Answer:
a)10.87
b)9.66
c)9.15
d)7.71
e) 5.56
f) 3.43
Explanation:
tep 1: Data given
Volume of 0.030 M NH3 solution = 30 mL = 0.030 L
Molarity of the HCl solution = 0.025 M
Step 2: Adding 0 mL of HCl
The reaction: NH3 + H2O ⇔ NH4+ + OH-
The initial concentration:
[NH3] = 0.030M [NH4+] = 0M [OH-] = OM
The concentration at the equilibrium:
[NH3] = 0.030 - XM
[NH4+] = [OH-] = XM
Kb = ([NH4+][OH-])/[NH3]
1.8*10^-5 = x² / 0.030-x
1.8*10^-5 = x² / 0.030
x = 7.35 * 10^-4 = [OH-]
pOH = -log [7.35 * 10^-4]
pOH = 3.13
pH = 14-3.13 = 10.87
Step 3: After adding 10 mL of HCl
The reaction:
NH3 + HCl ⇔ NH4+ + Cl-
NH3 + H3O+ ⇔ NH4+ + H2O
Calculate numbers of moles:
Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles
Moles HCl = 0.025 M * 0.010 L = 0.00025 moles
Moles NH4+ = 0 moles
Number of moles at the equilibrium:
Moles NH3 = 0.0009 -0.00025 =0.00065 moles
Moles HCl = 0
Moles NH4+ = 0.00025 moles
Concentration at the equilibrium:
[NH3]= 0.00065 moles / 0.040 L = 0.01625M
[NH4+] = 0.00625 M
pOH = pKb + log [NH4+]/[NH3]
pOH = 4.75 + log (0.00625/0.01625)
pOH = 4.34
pH = 9.66
Step 3: Adding 20 mL of HCl
Calculate numbers of moles:
Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles
Moles HCl = 0.025 M * 0.020 L = 0.00050 moles
Moles NH4+ = 0 moles
Number of moles at the equilibrium:
Moles NH3 = 0.0009 -0.00050 =0.00040 moles
Moles HCl = 0
Moles NH4+ = 0.00050 moles
Concentration at the equilibrium:
[NH3]= 0.00040 moles / 0.050 L = 0.008M
[NH4+] = 0.01 M
pOH = pKb + log [NH4+]/[NH3]
pOH = 4.75 + log (0.01/0.008)
pOH = 4.85
pH = 14 - 4.85 = 9.15
Step 4: Adding 35 mL of HCl
Calculate numbers of moles:
Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles
Moles HCl = 0.025 M * 0.035 L = 0.000875 moles
Moles NH4+ = 0 moles
Number of moles at the equilibrium:
Moles NH3 = 0.0009 -0.000875 =0.000025 moles
Moles HCl = 0
Moles NH4+ = 0.000875 moles
Concentration at the equilibrium:
[NH3]= 0.000025 moles / 0.065 L = 3.85*10^-4M
[NH4+] = 0.000875 M / 0.065 L = 0.0135 M
pOH = pKb + log [NH4+]/[NH3]
pOH = 4.75 + log (0.0135/3.85*10^-4)
pOH = 6.29
pH = 14 - 6.29 = 7.71
Step 5: adding 36 mL HCl
Calculate numbers of moles:
Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles
Moles HCl = 0.025 M * 0.036 L = 0.0009 moles
Moles NH4+ = 0 moles
Number of moles at the equilibrium:
Moles NH3 = 0.0009 -0.0009 =0 moles
Moles HCl = 0
Moles NH4+ = 0.0009 moles
[NH4+] = 0.0009 moles / 0.066 L = 0.0136 M
Kw = Ka * Kb
Ka = 10^-14 / 1.8*10^-5
Ka = 5.6 * 10^-10
Ka = [NH3][H3O+] / [NH4+]
Ka =5.6 * 10^-10 = x² / 0.0136
x = 2.76 * 10^-6 = [H3O+]
pH = -log(2.76 * 10^-6)
pH = 5.56
Step 6: Adding 37 mL of HCl
Calculate numbers of moles:
Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles
Moles HCl = 0.025 M * 0.037 L = 0.000925 moles
Moles NH4+ = 0 moles
Number of moles at the equilibrium:
Moles NH3 = 0.0009 -0.000925 =0 moles
Moles HCl = 0.000025 moles
Concentration of HCl = 0.000025 moles / 0.067 L = 3.73 * 10^-4 M
pH = -log 3.73*10^-4= 3.43
The pH of the solution in the titration of 30 mL of 0.030 M NH₃ with 0.025 M HCl, is:
a) pH = 10.86
b) pH = 9.66
c) pH = 9.15
d) pH = 7.70
e) pH = 5.56
f) pH = 3.43
Calculating the pH a) 0 mL
Initially, the pH of the solution is given by the dissociation of NH₃ in water.
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻ (1)
The constant of the above reaction is:
[tex] Kb = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]} = 1.76\cdot 10^{-5} [/tex] (2)
At the equilibrium, we have:
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻ (3)
0.030 M - x x x
[tex] 1.76\cdot 10^{-5}*(0.030 - x) - x^{2} = 0 [/tex]
After solving for x and taking the positive value:
x = 7.18x10⁻⁴ = [OH⁻]
Now, we can calculate the pH of the solution as follows:
[tex] pH = 14 - pOH = 14 + log(7.18\cdot 10^{-4}) = 10.86 [/tex]
Hence, the initial pH is 10.86.
b) 10 mL
After the addition of HCl, the following reaction takes place:
NH₃ + HCl ⇄ NH₄⁺ + Cl⁻ (4)
We can calculate the pH of the solution from the equilibrium reaction (3).
[tex] 1.76\cdot 10^{-5}(Cb - x) - (Ca + x)*x = 0 [/tex] (5)
Finding the number of moles of NH₃ and NH₄⁺
The number of moles of NH₃ (nb) and NH₄⁺ (na) are given by:
[tex] n_{b} = n_{i} - n_{HCl} [/tex] (6)
[tex] n_{b} = 0.030 mol/L*0.030 L - 0.025 mol/L*0.010 L = 6.5\cdot 10^{-4} moles [/tex]
[tex] n_{a} = n_{HCl} [/tex] (7)
[tex] n_{a} = 0.025 mol/L*0.010 L = 2.5 \cdot 10^{-4} moles [/tex]
Calculating the concentrations of NH₃ and NH₄⁺The concentrations are given by:
[tex] Cb = \frac{6.5\cdot 10^{-4} moles}{(0.030 L + 0.010 L)} = 0.0163 M [/tex] (8)
[tex] Ca = \frac{2.5 \cdot 10^{-4} mole}{(0.030 L + 0.010 L)} = 6.25 \cdot 10^{-3} M [/tex] (9)
Calculating the pHAfter entering the values of Ca and Cb into equation (5) and solving for x, we have:
[tex] 1.76\cdot 10^{-5}(0.0163 - x) - (6.25 \cdot 10^{-3} + x)*x = 0 [/tex]
x = 4.54x10⁻⁵ = [OH⁻]
Then, the pH is:
[tex] pH = 14 + log(4.54\cdot 10^{-5}) = 9.66 [/tex]
Hence, the pH is 9.66.
c) 20 mLWe can find the pH of the solution from the reaction of equilibrium (3).
Calculating the concentrations of NH₃ and NH₄⁺The concentrations are (eq 8 and 9):
[tex] Cb = \frac{0.030 mol/L*0.030 L - 0.025 mol/L*0.020 L}{(0.030 L + 0.020 L)} = 8.0\cdot 10^{-3} M [/tex]
[tex] Ca = \frac{0.025 mol/L*0.020 L}{(0.030 L + 0.020 L)} = 0.01 M [/tex]
Calculating the pHAfter solving the equation (5) for x, we have:
[tex] 1.76\cdot 10^{-5}(8.0\cdot 10^{-3} - x) - (0.01 + x)*x = 0 [/tex]
x = 1.40x10⁻⁵ = [OH⁻]
Then, the pH is:
[tex] pH = 14 + log(1.40\cdot 10^{-5}) = 9.15 [/tex]
So, the pH is 9.15.
d) 35 mLWe can find the pH of the solution from reaction (3).
Calculating the concentrations of NH₃ and NH₄⁺[tex] Cb = \frac{0.030 mol/L*0.030 L - 0.025 mol/L*0.035 L}{(0.030 L + 0.035 L)} = 3.85\cdot 10^{-4} M [/tex]
[tex] Ca = \frac{0.025 mol/L*0.035 L}{(0.030 L + 0.035 L)} = 0.0135 M [/tex]
Calculating the pHAfter solving the equation (5) for x, we have:
[tex] 1.76\cdot 10^{-5}(3.85\cdot 10^{-4} - x) - (0.0135 + x)*x = 0 [/tex]
x = 5.013x10⁻⁷ = [OH⁻]
Then, the pH is:
[tex] pH = 14 + log(5.013\cdot 10^{-7}) = 7.70 [/tex]
So, the pH is 7.70.
e) 36 mL Finding the number of moles of NH₃ and NH₄⁺[tex] n_{b} = 0.030 mol/L*0.030 L - 0.025 mol/L*0.036 L = 0 [/tex]
[tex] n_{a} = 0.025 mol/L*0.036 L = 9.0 \cdot 10^{-4} moles [/tex]
Since all the NH₃ reacts with the HCl added, the pH of the solution is given by the dissociation reaction of the NH₄⁺ produced in water.
At the equilibrium, we have:
NH₄⁺ + H₂O ⇄ NH₃ + H₃O⁺
Ca - x x x
[tex] Ka = \frac{x^{2}}{Ca - x} [/tex]
[tex] Ka(Ca - x) - x^{2} = 0 [/tex] (10)
Calculating the acid constant of NH₄⁺
We can find the acid constant as follows:
[tex] Kw = Ka*Kb [/tex]
Where Kw is the constant of water = 10⁻¹⁴
[tex] Ka = \frac{1\cdot 10^{-14}}{1.76 \cdot 10^{-5}} = 5.68 \cdot 10^{-10} [/tex]
Calculating the pH
The concentration of NH₄⁺ is:
[tex] Ca = \frac{9.0 \cdot 10^{-4} moles}{(0.030 L + 0.036 L)} = 0.0136 M [/tex]
After solving the equation (10) for x, we have:
x = 2.78x10⁻⁶ = [H₃O⁺]
Then, the pH is:
[tex] pH = -log(H_{3}O^{+}) = -log(2.78\cdot 10^{-6}) = 5.56 [/tex]
Hence, the pH is 5.56.
f) 37 mLNow, the pH is given by the concentration of HCl that remain in solution after reacting with NH₃ (HCl is in excess).
Calculating the concentration of HCl
[tex] C_{HCl} = \frac{0.025 mol/L*0.037 L - 0.030 mol/L*0.030 L}{(0.030 L + 0.037 L)} = 3.73 \cdot 10^{-4} M = [H_{3}O^{+}] [/tex]
Calculating the pH
[tex] pH = -log(H_{3}O^{+}) = -log(3.73 \cdot 10^{-4}) = 3.43 [/tex]
Therefore, the pH is 3.43.
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The metal tantalum becomes superconducting at temperatures below 4.483 K. Calculate the temperature at which tantalum becomes superconducting in degrees Celsius.
Answer:
The correct answer is "-268.667°C".
Explanation:
Given:
Temperature,
= 4.483 K (below)
Now,
The formula of temperature conversion will be:
⇒ [tex]T(^{\circ} C)=T(K)-273.15[/tex]
By putting the values, we get
⇒ [tex]=4.483-273.15[/tex]
⇒ [tex]=-268.667^{\circ} C[/tex]
Thus the above is the correct answer.
A student obtained an unknown metal sample that weighed 65.3 g and at a temperature of 99.8oC, he placed it in a calorimeter containing 43.7 g of water at 25.7oC. At equilibrium the temperature of the water and metal was 34.5oC. Knowing the specific heat of the water to be 4.18 J/goC, what is the specific heat of the metal
Answer:
0.377 J/gºC
Explanation:
From the question given above, the following data were obtained:
Mass of metal (Mₘ) = 65.3 g
Initial temperature of metal (Tₘ) = 99.8 °C
Mass of water (Mᵥᵥ) = 43.7 g
Initial temperature of water (Tᵥᵥ) = 25.7 °C
Equilibrium temperature (Tₑ) = 34.5 °C
Specific heat capacity of water (Cᵥᵥ) = 4.18 J/gºC
Specific heat capacity of metal (Cₘ) =?
The specific heat capacity of metal can be obtained as illustrated below:
Heat lost by metal = heat gained by water.
MₘCₘ(Tₘ – Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Cᵥᵥ)
65.3 × Cₘ (99.8 – 34.5) = 43.7 × 4.18 (34.5 – 25.7)
65.3Cₘ × 65.3 = 182.666 × 8.8
4264.09Cₘ = 1607.4608
Divide both side by 4264.09
Cₘ = 1607.4608 / 4264.09
Cₘ = 0.377 J/gºC
Therefore the specific heat capacity of the metal is 0.377 J/gºC
What type of a liquid will have a pH value equal to 12? (1 point)
Basic
Neutral
Strong acid
Weak aci
Answer: it will be basic
pH that ranges from 0-6 are acid
pH of EXACTLY 7 is neutral
pH greater than 7 are strongly basic or base
Consider the following equilibrium:
2H2(g)+S2(g)⇌2H2S(g)Kc=1.08×107 at 700 ∘C.
What is Kp?
Answer:
Consider the following equilibrium:
2H2(g)+S2(g)⇌2H2S(g)Kc=1.08×107 at 700 ∘C.
What is Kp?
Explanation:
Given,
[tex]Kc=1.08 * 10^7[/tex]
The relation between Kp and Kc is:
[tex]Kp=Kc * (RT)^d^e^l^t^a^(^n^)[/tex]
Where delta n represents the change in the number of moles.
For the given equation,
The Delta n = Number of moles of products - number of moles of reactants
(2-(2+1))
=-1.
Hence,
Kp=Kc/RT.
Thus,
[tex]Kp=1.08 * 10^7 / 8.314 J.K6-1.mol^-^1 x 973 K\\Kp=1335.06[/tex]
The answer is Kp=1335.06
The value of [tex]K_p[/tex] is [tex]1.35\times 10^5[/tex].
Explanation:
The relation between [tex]K_p \& K_c[/tex] is given by:
[tex]K_p=K_c(RT)^{\Delta n_g}[/tex]
Where:
[tex]K_c[/tex] = The equilibrium constant of reaction in terms of concentration
[tex]K_p[/tex] = The equilibrium constant of reaction in terms of partial pressure
R= The universal gas constant
T = The temperature of the equilibrium
[tex]n_g[/tex]= Change in gaseus moles
Given:
An equilibrium reaction, 700°C:
[tex]2H_2(g)+S_2(g)\rightleftharpoons 2H_2S(g),K_c=1.08\times 10^7[/tex]
To find:
The equilibrium constant in terms of partial pressure, [tex]K_p[/tex].
Solution:
The equilibrium constant of reaction in terms of concentration= [tex]K_c[/tex]
[tex]K_c=1.08\times 10^7[/tex]
The equilibrium constant of reaction in terms of partial pressure =[tex]K_p=?[/tex]
The gaseous moles of reactant side = [tex]n_r= 3[/tex]
The gaseous moles of product side = [tex]n_p= 2[/tex]
The temperature at which equilibrium is given = T
[tex]T = 700^oC+273.15 K=973.15K[/tex]
The change in gaseous mole = [tex]n_g=n_p-n_r=2-3 = -1[/tex]
[tex]K_p=1.08\times 10^7\times (0.0821 atm L/mol K\times 973.15 K)^{-1}\\K_p=1.35\times 10^5[/tex]
The value of [tex]K_p[/tex] is [tex]1.35\times 10^5[/tex].
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g Calculate the number of grams of aluminum that is produced in 1.00 h by the electrolysis of molten AlCl3 if the electrical current is 10.0A.
Answer:
3.36 grams Al°(s)
Explanation:
Given AlCl₃(s), determine the mass (grams) of Al°(s) produced from electrolysis of Aluminum Chloride at 10.0 amps for 1.00 hour.
AlCl₃(s) + 378.3°F (=192.4°C) => Al⁺³(l) + 3Cl⁻(l)
formula wt. Al° = 27g/mol
Faraday Constant (F°) = 96,500 amp·sec
? grams Al°(s) = 10.0amps x (1 mole e⁻/96,500amp-sec) x (1 mole Al°(s)/3 mole e⁻) x (27g Al°(s)/1 mole Al°(s)) x 3,600 sec = 3.36 grams Al°(s)
The 3.36 grams of aluminum are produced in 1 hour by the electrolysis of molten AlCl₃ when 10A current is passed.
What is electrolysis?Electrolysis is a process that uses an electrical current to break chemical compounds. The electric current is passed through the substance to bring the chemical change by gain or loss of electrons.
The electrolysis of the aluminum chloride in the molten state is represented as:
AlCl₃ → Al³⁺ + 3Cl⁻
At cathode: Al³⁺ + 3e⁻ → Al (s)
Given, the current. I = 10 A and t = 1 hr = 3600 s
We know that the current is calculated from the equation: I = q/t
q = I× t
q = (10A) × (3600s)
q = 36 × 10³ C
We know, 96500 C of the charge has electrons = 1 mol
36 × 10³ C of the charge has electrons = 0.373 mol
3 moles of electrons required to produce aluminum = 1 mol
0.373 mol of electrons will produce aluminum = 0.373/3 = 0.124 mol
We know that, the mass of one mole of Al = 27g
The mass of 0.124 mol of Al = 27 × 0.124 = 3.36 g
Therefore, the aluminum produced in 1 hour by the electrolysis of molten AlCl₃ is equal to 3.36 grams.
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A major component of gasoline is octane when octane is burned in air it chemically reacts with oxygen to produce carbon dioxide and water what mass of carbon dioxide is produced by the reaction of oxygen
gasoline is the chemical that is coming out of the air
Which of the following describes an organisms habitat?
A) where the organism lives
B) how the organism moves
C) what the organism eats
D) what eats the organism
Answer:
A) habitat
Explanation:
a habitat is essentially the organisms "home". also known as a "niche"
7.23 One equivalent of sodium methanethiolate is added to an electrophile that has two leaving groups. Which product will be major
The question is incomplete, the complete question is shown in the image attached to this answer
Answer:
A
Explanation:
We can see from the conditions of the reaction that the incoming nucleophile is -SCH3 and there are two possible leaving groups in the substrate.
First of all, we have to look at the conditions of the reaction. We can see that the reaction is taking place in DMF, a polar aprotoc solvent. This condition favours the SN2 synchronous mechanism over the SN1 ionic mechanism.
Hence, the nucleophile at the 1-position is preferentially substituted owing to the conditions of the reaction.
Thus, option A is the major product of the reaction.
In the titration of 82.0 mL of 0.400 M HCOOH with 0.150 M LiOH, how many mL of LiOH are required to reach the equivalence point
Answer:
218.7 mL
Explanation:
The reaction that takes place is:
HCOOH + LiOH → LiCOOH + H₂OFirst we calculate how many HCOOH moles reacted, using the given volume and concentration:
82.0 mL * 0.400 M = 32.8 mmol HCOOHAs 1 HCOOH mol reacts with 1 LiOH mol, 32.8 mmoles of LiOH are needed to react with 32.8 mmoles of HCOOH.
Finally we calculate how many mL of a 0.150 M solution would contain 32.8 mmoles:
32.8 mmol / 0.150 M = 218.7 mLWhich of the following is true for balancing equations?
A. There must be an equal number of atoms of each element on both sides of the equation.
B. The number of products should be equal to the number of reactants
C. The properties of products should be the same as the properties of the reactants
D. There must be an equal number of compounds on both sides of the equation
Answer:
A.
Explanation:
An equation with the equal amount and proportion of atoms of each element on both sides of the reaction is commonly referred to as a balanced chemical equation.
The law of conservation of matter asserts that no observable and empirical change in the amount of matter occurs within a conventional chemical process. As a result, each element in the product would have the same equal amount or numbers of atoms as the reactants.
What effect would a decrease in volume have on pressure, assuming that temperature (T) and moles of gas (n) are kept constant
Answer:
Pressure increases
Explanation:
Boyle's law states that; '' the volume of a given mass of ideal gas is inversely proportional to its pressure at constant temperature.
Hence, when the volume of a given mass of ideal gas is decreased, the molecules of the gas come closer together so they collide with each other and the walls of the container more frequently.
This implies that the pressure of the gas increases as volume decreases in accordance with Boyle's law.
What is the difference between conjugate acid-base pair?
a. a H atom. c. a mole water
b. a H+ ion d. a OH– ion
Answer:
b. a H+ ion
Explanation:
The concept of conjugate acid-base pair is related to Bronsted-Lowry acid-base theory and according to this theory, acid is a proton acceptor.
In short,
conjugate base is formed when an acid donates a proton.
conjugate acid is formed when a base accepts a proton.
A student performs an experiment similar to Experiment 1 using hydrochloric acid (HCl) and potassium hydroxide (KOH). The mass of the hydrochloric acid solution is 250.000 g. After combining the HCl and KOH, the final combined mass is 400.000 g. Given what you have learned about the conservation of mass in this experiment, what must have been the mass of the KOH solution
Answer:
150.000 g
Explanation:
The law of conservation of mass states that the mass of reactants and products of a reaction must be equal to one another.
In other words, for this case:
Mass of KOH + Mass of HCl = Mass of ProductsWe are given all required data to calculate the mass of the KOH solution:
Mass of KOH + 250.000 g = 400.000 gMass of KOH = 150.000 gWhat kind of light would an electron experiencing n=4 to n=2 drop emit?
Hi there!
[tex]\large\boxed{\text{Visible Light.}}[/tex]
According to the diagram and the arrows, a drop from level 4 (n = 4) to level 2 (n = 2) produces orange visible light.
[tex]\red\large{{}}[/tex]
A piece of solid tin is submerged in silver nitrate solution a reaction occurs producing tin(IV) nitrate solution and solid silver
Write a word equation write a skeleton equation write a balanced chemical equation
Answer:
Tin + silver trioxonitrate V -------->Tin IV nitrate + silver
Explanation:
The term word equation refers to an equation in which the reactants and products are written in words rather than chemical symbols.
Note than tin is above silver in the electrochemical series hence silver will be displaced by tin as follows;
Tin + silver trioxonitrate V -------->Tin IV nitrate + silver
Select the missing words to complete the definition of buffer capacity. Buffer capacity is the _____________ of acid or base a buffer can handle before pushing the _____________ outside of the buffer range.
Answer:
amount, pH value.
Explanation:
The buffer range is the pH range in which the buffer performs optimally, i.e., neutralizes even when a strong acid or base is introduced to it and resists any major change in its pH value.
The buffer capacity is the amount of acid or base that can be added before the pH of the buffer solution changes significantly.
Thus, the final statement becomes,
Buffer capacity is the amount of acid or base a buffer can handle before pushing the pH value outside of the buffer range.
Answer:
Amount
pH value
Buffer capacity is the amount of acid or base a buffer can handle before pushing the pH outside of the buffer range.
A buffer however consists of a weak acid and its conjugate base.Its major advantage is the ability to resist changes in pH when an acid or a base is added to the solution.
The human blood is also an example of a buffer solution as it is able to resist changes in pH when we eat or drink certain types of food.
An example of a buffer include acetic acid (HC₂H₃O₂) which is a weak acid and sodium acetate (NaC₂H₃O₂) which is a salt derived from that acid).
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43 mg = [?]g
A. 0.043 g
B. 4.3 g
C. 4300 g
D. 43,000 g
Answer:
Option A (0.043 g) is the correct answer.
Explanation:
Given:
= 43 mg
As we know,
[tex]1 \ mg = \frac{1}{1000} \ g[/tex]
then,
⇒ [tex]43 \ mg = \frac{43}{1000} \ g[/tex]
[tex]= 0.043 \ g[/tex]
Thus, the above is the correct alternative.
Which of the following two periods contain the same number of element?
A. 2 & 3
B.3 & 4
c. 4 & 6
D. 2 & 4
Answer:
4 and 6
Explanation:
Period 4 has 18 elements and so does period 6.
a laser emits light with a frequency of 4.69 x 10 to the 14th power s - 1 calculate the wavelength of this light.
Answer:
6.40x10^-7
Explanation:
answer with work is attached.
g A sample of chlorine gas starting at 681 mm Hg is placed under a pressure of 991 mm Hg and reduced to a volume of 513.7 mL. What was the initial volume, in mL, of the chlorine gas container if the process was performed at constant temperature?
Answer:
747.5 mL
Explanation:
Assuming ideal behaviour, we can solve this problem by using Boyle's law, which states that at constant temperature:
P₁V₁ = P₂V₂Where in this case:
P₁ = 681 mm HgV₁ = ?P₂ = 991 mm HgV₂ = 513.7 mLWe input the data given by the problem:
681 mm Hg * V₁ = 991 mm Hg * 513.7 mLAnd solve for V₁:
V₁ = 747.5 mLA student needed to make a 3 g/L NaCl solution. The student weighed 3 g of NaCl in a beaker and measured 1 L of water in a 1L volumetric flask that was labeled TC. The student then added the water to the beaker containing the NaCl. What errors did this student make? Describe how this can be performed properly.
Answer:
The answer is in the explanation.
Explanation:
A solution is defined as the homogeneous mixture of a solute (In this case, NaCl) and the solvent (water).
To prepare 1L of the solution, the student can weigh the 3g of NaCl in the volumetric flask but need to add slowly water to dissolve the NaCl (That is very soluble in water). When all NaCl is dissolved the student must transfer the solution to the 1L volumetric flask. Then, you must add more water to the beaker until "Clean" all the solute of the beaker to transfer it completely to the volumetric flask.
explain why the melting point of a solid is equal to the freezing point of it's liquid.
Explanation:
Because melting point and freezing point describe the same transition of matter, in this case from liquid to solid (freezing) or equivalently, from solid to liquid (melting). It is stuck on 0 ∘C during the entire melting or freezing process. None except melting is when you heat up and freezing when you cool down.hope it helps.stay safe healthy and happy.A 0.50 mol sample of COBr2 is transferred to a 9.50-L flask and heated until equilibrium is attained. Calculate the equilibrium concentrations of each species.
Answer:
Equlibrium concentration for each species ae as follows:
[CO] = 0.043 mol/L
[Br₂] = 0.043 mol/L
[COBr₂] = 0.01 mol/L
Explanation:
Let take a look at the chemical equation taking place at equilibrium
COBr2(g) ⇄ CO(g) + Br2(g)
The concentration of COBr2 i.e.
[COBr2] = no of moles/volume
= 0.50 mol/9.50 L
[COBr2] = 0.0530 mol/L
At standard conditions
Kc for COBr2 = 0.190
Now, the ICE table for the above reaction can be computed as follows:
COBr2(g) ⇄ CO(g) + Br2(g)
Initial 0.053 0 0
Change -x +x +x
Equilibrium (0.053 - x) x x
[tex]\mathsf{K_c = \dfrac{[CO][Br_2]}{[COBr_2]}}[/tex]
[tex]K_c = \dfrac{(x) (x)}{(0,053 -x)}[/tex]
[tex]0.190= \dfrac{x^2}{(0.053 -x)}[/tex]
x² = 0.190(0.053 - x)
x² = 0.01007 - 0.190x
x² + 0.190x - 0.01007 = 0
Using quadratic formula:
x ≅ 0.043 mol/L
SInce: x = [CO][Br₂] = 0.043 mol/L
[COBr₂] = 0.053 - x
[COBr₂] = 0.053 - 0.043 mol/L
[COBr₂] = 0.01 mol/L
What is different between margerine and butter in term of organic chemistry
Answer:
The most important difference between the two is that butter is derived from dairy and is rich in saturated fats, whereas margarine is made from plant oils. ... If the margarine contains partially hydrogenated oils, it will contain trans fat, even if the label claims that it has 0 g.
Explanation:
(⌒_⌒;)
If you could travel at the speed of light, how long would it take to travel from one side of the Milky Way galaxy to the other?
Answer:
It would take 200,000 years for a spaceship traveling at the speed of light to go across the entire galaxy.
EXTRACTION OF CAFFEINE 1.Explain the reason each step of the separation is performed with three portions of the solvent rather than with a single portion of solvent. (e.g. Done in triplicate.) (0.5 pt) 2.Why should a separatory funnel not be vigorously shaken
Answer:
Throughout the explanations section below you will find a description of the question.
Explanation:
(1)
Whether a solution would be positioned inside a separative funnel, combined water, as well as solvent, disintegrate particulate caffeine. In every stage, the caffeine content incorporated relies upon the coefficient of caffeine partitioning throughout the combination of water as well as fluid.Thus, increasingly caffeine is taken from the solvent whenever the moment you bring additional solvent. Consequently, we separate the solvent from the single component.(2)
For compounds to be mixed thoroughly and separated into different layers, a shuddering mixture within the dividing funnel would be essential.However, it vibrates the separation funnel forcefully, restricts airflow within the funnel, which can also induce the fluid under it to burst or causing fluid to fire.Which of the following molecules can be used in catabolic reactions to generate the carbon backbones required for gluconeogenesis?
a. glutathione, a short peptide containing glutamate, serine and histidine
b. butyrate, a short chain fatty acid
c. fructose, a monosaccharide
d. starch, a polysaccharide
Answer:
The correct option is A
Explanation:
Some amino acids, called glucogenic amino acids, when catabolized convert there carbon backbones to tricarboxylic acid (TCA) cycle intermediates. These intermediates can be subsequently metabolized into carbon dioxide and water with the release of ATP or the formation of glucose (known as gluconeogenesis.
All amino acids (with the exception of leucine and lysine) are glucogenic and can thus generate the carbon backbones required for gluconeogenesis. Thus, the correct option is a.
is -2 degrees Fahrenheit warmer or is -17 degrees fahrenheit?
Assign priorities in the following set of substituents according to Cahn-Ingold-Prelog rules.
-OCH3 -Br -Cl -CH2OH
А B C D
(Provide your ranking through a string like abcd, starting with the one with the highest priority).
Answer:
Assign priorities in the following set of substituents according to Cahn-Ingold-Prelog rules.
-OCH3 -Br -Cl -CH2OH
Explanation:
To give priorities for the substituents that are attached to chiral carbon and to assign either R or S-configuration the following rules were proposed:
1. The atom with the highest atomic number is given first priority.
2. If the Groups attached to chiral carbon are having the same first atom, then check for the atomic number of the second atom.
Among the given groups,
-Br has the highest atomic number, so it is given first priority.
Then, -Cl.
Then, -OCH3
and the last one is -CH2OH.
Hence, the order is :
BCAD.
Calculate the displacement (the total volume of the cylinder through which the piston move) of a 5.70L automobile engine in cubic inches, (1inch=2.54cm)
Answer:
348 inches³
Explanation:
From our previous knowledge of units conversion:
We know that 1000 cm³ makes 1 Liter.
Thus, for a 5.70 L automobile engine in cubic meters will be:
= 5.70 × 1000 cm³
= 5700 cm³
Now, the displacement of the automobile in cubic inches provided that 1 inch = 2.534 cm is:
⇒ 5700× (1/ (2.54)³) in³
= 5700×0.0610 in³
= 347.7 in³
≅ 348 inches³