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.
To find the concentration in ppm just downstream, we calculate the mass of the pollutant using the flow rate and concentration. The concentration in ppm just downstream is 50 ppm. To find the number of pounds of substance per day passing a given spot downstream, we convert the mass from grams to pounds.
Explanation:In order to find the concentration in ppm just downstream, we need to calculate the mass of pollutant that is being released into the stream. We can use the following formula:
Mass = Flow rate x Concentration
Using the given flow rates and concentrations:
Upstream Mass = 10 MGD x 3.0 mg/L = 30 mg/L
Downstream Mass = 5 MGD x 10.0 mg/L = 50 mg/L
So the concentration in ppm just downstream is:
50 mg/L x (1 ppm/1 mg/L) = 50 ppm
To find the number of pounds of substance per day passing a given spot downstream, we will need to calculate the mass in pounds. We can use the following conversions:
1 lbm = 2.2 kg
1 kg = 1000 g
1 g = 1000 mg
Using the given flow rate and concentration:
Downstream Mass (in g/day) = 5 MGD x 10.0 mg/L = 50,000 g/day
Downstream Mass (in kg/day) = 50,000 g/day x (1 kg/1000 g) = 50 kg/day
Downstream Mass (in lbm/day) = 50 kg/day x (1 lbm/2.2 kg) = 22.7 lbm/day
So the number of pounds of substance per day passing a given spot downstream is approximately 22.7 pounds.
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
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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What is the pH of a 0.18 M CH3NH3+Cl– aqueous solution? The pKb ofCH3NH2 is 3.44
Answer:
5.65
Explanation:
Given that:
[tex]pK_{b}\ of\ CH_3NH_2=3.44[/tex]
[tex]K_{b}\ of\ CH_3NH_2=10^{-3.44}=3.6308\times 10^{-4}[/tex]
[tex]K_a\ of\ CH_3NH_3^+Cl^-=\frac {K_w}{K_b}=\frac {10^{-14}}{3.6308\times 10^{-4}}=2.7542\times 10^{-11}[/tex]
Concentration = 0.18 M
Consider the ICE take for the dissociation as:
[tex]CH_3NH_3^+[/tex] ⇄ H⁺ + [tex]CH_3NH_2[/tex]
At t=0 0.18 - -
At t =equilibrium (0.18-x) x x
The expression for dissociation constant of acetic acid is:
[tex]K_{a}=\frac {\left [ H^{+} \right ]\left [CH_3NH_2 \right ]}{[CH_3NH_3^+]}[/tex]
[tex]2.7542\times 10^{-11}=\frac {x^2}{0.18-x}[/tex]
x is very small, so (0.18 - x) ≅ 0.18
Solving for x, we get:
x = 0.2227×10⁻⁵ M
pH = -log[H⁺] = -log(0.2227×10⁻⁵) = 5.65
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]
Trinitrotoluene (C-H5N306, 227.1 g/mol) is easily detonated. How many grams of carbon are in 57.6 grams of TNT? Avogadro's Number: 1 mole = 6.02 x 1023 species A. 403 g B. 57.6 g C. 21.3 g D. 1.78 g E. None of the above
Answer: The correct answer is Option C.
Explanation:
The chemical formula for trinitrotoluene is [tex]C_7H_5N_3O_6[/tex]
In 1 mole of TNT, 7 moles of carbon atom, 5 moles of hydrogen atom, 3 moles of nitrogen atom and 6 moles of oxygen atom are present.
We know that:
Mass of trinitrotoluene = 227.1 g/mol
Mass of carbon = 12.01 g/mol
We are given:
Mass of TNT = 57.6 grams
To calculate the mass of carbon in given amount of TNT, we apply unitary method:
In 227.1 grams of TNT, amount of carbon present is = [tex](7\times 12.01)=84.07g[/tex]
So, in 57.6 grams of TNT, the amount of carbon present is = [tex]\frac{84.07g}{227.1g}\times 57.6g=21.3g[/tex]
Hence, the correct answer is Option C.
Which of the following gives the net ionic reaction for the reaction used in this experiment?
H+(aq) + OH-(aq) → H2O(l)
no net reaction
Ba2+(aq) + SO42-(aq) → BaSO4(s)
Ba2+(aq) + 2 OH-(aq) + 2 H+(aq) + SO42-(aq) → BaSO4(s) + 2 H2O(l)
Answer: Option (d) is the correct answer.
Explanation:
The given reaction will be as follows.
[tex]Ba(OH)_{2}(aq) + H_{2}SO_{4}(aq) \rightarrow BaSO_{4}(s) + 2H_{2}O(l)[/tex]
In the ionic form, the equation will be as follows.
[tex]Ba^{2+}(aq) + SO^{2-}_{4}(aq) \rightarrow BaSO_{4}(s)[/tex] ........ (1)
[tex]H^{+}(aq) + OH^{-}(aq) \rightarrow H_{2}O(l)[/tex] ............ (2)
Hence, for the net ionic equation we need to add both equation (1) and (2). Therefore, the net ionic equation will be as follows.
[tex]Ba^{2+}(aq) + SO^{2-}_{4}(aq) + H^{+}(aq) + OH^{-}(aq) \rightarrow BaSO_{4}(s) + H_{2}O(l)[/tex]
Now, balancing the atoms on both the sides we get the net ionic equation as follows.
[tex]Ba^{2+}(aq) + SO^{2-}_{4}(aq) + 2H^{+}(aq) + 2OH^{-}(aq) \rightarrow BaSO_{4}(s) + 2H_{2}O(l)[/tex]
The correct net ionic reaction for the experiment is H+(aq) + OH-(aq) → H2O(l), an example of an acid-base neutralization reaction. The other options included the formation of a precipitate, which is not part of the net ionic reaction.
Explanation:Based on the provided options, the correct net ionic reaction for the experiment seems to be H+(aq) + OH-(aq) → H2O(l). This reaction is a classic example of an acid-base neutralization reaction, where an acid (H+) and a base (OH-) react to form water. The other two reactions involve the formation of a precipitate (BaSO4), but the full reaction is simplified to leave out the precipitate ions on either side. This does not occur in the first reaction. Hence, the first reaction is the correct net ionic reaction for this experiment.
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which higher heat transfer ?
a)Fan with air
b)boiling water
c)cooled liquid
Answer:
b)boiling water
Explanation:
High heat is transferred with the thermal conduction process, also called diffusion. Always occurs from a region of higher temperatures to a region of lower temperatures. The flow of fluid may be forced into external processes. As the particles rapidly move and vibrate and the transfer of heat from one state to the other by the interaction of particles and rate of heat lost has to be the same as absorbed heat during vaporization at the same fluid pressure. A component of heat transfer can be seen as a heat sink in which transfers the heat generated within any medium. Thus evaporative cooling happens after the vapor is added to the surrounding air.How many atoms are found in 3.45g of CO2?
Answer: The number of carbon and oxygen atoms in the given amount of carbon dioxide is [tex]4.72\times 10^{22}[/tex] and [tex]9.44\times 10^{22}[/tex] respectively
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of carbon dioxide gas = 3.45 g
Molar mass of carbon dioxide gas = 44 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of carbon dioxide gas}=\frac{3.45g}{44g/mol}=0.0784mol[/tex]
1 mole of carbon dioxide gas contains 1 mole of carbon and 2 moles of oxygen atoms.
According to mole concept:
1 mole of a compound contains [tex]6.022\time 10^{23}[/tex] number of molecules
So, 0.0784 moles of carbon dioxide gas will contain [tex]1\times 0.0784\times 6.022\times 10^{23}=4.72\times 10^{22}[/tex] number of carbon atoms and [tex]2\times 0.0784\times 6.022\times 10^{23}=9.44\times 10^{22}[/tex] number of oxygen atoms
Hence, the number of carbon and oxygen atoms in the given amount of carbon dioxide is [tex]4.72\times 10^{22}[/tex] and [tex]9.44\times 10^{22}[/tex] respectively
Be sure to answer all parts. The thermal decomposition of phosphine (PH3) into phosphorus and molecular hydrogen is a first-order reaction: 4PH3(g) → P4(g) + 6H2(g) The half-life of the reaction is 35.0 s at 680°C. Calculate the first-order rate constant for the reaction: s−1 Calculate the time required for 77.0 percent of the phosphine to decompose: s
Answer:
k = 0.0198 s⁻¹
t = 74.25 seconds
Explanation:
Given that:
Half life = 35.0 s
[tex]t_{1/2}=\frac {ln\ 2}{k}[/tex]
Where, k is rate constant
So,
[tex]k=\frac {ln\ 2}{t_{1/2}}[/tex]
[tex]k=\frac {ln\ 2}{35.0}\ s^{-1}[/tex]
The rate constant, k = 0.0198 s⁻¹
Using integrated rate law for first order kinetics as:
[tex][A_t]=[A_0]e^{-kt}[/tex]
Where,
[tex][A_t][/tex] is the concentration at time t
[tex][A_0][/tex] is the initial concentration
Given:
77.0 % is decomposed which means that 0.77 of [tex][A_0][/tex] is decomposed. So,
[tex]\frac {[A_t]}{[A_0]}[/tex] = 1 - 0.77 = 0.23
t = ?
[tex]\frac {[A_t]}{[A_0]}=e^{-k\times t}[/tex]
[tex]0.23=e^{-0.0198\times t}[/tex]
t = 74.25 seconds
When in a reaction heat is a reactant then the reaction is called thermal decomposition reaction. The first-order rate constant is 0.0198 per second and the time required is 74.25 seconds.
What is a rate constant?The rate constant is the specific rate that is the proportionality constant in a reaction and depicts the relation between the chemical reaction rate and the concentration of the reactants.
Rate constant can be given as,
[tex]\rm t\dfrac{1}{2} = \dfrac{ln 2}{k}[/tex]
The half-life of the reaction is 35 seconds.
Substituting values in the above equation:
[tex]\begin{aligned}\rm k &= \rm \dfrac{ln\;2}{t\frac{1}{2}}\\\\&= \dfrac{\rm ln\2}{35}\\\\&= 0.0198 \;\rm s^{-1}\end{aligned}[/tex]
The time taken can be calculated by the rate law for first order:
[tex]\rm [A_{t}] = [A_{o}]e^{-kt}[/tex]
Here,
Concentration at time t = [tex]\rm [A_{t}][/tex]The initial concentration = [tex]\rm [A_{o}][/tex]Solving for time (t):
[tex]\begin{aligned}\rm \dfrac {[A_{t}]}{[A_{o}]} &= 1 - 0.77\\\\0.23 &= \rm e ^{-0.0198 \times t}\\\\\rm t &= 74.25\;\rm seconds\end{aligned}[/tex]
Therefore, the rate constant is 0.0198 per second and the time taken is 74.25 seconds.
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Which of the following regions of the periodic table tends to prefer a –1 charge and occupies Group 17?
a. Alkaline earth metals.
b. Halogens.
c. Noble gases.
Answer:
b. halogens
Explanation:
The elements of group 17 are called halogens. These are six elements Fluorine, Chlorine, Bromine, Iodine, Astatine. Halogens are very reactive these elements cannot be found free in nature. Their chemical properties are resemble greatly with each other. As we move down the group in periodic table size of halogens increases that's way fluorine is smaller in size as compared to other halogens elements. Their boiling points also increases down the group which changes their physical states.
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.
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
Why are the electronegativities of the 5d elements so high?
Explanation:
The electronegativity values of the 5d series are very high because the size of the 5d orbitals are much larger as compared to the size of the 3d and 4d orbitals. As a consequence of this, the shielding capacity of 5d elements are low or these elements are not effective at shielding the nuclear charge . Thus, this causes the increase in the effective nuclear charge which makes the electronegativity values to increase steeply from the Lutetium (1.27) to the Gold (2.54).
The weight of the body in the air is .... the weight of the submerged body
a) equal
b) more than
c) lower than
d) Not related
Answer:
the correct answer is option 'b': More than
Explanation:
The 2 situations are represented in the attached figures below
When an object is placed in air it is acted upon by force of gravity of earth which is measured as weight of the object.
While as when any object is submerged partially or completely in any fluid the fluid exerts a force in upward direction and this force is known as force of buoyancy and it's magnitude is given by Archimedes law as equal to the weight of the fluid that the body displaces, hence the effective force in the downward direction direction thus the apparent weight of the object in water decreases.
Define ""point source pollution"" and ""non-point pollution"".
Answer:
Point source pollution:
If pollution comes from a fix source then is called point source pollution.It have a specific location where pollution comes.
Ex: Air pollution ,water pollution.
Non point source pollution:
If pollution comes from number of sources then is called point source pollution.This pollution does not have a specific point of source.
Ex: Spills ,leaks ,Sewage over flow etc.
Answer:
Defined as under.
Explanation:
A point source pollution is a single identifiable source of pollutants like air, water, thermal and light. And has a negligible extent, distinguishes from other geometric or area sources of nonpoint, etc. While nonpoint source pollution is derived from various sources. Such as results of land runoff. atmospheric drainage and seepage they may affect a body of water such as from the agricultural areas draining to a river. Various pathogens and viruses found in the body of water, also the presence of poorly managed livestock, pets, and other these problems is also elated to the urban and rural asphalt and concrete, highway runoff and mining activities.What should be done in case of acid spills and mercury spills? Explain
Explanation:
Acid spills
These types of spills should be neutralized with base like sodium bicarbonate and then must be cleaned up by using paper towel or sponge. Strong base like sodium hydroxide must not be used to neutralize. Best base to use is sodium bicarbonate which has much less chance of the injury.
Mercury spills
Mercury is found commonly in the thermometers. If one have mercury spill, it must be cleaned up immediately with either commercial product like Hg Aborb ™ or by using elemental sulfur. Mercury sponges can also be purchased that form amalgam with liquid mercury and thus trapping it on surface of sponge.
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].
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)
What is the pH of a solution containing 0.049 M of formic acid and 0.055 M of sodium formate?
Answer:
pH of solution is 3.80
Explanation:
Formic acid an weak acid and formate is conjugate base of formic acidHence solution containing formic acid and formate acts as a buffer.According to Henderson-Hasselbalch equation for a buffer consist of an weak acid (formic acid) and it's conjugate base (formate)-[tex]pH=pK_{a}(formic acid)+log(\frac{C_{formate}}{C_{formic acid}})[/tex]
where, C stands for concentration
[tex]pK_{a}[/tex] of formic acid 3.75So, [tex]pH=3.75+log(\frac{0.055}{0.049})=3.80[/tex]
7. For the system PCls(g) → PC13(g) + Cl2(g) Kis 26 at 300°C. In a 5.0-L flask, a gaseous mixture consists of all three gases with partial pressure as follows: Ppcis = 0.012 atm, Pc2=0.45 atm, Ppci3 -0.90 atm. a) Is the mixture at equilibrium? Explain. b) If it is not at equilibrium, which way will the system shift to establish equilibrium?
Answer:
a) Reaction is not at equilibrium
b) Reaction will move towards backward direction
Explanation:
[tex]PCl_5(g) \rightarrow PCl_3(g) + Cl_2(g)[/tex]
Equilibrium constant = 26
[tex]Reaction\ quotient (Q) = \frac{[p_{PCl_3}]\times [p_{Cl_2}]}{[p_{PCl_5}]}[/tex]
[tex][p_{PCl_5}] = 0.012 atm[/tex]
[tex][p_{PCl_3}]= 0.90 atm[/tex]
[tex][p_{Cl_2}]= 0.45 atm[/tex]
[tex]Reaction\ quotient (Q) = \frac{[p_{PCl_3}]\times [p_{Cl_2}]}{[p_{PCl_5}]}[/tex]
[tex]Reaction quotient (Q) =\frac{0.90\times 0.45} {0.012} = 33.75[/tex]
As reaction quotient (Q) is more than equilibrium constant, so reaction is not at equilibrium and reaction will move towards backward direction.
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 )
For which of these is there an increase in entropy? KCI(aq)+AgNO3(aq)KNO3(aq)+AgCI(s) NaCl(s)NaCl(aq) 2NaOH(aq)+CO2(g)Na2CO3(aq)+H20(1) C2H5OH(g)C2H5OH(I)
Answer: NaCl (s) → NaCl (aq)
Explanation:
Entropy is often associated with the disorder or randomness of a system. Therefore, in each reaction, it is necessary to evaluate if the disorder increases or decreases to understand what happens to the entropy:
1) KCl (aq) + AgNO₃ (aq) → KNO₃ (aq) + AgCl (s) - In this reaction, we have only aqueous reactants in the beginning and in the product we have a precipitate. The solid state is more organised than the liquid, consequently, the entropy decreases.
2) NaCl (s) → NaCl (aq) - In this case, oposite to the first one, we go from a solid state to an aqueous state. The solvation of the ions Na⁺ and Cl⁻ is random while the solid state is very organised. Therefore, in this reaction the entropy increases.
3) 2NaOH (aq) + CO₂ (g) → Na₂CO₃ (aq) + H₂O (l) - In this reaction, the reactants have higher entropy because of the gas CO₂. Therefore, the entropy decreases.
4) C₂H₅OH (g) → C₂H₅OH (l) - In this reaction, the reactant is a gas and the product a liquid. Therefore, the entropy decreases.
Blood substitute. As noted in this chapter, blood contains a total concentration of phosphate of approximately 1 mM and typically has a pH of 7.4. You wish to make 100 liters of phosphate buffer with a pH of 7.4 from NaH 2 PO 4 (molecular weight, 119. 98 g mol 1) and Na 2 HPO 4 (molecular weight, 141. 96 g mol 1). How much of each (in grams) do you need? Berg, Jeremy M.. Biochemistry (p. 25). W. H. Freeman. Kindle Edition.
Answer:
Mass of NaH₂PO₄ = 4.707 gMass of Na₂HPO₄ = 8.627 gExplanation:
The equilibrium relevant for this problem is:
H₂PO₄⁻ ↔ HPO₄⁻² + H⁺
The Henderson–Hasselbalch (H-H) equation is needed to solve this problem:
pH= pka + [tex]log\frac{[A^{-} ]}{[HA]}[/tex]
In this case, [A⁻] = [HPO₄⁻²], [HA] = [H₂PO₄⁻], pH = 7.4; from literature we know that pka=7.21.
We use the H-H equation to describe [HPO₄⁻²] in terms of [H₂PO₄⁻]:
[tex]7.4=7.21+log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]} \\0.19=log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]} \\10^{0.19}= \frac{[HPO4^{-2} ]}{[H2PO4^{-} ]} \\1.549*[H2PO4^{-} ]=[HPO4^{-2} ][/tex]
The problem tells us that the concentration of phosphate is 1 mM, which means:
[HPO₄⁻²] + [H₂PO₄⁻] = 1 mM = 0.001 M
In this equation we can replace [HPO₄⁻²] with the term expressed in the H-H eq:
1.549 * [H₂PO₄⁻] + [H₂PO₄⁻] = 0.001 M
2.549 * [H₂PO₄⁻] = 0.001 M
[H₂PO₄⁻] = 3.923 * 10⁻⁴ M
With the value of [H₂PO₄⁻] we can calculate [HPO₄⁻²]:
[HPO₄⁻²] + 3.923 * 10⁻⁴ M = 0.001 M
[HPO₄⁻²] = 6.077 * 10⁻⁴ M
With the concentrations, the molecular weight, and the volume, we calculate the mass of each reagent:
Mass of NaH₂PO₄ = 3.923 * 10⁻⁴ M * 100 L * 119.98 g/mol = 4.707 gMass of Na₂HPO₄ = 6.077 * 10⁻⁴ M * 100 L * 141.96 g/mol = 8.627 gCalculate 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.
Calculate the freezing point of a solution made from 220g of octane (C Hua), molar mass = 114,0 gmol dissolved in 1480 g of benzene. Benzene freezes at 5.50"C and its Kvalue is 5.12C/m. -1.16°C 0.98°C 666"C 12 2°C 5.49°C 10 12 AM A A 2019 Backspace yuo Pill но кL
Answer: Freezing point of a solution will be [tex]-1.16^0C[/tex]
Explanation:
Depression in freezing point is given by:
[tex]\Delta T_f=i\times K_f\times m[/tex]
[tex]\Delta T_f=T_f^0-T_f=(5.50-T_f)^0C[/tex] = Depression in freezing point
i= vant hoff factor = 1 (for non electrolyte)
[tex]K_f[/tex] = freezing point constant = [tex]5.12^0C/m[/tex]
m= molality
[tex]\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]
Weight of solvent (benzene)= 1480 g =1.48 kg
Molar mass of solute (octane) = 114.0 g/mol
Mass of solute (octane) = 220 g
[tex](5.50-T_f)^0C=1\times 5.12\times \frac{220g}{114.0 g/mol\times 1.48kg}[/tex]
[tex](5.50-T_f)^0C=6.68[/tex]
[tex]T_f=-1.16^0C[/tex]
Thus the freezing point of a solution will be [tex]-1.16^0C[/tex]
You make 1 Liter of an aqueous solution containing 9.20 ml of 57.8 mM acetic acid and 56.2 mg of sodium acetate (MW = 82.0 g/mole). a. How many moles of acetic acid did you add? b. How many moles of sodium acetate did you add? c. What is the appropriate formula for calculating the pH of the above solution? d. What is the pH of the above solution? Is it acidic or basic? Slightly or strongly so? (Use the same scale as on HW 1)
Answer:
a) 5,3176x10⁻⁴ moles
b) 6,85x10⁻⁴ moles
c) The appropriate formula to calculate is Henderson-Hasselbalch.
d) pH = 4,86. Acidic solution but slighty
Explanation:
a) moles of acetic acid:
9,20x10⁻³L × 57,8x10⁻³M = 5,3176x10⁻⁴ moles
b) moles of sodium acetate:
56,2x10⁻³g ÷ 82,0 g/mole = 6,85x10⁻⁴ moles
c) The appropriate formula to calculate is Henderson-Hasselbalch:
pH= pka + log₁₀ [tex]\frac{[A^-]}{[HA]}[/tex]
d) pH= 4,75 + log₁₀ [tex]\frac{[6,85x10_{-4}]}{[5,3176x10_{-4}]}[/tex]
pH = 4,86
3 < pH < 7→ Acidic solution but slighty
I hope it helps!
A gram of gasoline produces 45.0kJ of energy when burned. Gasoline has a density of 0.77/gmL. How would you calculate the amount of energy produced by burning 5.2L of gasoline? Set the math up. But don't do any of it. Just leave your answer as a math expression. Also, be sure your answer includes all the correct unit symbols.
Answer:
Math expression: [tex]=\frac{0.77\ g/ml*5200\ ml}{1\ g} *45.0\ kJ[/tex]
Explanation:
Given:
Energy produced per gram of gasoline = 45.0 kJ
Density of gasoline = 0.77 g/ml
Volume of gasoline = 5.2 L=5200 ml
To determine:
The amount of energy produced by burning 5.2 L gasoline
Calculation set-up:
1. Calculate the mass (m) of gasoline given the density (d) and volume (v)
[tex]m = d*v\\\\m = 0.77 g/ml*5200 ml[/tex]
2. Calculate the amount of energy produced
[tex]=\frac{0.77\ g/ml*5200\ ml}{1\ g} *45.0\ kJ=180180 kJ[/tex]
Various members of a class of compounds, alkenes, react with hydrogen to produce a corresponding alkane. Termed hydrogenation, this type of reaction is used to produce products such as margarine. A typical hydrogenation reaction is C10H20() + H2(g) → C10H22(5) Decene Decane How much decane can be produced in a reaction of excess decene with 2.45 g hydrogen? Give your answer in scientific notation. O *10 g decane
Answer: The mass of decane produced is [tex]1.743\times 10^2g[/tex]
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
Mass of hydrogen gas = 2.45 g
Molar mass of hydrogen gas = 2 g/mol
Putting values in equation 1:, we get:
[tex]\text{Moles of }H_2=\frac{2.45g}{2g/mol}=1.225mol[/tex]
The chemical equation for the hydrogenation of decene follows:
[tex]C_{10}H_{20}(l)+H_2(g)\rightarrow C_{10}H_{22}(s)[/tex]
As, decene is present in excess. So, it is considered as an excess reagent.
Thus, hydrogen gas is a limiting reagent because it limits the formation of products.
By Stoichiometry of the reaction:
1 mole of hydrogen gas produces 1 mole of decane.
So, 1.225 moles of hydrogen gas will produce = [tex]\frac{1}{1}\times 1.225=1.225mol[/tex] of decane
Now, calculating the mass of decane by using equation 1, we get:
Moles of decane = 1.225 mol
Molar mass of decane = 142.30 g/mol
Putting values in equation 1, we get:
[tex]1.225mol=\frac{\text{Mass of decane}}{142.30g/mol}\\\\\text{Mass of carbon dioxide}=(1.225mol\times 142.30g/mol)=174.3g=1.743\times 10^2g[/tex]
Hence, the mass of decane produced is [tex]1.743\times 10^2g[/tex]
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
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