Answer:
C H2SO4 = 9.79 M
Explanation:
molarity (M) ≡ # dissolved species / V slnH2SO4 ↔ H3O+ + SO4-∴ %w/w H2SO4 = 960% = g H2SO4 / g sln * 100
⇒ 9.6 = g H2SO4 / g sln
calculation base: 1000 g sln
⇒ g H2SO4 = 9600g
⇒ mol H2SO4 = 9600 g H2SO4 * ( mol H2SO4/ 980g H2SO4 ) = 9.796 mol H2SO4
⇒ V sln = 1000g sln / 1000g/L = 1 L sln
∴ ρ H20 ≅ 1000 Kg/m³ = 1000 g/L
⇒ C H2SO4 = 9.796 mol H2SO4 / 1 L sln
⇒ C H2SO4 = 9.796 M
Diet cola drinks have a pH of about 3.0, while milk has a pHof
about 7.0. How many times greater is theH3O+
concentration in diet cola than inmilk?
A. 2.3 times higher in diet cola than in milk.
B. 400 times higher in diet cola than in milk.
C. 0.43 times higher in diet cola than in milk.
D. 1,000 times higher in diet cola than in milk.
E. 10,000 times higher in diet cola than in milk.
Answer:
[tex]H_{3}O^{+}[/tex] concentration is 10000 times higher in diet cola than milk
Explanation:
pH is negative logarithm of concentration of [tex]H_{3}O^{+}[/tex] in molarity. It is expressed as: [tex]pH=-log[H_{3}O^{+}][/tex]So, [tex][H_{3}O^{+}]=10^{-pH}[/tex](M)For diet cola, [tex][H_{3}O^{+}]=10^{-3.0}[/tex] (M)For milk, [tex][H_{3}O^{+}]=10^{-7.0}[/tex] (M)So, [tex]\frac{[H_{3}O^{+}](diet cola)}{[H_{3}O^{+}](milk)}=\frac{10^{-3.0}}{10^{-7.0}}=10^{4}=10000[/tex]Hence [tex]H_{3}O^{+}[/tex] concentration is 10000 times higher in diet cola than milkFinal answer:
The concentration of hydronium ions (H3O+) is 10,000 times higher in diet cola with a pH of 3.0 than in milk with a pH of 7.0, due to the logarithmic nature of the pH scale where each pH unit represents a tenfold change in H3O+ concentration.
Explanation:
The pH scale is logarithmic, meaning that each whole number on the pH scale represents a tenfold difference in hydronium ion concentration. Therefore, a change in pH unit corresponds to a change of a factor of 10 in the H3O+ concentration. Diet cola has a pH of about 3.0, and milk has a pH of about 7.0, a difference of 4 pH units. This implies that the hydronium ion concentration in diet cola is 104 times greater than in milk, as each unit difference accounts for a tenfold increase. Thus, the H3O+ concentration is 10,000 times higher in diet cola than in milk.
Alyssa is trying to balance the following equation: Pb(OH)2 + 2H2SO4 => PbSO4 + 2H20 What number should be used to replace the question mark (?) shown in the equation above to balance it? O a. 3 O b. 1 Oco O d. 2 Oe. 4
Answer:
2 Pb(OH)2 + 2H2SO4 => 2 PbSO4 + 4 H20
Explanation:
Since there's no "?" shown in the equation, let's balance it and solve it entirely.
Pb(OH)2 + 2H2SO4 => PbSO4 + 2H20
1Pb + 10O + 6H + 2S ≠ 1Pb + 6O + 4H + 1S → it needs to be balanced.
To do this, let's start by looking at the elements that are only presnet once on each side:
On the products half, S is only present in PbSO4 → if we look at the reagents half, we can see it needs a "2" → then Pb is multiplied by 2 too → so Pb(OH)2 on the reagents half will also need a "2" → final count on O and H on the reagents side: 12O and 8H → to balance it, you need 4 water molecules on the products side.
Calculate the number of moles of a gas that is present in a 7.55 L container at 45°C, if the gas exerts a pressure of 725mm Hg. Enter your answer in the box provided. mol
Answer: The number of moles of gas present is 0.276 moles
Explanation:
To calculate the number of moles of gas, we use the equation given by ideal gas:
PV = nRT
where,
P = Pressure of the gas = 725 mm Hg
V = Volume of the gas = 7.55 L
n = number of moles of gas = ?
R = Gas constant = [tex]62.3637\text{ L.mmHg }mol^{-1}K^{-1}[/tex]
T = Temperature of the gas = [tex]45^oC=(45+273)K=318K[/tex]
Putting values in above equation, we get:
[tex]725mmHg\times 7.55L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 318K\\\\n=0.276mol[/tex]
Hence, the number of moles of gas present is 0.276 moles
Calculate the terminal velocity of a droplet (radius =R, density=\rho_d) when its settling in a stagnant fluid (density=\rho_f).
Answer:
[tex]V=\dfrac{2}{9\ \mu}R^2g(\rho_d-\rho_f)[/tex]
Explanation:
Given that
Radius =R
[tex]Density\ of\ droplet=\rho_d[/tex]
[tex]Density\ of\ fluid=\rho_f[/tex]
When drop let will move downward then so
[tex]F_{net}=F_{weight}-F_{b}-F_d[/tex]
Fb = Bouncy force
Fd = Drag force
We know that
[tex]F_b=\dfrac{4\pi }{3}R^3\ \times \rho_f\times g[/tex]
[tex]F_{weight}=\dfrac{4\pi }{3}R^3\ \times \rho_d\times g[/tex]
[tex]F_{d}=6\pi \mu\ R\ V[/tex]
μ=Dynamic viscosity of fluid
V= Terminal velocity
So at the equilibrium condition
[tex]F_{net}=F_{weight}-F_{b}-F_d[/tex]
[tex]0=F_{weight}-F_{b}-F_d[/tex]
[tex]F_{weight}=F_{b}+F_d[/tex]
[tex]\dfrac{4\pi }{3}R^3\ \times \rho_d\times g=\dfrac{4\pi }{3}R^3\ \times \rho_f\times g+6\pi \mu\ R\ V[/tex]
So
[tex]V=\dfrac{2}{9\ \mu}R^2g(\rho_d-\rho_f)[/tex]
This is the terminal velocity of droplet.
Question 10 0 / 3.5 points Many high temperature studies have been carried out on the equilibrium of the reaction: 2SO2(g) + O2(g) = 2 SO3(g) In one study the reaction vessel initially contained (5.000x10^-3) M SO2, (2.50x10^-3) MO2, and no SO3. If it was determined that at equilibrium the SO2 concentration was (2.8x10^-3) M, determine Kc at this temperature for the reaction as written. • Answers must be written in scientific notation • Write your answer using ONE decimal place (TWO significant figures), even if this is not the correct number of significant figures (e.g., 3.4E-6 or 3.4 x 10-6). • Do NOT use spaces. • Do NOT include units.
Answer: [tex]4.4\times 10^{2}[/tex]
Explanation:
The chemical reaction follows the equation:
[tex]2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)[/tex]
t = 0 [tex]5.000\times 10^{-3}[/tex] [tex]2.50\times 10^{-3}[/tex] 0
At eqm [tex](5.000\times 10^{-3}-2x)[/tex] [tex](2.50\times 10^{-3}-x)[/tex] (2x)
The expression for [tex]K_c[/tex] for the given reaction follows:
[tex]K_c=\frac{[SO_3]^2}{[SO_2]^2[O_2]}[/tex]
[tex]K_c=\frac{[2x]^2}{[5.000\times 10^{-3}-2x]^2[2.50\times 10^{-3}-x]}[/tex]
Given : [tex][SO_2]_{eqm}=2.80\times 10^{-3}[/tex]
[tex]5.000\times 10^{-3}-2x=2.80\times 10^{-3}[/tex]
[tex]x=1.1\times 10^{-3}[/tex]
Putting the values we get:
[tex]K_c=\frac{[2\times 1.1\times 10^{-3}]^2}{[5.000\times 10^{-3}-2\times 1.1\times 10^{-3}]^2[2.50\times 10^{-3}-1.1\times 10^{-3}]}[/tex]
[tex]K_c=4.4\times 10^2[/tex]
Therefore, the equilibrium concentration [tex]4.4\times 10^{2}[/tex]
Which statement about protons is false?
1. Protons have the same magnitude of charge as electrons but are opposite in sign.
2. Protons have about the same mass as neutrons.
3. All atoms have protons.
4. Protons have about the same mass as electrons.
Only one can be selected.
Answer: The correct answer is Option 4.
Explanation:
There are three sub-atomic particles present in an atom. They are: electrons, protons and neutrons.
Protons constitute in each and every atom.
The charge on proton is of equal magnitude as that of electron but having opposite sign. Proton carry a positive charge and electron carry a negative charge.
Protons and neutrons, both determine the mass of an atom.
Mass of 1 proton = 1.007276 u
Mass of 1 neutron = 1.008664 u
Mass of 1 electron = 0.00054858 u
Mass of proton is almost same as that of neutron but is more than the mass of electron.
Hence, the correct answer is Option 4.
Final answer:
The false statement about protons is that they have about the same mass as electrons. In reality, a proton's mass is approximately 1836 times greater than an electron's mass, making this statement incorrect.
Explanation:
The statement about protons that is false is: Protons have about the same mass as electrons.
Let's review the statements to identify the false one:
Protons have the same magnitude of charge as electrons but are opposite in sign. This statement is true since protons and electrons have charges with equal magnitude but opposite signs, with protons having a +1 charge and electrons a -1 charge.Protons have about the same mass as neutrons. This statement is true as protons and neutrons are similar in mass, both approximately 1.67 × 10⁻²⁴ grams, or one atomic mass unit (amu).All atoms have protons. This statement is true as protons are a fundamental component of all atomic nuclei.Protons have about the same mass as electrons. This statement is false because the mass of a proton is about 1836 times greater than that of an electron. Therefore, the proton has significantly more mass compared to the electron.Understanding the basic properties of subatomic particles, such as their mass and charge, is essential in chemistry.
a sample of an oxide of antimony (sb) contain 19.75 g of antimony combine with 6.5 g of oxygen . what is the simplest formula for the oxide
Explanation:
The given data is as follows.
Mass of antimony = 19.75 g
Molar mass of Sb = 121.76 g/mol
Therefore, calculate number of moles of Sb as follows.
Moles of Sb = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{19.75 g}{121.76 g/mol}[/tex]
= 0.162 mol
Mass of oxygen given is 6.5 g and molar mass of oxygen is 16 g/mol. Hence, moles of oxygen will be calculated as follows.
Moles of oxygen = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{6.5 g}{16 g/mol}[/tex]
= 0.406 mol
Hence, ratio of moles of Sb and O will be as follows
Sb : O
[tex]\frac{0.162}{0.162} : \frac{0.406}{0.162}[/tex]
1 : 2.5
We multiply both the ratio by 2 in order to get a whole number. Therefore, the ratio will be 2 : 5.
Thus, we can conclude that the empirical formula of the given oxide is [tex]Sb_{2}O_{5}[/tex].
Not only due trees "fix" carbon but so do green vegetables. Photosynthesis in spinach leaves produces glucose via the Calvin cycle which involves the fixation of CO2 with ribulose 1-5 bisphosphate to form 3-phosphoglycerate via C3H8P2011(aq) + H2O(aq) + CO2(g) → 2 CzH4PO3(aq) + 2 H+(aq) If 15.0 g of 3-phosphoglycerate is formed by this reaction at T = 298 K and P = 1.00 atm what volume of CO2 is fixed? [1.00 L]
Answer : The volume of [tex]CO_2[/tex] gas is 1.00 L
Explanation :
First we have to determine the moles of [tex]C_3H_4PO_7^{3-}[/tex].
Molar mass of [tex]C_3H_4PO_7^{3-}[/tex] = 182.9 g/mole
[tex]\text{ Moles of }C_3H_4PO_7^{3-}=\frac{\text{ Mass of }C_3H_4PO_7^{3-}}{\text{ Molar mass of }C_3H_4PO_7^{3-}}=\frac{15.0g}{182.9g/mole}=0.0820moles[/tex]
Now we have to calculate the moles of [tex]CO_2[/tex].
The given balanced chemical reaction is:
[tex]C_5H_8P_2O_{11}^{4-}(aq)+H_2O(aq)+CO_2(g)\rightarrow 2C_3H_4PO_7^{3-}(aq)+2H^+(aq)[/tex]
From the reaction we conclude that,
As, 2 moles of [tex]C_3H_4PO_7^{3-}[/tex] produce from 1 mole of [tex]CO_2[/tex]
So, 0.0820 moles of [tex]C_3H_4PO_7^{3-}[/tex] produce from [tex]\frac{0.0820}{2}=0.041moles[/tex] of [tex]CO_2[/tex]
Now we have to calculate the volume of [tex]CO_2[/tex] gas.
Using ideal gas equation:
[tex]PV=nRT[/tex]
where,
P = pressure of gas = 1.00 atm
V = volume of gas = ?
T = temperature of gas = 298 K
n = number of moles of gas = 0.041 mole
R = gas constant = 0.0821 L.atm/mole.K
Now put all the given values in the ideal gas equation, we get:
[tex](1.00atm)\times V=(0.041mole)\times (0.0821L.atmK^{-1}mol^{-1})\times (298K)[/tex]
[tex]V=1.00L[/tex]
Therefore, the volume of [tex]CO_2[/tex] gas is 1.00 L
Five million gallons per day (MGD) of wastewater, with a concentration of 10.0 mg/L of a conservative pollutant, is released into a stream having an upstream flow of 10 MGD and pollutant concentration of 3.0 mg/L. What is the concentration in ppm just downstream of the release point? How many pounds of substance per day pass a given spot downstream?
Answer:
a) The concentration in ppm (mg/L) is 5.3 downstream the release point.
b) Per day pass 137.6 pounds of pollutant.
Explanation:
The first step is to convert Million Gallons per Day (MGD) to Liters per day (L/d). In that sense, it is possible to calculate with data given previously in the problem.
Million Gallons per day [tex]1 MGD = 3785411.8 litre/day = 3785411.8 L/d[/tex]
[tex]F_1 = 5 MGD (\frac{3785411.8 L/d}{1MGD} ) = 18927059 L/d\\F_2 =10 MGD (\frac{3785411.8 L/d}{1MGD} )= 37854118 L/d [/tex]
We have one flow of wastewater released into a stream.
First flow is F1 =5 MGD with a concentration of C1 =10.0 mg/L.
Second flow is F2 =10 MGD with a concentration of C2 =3.0 mg/L.
After both of them are mixed, the final concentration will be between 3.0 and 10.0 mg/L. To calculate the final concentration, we can calculate the mass of pollutant in total, adding first and Second flow pollutant, and dividing in total flow. Total flow is the sum of first and second flow. It is shown in the following expression:
[tex]C_f = \frac{F1*C1 +F2*C2}{F1 +F2}[/tex]
Replacing every value in L/d and mg/L
[tex]C_f = \frac{18927059 L/d*10.0 mg/L +37854118 L/d*10.0 mg/L}{18927059 L/d +37854118 L/d}\\C_f = \frac{302832944 mg/d}{56781177 L/d} \\C_f = 5.3 mg/L[/tex]
a) So, the concentration just downstream of the release point will be 5.3 mg/L it means 5.3 ppm.
Finally, we have to calculate the pounds of substance per day (Mp).
We have the total flow F3 = F1 + F2 and the final concentration [tex]C_f[/tex]. It is required to calculate per day, let's take a time of t = 1 day.
[tex]F3 = F2 +F1 = 56781177 L/d \\M_p = F3 * t * C_f\\M_p = 56781177 \frac{L}{d} * 1 d * 5.3 \frac{mg}{L}\\M_p = 302832944 mg[/tex]
After that, mg are converted to pounds.
[tex]M_p = 302832944 mg (\frac{1g}{1000 mg} ) (\frac{1Kg}{1000 g} ) (\frac{2.2 lb}{1 Kg} )\\M_p = 137.6 lb[/tex]
b) A total of 137.6 pounds pass a given spot downstream per day.
part A;
The equation for molarity, M, is
M=n/V
where V is volume and n is the number of moles of solute.
A certain solution has a molarity of M = 2.73 mol/L and has a volume of V = 0.650 L . What is the value of n?
Express your answer numerically in moles.
part B;
The equation for photon energy, E, is
E=hcλ
where h = 6.626×10−34 J⋅s (Planck's constant) and c = 2.99×108 m/s (the speed of light).
What is the wavelength, λ, of a photon that has an energy of E = 3.98×10−19 J ?
Express your answer numerically in meters.
part C;
The ideal gas equation is
PV=nRT
where P is pressure, V is volume, n is the number of moles, R is a constant, and T is temperature.
You are told that a sample of gas has a pressure of P = 899 torr , a volume of V = 3280 mL, and a temperature of T = 307 K . If you use R = 8.206×10−2 L⋅atm/(K⋅mol) , which of the following conversions would be necessary before you could find the number of moles of gas, n, in this sample?
Check all that apply.
View Available Hint(s)
Check all that apply.
Convert the pressure to atmospheres (atm).
Convert the pressure to pascals (Pa).
Convert the volume to cubic meters (m3).
Convert the volume to liters (L).
Convert the temperature to degrees Celsius (∘C).
Convert the temperature to degrees Fahrenheit (∘F).
Answer:
Part A : n = 1.77 moles
Part B : λ = [tex]2.01*10^{6} m^{-1}[/tex]
Part C : n = 0.154 moles
Explanation:
Part A
The problem gives you the equation for molarity M:
[tex]M=\frac{n}{V}[/tex]
n is the number of moles of solute and V is the volume
Then the problem gives you the molarity of a substance [tex]M=2.73\frac{mol}{L}[/tex] and the volume V = 0.650L, so you need to solve the equation for n:
[tex]M=\frac{n}{V}[/tex]
as the V is dividing it passes to multiply the M:
n = M*V
and you should replace the values:
[tex]n = 2.73\frac{mol}{L}*0.650L[/tex]
n = 1.77 moles
Part B
This time you have to solve the equation E = hcλ for λ that is the unknown information, so you have:
E = hcλ
h and c are multiplying so they pass to divide the E:
λ = [tex]\frac{E}{hc}[/tex]
and replacing the values:
λ = [tex]\frac{3.98*10^{-19}J}{(6.626*10^{-34}J.s)(2.99*10^{8}\frac{m}{s})}[/tex]
λ = [tex]2.01*10^{6} m^{-1}[/tex]
PartC
In this part the problem gives you the equation PV=nRT and the first thing you should do is to verify that all the quantities are in consistent units so:
[tex]R=8.206*10^{-2} \frac{L.atm}{K.mol}[/tex] so you need to convert the pressure to atmospheres and convert the volume to liters.
- Convert the pressure to atmospheres:
[tex]P=899torr*\frac{0.00131579atm}{1torr}[/tex]
P = 1.18 atm
- Convert the volume to liters:
[tex]V=3280mL*\frac{1L}{1000mL}[/tex]
V = 3.28L
To find the number of moles n, you should solve the equation for n:
Pv = nRT
As R and T are multiplying the n, they pass to divide to the other side of the equation:
[tex]n=\frac{PV}{RT}[/tex]
And finally you should replace the values:
[tex]n=\frac{(1.18atm)(3.28L)}{(8.206*10^{-2}\frac{L.atm}{K.mol})(307K)}[/tex]
n = 0.154 moles
Which statement about entropy is true? O Non-spontaneous reactions sometimes will happen all by themselves. O Entropy stops when pure substances are in solid state The entropy of the universe is increasing. o During a change of state there is no change in entropy.
Answer:
Explanation:
Entropy stops when pure substances are in solid state: false. The entropy is 0 when the temperature is 0K, not by the solid state.The entropy of the universe is increasing. true. The second law of thermodynamics says: The level of disorder in the universe is steadily increasing.During a change of state there is no change in entropy. false. A change of state occurs with a change of the energy of the system and a change of energy always will change the entropy of the systemNon-spontaneous reactions sometimes will happen all by themselves false. Non-spontaneous reactions never occurs by themselves. Also, this have no immediate relation between non-spontaneous reactions and entropy. Also you have to study the enthalpy of the process.I hope it helps!
Enter your answer in the provided box. Calculate the composition of the following solution in mass/volume % 0.600 g NaOH per 2.00 mL % NaOH
Answer:
30 m/v %
Explanation:
First, we need to understand the definition of mass/volume percentage (m/v %). This is a way of expressing concentration of a substance and it means grams of a determined compound per 100 mL of solution:
m/v% = g of X substance/100 mL of solution
Having said that, if we have 0.600 g of NaOH per 2.00 mL of solution, then in 100 mL of solution we will have:
2.00 mL solution ---- 0.600 g of NaOH
100 mL solution---- x=(100 mL × 0.600 g NaOH)/ 2.00 mL = 30 g = 30 %m/v
So, the concentration of a solution containing 0.600 g of NaOH per 2.00 mL is a 30 m/v% solution.-
Write 605,400.34 in Scientific Notation with 4 significant figures.
Answer : The correct answer is, [tex]6.054\times 10^5[/tex]
Explanation :
Scientific notation : It is the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.
For example :
5000 is written as [tex]5.0\times 10^3[/tex]
889.9 is written as [tex]8.899\times 10^{-2}[/tex]
In this examples, 5000 and 889.9 are written in the standard notation and [tex]5.0\times 10^3[/tex] and [tex]8.899\times 10^{-2}[/tex] are written in the scientific notation.
[tex]8.89\times 10^{-2}[/tex] this is written in the scientific notation and the standard notation of this number will be, 0.00889.
If the decimal is shifting to right side, the power of 10 is negative and if the decimal is shifting to left side, the power of 10 is positive.
As we are given the 605,400.34 in standard notation.
Now converting this into scientific notation, we get:
[tex]\Rightarrow 605,400.34=6.054\times 10^5[/tex]
As, the decimal point is shifting to left side, thus the power of 10 is positive.
Hence, the correct answer is, [tex]6.054\times 10^5[/tex]
A 25.0-mL solution of 0.100 M CH3COOH
istitrated with a 0.200 M KOH solution. Calculate thepH
after the following additions of the KOH solution : (a) 0.0 mL,(b)
5.0 mL, (c) 10.0 mL, (d) 12.5 mL, (e) 15.0 mL.
Answer:
a) pH = 2,88
b) pH = 4,58
c) pH = 5,36
d) pH = 8,79
e) pH = 12,10
Explanation:
In a titration of a strong base (KOH) with a weak acid (CH₃COOH) the reaction is:
CH₃COOH + KOH → CH₃COOK + H₂O
a) Here you have just CH₃COOH, thus:
CH₃COOH ⇄ CH₃COO⁻ + H⁺ where ka =1,74x10⁻⁵ and pka = 4,76
When this reaction is in equilibrium:
[CH₃COOH] = 0,100 -x
[CH₃COO⁻] = x
[H⁺] = x
Thus, equilibrium equation is:
1,74x10⁻⁵ = [tex]\frac{[x][x] }{[0,100-x]}[/tex]
The equation you will obtain is:
x² + 1,74x10⁻⁵x - 1,74x10⁻⁶ = 0
Solving:
x = -0,0013278193 ⇒ No physical sense. There are not negative concentrations
x = 0,0013104193
As x = [H⁺] and pH = - log [H⁺]
pH = 2,88
b) Here, it is possible to use:
CH₃COOH + KOH → CH₃COOK + H₂O
With adition of 5,0 mL of 0,200M KOH solution the initial moles are:
CH₃COOH = [tex]0,025 L.\frac{0,100 mol}{L} =[/tex] = 2,5x10⁻³ mol
KOH = [tex]0,005 L.\frac{0,200 mol}{L} =[/tex] = 1,0x10⁻³ mol
CH₃COOK = 0.
In equilibrium:
CH₃COOH = 2,5x10⁻³ mol - 1,0x10⁻³ mol = 1,5x10⁻³ mol
KOH = 0 mol
CH₃COOK = 1,0x10⁻³ mol
Now, you can use Henderson–Hasselbalch equation:
pH = 4,76 + log [tex]\frac{1,0x10^{-3} }{1,5x10^{-3} }[/tex]
pH = 4,58
c) With adition of 10,0 mL of 0,200M KOH solution the initial moles are:
CH₃COOH = [tex]0,025 L.\frac{0,100 mol}{L} =[/tex] = 2,5x10⁻³ mol
KOH = [tex]0,010 L.\frac{0,200 mol}{L} =[/tex] = 2,0x10⁻³ mol
CH₃COOK = 0.
In equilibrium:
CH₃COOH = 2,5x10⁻³ mol - 2,0x10⁻³ mol = 0,5x10⁻³ mol
KOH = 0 mol
CH₃COOK = 2,0x10⁻³ mol
Now, you can use Henderson–Hasselbalch equation:
pH = 4,76 + log [tex]\frac{2,0x10^{-3} }{0,5x10^{-3} }[/tex]
pH = 5,36
d) With adition of 12,5 mL of 0,200M KOH solution the initial moles are:
CH₃COOH = [tex]0,025 L.\frac{0,100 mol}{L} =[/tex] = 2,5x10⁻³ mol
KOH = [tex]0,0125 L.\frac{0,200 mol}{L} =[/tex] = 2,5x10⁻³ mol
CH₃COOK = 0.
Here we have the equivalence point of the titration, thus, the equilibrium is:
CH₃COO⁻ + H₂O ⇄ CH₃COOH + OH⁻ kb = kw/ka where kw is equilibrium constant of water = 1,0x10⁻¹⁴; kb = 5,75x10⁻¹⁰
Concentrations is equilibrium are:
[CH₃COOH] = x
[CH₃COO⁻] = 0,06667-x
[OH⁻] = x
Thus, equilibrium equation is:
5,75x10⁻¹⁰ = [tex]\frac{[x][x] }{[0,06667-x]}[/tex]
The equation you will obtain is:
x² + 5,75x10⁻¹⁰x - 3,83x10⁻¹¹ = 0
Solving:
x = -0.000006188987⇒ No physical sense. There are not negative concentrations
x = 0.000006188
As x = [OH⁻] and pOH = - log [OH⁻]; pH = 14 - pOH
pOH = 5,21
pH = 8,79
e) The excess volume of KOH will determine pH:
With 12,5mL is equivalence point, the excess volume is 15,0 -12,5 = 2,5 mL
2,5x10⁻³ L × [tex]\frac{0,200 mol}{1L}[/tex] ÷ 0,040 L = 0,0125 = [OH⁻]
pOH = - log [OH⁻]; pH = 14 - pOH
pOH = 1,90
pH = 12,10
I hope it helps!
Answer:
The pH of solution on addition of 0.0, 5.0, 10.0 12.5 and 15 ml of KOH will be 2.87, 4.56, 5.34, 4.74, and 12.09 respectively.
Explanation:
pH can be calculated by the evaluation of Hydronium ions in the solution.
[tex]\rm K_a[/tex] of [tex]\rm CH_3COOH[/tex] is 1.8 [tex]\rm \times10^-^5[/tex]
(a) Hydrogen ion concentration = [tex]\rm \sqrt{K_a\;\times\;CH_3COOH\;concentartion}[/tex]
Hydrogen ion concentration = [tex]\rm \sqrt{1.8\;\times\;10^-^5\;\times\;0.1}[/tex]
Hydrogen ion concentration = 1.34 [tex]\times\;10^-^3[/tex] M
pH of solution = log [Hydrogen ion concentration]
pH = log [[tex]\rm 1.34\;\times10^-^5[/tex]]
pH = 2.87
(b) On addition of 5 ml of KOH of 0.200 M, the moles of KOH added are:
moles of KOH = [tex]\rm \frac{volume\;(ml)}{1000}\;\times\;\frac{molarity}{L}[/tex]
moles of KOH = [tex]\rm \frac{5}{1000}\;\times\;\frac{0.200}{L}[/tex]
moles of KOH = [tex]\rm 1\;\times\;10^-^3[/tex] M
The initial moles of [tex]\rm CH_3COOH[/tex] are:
moles of [tex]\rm CH_3COOH[/tex] = [tex]\rm \frac{25}{1000}\;\times\;\frac{0.100}{L}[/tex]
moles of [tex]\rm CH_3COOH[/tex] = [tex]\rm 2.5\;\times\;10^-^3[/tex]
At equilibrium, the concentration of [tex]\rm CH_3COOH[/tex] = [tex]\rm 2.5\;\times\;10^-^3[/tex] - [tex]\rm 1\;\times\;10^-^3[/tex] mol
concentration of [tex]\rm CH_3COOH[/tex] = [tex]\rm 1.5\;\times\;10^-^3[/tex] moles.
The concentration of [tex]\rm CH_3COOK[/tex] = [tex]\rm 1\;\times\;10^-^3[/tex] moles
pH = [tex]\rm pK_a\;+\;log\;\frac{salt}{acid}[/tex]
pH = 4.76 + log [tex]\rm \frac{1\;\times\;10^-^3}{1.5\;\times\;10^-^3}[/tex]
pH = 4.58
(c) On addition of 10 ml of KOH,
moles of KOH = [tex]\rm \frac{10}{1000}\;\times\;\frac{0.200}{L}[/tex]
moles of KOH = [tex]\rm 2\;\times\;10^-^3[/tex] moles
moles of [tex]\rm CH_3COOH[/tex] = [tex]\rm 2.5\;\times\;10^-^3[/tex] moles
moles of [tex]\rm CH_3COOK[/tex] = [tex]\rm 2\;\times\;10^-^3[/tex] moles
pH = 4.76 + log [tex]\rm \frac{2\;\times\;10^-^3}{0.5\;\times\;10^-^3}[/tex]
pH = 5.36
(d) On addition of 12.5 ml of KOH,
moles of KOH = [tex]\rm \frac{12.5}{1000}\;\times\;\frac{0.200}{L}[/tex]
moles of KOH = [tex]\rm 2.5\;\times\;10^-^3[/tex]
moles of [tex]\rm CH_3COOH[/tex] = [tex]\rm 2.5\;\times\;10^-^3[/tex] moles
moles of [tex]\rm CH_3COOK[/tex] = [tex]\rm 2.5\;\times\;10^-^3[/tex] moles
pH = 4.76 + log [tex]\rm \frac{2\;\times\;10^-^3}{0}[/tex]
pH = 4.74
(e) On addition of 15.0 ml of KOH,
The excess point is reached after the addition of 12.5 ml of KOH. After further addition of KOH, the pOH will be there.
The OH concentration on addition of KOH = [tex]\rm \frac{(\frac{15}{1000}\;\times0.200\;moles)\;-\;(\frac{25}{1000}\;\times0.100\;moles) }{\frac{25}{1000}\;+\;\frac{15}{1000} }[/tex]
= 0.0125 M
pOH = -log [OH concentration]
pOH = -log [0.0125]
pOH = 1.903
pH = 14 - pOH
pH = 14 - 1.903
pH = 12.097
The pH of solution on addtion of KOH will be :
O ml KOH = 2.87
5 ml KOH= 4.56
10 ml KOH = 5.36
12.5 ml KOH = 4.74
15 ml KOH = 12.907.
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DrugDigest is a website that is owned and hosted by a. Food and Drug Administration c. Express Scripts b. National Library of Medicine d. Blue cross/Blue Shield
Answer:
The answer is c. Express Scripts Company
Explanation:
DrugDigest is a website tool that helps you to find information about medicaments you consume or you see for example you introduce the first three letters of the drug in the website and you will have a detailed description of brands, supplied information eg. if the medicament comes by pills or by syrup and also the concentration of the medicament and recommendations ( under which case you can take the medicine, effects, side effects)
Which of the following is (are) generally property of metals? 2. a) Electrical insulating b) Thermally conductive c) Mechanically brittle d) Typically gases or liquids
Answer:
b) Thermally conductive
Explanation:
Few properties of metals are:
Metals are hard except metals, sodium and potassium.Metals are good conductors of heat and electricity as they have free electrons.All metals are solids at the room temperature except mercury.Metals have high malleability and ductility.Hence, correct options is B. The rest are not properties of metals.
Enter your answer in the box provided. How many grams of helium must be added to a balloon containing 6.24 g helium gas to double its volume? Assume no change in temperature or pressure. g
Answer : The mass of helium gas added must be 12.48 grams.
Explanation : Given,
Mass of helium (He) gas = 6.24 g
Molar mass of helium = 4 g/mole
First we have to calculate the moles of helium gas.
[tex]\text{Moles of }He=\frac{\text{Mass of }He}{\text{Molar mass of }He}=\frac{6.24g}{4g/mole}=1.56moles[/tex]
Now we have to calculate the moles of helium gas at doubled volume.
According to the Avogadro's law, the volume of gas is directly proportional to the number of moles of gas at same pressure and temperature. That means,
[tex]V\propto n[/tex]
or,
[tex]\frac{V_1}{V_2}=\frac{n_1}{n_2}[/tex]
where,
[tex]V_1[/tex] = initial volume of gas = V
[tex]V_2[/tex] = final volume of gas = 2V
[tex]n_1[/tex] = initial moles of gas = 1.56 mole
[tex]n_2[/tex] = final moles of gas = ?
Now we put all the given values in this formula, we get
[tex]\frac{V}{2V}=\frac{1.56mole}{n_2}[/tex]
[tex]n_2=3.12mole[/tex]
Now we have to calculate the mass of helium gas at doubled volume.
[tex]\text{Mass of }He=\text{Moles of }He\times \text{Molar mass of }He[/tex]
[tex]\text{Mass of }He=3.12mole\times 4g/mole=12.48g[/tex]
Therefore, the mass of helium gas added must be 12.48 grams.
Final answer:
To double the volume of a helium balloon at constant temperature and pressure, an additional 6.24 grams of helium must be added, resulting in a total of 12.48 grams of helium in the balloon.
Explanation:
The question involves the concept of the molar volume of a gas, which is a central principle in chemistry, particularly in the context of gases under the ideal gas law. To double the volume of helium in a balloon at constant temperature and pressure, the amount of helium in moles must be doubled as well. If initially there are 6.24 grams of helium in the balloon, we first convert this mass to moles using the molar mass of helium (4.00 g/mol) and then calculate the additional moles (and hence grams) of helium needed to double the volume.
Initial moles of helium (n1): 6.24 g / 4.00 g/mol = 1.56 moles
To double the volume, we need another 1.56 moles of helium. The mass of additional helium required is:
Additional mass (Δm): 1.56 moles x 4.00 g/mol = 6.24 g
Therefore, to double the volume of the helium balloon, an additional 6.24 grams of helium gas must be added.
Liquid X has a density of 0.834 g/mL. What is the volume, in ml, of 205 g of liquid X? Please report your answer to the nearest whole mL.
Answer: The volume of liquid X is 246 mL
Explanation:
To calculate volume of a substance, we use the equation:
[tex]\text{Density of a substance}=\frac{\text{Mass of a substance}}{\text{Volume of a substance}}[/tex]
We are given:
Mass of liquid X = 205 g
Density of liquid X = 0.834 g/mL
Putting values in above equation, we get:
[tex]0.834g/mL=\frac{205g}{\text{Volume of liquid X}}\\\\\text{Volume of liquid X}=246mL[/tex]
Hence, the volume of liquid X is 246 mL
What is the mass, in g, of 27.2 mol of N? Be sure to answer all parts. Enter your answer in scientific notation. x 10 (select) AgN
Explanation:
It is known that number of moles of a substance equal to mass divided by its molar mass.
Mathematically, No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
It is given that no. of oles present are 27.2 mol and molar mass of N is 14 g/mol.
Therefore, calculate mass of N as follows.
No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
27.2 mol = [tex]\frac{mass}{14 g/mol}[/tex]
mass = 380.8 g
Thus, we can conclude that mass of N is 380.8 g.
The equation for density, d, is
d=m/V
where m is mass and V is volume.
What is the density, d, of a substance with a volume of V = 18.2 cm3 and a mass of m = 61.6 g ?
Express your answer numerically in grams per cubic centimeter.
Answer: The density of substance is [tex]3.38g/cm^3[/tex]
Explanation:
To calculate density of a substance, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
We are given:
Mass of substance = 61.6 g
Volume of substance = [tex]18.2cm^3[/tex]
Putting values in above equation, we get:
[tex]\text{Density of substance}=\frac{61.6g}{18.2cm^3}\\\\\text{Density of substance}=3.38g/cm^3[/tex]
Hence, the density of substance is [tex]3.38g/cm^3[/tex]
Final answer:
The density of the substance with a mass of 61.6 g and a volume of 18.2 [tex]cm^3[/tex] is 3.385 g/[tex]cm^3[/tex], calculated by dividing mass by volume using the density formula.
Explanation:
The density of a substance is calculated using the formula d = m/V, where d is the density, m is the mass, and V is the volume.
To find the density of a substance with a given mass and volume, you simply divide the mass by the volume. For a substance with a mass of m = 61.6 g and a volume of V = 18.2 [tex]cm^3[/tex], the density d would be calculated as follows:
d = m/V
d = 61.6 g / 18.2 [tex]cm^3[/tex]
d \3.385 g/[tex]cm^3[/tex]
This means that the density measurement of the substance is 3.385 grams per cubic centimeter, expressed as g/[tex]cm^3[/tex].
carbon atoms stick to catalyst surface in craking processes ?
1- sintring
2- poising
3- aging
4- cocking or fouling
Answer:
option D - coking or fouling
Explanation:
Coking (not cocking) is the process involving the deposition of small carbon particles (created by simply putting carbon atoms) on a catalyst's accessible surface, leading in a reduction in the area accessible for catalytic activity. This is also sometimes related to fouling catalysts or merely fouling them.
When iron(II) chloride reacts with silver nitrate, iron(II) nitrate and silver chloride are produced. The balanced equation for this reaction is: FeCl2 (aq) + 2AgNO3(aq) --> Fe(NO3)2(aq) + 2AgCl(s) Suppose 2.86 moles of iron(II) chloride react. The reaction consumes moles of silver nitrate. The reaction produces moles of iron(II) nitrate and moles of silver chloride. Submit Answer & Next
Answer: The moles of silver nitrate reacted is 5.72 moles, moles of iron (II) nitrate produced is 2.86 moles and moles of silver chloride produced is 5.72 moles
Explanation:
We are given:
Moles of iron (II) chloride = 2.86 moles
For the given chemical equation:
[tex]FeCl_2(aq.)+2AgNO_3(aq.)\rightarrow Fe(NO_3)_2(aq.)+2AgCl(s)[/tex]
For silver nitrate:By Stoichiometry of the reaction:
1 mole of iron (II) chloride reacts with 2 moles of silver nitrate.
So, 2.86 moles of iron (II) chloride will react with = [tex]\frac{2}{1}\times 2.86=5.72mol[/tex] of silver nitrate
Moles of silver nitrate reacted = 5.72 moles
For iron (II) nitrate:By Stoichiometry of the reaction:
1 mole of iron (II) chloride produces 1 mole of iron (II) nitrate
So, 2.86 moles of iron (II) chloride will produce = [tex]\frac{1}{1}\times 2.86=2.86mol[/tex] of iron (II) nitrate
Moles of iron (II) nitrate produced = 2.86 moles
For silver chloride:By Stoichiometry of the reaction:
1 mole of iron (II) chloride produces 2 moles of silver chloride
So, 2.86 moles of iron (II) chloride will produce = [tex]\frac{2}{1}\times 2.86=5.72mol[/tex] of silver chloride
Moles of silver chloride produced = 5.72 moles
Hence, the moles of silver nitrate reacted is 5.72 moles, moles of iron (II) nitrate produced is 2.86 moles and moles of silver chloride produced is 5.72 moles
When 2.86 moles of FeCl2 react, they consume 5.72 moles of AgNO3, produce 2.86 moles of Fe(NO3)2, and produce 5.72 moles of AgCl.
Explanation:In the balanced equation FeCl2 (aq) + 2AgNO3(aq) --> Fe(NO3)2(aq) + 2AgCl(s), each FeCl2 molecule reacts with 2 AgNO3 molecules. This means that for every mole of FeCl2, 2 moles of AgNO3 are consumed. Therefore, if we have 2.86 moles of FeCl2, it will consume 2 x 2.86 = 5.72 moles of AgNO3.
In this reaction, for every mole of FeCl2 consumed, 1 mole of Fe(NO3)2 is produced. Therefore, if we have 2.86 moles of FeCl2, 2.86 moles of Fe(NO3)2 will be produced.
In addition, for every mole of FeCl2 consumed, 2 moles of AgCl are produced. Therefore, if we have 2.86 moles of FeCl2, 2 x 2.86 = 5.72 moles of AgCl will be produced.
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weak acid. d. weak base. An amphoteric species is one that reacts as a. acid only. b. base only. a(n) c. acid or base. None of the above
Answer: The nature of an amphoteric substance is amphoprotic; it has ability to donate proton or gain proton. So the amphotric substances are those which can react as both acid or base in a reaction. Many oxides of metals when subjected to a reaction mixture can acts as both an acid or either a base, they are called amphoteric oxides. some of the examples of these amphoteric oxide is zinc oxide and lead oxide.
Hence, the correct option here is (c )
Use the References to access important values if needed for this question. The radius of a potassium atom is 231 pm. How many potassium atoms would have to be laid side by side to span a distance of 2.91 mm? atoms Submit Answer Try Another Version 1 item attempt remaining
Answer:
6298702 potassium atoms would have to be laid side by side to span a distance of 2,91 mm
Explanation:
If the radius of a potassium atom is 231 pm, then the diameter would be:
d = 2r = 2*(231) = 462 pm
So each potassium atom occupies a space of 462 pm, we can express this relationship as follows:
[tex]\frac{1 potassium atom}{462 pm}[/tex]
To solve this problem, we'll use the following conversion factors:
1 pm = 1×E−12 m
1000 mm = 1 m
We begin to accommodate all our relationships starting from that numerical expression that is not written as a relationship (usually the one in the question), and in such a way that the units are eliminated between them.
[tex]2, 91 mm * \frac{1 m}{1000 mm}*\frac{ 1 pm}{1*10^{-12}m }*\frac{1 potassium atom}{462 pm}= 6298701.3[/tex] potassium atoms
So, we'll need 6298702 potassium atoms to span a distance of 2,91 mm
A double pipe heat exchanger is to be designed to heat 1 kg/s of a cold fluid from 30°C to 60°C using 2 kg/s of a hot fluid at 100°C. The two streams have equal specific heat capacities and overall heat transfer coefficient. Calculate the ratio of heat transfer areas of counter current to co-current. a) 1.142 b) 0.875 c) 0.927 d) 1.077
Answer:
big question
Explanation:
QUESTION 1 1.041 points If a 50.00 ml aliquot of a 0.12 M NaCl solution is added to 30.00 mL of a 0.18 M CaCl solution, what is the concentration of calcium ion in the mixture? 0.10M 0.086 M 0,068 M 0.36 M 0.090 M
Answer:
[Ca²⁺] = 0.068 M
Explanation:
The concentration of the calcium ion will be equal to the amount of calcium in CaCl₂, divided by the total volume:
C = n/V
When CaCl₂ dissociates in water, one mole of calcium ion is produced for every mole of CaCl₂, so the molar ratio of CaCl₂ to Ca²⁺ is 1:1. The moles of Ca²⁺ are calculated as follows:
(0.18 mol/L)(30.00 mL) = 5.4 mmol CaCl₂ = 5.4 mmol Ca²⁺
The total volume is (50.00 mL + 30.00 mL) = 80.00 mL
Thus, the concentration of Ca²⁺ is:
C = n/V = (5.4 mmol)/(80.00 mL) = 0.068 M
How do you know when the central atom in a lewis structure will have more than 8 electrons?
One particular example in mind is BrO3F. The structure should have 32 electrons, and it does when Br has a single bond with the other atoms and their formal charge cancels out to have a formal charge of zero.
Answer:
To know when a central atom in a lewis structure will have more than 8 electrons it is important to know where is the element at the periodic table and the orbital configuration of the atom.
Explanation:
The octet rule establishes that an atom could win, lose, or share electrons with other atoms till every atom have eight valence electrons. However exist exceptions to the octet rule. Some elements like Be, B and Al are stable with only six valence electrons because these three elements are small and with low electronegativity. On the other hand, some elements from the 3d orbital could expand their octet and still be stable, like S in SF6. This happens because elements from 3d orbital have enough space to suit more electrons.
The steam generator in a sodium-cooled reactor power plant receives sodium at the rate of 10 x 106 lbm/hr. The sodium temperatures are 600 and | 1000°F. 800°F steam is generated at 1,000 psia from feedwater at 340°F necessary reactor output in Mw(t) Assuming no heat losses, find the temperature difference at the pinch point in the generator. (ep for Na = 0.3.)
Explanation:
It is known that pinch point temperature difference is defined as the difference between the temperature of exhaust existing the evaporator and the temperature of water evaporation.
[tex]\Delta T_{p.p}[/tex] = Exhaust existing the evaporator temperature - Temperature of water evaporation
Since, it is given that exhaust existing the evaporator temperature = [tex]1000^{o}F[/tex]
Temperature of water evaporation = [tex]800^{o}F[/tex]
Hence, calculate [tex]\Delta T_{p.p}[/tex] using the above formula as follows.
[tex]\Delta T_{p.p}[/tex] = Exhaust existing the evaporator temperature - Temperature of water evaporation
= [tex]1000^{o}F[/tex] - [tex]800^{o}F[/tex]
= [tex]200^{o}F[/tex]
Thus, we can conclude that the temperature difference at the pinch point in the generator is [tex]200^{o}F[/tex].
A protein subunit from an enzyme is part of a research study and needs to be characterized. A total of 0.150 g of this subunit was dissolved in enough water to produce 2.00 mL of solution. At 28 ∘C the osmotic pressure produced by the solution was 0.138 atm. What is the molar mass of the protein?
To find the molar mass of the protein subunit, use the formula for osmotic pressure. Rearrange the formula to solve for M and plug in the given values. Convert the mass of the protein to moles using the molar mass.
Explanation:To find the molar mass of the protein subunit, we can use the formula for osmotic pressure, II = MRT. Given that the osmotic pressure is 0.138 atm and the temperature is 28 °C (which is 310 K), we can rearrange the formula to solve for M. The value of R is 0.08206 L atm/mol K. Plugging in the values, we get:
0.138 atm = M * (0.08206 L atm/mol K) * (310 K)
Solving for M gives:
M = 0.138 atm / (0.08206 L atm/mol K * 310 K)
Calculating this expression will give us the molarity of the solution. Since we dissolved 0.150 g of the protein subunit in enough water to make 2.00 mL of solution, we can convert this to moles using the molar mass of the protein. The molar mass is given by:
Molar mass = Mass / Moles
Therefore, we can rearrange the equation to solve for the molar mass:
Molar mass = Mass / Moles = (0.150 g / M) * (1 mol / 1000 g)
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To convert from liters/second to cubic gallons/minute, multiply the number of liters/second by 15.850 0 0.0353 00.2642 0 60
Answer: 15.850
Explanation:
The conversion used from liters to gallons is:
1 L = 0.264172 gallon
The conversion used from sec to min is:
60 sec = 1 min
1 sec =[tex]\frac{1}{60}\times 1=0.017min[/tex]
We are asked: liters/sec = gallons/min
[tex]liters/sec=\frac{0.264172}{0.017}=15.850gallons/min[/tex]
Therefore, to convert from liters/second to gallons/minute, multiply the number of liters/second by 15.850.