Answers
Chemical reaction: CH₃NH₂(aq)+ H₂O(l) ⇌ CH₃NH₃⁺(aq) + OH⁻(aq).
Kb(CH₃NH₂) = 4,4·10⁻⁴.
c₀(CH₃NH₂) = 0,57 M.
c(CH₃NH₃⁺) = c(OH⁻) = x.
c(NH₂OH) = 0,57 M - x.
Kb = c(CH₃NH₃⁺) · c(OH⁻) / c(CH₃NH₂).
0,00044 = x² / (0,57 M - x).
Solve quadratic equation: x = c(OH⁻) = 0,0156 mol/L.
pOH = -log(0,0156 mol/L.) = 1,80.
pH = 14 - 1,80 = 12,2.
The pH value is 12.2
Further explanationGiven:
0.57 M methylamine (CH₃NH₂)[tex]K_b = 4.4 \times 10^{-4}[/tex]Question:
The pH value of methylamine
The Process:
Methylamine is a weak base. When a weak base reacts with water, it produces its conjugate acid and hydroxide ions.
[tex]\boxed{ \ CH_3NH_2_{(aq)} + H_2O_{(l)} \rightleftharpoons CH_3NH_3_{(aq)} + OH^-_{(aq)} \ }[/tex]
CH₃NH₂ is the conjugate acid of CH₃NH₂.The concentration of hydroxide ions is needed to calculate pH.Let's prepare the equilibrium system to calculate the concentration of hydroxide ions. In chemical equilibrium, the liquid phase has no effect.
Initial concentration (in molars): [tex]\boxed{ \ [CH_3 NH_2] = 0.57 \ }[/tex]Change (in molars): [tex]\boxed{ \ [CH_3NH_2] = -x \ } \boxed{ \ [CH_3NH_3] = +x \ } \boxed{ \ [OH^-] = +x \ }[/tex]Equilibrium (in molars): [tex]\boxed{ \ [CH_3NH_2] = 0.57 - x \ } \boxed{ \ [CH_3NH_3] = x \ } \boxed{ \ [OH^-] = x \ }[/tex][tex]\boxed{ \ K_b = \frac{ [CH_3NH_3] [OH^-] }{ [CH_3NH_2] } \ }[/tex]
Here Kb acts as Kc or equilibrium constant.
[tex]\boxed{ \ 4.4 \times 10^{-4} = \frac{ x \cdot x }{ 0.57 - x } \ }[/tex]
[tex]\boxed{ \ 4.4 \times 10^{-4} = \frac{x^2}{0.57 - x} \ }[/tex]
[tex]\boxed{ \ 2.508 \times 10^{-4} - 4.4 \times 10^{-4}x = x^2 \ }[/tex]
[tex]\boxed{ \ x^2 + 4.4 \times 10^{-4}x - 2.508 \times 10^{-4} = 0 \ }[/tex]
The solution is obtained through the formula of quadratic equations, i.e., [tex]\boxed{ \ x = [OH^-] = 0.0156 \ M \ }[/tex]
Next, we calculated the pOH value followed by the pH value.
[tex]\boxed{ \ pOH = -log [OH^-] \ }[/tex]
[tex]\boxed{ \ pOH = -log [0.0156] \ }[/tex]
We get [tex]\boxed{ \ pOH = 1.81 \ }[/tex]
[tex]\boxed{ \ pH + pOH = 14 \ }[/tex]
[tex]\boxed{ \ pH = 14 - pOH \ }[/tex]
[tex]\boxed{ \ pH = 14 - 1.81 \ }[/tex]
Thus [tex]\boxed{\boxed{ \ pH = 12.19 \ rounded \ to \ 12.2 \ }}[/tex]
- - - - - - -
Quick Steps
0.57 M methylamine (CH₃NH₂)
[tex]K_b = 4.4 \times 10^{-4}[/tex]
We immediately use the formula to calculate the concentration of hydroxide ions for weak bases.
[tex]\boxed{\boxed{ \ [OH^-] = \sqrt{K_b \times base \ concentration} \ }}[/tex]
[tex]\boxed{ \ [OH^-] = \sqrt{4.4 \times 10^{-4} \times 0.57} \ }[/tex]
[tex]\boxed{ \ [OH^-] = 0.0158 \ M \ }[/tex]
Like the steps above, we calculated the pOH value followed by the pH value.
[tex]\boxed{ \ pOH = -log [OH^-] \ }[/tex]
[tex]\boxed{ \ pOH = -log [0.0158] \ }[/tex]
[tex]\boxed{ \ pOH = 1.8 \ }[/tex]
[tex]\boxed{ \ pH = 14 - pOH \ }[/tex]
[tex]\boxed{ \ pH = 14 - 1.8 \ }[/tex]
Thus [tex]\boxed{\boxed{ \ pH = 12.2 \ }}[/tex]
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Hydrogen is used as a rocket fuel because it is very light and reacts explosively and completely with oxygen. for the combustion reaction 2h2(g)+o2(g)⇌2h2o(g) what is the likely magnitude of the equilibrium constant k?
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The likely magnitude of the equilibrium constant k for the reaction btween hydrogen and oxygen is k>10³.
The equilibrium constant k for the reaction 2H₂(g) + O₂(g)⇄ 2H₂O(g) can be expressed as follows:
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The reaction tends to completion hence K should be in the order of K>103.
The equilibrium constant is a number that indicates the extent to which reactants are converted to products in a chemical reaction. A high value of equilibrium constant indicates that the system contains mostly products and few reactants at equilibrium. A low equilibrium constant indicates that the concentration of reactants is greater than the concentration of products at equilibrium.
For the reaction; 2H2(g) + O2(g) ⇌ 2H2O(g), we know the reaction to be explosive and tend to completion since it is a combustion reaction. This means that the reactants are mostly converted to products and the equilibrium constant will be large. Hence K should be in the order of K>103.
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Hydrogen is used as a rocket fuel because it is very light and reacts explosively and completely with oxygen. For the combustion reaction 2H2(g)+O2(g)⇌2H2O(g) what is the likely magnitude of the equilibrium constant K?
1. K<10−3
2. 10−3
3. K=0
4. K>103
Determine the mass of lithium hydroxide produced when 0.83g of lithium nitride reacts with water
Answers
The mass of Lithium hydroxide that is produced when 0.83g of lithium nitride reacts with water is 1,7122 g.
The reaction between water and lithium nitride is the following:
Li₃N(s) + 3H₂O(l) → NH₃(g) + 3LiOH(aq)
From this reaction we can calculate the mass of lithium hydroxide produced when 0,83 g of Lithium Nitride reacts with water, using the following conversion factor to go from grams of Li₃N to grams of LiOH using the reaction coefficients and molar masses:
[tex]0,83gLi_3N* \frac{1 mol Li_3N}{34,83 g Li_3N}* \frac{3 mol LiOH}{1 mol Li_3N}* \frac{23,95 g LiOH}{1 mol LiOH}= 1,7122 g LiOH [/tex]
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When 0.83g of lithium nitride reacts with water, 1.7g of lithium hydroxide is produced according to the balanced chemical equation and stoichiometry.
Explanation:To determine the mass of lithium hydroxide (LiOH) produced, we first need to understand the balanced chemical equation of the reaction.
Lithium nitride (Li₃N) reacts with water (H₂O) to form lithium hydroxide and ammonia (NH₃), which can be represented as:
Li₃N + 3H₂O → 3LiOH + NH₃
From the equation, we can see that 1 mole of lithium nitride reacts with 3 moles of water to produce 3 moles of lithium hydroxide.
Next, we need to convert the mass of lithium nitride to moles using the molar mass of lithium nitride, which is approximately 34.83 g/mol.
0.83 g Li₃N * (1 mol/34.83 g) = 0.0238 mol
Since the molar ratio of Li₃N to LiOH is 1:3, we multiply this mole by 3 to get the moles of lithium hydroxide produced.
Moles of LiOH = 0.0238 mol * 3 = 0.0714 mol
Finally, we convert the moles of LiOH to grams using the molar mass of LiOH, which is 23.95 g/mol.
Mass of LiOH = 0.0714 mol * 23.95 g/mol = 1.7 g
So, when 0.83g of lithium nitride reacts with water, 1.7g of lithium hydroxide is produced.
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What is the empirical formula of a substance that contains 5.28 g of c, 0.887 g of h, and 3.52 g of o?
Answers
Final answer:
The empirical formula of the compound is CH2O.
Explanation:
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Step 2: Divide each element's mole value by the smallest mole value to get the simplest ratio. In this case, the smallest mole value is 0.22 mol O. Dividing the other mole values by 0.22 gives us 0.44 mol C, 0.877 mol H, and 1 mol O.
The empirical formula of the compound is CH2O.
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Answer:
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and I think the second skill should be B or C
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Answers
find the mole of each molecule
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The thermodynamic law that explains the movement of heat energy during the refrigeration cycle is the first/zeroth/second.
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is the answer to this queston
Given the balanced equation 3H2(g)+N2(g)=2NH3(g), calculate the mass of NH3 produced by the complete reaction of 2.55g of H2
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The mass of NH₃ produced by the complete reaction of 2,55 g of H₂ is 14,3622 g
To calculate the mass of NH₃ produced in the chemical reaction we need to use the following conversion factor, to go from the mass of H₂ to the mass of NH₃, using the reaction coefficients for the reagents and the products, and the molar masses of each compound:
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The mass of NH3 produced from the complete reaction of 2.55g of H2 is calculated to be 14.45 g, based on the molar masses of H2 and NH3 and the stoichiometry of the given balanced chemical equation.
Explanation:To calculate the mass of NH3 produced by the complete reaction of 2.55g of H2, we first need to determine the molar mass of H2. Knowing that the atomic mass of hydrogen is 1.0, we can say that the molar mass of H2 is 2.0 g/mol. Next, we need to find out how many moles of H2 are in 2.55 g:
Number of moles of H2 = mass (g) / molar mass (g/mol) = 2.55 g / 2.0 g/mol = 1.275 mol
Using the stoichiometry of the balanced equation (N2(g) + 3H2(g) → 2NH3(g)), we can see that 3 moles of H2 produce 2 moles of NH3. Therefore, we can set up a proportion to find the number of moles of NH3 that would be produced from 1.275 moles of H2:
(1.275 mol H2) * (2 mol NH3 / 3 mol H2) = 0.85 mol NH3
Now, to find the mass of NH3, we need to know its molar mass. The atomic mass of nitrogen is 14.0 and hydrogen is 1.0, so the molar mass of NH3 is 14.0 + (3 * 1.0) = 17.0 g/mol. Finally, we can calculate the mass of NH3 produced:
Mass of NH3 = number of moles * molar mass = 0.85 mol * 17.0 g/mol = 14.45 g
If the equilibrium constant kc for the reaction below is 1.71 × 10–1, what will be the equilibrium pressure of no if the initial partial pressures of the three gases are all 1.96 × 10–3 atm? $$
Answers
N₂ + O₂ ⇄ 2NO
We can write an expression for the equilibrium constant. We are given the value of Kc, but we must use Kp if we are dealing with partial pressures. However, since 2 moles of gas are formed from 2 moles of gas, the value of Kc is equal to Kp so we do not need to change anything.
Kc = (Pno)²/(Pn2)(Po2) = 1.71 x 10⁻¹
We must first use the initial pressures to find the reaction quotient to decide which way the equilibrium will shift:
Q = (1.96 x 10⁻³)²/(1.96 x 10⁻³)(1.96 x 10⁻³) = 1
Since Q > Kp, the equilibrium will shift to the left. Now we can prepare an ICE table to find the equilibrium pressures:
N₂ O₂ NO
I 1.96 x 10⁻³ 1.96 x 10⁻³ 1.96 x 10⁻³
C +x +x -2x
E 0.00196 + x 0.00196 + x 0.00196 - 2x
We enter these equilibrium pressures into the equilibrium expression and solve for x:
0.171 = (0.00196 - 2x)²/(0.00196 + x)²
Expand the equation and simplify to a quadratic function:
3.829x² - 0.00851x + 3.183·10⁻⁶ = 0
x = ((0.00851) +/- sqrt((0.00851)² - 4(3.829)(3.183 x 10⁻⁶)))/2(3.829)
x = 0.000476 atm
Now we can solve for the equilibrium partial pressures:
Pno = 0.00196 - 2(0.000476) = 0.00101 atm
Pn2 = 0.00196 + 0.000476 = 0.00244 atm
Po2 = 0.000196 + 0.000476 = 0.00244 atm
Therefore, the equilibrium partial pressure of NO is 0.00101 atm.
Final answer:
To determine the equilibrium pressure of NO, set up an ICE table using the values of the equilibrium constant (Kc) and initial partial pressures. Assume the increase in NO pressure is 'x' and the decrease in N2O and O2 pressures is '2x' and 'x', respectively. Calculate the equilibrium pressures by adding the initial pressures to their corresponding changes.
Explanation:
To determine the equilibrium pressure of NO, we need to set up an ICE table and use the values of the equilibrium constant (Kc) and initial partial pressures of the three gases. Let's assume the increase in the pressure of NO is 'x'. Since the stoichiometric coefficient of NO is 2, the decrease in the pressures of N2O and O2 will be '2x' and 'x', respectively. The equilibrium partial pressures of the three gases can be calculated by adding the initial pressures to their corresponding changes.
Initial pressures:
N2O: 1.96 × 10-3 atmO2: 1.96 × 10-3 atmNO: 1.96 × 10-3 atmChange in pressures:
N2O: -2xO2: -xNO: xEquilibrium pressures:
N2O: (1.96 × 10-3 - 2x) atmO2: (1.96 × 10-3 - x) atmNO: (1.96 × 10-3 + x) atmSince the equilibrium constant (Kc) is given as 1.71 × 10-1, which represents the ratio of the equilibrium concentrations of products to reactants, we can set up the following equation:
Kc = [NO]2 / ([N2O] * [O2])
Substituting the equilibrium pressures into this equation:
1.71 × 10-1 = ([1.96 × 10-3 + x]2 / ([1.96 × 10-3 - 2x] * [1.96 × 10-3 - x]))
Now, solve this equation to find the value of 'x', which represents the equilibrium increase in the pressure of NO.
Once 'x' is determined, substitute it back into the equations for the equilibrium pressures to calculate the final equilibrium pressures of N2O, O2, and NO.
A car travels 31.2 miles in 36.0 minutes. What is the average speed of the car during this trip, in miles per hour (mph)?
Answers
in this instance, distance traveled - 31.2 miles
time taken to travel this distance - 36 minutes
time is given in minutes, we need to convert minutes to hours
60 minutes - 1 hour
Therefore 1 minute - 1/60 hr
36 minutes - 1/60 hr/minutes x 36 minutes
- 0.6 hr
speed = distance travelled / time taken
= 31.2 miles / 0.6 hr
= 52 mph (miles per hour)
average speed means that on average the car has been travelling at 52 mph speed. within one hour on average the car can cover 52 miles distance.
Answer:
0.8 mph
Explanation:
divide distance by time
31.2/ 36 = 0.866..
Draw one of the isomeric c5h12o alcohols that can be prepared by lithium aluminum hydride reduction of a ketone.
Answers
The Lithium Aluminum hydride (LiAlH₄) reduction of a ketone is a process in which the ketone is reduced and the Lithium Aluminum hydride is oxidized. The general chemical reaction for this kind of reduction is:
1)LiAlH₄
R(C=O)R ---------→R(C-OH)R
2)H₃O⁺
The structure for one of the isomeric C₅H₁₂O alcohols that can be prepared by lithium aluminum hydride reduction of a ketone is shown in the attached image, the compound is called 3-Pentanol:
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Helium has a density of 1.79 x 10-4 g/mL at standard temperature and pressure. A balloon has a volume of 6.3 liters. Calculate the mass of helium that it would take to fill the balloon
Answers
In other words in 1 mL there's 1.79 x 10⁻⁴ g of He.
To fill a volume of 6.3 L the mass of He required
= 1.79 x 10⁻⁴ g/mL * 6300 mL
= 11 277 * 10⁻⁴ g
Therefore mass of He required = 1.1277 g of He
Answer:
1.1 x 10-3 g
Explanation:
Consider this reaction: 2al(s) + 3 cucl2(aq) → 2alcl3(aq) + 3 cu(s) if the concentration of cucl2 drops from 1.000 m to 0.655 m in the first 30.0 s of the reaction, what is the average rate of reaction over this time interval?
Answers
The average rate of reaction for CuCl₂ is found by dividing the concentration change (0.345 M) by the time interval (30.0 s), resulting in an average reaction rate of 0.0115 M/s for both the disappearance of CuCl₂ and the formation of Cu.
The student is asking about calculating the average rate of reaction using changes in the concentration of a reactant over a given time interval. The average rate of reaction can be calculated by dividing the change in concentration of a reactant by the time period over which the change occurred. In this case, the concentration of CuCl₂ drops from 1.000 M to 0.655 M over 30.0 seconds.
To find the average rate at which CuCl₂ has reacted, we can use the formula:
Calculate the change in concentration of CuCl₂:(Initial concentration) - (Final concentration) = 1.000 M - 0.655 M = 0.345 M. Divide the change in concentration by the time interval:
0.345 M / 30.0 s = 0.0115 M/s.
The average rate of reaction for the disappearance of CuCl₂ is 0.0115 M/s. Since the reaction stoichiometry shows 3 moles of CuCl₂ produces 3 moles of Cu, the average rate of formation of Cu is also 0.0115 M/s.
Consider the unbalanced equation for the combustion of hexane: αc6h14(g)+βo2(g)→γco2(g)+δh2o(g) part a balance the equation. give your answer as an ordered set of numbers α, β, γ, ... use the least possible integers for the coefficients. α, β, γ, δ = request answer part b determine how many moles of o2 are required to react completely with 6.4 moles c6h14. express your answer using two significant figures. n = mol
Answers
Answer :
Part A : The value of [tex]\alpha, \beta, \gamma \text{ and }\delta[/tex] are 2, 19, 12 and 14 respectively.
Part B : [tex]6.0\times 10^1moles[/tex] of [tex]O_2[/tex] are required to react with 6.4 moles of [tex]C_6H_{14}[/tex].
Solution :
Part A :
The given unbalanced equation is,
[tex]\alpha C_6H_{14}(g)+\beta O_2(g)\rightarrow \gamma CO_2(g)+\delta H_2O(g)[/tex]
The balanced equation is,
[tex]2C_6H_{14}(g)+19O_2(g)\rightarrow 12CO_2(g)+14H_2O(g)[/tex]
Part B :
From the balanced equation, we conclude that
2 moles of [tex]C_6H_{14}[/tex] react with 19 moles of [tex]O_2[/tex]
6.4 moles of [tex]C_6H_{14}[/tex] react with [tex]\frac{19moles}{2moles}\times 6.4moles=60.8moles=6.0\times 10^1moles[/tex] of [tex]O_2[/tex]
Therefore, [tex]6.0\times 10^1moles[/tex] of [tex]O_2[/tex] are required to react with 6.4 moles of [tex]C_6H_{14}[/tex].
The balanced equation is 2C3H7(g) + 9O2(g) → 6CO2(g) + 7H2O(g), therefore the coefficients α, β, γ, δ are 2,9,6,7, respectively. And 29 moles of O2 are required to react completely with 6.4 moles of C6H14.
Explanation:For part a, to balance the equation for the combustion of hexane, we have to adjust the coefficients in the equation C6H14(g) + αO2(g) → βCO2(g) + γH2O(g) such that the number of atoms of each element on both sides of the equation are equal. The balanced equation becomes: 2C3H7(g) + 9O2(g) → 6CO2(g) + 7H2O(g) . It means that α, β, γ, δ = 2, 9, 6, 7 respectively.
For part b, according to the balanced equation, 9 moles of O2 are required to react completely with 2 moles of C6H14. So, if there are 6.4 moles of C6H14, you will need: (6.4 moles C6H14 * 9 moles O2) / 2 moles C6H14 = 28.8 ≈ 29 moles of O2. In this case 'n' equals 29 moles.
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Write the atomic cores for potassium and chloride ions
Answers
The target is to determine the Lewis structer for the ionic compound KCl.
The atomic cores are the electrons but the outermost shell (i.e. valence electrons).
Potassium has atomic number (Z) 19.
So, its electron distribution is 1s2 2s2 2p6 3s2 3p6 4s
The valence electron is 4s, so the atomic core is 1s2 2s2 2p6 3s2 3p6.
Potassium ion, K+, has lost one electron, the outermost electron, so it has the same electrons as the atomic core of K.
Chlorine has atomic number 17.
The electron distribution of chlorine is 1s2 2s2 2p6 3s2 3p5.
The valence eletrons are 3s2 3p5, so the atomic core of chlorine is 1s2 2s2 2p6.
The ion Cl- has gained one electron so its electron configuration is 1s2 2s2 2p6 3s2 3p6.
To represent the Lewis structure you only use the valence electrons, so the representation is:
**
K(+) +*Cl(-) **
**
Where the * marks represent the 7 valence electrons of chloride atom and the + mark represent the electron gained by chloride and released by potassium.
So, the ion K(+) has lost its only valence electron and the ion Cl(-) has increased its valence electrons to 8. In this way both have the stable configuration of a noble gas.
The atomic core for a potassium ion (K+) is [Ar]4s¹ without the one electron it loses to become an ion. The atomic core for a chloride ion (Cl-) is [Ne]3s² 3p⁵ with an extra electron gained to fill its 3p orbital.
Explanation:The atomic cores for potassium and chloride ions are represented by their respective electron configurations and their charges after ion formation. Potassium (K) has an atomic number of 19, which means it has 19 electrons. The electron configuration is [Ar]4s¹. When potassium forms an ion, it loses one electron to achieve a noble gas configuration, making it a potassium ion with a charge of +1, represented as K⁺.
Chlorine (Cl) has an atomic number of 17, which means it has 17 electrons. The electron configuration is [Ne]3s² 3p⁵. When chlorine forms an ion, it gains one electron to fill its 3p orbital, becoming a chloride ion with a charge of -1, represented as Cl⁻.
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During respiration, a plant uses glucose (sugar) and oxygen to give it the energy it needs to survive. During this process ______ and _____ are released.
Question 6 options:
hydrogen and oxygen
sunlight and water
water and carbon dioxide
nitrogen and carbon dioxide
Answers
Answer:
water carbon dioxide
Explanation:
The part of the flower responsible for producing pollen (sperm) is the _______.
filament
anther
stigma
style
Answers
Here is your answer:
The proper answer to this question is option C "stigma".
Here is how:
The stigma is responsible for producing pollen in a plant.
Your answer is C.
If you need anymore help feel free to ask me!
Hope this helps!
how many moles are there in 78.3g of CO2?
Answers
CO2 — 12 + (16x2) = 12+ 32 = 44 g
Therefore molar mass of CO2 is 44 g/mol
In 44 g of CO2 there’s 1 mol of CO2
Then 1 g of CO2 there’s 1/44 mol of CO2
Therefore in 78.3 g of CO2 there’s — 1/44 x 78.3 =1.78 mol of CO2
What is the number “4” in SiCl4?
Answers
A subscript is the answer.
Hope this answer has helped you.
Determine the molecular formula of a compound that has a molar mass of 183.2 g/mol and an empirical formula of c2h5o2.
Answers
Good luck and don't forget to rate or mark Brainliest :)
Answer:
[tex]C_6H_{15}O_6[/tex]
Explanation:
Hello,
In this case, one computes the molar mass for the given empirical formula as follows:
[tex]M_{empirical}=12*2+1*5+16*2=61g/mol[/tex]
In such a way, one computes how many times the molecular formula's molecular mass is contained into the empirical formula's molecular mass in order to determine the whole number relating them as shown below:
[tex]\frac{183.2g/mol}{61g/mol}=3[/tex]
Finally, the molecular formula is three times the empirical formula, hence:
[tex]C_6H_{15}O_6[/tex]
Best regards.
Write a net ionic equation to show that acetylsalicylic acid (aspirin), hc9h7o4, behaves as a brønsted-lowry acid in water.
Answers
HC₉H₇O₄(aq) + H₂O(l) ⇄ C₉H₇O₄⁻(aq) + H₃O⁺(aq).
In this chemical reaction aspirin(HC₉H₇O₄) is an acid because it donates a proton (H⁺) to water and becomes its conjugate base (C₉H₇O₄⁻) and water is a base because it accepts a proton from aspirin and becomes its conjugate acid, the hydronium ion, (H₃O⁺).
Explanation:
According to the Bronsted-Lowry conjugate acid-base theory:
An acid is defined as a substance which looses donates protons and thus forming conjugate base.A base is defined as a substance which accepts protons and thus forming conjugate acid.Acetylsalicylic acid when dissolved in water donates its proton to form conjugate base and water gains the proton to form conjugate acid.The net ionic equation is given as:
[tex]HC_9H_7O_4(0+H_2O\rightarrow (C_9H_7O_4)^{-}+H_3O^+[/tex]
Determine the limiting reactant. 2al(s)+3br2(g)→2albr3(s)
Answers
m(Al) = 25 g.
n(Al) = m(Al) ÷ M(Al).
n(Al) = 25 g ÷ 27 g/mol.
n(Al) = 0,926 mol.
m(Br₂) = 25 g.
n(Br₂) = 25 g ÷ 160 g/mol.
n(Br₂) = 0,156 mol.
Bromine is the limiting reactant, because it has less amount of substance.
The limiting reactant in a chemical reaction is determined by calculating the amount of product (AlBr3 in this case) that would be produced from the complete reaction of each of the reactants separately. The reactant yielding the lesser amount of product is the limiting reactant.
Explanation:To determine the limiting reactant in a chemical reaction, you can use a method where you compare the amount of each reactant to the amount of product expected from its complete reaction. Taking the given equation, 2Al(s)+3Br2(g)→2AlBr3(s), we would calculate the amount of the product AlBr3 that would be formed from the complete reaction of each reactant, Al and Br2, separately. The reactant that yields the lesser amount of the product AlBr3 is the limiting reactant. This concept can be understood using a simple analogy. Let's think of making a sandwich: 2 slices of bread + 1 slice of cheese yields 1 sandwich. If you have an excess of bread but only one slice of cheese, you can only make one sandwich. The cheese is the limiting reactant as it limits the yield of sandwich and the amount of bread left unreacted is the excess reactant. Similarly, in the chemical reaction given above, the reactant that produces less amount of product (AlBr3) acts as the limiting reactant.
Learn more about Limiting Reactant here:https://brainly.com/question/33417913
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The volume of an automobile air bag was 66.8 l when inflated at 25 °c with 77.8 g of nitrogen gas. what was the pressure in the bag in kpa
Answers
PV = nRT
P - pressure
V - volume - 66.8 / 1000 m³
n - number of moles - 77.8 g/ 28 g/mol = 2.77 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - 25 °C + 273 = 298 K
substituting these values in the equation
P x 0.0668 m³ = 2.77 mol x 8.314 Jmol⁻¹K⁻¹ x 298 K
P = 102.7 kPa