO₄ + KOH → K₂SO₄ + H₂O#) Balanced chemical equation (including phases)
H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l) ← answerQuestion 2:
1) Mol ratio:
Using the stoichiometric coefficients of the balanced chemical equation you get the mol ratio:
1 mol H₂SO₄ (aq) : 2 mol KOH (aq) : 1 mol K₂SO₄ (aq) : mol 2H₂O (l)2) Moles of H₂SO₄:
V = 0.750 literM = 0.480 mol/literM = n/V ⇒ n = M×V = 0.480 mol/liter × 0.750 liter = 0.360 mol3) Moles of KOH:
V = 0.700 literM = 0.290 mol/literM = n/V ⇒ n = M × V = 0.290 mol/liter × 0.700 liter = 0.203 mol4) Determine the limiting reagent:
a) Stoichiometric ratio:
1 mol H₂SO₄ / 2 mol NaOH = 0.500 mol H₂SO4 / mol NaOH
b) Actual ratio:
0.360 mol H₂SO4 / 0.203 mol NaOH = 1.77 mol H₂SO₄ / mol NaOH
Since hte actual ratio of H₂SO₄ is greater than the stoichiometric ratio, you conclude that H₂SO₄ is in excess.
5) Amount of H₂SO₄ that reacts:
Since, KOH is the limiting reactant, using 0.203 mol KOH and the stoichiometryc ratio 1 mol H₂SO₄ / 2 mol KOH, you get:x / 0.203 mol KOH = 1 mol H₂SO₄ / 2 mol KOH ⇒
x = 0.203 / 2 = 0.0677 mol of H₂SO₄
6) Concentration of H₂SO₄ remaining:
Initial amount - amount that reacted = 0.360 mol - 0.0677 mol = 0.292 molTotal volume = 0.700 liter + 0.750 liter = 1.450 literConcetration = MM = n / V = 0.292 mol / 1.450 liter = 0.201 M ← answer
The balanced neutralization reaction between H2SO4 and KOH is H2SO4 + 2KOH → K2SO4 + 2H2O. To calculate the concentration of sulfuric acid remaining after neutralization, use the moles of KOH reacted and the mole ratio in the balanced equation.
Explanation:(a) The neutralization reaction between H2SO4 (sulfuric acid) and KOH (potassium hydroxide) in aqueous solution can be represented as:
H2SO4 + 2KOH → K2SO4 + 2H2O
(b) To determine the concentration of sulfuric acid remaining after neutralization, we need to calculate the moles of KOH reacted using the given volumes and concentrations. Since H2SO4 has a 1:2 mole ratio with KOH in the equation, half of the moles of KOH reacted will be the moles of sulfuric acid neutralized. Therefore, the concentration of sulfuric acid remaining can be calculated by subtracting the moles of sulfuric acid neutralized from the initial concentration.
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Of the following gases, ________ will have the greatest rate of effusion at a given temperature. Of the following gases, ________ will have the greatest rate of effusion at a given temperature. NH3 HCl CH4 Ar HBr
Answers
Answer: From the given gases, the greatest rate of effusion is of [tex]CH_4[/tex]
Explanation:
Rate of effusion of a gas is determined by a law known as Graham's Law.
This law states that the rate of effusion or diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:
[tex]\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}[/tex]
It is visible that molar mass is inversely related to rate of effusion. So, the gas having lowest molar mass will have the highest rate of effusion.
For the given gases:
Molar mass of [tex]NH_3=17g/mol[/tex]
Molar mass of [tex]HCl=36.5g/mol[/tex]
Molar mass of [tex]CH_4=16g/mol[/tex]
Molar mass of [tex]Ar=40g/mol[/tex]
Molar mass of [tex]HBr=81g/mol[/tex]
The molar mass of methane gas is the lowest. Thus, it will have the greatest rate of effusion.
Hence, the greatest rate of effusion is of [tex]CH_4[/tex]
Answer:
CH4
CH4
Explanation:
I need help knowing how to find the name of the element. I am able to do the calculations to find the relative atomic mass of the element. Please help.
Isotope ?-35 has an atomic mass of 35.
75% of all atoms of element ? are of this form.
Isotope ?-37 has an atomic mass of 37.
25% of all atoms of element ? are of this form.
What is the relative atomic mass of the element?
Now, what is this element's name?
Answers
Answer: Chlorine, amu 35.45
Calculate the relative atomic mass by multiplying the mass of each isotope by its percentage, then sum. The relative atomic mass is 35.5, which corresponds to the element chlorine.
Explanation:The relative atomic mass of an element is the weighted average mass of the atoms in a naturally occurring sample of the element, taking into account the percentages of each isotope present. You seem to have two isotopes of the same element present: Isotope-35 and Isotope-37.
To calculate the relative atomic mass, you multiply the mass of each isotope by the percentage (in decimal form) in which it is found, and then sum those values. Let's see how this works:
Isotope-35: 35 * 0.75 = 26.25 Isotope-37: 37 * 0.25 = 9.25
Add these two values together (26.25 + 9.25) to find the relative atomic mass: 35.5. The element with a relative atomic mass of approximately 35.5 is chlorine (Cl).
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Given the following equation: |4 NH3 (g) + 5 О2 (g) —> 4 NO (g) + 6 H20 () How many grams of H20 is produced if 21.1 grams of NH3 reacts with 73.9 grams of O2?
Answers
Answer: The mass of water that can be formed are 33.48 g.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ....(1)
For ammonia gas:Given mass of ammonia gas = 21.1 g
Molar mass of ammonia gas = 17 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of ammonia}=\frac{21.1g}{17g/mol}=1.24mol[/tex]
For oxygen gas:Given mass of oxygen gas = 73.9 g
Molar mass of oxygen gas = 32 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of oxygen gas}=\frac{73.9g}{32g/mol}=2.30mol[/tex]
For the given chemical equation:
[tex]4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(l)[/tex]
By Stoichiometry of the reaction:
4 moles of ammonia gas reacts with 5 moles of oxygen gas.
So, 1.24 moles of ammonia gas will react with = [tex]\frac{5}{4}\times 1.24=1.55moles[/tex] of oxygen gas.
As, given amount of oxygen gas is more than the required amount. Thus, it is considered as an excess reagent.
So, ammonia gas is considered as a limiting reagent because it limits the formation of products.
By Stoichiometry of the above reaction:
4 moles of ammonia gas is producing 6 moles of water.
So, 1.24 moles of ammonia gas will produce = [tex]\frac{6}{4}\times 1.24=1.86moles[/tex] of water.
Now, calculating the mass of water by using equation 1, we get:
Moles of water = 1.86 moles
Molar mass of water = 18 g/mol
Putting all the values in equation 1, we get:
[tex]1.86mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=33.48g[/tex]
Hence, the mass of water that can be formed are 33.48 g
If 100.0 mL of 0.453 M Na2SO4 are added to 100.0 mL of 0.907 M Pb(NO3)2, how many grams of PbSO4 can be produced?
Na2SO4(aq)+Pb(NO3)2(aq)⟶2NaNO3(aq)+PbSO4(s)
Answers
The mass of PbSO₄ that could be produced is 13.74 g
From the question,
We are to determine the mass of PbSO₄ that could be produced from the reaction
The given balanced chemical equation for the reaction is
Na₂SO₄(aq) + Pb(NO₃)₂(aq) ⟶ 2NaNO₃(aq) + PbSO₄(s)
This means
1 mole of Na₂SO₄ reacts with 1 mole of Pb(NO₃)₂ to produce 2 moles of NaNO₃ and 1 mole of PbSO₄
Now, we will determine the number of moles of each reactant present
For Na₂SO₄Volume = 100.0 mL = 0.1 L
Concentration = 0.453 M
Using the formula
Number of moles = Concentration × Volume
∴ Number of moles of Na₂SO₄ = 0.453 × 0.1
Number of moles of Na₂SO₄ = 0.0453 mol
For Pb(NO₃)₂Volume = 100.0 mL = 0.1 L
Concentration = 0.907 M
∴ Number of moles of Pb(NO₃)₂ = 0.907 × 0.1
Number of moles of Pb(NO₃)₂ = 0.0907 mol
Since, 1 mole of Na₂SO₄ reacts with 1 mole of Pb(NO₃)₂
Then,
0.0453 mole of Na₂SO₄ will react with 0.0453 mole of Pb(NO₃)₂
Now, from the balanced chemical equation
1 mole of Na₂SO₄ reacts with 1 mole of Pb(NO₃)₂ to produce 1 mole of PbSO₄
Then,
0.0453 mole of Na₂SO₄ will react with 0.0453 mole of Pb(NO₃)₂ to produce 0.0453 mole of PbSO₄
∴ The number of moles of PbSO₄ produced is 0.0453 mole
Now, for the mass of PbSO₄ produced
From the formula
Mass = Number of moles × Molar mass
Molar mass of PbSO₄ = 303.26 g/mol
∴ Mass of PbSO₄ produced = 0.0453 × 303.26
Mass of PbSO₄ produced = 13.737678 g
Mass of PbSO₄ produced ≅ 13.74 g
Hence, the mass of PbSO₄ that could be produced is 13.74 g
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The question involves the stoichiometric calculation during a chemical reaction. In the reaction, Pb(NO₃)₂ and Na₂SO₄ react to produce PbSO₄. Given the volumes and molarities, we determined Na₂SO₄ to be the limiting reactant. Therefore, approximately 13.72 g of PbSO₄ can be produced.
Explanation:This question is trying to understand a stoichiometric calculation involved in a chemical reaction. We start by determining the limiting reactant. To do this, divide the number of moles of each reactant by the stoichiometric coefficient in the balanced chemical equation. The reactant that returns the smallest number is the limiting reactant.
In this case, the reaction uses one mole of Pb(NO₃)₂ and one mole of Na₂SO₄ to produce PbSO₄, therefore the molar ratio is 1:1. Pb(NO₃)₂ has a higher molarity, hence it will not be the limiting factor. Therefore, Na₂SO₄ is the limiting reactant. Considering this, we calculate that (0.453 moles/L × 0.1 L = 0.0453 moles of Na₂SO₄) are present.
To find the mass of PbSO₄ that can be produced, we use the molar mass of PbSO₄ (303.26 g/mol)—based on the atomic masses of lead, sulfur, and oxygen. From this we can calculate the mass as follows: 0.0453 moles × 303.26 g/mol = 13.72398 g of PbSO₄.
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The vapor pressure of water at 65oC is 187.54 mmHg. What is the vapor pressure of a ethylene glycol (CH2(OH)CH2(OH)) solution made by dissolving 22.37 g of ethylene glycol in 82.21 g of water?
Answers
Answer:
173.83 mmHg is the vapor pressure of a ethylene glycol solution.
Explanation:
Vapor pressure of water at 65 °C=[tex]p_o= 187.54 mmHg[/tex]
Vapor pressure of the solution at 65 °C= [tex]p_s[/tex]
The relative lowering of vapor pressure of solution in which non volatile solute is dissolved is equal to mole fraction of solute in the solution.
Mass of ethylene glycol = 22.37 g
Mass of water in a solution = 82.21 g
Moles of water=[tex]n_1=\frac{82.21 g}{18 g/mol}=4.5672 mol[/tex]
Moles of ethylene glycol=[tex]n_2=\frac{22.37 g}{62.07 g/mol}=0.3603 mol[/tex]
[tex]\frac{p_o-p_s}{p_o}=\frac{n_2}{n_1+n_2}[/tex]
[tex]\frac{187.54 mmHg-p_s}{187.54 mmHg}=\frac{0.3603 mol}{0.3603 mol+4.5672 mol}[/tex]
[tex]p_s=173.83 mmHg[/tex]
173.83 mmHg is the vapor pressure of a ethylene glycol solution.
Phosgene (carbonyl chloride), COCl2, is an extremely toxic gas that is used in manufacturing certain dyes and plastics. Phosgene can be produced by reacting carbon monoxide and chlorine gas at high temperatures: CO(g) Cl2(g)⇌COCl2(g) Carbon monoxide and chlorine gas are allowed to react in a sealed vessel at 477 ∘C . At equilibrium, the concentrations were measured and the following results obtained: Gas Partial Pressure (atm) CO 0.830 Cl2 1.30 COCl2 0.220 What is the equilibrium constant, Kp, of this reaction
Answers
The equilibrium constant for the phosgene formation reaction, given the partial pressures of carbon monoxide, chlorine gas, and phosgene at equilibrium, is approximately 0.204.
The student has asked to calculate the equilibrium constant (Kp) for the formation of phosgene (COCl₂) from carbon monoxide (CO) and chlorine gas (Cl₂) at a given temperature. The equilibrium constant for a reaction can be determined using the partial pressures of the gases at equilibrium. The equation for Kp in this reaction is Kp = PCOCl₂ / (PCO * PCl₂). Given the partial pressures at equilibrium (PCO = 0.830 atm, PCl₂ = 1.30 atm, PCOCl₂ = 0.220 atm), we can calculate Kp as follows:
Kp = 0.220 atm / (0.830 atm * 1.30 atm) = 0.220 / 1.079 = 0.204
Therefore, the equilibrium constant, Kp, for the reaction under the given conditions is approximately 0.204.
Calcium hydride (CaH2) reacts with water to form hydrogen gas: CaH2(s) + 2H2O(l) → Ca(OH)2(aq) + 2H2(g) How many grams of CaH2 are needed to generate 48.0 L of H2 gas at a pressure of 0.995 atm and a temperature of 32 °C?
Answers
Answer:
40.g CaH2
Explanation:
1. ideal gas law(PV = nRT) → use ideal gas law first when volume is given
P = 0.995atm
V = 48.0L H2
n = ?
R = 0.0821L atm/molK
T = 32 + 273 = 305K
n = (0.995atm x 48.0L H2)/(0.0821L atm/molK x 305K) → do not simplify as small decimals might change the answer
2. Conversions
2 H2 and 1 CaH2 → 1/2
(mole of H2) x 1/2 x (molar mass of CaH2)
(0.995atm x 48.0L H2)/(0.0821L atm/molK x 305K) x 1/2 x (40.08 + 2.02) = 40.g CaH2
the longer answer will be 40.14887882 but as the minimum sigfig given in the question is 2, it is 40.g CaH2.
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If 3.50 g of the unknown compound contained 0.117 mol of C and 0.233 mol of H, how many moles of oxygen, O, were in the sample? Express your answer to three significant figures and include the appropriate units.
Answers
Answer:
0.116 g.
Explanation:
Firstly, we can find the mass of C and H in the unknown compound:mass of C = (no. of moles of C)(atomic mass of C) = (0.117 mol)(12.01 g/mol) = 1.405 g.
mass of H = (no. of moles of H)(atomic mass of H) = (0.233 mol)(1.01 g/mol) = 0.235 g.
∴ mass of O = mass of unknown sample - (mass of C + mass of H) = 3.50 g - (1.405 g + 0.235) = 1.86 g.
∴ no. of moles of O = (mass of O)/(atomic mass of O) = (1.86 g)/(16.0 g/mol) = 0.116 g.
The vapor pressure of liquid acetone, CH3COCH3, is 100 mm Hg at 281 K. A 6.06E-2 g sample of liquid CH3COCH3 is placed in a closed, evacuated 360. mL container at a temperature of 281 K. Calculate what the ideal gas pressure would be in the container if all of the liquid acetone evaporated.
Answers
Final answer:
The ideal gas pressure in the container if all the liquid acetone evaporated is calculated by converting the sample mass to moles, ensuring all units are consistent, and applying the Ideal Gas Law. The result is an ideal gas pressure of approximately 0.064 atm.
Explanation:
To calculate the ideal gas pressure in the container if all of the liquid acetone evaporated, we need to determine the amount of acetone in moles and use the Ideal Gas Law. Given that the vapor pressure of acetone is 100 mm Hg at 281 K, we will first convert the 6.06E-2 g of acetone into moles using acetone's molar mass.
The molar mass of acetone (CH3)2CO is approximately 58.08 g/mol. By dividing the mass of the acetone sample by its molar mass, we obtain:
Number of moles = mass/molar mass = 6.06E-2 g / 58.08 g/mol ≈ 1.04E-3 mol
Next, we apply the Ideal Gas Law which is PV=nRT. To use this law, we must ensure all units are consistent. The vapor pressure must be in atmospheres, temperature in Kelvin, and volume in liters. The conversion from mm Hg to atm is 1 atm = 760 mm Hg, so 100 mm Hg = 100/760 atm. Also, convert the volume from mL to L by dividing by 1000.
Using these conversions and the Ideal Gas Law:
P = (nRT)/V = (1.04E-3 mol * 0.0821 L atm/(K mol) * 281 K) / (360 mL * 1 L/1000 mL) ≈ 0.064 atm
This is the ideal gas pressure if all of the acetone vaporizes at 281 K in the 360 mL container.
Classify each of these reactions. Ba(ClO3)2(s)⟶BaCl2(s)+3O2(g)Ba(ClO3)2(s)⟶BaCl2(s)+3O2(g) acid–base neutralization precipitation redox none of the above 2NaCl(aq)+K2S(aq)⟶Na2S(aq)+2KCl(aq)2NaCl(aq)+K2S(aq)⟶Na2S(aq)+2KCl(aq) acid–base neutralization precipitation redox none of the above CaO(s)+CO2(g)⟶CaCO3(s)CaO(s)+CO2(g)⟶CaCO3(s) acid–base neutralization precipitation redox none of the above KOH(aq)+AgNO3(aq)⟶KNO3(aq)+AgOH(s)KOH(aq)+AgNO3(aq)⟶KNO3(aq)+AgOH(s) acid–base neutralization precipitation redox none of the above
Answers
The first reaction is a redox reaction, the second and third reactions are none of the above, and the fourth reaction is a precipitation reaction.
Explanation:The reactions asked in the question can be classified as follows:
Ba(ClO₃)₂(s)⟶BaCl₂(s)+3O₂(g): This is a redox reaction. It involves a transfer of electrons which is characterized by changes in oxidation states. Here, chlorine is reduced from +5 in ClO₃⁻ to -1 in Cl⁻, and oxygen is oxidized from -2 in ClO₃⁻ to 0 in O₂.2NaCl(aq)+K₂S(aq)⟶Na₂S(aq)+2KCl(aq): This is a type of double displacement reaction known as 'metathesis', but it can't be classified as acid-base neutralization, redox, or precipitation, so it would fall under none of the above.CaO(s)+CO₂(g)⟶CaCO₃(s): It's a combination reaction resulting in the formation of a single product, calcium carbonate. Given the options, this reaction would also be classified as none of the above.KOH(aq)+AgNO₃(aq)⟶KNO₃(aq)+AgOH(s): This is a precipitation reaction where soluble ions in solution react to form an insoluble product, AgOH(s), which precipitates out of solution.Learn more about Chemical reaction classification here:https://brainly.com/question/8117294
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The first reaction is an oxidation-reduction (combustion) reaction, the second reaction is a precipitation reaction, the third reaction is an acid-base neutralization reaction, and the fourth reaction is also an acid-base neutralization reaction.
Explanation:The first reaction, Ba(ClO3)2(s)⟶BaCl2(s)+3O2(g), is a decomposition reaction also known as oxidation-reduction (combustion). The solid compound breaks down into a solid product and a gas.
The second reaction, 2NaCl(aq)+K2S(aq)⟶Na2S(aq)+2KCl(aq), is a precipitation reaction. The combination of two aqueous solutions forms an insoluble product.
The third reaction, CaO(s)+CO2(g)⟶CaCO3(s), is an acid-base neutralization reaction. The solid oxide reacts with a gas to form a solid carbonate.
The fourth reaction, KOH(aq)+AgNO3(aq)⟶KNO3(aq)+AgOH(s), is also an acid-base neutralization reaction. The aqueous solutions react to form a solid hydroxide and an aqueous salt.
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Calculate the change in pH when 5.00 mL of 0.100 M HCl is added to 75.0 mL of a buffer solution that is 0.100 M in NH3 and 0.100 M in NHaCI.
Answers
The first part of the answer is in the photo attached. I hope i can attach a second photo...
To find the change in pH, use the Henderson-Hasselbalch equation with the initial concentrations of NH3 and NH4+ to find the initial pH. Adjust concentrations after addition of HCl to find the final pH. Subtract initial pH from final to find change.
Explanation:The question asks us to calculate the change in pH when 5.00 ml of 0.100 M HCl is added to 75.0 ml of a buffer solution that is 0.100 M in NH3 and 0.100 M in NH4Cl. This is a problem related to acid-base chemistry. We have a strong acid (HCl) being added to a buffer solution composed of NH3 (a weak base) and NH4Cl (the conjugate acid of the base).
First, it's important to use the Henderson-Hasselbalch equation, pH = pKa + log([A-]/[HA]), where pKa is the -log of the Ka, [A-] is the concentration of base (NH3 in this case) and [HA] is the concentration of conjugate acid (NH4+ in this case). Plug the initial concentrations into the equation to find initial pH. Then, calculate the moles of HCl being added and adjust the concentrations of NH4+ and NH3 accordingly. Using the new concentrations in Henderson-Hasselbalch equation will give the final pH. The difference between the initial and final pH is the change in pH.
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The value of ΔH° for the reaction below is -126 kJ. The amount of heat that is released by the reaction of 10.0 g of Na2O2 with water is ________ kJ. 2Na2O2 (s) + 2H2O (l) → 4NaOH (s) + O2 (g)
Answers
Approximately -8 kJ of heat is released when 10 grams of Na2O2 react with water, based on the provided ΔH° value for the reaction.
Explanation:The value of ΔH° (-126 kJ) is the enthalpy change of the reaction when 2 moles of Na2O2 react with water to form 4 moles of NaOH and O2. The molar mass of Na2O2 is about 78.0 g/mol. Therefore, 10.0 g of Na2O2 is roughly equivalent to 0.128 moles. Given that the ΔH° value provided is for 2 moles, the heat released by the reaction of 0.128 moles would be (0.128/2) x -126 kJ which is approximately -8 kJ. So, the amount of heat that is released by the reaction of 10.0 g of Na2O2 with water is roughly -8 kJ.
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A monoprotic acid is an acid that donates a single proton to the solution. Suppose you have 0.140 g of a monoprotic acid dissolved in 35.0 mL of water. This solution is then neutralized with 14.5 mL of 0.110 M NaOH. What is the molar mass of the acid?
Answers
Answer:
molar mass HA = 87.8 g/mole
Explanation:
Given 0.140g HA + 14.5ml(0.110M NaOH) => NaA + H₂O
Rxn is a 1:1 rxn ration => moles HA neutralized = moles NaOH used
=> 0.140g/molar mass of HA = (0.110M)(0.0145L)
=> molar mass of HA = (0.140g)/[(0.110moles/L)(0.0145L)] = 87.8 g/mole
Answer: The molar mass of monoprotic acid is 87.72 g/mol
Explanation:
To calculate the concentration of acid, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of monoprotic acid
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.
We are given:
[tex]n_1=1\\M_1=?M\\V_1=35.0mL\\n_2=1\\M_2=0.110M\\V_2=14.5mL[/tex]
Putting values in above equation, we get:
[tex]1\times M_1\times 35.0=1\times 0.110\times 14.5\\\\x=\frac{1\times 0.110\times 14.5}{1\times 35.0}=0.0456M[/tex]
To calculate the molecular mass of acid, we use the equation used to calculate the molarity of solution:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]
We are given:
Molarity of solution = 0.0456 M
Given mass of acid = 0.140 g
Volume of solution = 35.0 mL
Putting values in above equation, we get:
[tex]0.0456M=\frac{0.140\times 1000}{\text{Molar mass of acid}\times 35}\\\\\text{Molar mass of acid}=\frac{0.140\times 1000}{0.0456\times 35}=87.72g/mol[/tex]
Hence, the molar mass of monoprotic acid is 87.72 g/mol