Answer:
0,040 M
Explanation:
The global reaction of the problem is:
Al(OH) (s) + OH⁻ ⇄ Al(OH)₂⁻(aq) K= 40
The equation of equilibrium is:
K = [tex]\frac{[Al(OH)_{2} ^-]}{[Al(OH)][OH^-]}[/tex]
The concentration of OH⁻ is:
pOH = 14 - pH = 3
pOH = -log [OH⁻]
[OH⁻] = 1x10⁻³
Thus:
40 = [tex]\frac{[Al(OH)_{2} ^-]}{[Al(OH)][1x10^{-3}]}[/tex]
0,04M = [tex]\frac{[Al(OH)_{2} ^-]}{[Al(OH)]}[/tex]
This means that 0,04 M are the number of moles that the solvent can dissolve in 1L, in other words, solubility.
I hope it helps!
A system that had work done on it but which receives or loses no heat from or to the surroundings has
w < 0, ΔE > 0
w = - ΔE
w > 0, ΔE < 0
w = ΔE
Answer:
w >0, ΔE < 0
Explanation:
As per the first law of thermodynamics,
ΔE = Q - W
Where,
ΔE = Change in internal energy
Q = Heat receive or heat loss
W = work done
Work done by the system is always -ve.
Work done on the system is always +ve.
It is given that work done on the system.
W = +ve or W > 0
As, there is no heat receive or heat loss
So, Q = 0
Now, as per the first law of thermodynamics.
ΔE = Q - W
Q = 0
ΔE = - W
or ΔE < 0
So, answer would be w > 0, ΔE < 0
What is the pH at each of the points in the titration of 25.00 mL of 0.2000 M
HCl by 0.2000 M NaOH:
i) Before adding NaOH
ii) After adding 24.00 mL NaOH
Answer:
i) pH = 0.6990
ii) pH = 2.389
Explanation:
i) Before adding aqueous NaOH, there are 25.00 mL of 0.2000 M HCl. HCl reacts with the water in the aqueous solution as follows:
HCl + H₂O ⇒ H₃O⁺ + Cl⁻
The HCl and H₃O⁺ are related to each other through a 1:1 molar ratio, so the concentration of H₃O⁺ is equal to the HCl concentration.
The pH is related to the hydronium ion concentration as follows:
pH = -log([H₃O⁺]) = -log(0.2000) = 0.699
ii) Addition of NaOH causes the following reaction:
H₃O⁺ + NaOH ⇒ 2H₂O + Na⁺
The H₃O⁺ and NaOH react in a 1:1 molar ratio. The amount of NaOH added is calculated:
n = CV = (0.2000 mol/L)(24.00 mL) = 4.800 mmol NaOH
Thus, 4.800 mmol of H₃O⁺ were neutralized.
The initial amount of H₃O⁺ present was:
n = CV = (0.2000 mol/L)(25.00 mL) = 5.000 mmol H₃O⁺
The amount of H₃O⁺ that remains after addition of NaOH is:
(5.000 mmol) - (4.800 mmol) = 0.2000 mmol
The concentration of H₃O⁺ is the amount of H₃O⁺ divided by the total volume. The total volume is (25.00 mL) + (24.00 mL ) = 49.00 mL
C = n/V = (0.2000 mmol) / (49.00 mL) = 0.004082 M
The pH is finally calculated:
pH = -log([H₃O⁺]) = -log(0.004082) = 2.389
The solubility of acetanilide is 0.53 g in 100 mL of ice‑cold water, and 5.50 g in 100 mL of boiling water. What is the maximum percent recovery that can be achieved for the recrystallization of acetanilide from water?
Answer:
MPR=90,36%
Explanation:
The recrystallization is a purification process where the solid to purify is dissolved in an appropriated dissolvent and then, changing the conditions the solubility changes and that solid (that was in solution before) precipitates and form crystals.
In this case, for boiling water 5.50 g of acetanilide could be dissolved and then cold the water, so the mass of crystals formed will be
[tex]m_{crystal}=\left(5.5-0.53\right)g=4.97g[/tex]
The maximum percent recovery is then
[tex]MPR=\frac{mass\ of\ solid\ recovered}{mass\ of\ solid\ disolved}\times 100\% = \frac{4.97g}{5.5g}\times 100\%=90,36\%[/tex]
A 36.5 lb child has a Streptococcus infection. Amoxicillin is prescribed at a dosage of 25 mg per kg of body weight b.id What is the meaning of the Latin abbreviation b.i.d.? O every other day O as needed O twice daily O once daily How many hours should pass between each administration? number of hours: 413 How many milligrams of amoxicillin should be given at each administration? mass of amoxicillin: 413 Amoxicillin should be stored between 0 °C and 20 °C. Should the amoxicillin be stored in the freezer or the refrigerator?
Answer:
a) b.i.d: twice daily.
b) 12 hour between each administration.
c) mg amoxicilin/administration = 413.9 mg/administration.
d) should be stored in the refrigerator
Explanation:
mass child = 36.5 Lb * ( 453.592 g/Lb ) = 16556.11 g = 16.556 Kgdosage: 25mg/kg body b.i.d.∴ b.i.d. : refer to twice a day; so 12 hours will pass between each administration of the medication.
⇒ mg amoxicilin/administration = 25 mg/ kg * 16.556 Kg = 413.9 mg amoxicilin.
Amoxicillin should be stored in the refrigerator, since in this section the temperature is kept within the storage range
Explain how a Buffer acts to resist pH change in a system upon addition of an acid or a base
Answer:
Explanation has been given below.
Explanation:
A buffer consists of either of a weak acid along with it's conjugate base or a weak base along with it's conjugate acid.Let's consider a buffer consists of a weak acid along with it's conjugate baseIf we add an acid to this buffer then conjugate base gets protonated and converted to corresponding weak acid. So effect of addition of acid gets neutralized by forming weak acid rather than increase in concentration of proton in solution.If we add a base to this buffer then weak acid gets converted to corresponding conjugate base. So effect of addition of base gets neutralized by forming conjugate base rather than in crease in concentration of hydroxyl ion in solution.Find the percent by mass of tin in the original sample, assuming that it contains no other reducing agents. A sample of impure tin of mass 0.528 g is dissolved in strong acid to give a solution of Sn2+. The solution is then titrated with a 0.0448 M solution of NO3−, which is reduced to NO(g). The equivalence point is reached upon the addition of 4.03×10−2 L of the NO3− solution.
Final answer:
To find the mass percent of chloride in the original dry sample, you can use the formula: Mass percent of chloride = (mass of chloride / mass of original sample) x 100%. Use the volume of AgNO3 solution used in the titration, the solution's molarity, and the molar mass of chloride to calculate the mass of chloride.
Explanation:
The percent by mass of chloride in the original dry sample can be calculated using the following formula:
Mass percent of chloride = (mass of chloride / mass of original sample) x 100%
In this case, the mass of chloride can be determined by multiplying the volume of AgNO3 solution used in the titration (28 mL) by the molarity of the solution (0.1 M) and the molar mass of chloride (35.453 g/mol).
Then, using the mass of chloride and the mass of the original sample (0.200 g), the percent by mass of chloride in the original dry sample can be calculated.
A technician tares a 100.0 mL volumetric flask at 150.00 g. After adding sodium chloride to the flask it then weighs 158.84 g. Assuming an error of 0.2 mL in the volumetric volume and 0.005 g in the weight, calculate the molar concentration of sodium chloride and its associated standard deviation.
To find the molar concentration of NaCl, subtract the tare weight from the total weight to get the mass of NaCl, calculate moles of NaCl, and divide by the solution volume. To estimate the standard deviation, propagate the errors from the mass and volume measurements according to the rules of error propagation. Specific numerical values for the standard deviation cannot be provided without exact formulas.
Explanation:The question pertains to calculating the molar concentration of sodium chloride (NaCl) in a solution, and its associated standard deviation, given certain experimental measurements and potential error margins. First, the mass of NaCl added to the solution is found by subtracting the tare weight of the volumetric flask from the total weight after NaCl was added, yielding 8.84 g of NaCl. The molecular weight of NaCl is 58.44 g/mol, which allows determination of the moles of NaCl present.
To find the molar concentration, divide the moles of NaCl by the volume of the solution in liters (assuming the 100.0 mL flask volume as ideal, the error in volume would be considered in calculating the standard deviation, not the concentration itself). Then, to address the error margins, propagate the errors from the mass and volume measurements to estimate the standard deviation of the calculated concentration.
Note: Without specific formulas for error propagation and the exact calculation method for standard deviation provided in the question, a detailed numerical solution including the standard deviation calculation cannot be accurately presented. However, this process typically involves the square root of the sum of squared fractional uncertainties of the measurements involved.
Calculate the mass of 1.0 L of helium (He), 1.0 L of chlorine gas (Cl2), and 1.0 L of air (79% N2, 21% O2 by volume) at 25°C and 1 atm total pressure. Explain why a balloon filled with helium rises and why leaks of chlorine gas can be dangerous.
To calculate the mass we use the following formulas:
PV=nRT (1)
and
n = m / M (2)
where:
P - pressure (atm)
V - volume (L)
n - moles
R - gas constant = 0.082 (L × atm) / (mol × K)
T - temperature (°K) (25°C + 273 = 298°K)
m - mass (g)
M - molecular mass (g/mole)
Now we rewrite equation (1):
n = PV / RT
And replace n with m / M from equation (2):
m / M = PV / RT
m = (P × V × M) / (R ×T)
1 L of He will have a mass of:
m = (1 × 1 × 4) / (0.082 × 298) = 0.1637 g
1 L of Cl₂ will have a mass of:
m = (1 × 1 × 71) / (0.082 × 298) = 2.9055 g
1.0 L of air will contain 0.79 L of N₂ and 0.21 L of O₂
0.79 L of N₂ will have a mass of:
m = (1 × 0.79 × 28) / (0.082 × 298) = 0.9052 g
0.21 L of O₂ will have a mass of:
m = (1 × 0.21 × 32) / (0.082 × 298) = 0.2750 g
mass of air = mass of N₂ + mass of O₂
mass of air = 0.9052 + 0.2750 = 1.1802 g
A balloon filed with helium will rise because as you see 1 L of helium is lighter than 1 L of air.
Chlorine gas is dangerous because chlorine is very toxic for human life and more of that is heavier than the air so will diffuse very hard from the area where the leak appeared.
Glycerol is a syrupy liquid often used in cosmetics and soaps. A 3.25-L sample of pure glycerol has a mass of 4.10 x 10 g.
What is the density of glycerol in g/cm"? Express your answer in grams per cubic centimeter.
Explanation:
It is known that density is the amount of mass present in liter of solution or substance.
Mathematically, Density = [tex]\frac{mass}{volume}[/tex]
It is given that volume is 3.25 L and mass is [tex]4.10 \times 10^{3} g[/tex]. Hence, calculate the density of glycerol as follows.
Density = [tex]\frac{mass}{volume}[/tex]
= [tex]\frac{4.10 \times 10^{3} g}{3.25 L}[/tex]
= [tex]1.26 \times 10^{3} g/L[/tex]
As, 1 L = 1000 [tex]cm^{3}[/tex].
So, [tex]1.26 \times 10^{3} g/L \times \frac{1000 cm^{3}}{1 L}[/tex]
= [tex]1260 \times 10^{6} g/cm^{3}[/tex]
Thus, we can conclude that the density of glycerol is [tex]1260 \times 10^{6} g/cm^{3}[/tex].
Question 3 (1 point) The extinction coefficient for copper sulfate in aqueous solution is 12 M-1.cm-1 at 800 nm. If the absorbance of the copper sulfate solution in 0.50 cm cuvette is 0.50 at 800 nm, the concentration of copper sulfate in this solution is: O3M O 83 mm 8.3 mm 12 M-cm 0.042 M
Answer:
The concentration of the copper sulfate solution is 83 mM.
Explanation:
The absorbance of a copper sulfate solution can be calculated using Beer-Lambert Law:
A = ε . c . l
where
ε is the extinction coefficient of copper sulfate (ε = 12 M⁻¹.cm⁻¹)
c is its molar concentration (what we are looking for)
l is the pathlength (0.50 cm)
We can use this expression to find the molarity of this solution:
[tex]c=\frac{A}{\epsilon.l } =\frac{0.5}{12M^{-1}cm^{-1}0.50cm } =0.083M=83mM[/tex]
The concentration of copper sulfate in the solution is calculated using Beer's law and the given values, resulting in a concentration of 0.042 M.
Explanation:The concentration of copper sulfate in solution can be calculated using Beer's law, which is the relation A = εbc where A is absorbance, ε is the extinction coefficient, b is the path length, and c is the concentration. Given the question, the extinction coefficient for copper sulfate is 12 M-1.cm-1, the absorbance is 0.50, and the cuvette path length is 0.50 cm. To find the concentration c, rearrange the equation to c = A / (εb) and substitute into that the values provided, resulting in a concentration of 0.042 M for copper sulfate in the solution.
What are the names of the following compounds: FeCl HNO NaSO SO
Answer:
FeCl: Ferric Chloride (also called iron chloride), comes from Fe (ferrum, or iron), and Cl (Chlorine)
HNO: Nitroxyl, from N (Nitrogen), and the acidic nature of a radical ending in -yl.
NaSO: Sodium sulfate, Na (Sodium), S (Sulfur), O (Oxygen).
SO: Sulfur monoxide (Mono-One), O (Oxygen) and S (Sulfur).
A material has an ASTM grain size number of 7. Determine the magnification, if the number of grains per square inch observed is: i. 64 grains/in ii. 500 grains/in? iii. 20 grains/in-
Answer:
A) M = 100X
B) M = 36X
C) M = 178.88X
Explanation:
Given data:
ASTM grain size number 7
a) total grain per inch^2 - 64 grain/inch^2
we know that number of grain per square inch is given as
[tex]Nm = 2^{n-1} (\frac{100}{M})^2[/tex]
where M is magnification, n is grain size
therefore we have
[tex]64 = 2^{7-1}(\frac{100}{M})^2[/tex]
solving for M we get
M = 100 X
B) total grain per inch^2 = 500 grain/inch^2
we know that number of grain per square inch is given as
[tex]Nm = 2^{n-1} (\frac{100}{M})^2[/tex]
where M is magnification, n is grain size
therefore we have[tex]500 = 2^{7-1}(\frac{100}{M})^2[/tex]
solving for M we get
M = 36 X
C) Total grain per inch^2 = 20 grain/inch^2
we know that number of grain per square inch is given as
[tex]Nm = 2^{n-1} (\frac{100}{M})^2[/tex]
where M is magnification, n is grain size
therefore we have[tex]20 = 2^{7-1}(\frac{100}{M})^2[/tex]
solving for M we get
M = 178.88 X
A tank with a height of 100 feet and a constant cross sectional area of 10 ft has a constant input flow of 15 ft /hour of pulp stock at 1% consistency and has a screen on the exit flow that only allows water to be removed and keeps all of the fiber in the tank. The tank is well mixed and completely full (i.e, an overflow type condition where flow in equals flow out). What is the consistency as a function of time if the tank contents starts off at 0% consistency. Plot it using excel or another spreadsheet tool, see below. The tank will become clogged when the consistency reaches 6%, when will this happen?
Answer:
(a) The consistency as a function of time is C=0.15*t.
(b) The tank will become clogged in 24 minutes.
Explanation:
The rate of accumulation of the pulp stock can be defined as
[tex]\frac{dC}{dt}=Q_{i}*C_{i}-Q_{o}*C_{o}[/tex]
In this case, Co is 0, because the exit flow is only water and 0% fiber.
[tex]frac{dC}{dt}=Q_{i}*C_{i}-Q_{o}*0=Q_{i}*C_{i}[/tex]
Rearranging adn integrating
[tex]dC = (Q_{i}*C_{i})dt\\\int dC = \int (Q_{i}*C_{i})dt\\C=(Q_{i}*C_{i})*t+constant[/tex]
At t=0, C=0,
[tex]C=(Q_{i}*C_{i})*t+constant\\0=(Q_{i}*C_{i})*0+constant\\0=constant\\\\C=(Q_{i}*C_{i})*t[/tex]
[tex]C=(15*0.01)*t=0.15*t[/tex]
(b) The time at when the concentration reaches 6% is 0.4 hours or 24 minutes.
[tex]C=0.15*t\\0.06=0.15*t\\t=0.06/0.15=0.4[/tex]
The consistency of the pulp stock in the tank can be calculated using the equation of continuity. The consistency as a function of time can be determined by calculating the flow rate, density, mass flow rate, and the ratio of fiber mass to pulp stock mass. The graph of consistency as a function of time can be plotted using time intervals and consistency values.
Explanation:The equation of continuity states that the mass flow rate into a volume has to equal the mass flow rate out of the volume. In this case, the tank is well-mixed and completely full, so the flow in equals the flow out. We can use this equation to calculate the consistency of the pulp stock in the tank as a function of time.
Step 1: Calculate the flow rate
The flow rate can be calculated by multiplying the input flow rate (15 ft/hr) by the cross-sectional area of the tank (10 ft2). This gives us a flow rate of 150 ft3/hr.
Step 2: Calculate the density
The density of the pulp stock is given as 1% consistency. To convert this to density, we need to know the density of water, which is 62.4 lb/ft3. The density of the pulp stock can then be calculated as (0.01)(62.4 lb/ft3).
Step 3: Calculate the mass flow rate
The mass flow rate can be calculated by multiplying the flow rate by the density. This gives us a mass flow rate of (150 ft3/hr)(0.01)(62.4 lb/ft3).
Step 4: Calculate the consistency
The consistency is the mass of fiber in the tank divided by the mass of the pulp stock in the tank. Since all of the fiber is kept in the tank, the consistency is equal to the mass of fiber divided by the mass flow rate. This can be calculated as 100 ft (the height of the tank) divided by the mass flow rate calculated in step 3.
Step 5: Plot the consistency as a function of time
To plot the consistency as a function of time, you can create a table with time intervals and calculate the consistency at each interval using the formula calculated in step 4. Then, plot the time intervals on the x-axis and the consistency values on the y-axis.
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You want to make 500 ml of a 1 N solution of sulfuric acid (H2SO4, MW: 98.1). How many grams of sulfuric acid do you need?
a.12.3 g
b.44.0 g
c.24.5 g
d.88.0 g
Please show all steps, so I can understand how to do this, thanks!
Answer:
24.525 g of sulfuric acid.
Explanation:
Hello,
Normality (units of eq/L) is defined as:
[tex]N=\frac{eq_{solute}}{V_{solution}}[/tex]
Since the sulfuric acid is the solute, and we already have the volume of the solution (500 mL) but we need it in liters (0.5 L, just divide into 1000), the equivalent grams of solute are given by:
[tex]eq_{solute}=N*V_{solution}=1\frac{eq}{L}*0.5L=0.5 eq[/tex]
Now, since the sulfuric acid is diprotic (2 hydrogen atoms in its formula) 1 mole of sulfuric acid has 2 equivalent grams of sulfuric acid, so the mole-mass relationship is developed to find its required mass as follows:
[tex]m_{H_2SO_4}=0.5eqH_2SO_4(\frac{1molH_2SO_4}{2 eqH_2SO_4}) (\frac{98.1 g H_2SO_4}{1 mol H_2SO_4} )\\m_{H_2SO_4}=24.525 g H_2SO_4[/tex]
Best regards.
Calculate: (a) the weight (in lbf) of a 30.0 lbm object. (b) the mass in kg of an object that weighs 44N. (c) the weight in dynes of a 15-ton object (not metric tons)
Answer:
a) 965,1 lbf
b) 4,5 kg
c) 1,33 * 10^6 dynes
Explanation:
Mass of an object refers to the amount of mattter it cotains, it can be expressed it gr, kg, lbm, ton, etc.
Weight of an object refers to a force, and is the measurement of the pull of gravitiy on an object. It may be definide as the mass times the acceleration of gravity.
w=mg
In Planet Earth, the nominal "average" value for gravity is 9,8 m/s² (in the International System) or 32,17 ft/s² (in the FPS system).
To solve this problem we'll use the following conversion factors:
1 lbf = 1 lbm*ft/s²
1 N = 1 kg*m/s²
1 dyne = 1 gr*cm/s² and 1 N =10^5 dynes
1 ton = 907,18 kg
1 k = 1000 gr
a) m = 30 lbm
[tex]w = 30 lbm * 32,17 ft/s^{2} = 965, 1 \frac{lmb*ft}{s^{2} } = 965,1 lbf[/tex]
b) w = 44 N
First, we clear m of the weight equation and then we replace our data.
[tex]m = \frac{w}{g} = \frac{44 N}{9,8 \frac{m}{{s}^{2}} } = 4,5 kg[/tex]
c) m = 15 ton
[tex]m=15 ton * \frac{907,18 kg}{1 ton} = 13607,7 kg \\ w = mg = 13607,7 kg * 9,8 m/s2 = 133355,5 N * \frac{10^{5} dynes }{1 N} = 1,33 * 10^{6}dynes [/tex]
The weight of a 30 lbm object is 966 lbf. The mass of an object weighing 44N is 4.49 kg. And, the weight of a 15-ton object is 1.34 x 10¹² dynes.
Explanation:To Calculate: (a) the weight (in lbf) of a 30.0 lbm object we need to use the fact that 1 lbm equals 32.2 lb force (lbf). Therefore, a 30.0 lbm object weight would be 30 lbm * 32.2 = 966 lbf.
In (b), the mass in kg of an object that weighs 44N can be calculated by dividing the weight by Earth's gravity (approximately 9.8m/s²). 44N / 9.8m/s² = 4.49 kg.
Lastly, in (c), to find the weight in dynes of a 15-ton object we first convert the weight to pounds since 1 ton equals 2000 lbs. Then we convert pounds to Newtons (1 lb = 4.44822 N) and finally Newtons to dynes (1 N = 1,000,000 dynes). So, 15 ton * 2000 = 30000lb * 4.44822 N/lb * 1,000,000 dynes/N = 1.34 x 10¹² dynes.
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If volumes are additive and 253 mL of 0.19 M potassium bromide is mixed with 441 mL of a potassium dichromate solution to give a new solution in which [K+] is 0.846 M, what is the concentration of the potassium dichromate used to make the new solution?
Answer:
The concentration of the Potassium Dichromate solution is 0.611 M
Explanation:
First of all, we need to understand that in the final solution we'll have potassium ions coming from KBr and also K2Cr2O7, so we state the dissociation equations of both compounds:
KBr (aq) → K+ (aq) + Br- (aq)
K2Cr2O7 (aq) → 2K+ (aq) + Cr2O7 2- (aq)
According to these balanced equations when 1 mole of KBr dissociates, it generates 1 mole of potassium ions. Following the same thought, when 1 mole of K2Cr2O7 dissociates, we obtain 2 moles of potassium ions instead.
Having said that, we calculate the moles of potassium ions coming from the KBr solution:
0.19 M KBr: this means that we have 0.19 moles of KBr in 1000 mL solution. So:
1000 mL solution ----- 0.19 moles of KBr
253 mL solution ----- x = 0.04807 moles of KBr
As we said before, 1 mole of KBr will contribute with 1 mole of K+, so at the moment we have 0.04807 moles of K+.
Now, we are told that the final concentration of K+ is 0.846 M. This means we have 0.846 moles of K+ in 1000 mL solution. Considering that volumes are additive, we calculate the amount of K+ moles we have in the final volume solution (441 mL + 253 mL = 694 mL):
1000 mL solution ----- 0.846 moles K+
694 mL solution ----- x = 0.587124 moles K+
This is the final quantity of potassium ion moles we have present once we mixed the KBr and K2Cr2O7 solutions. Because we already know the amount of K+ moles that were added with the KBr solution (0.04807 moles), we can calculate the contribution corresponding to K2Cr2O7:
0.587124 final K+ moles - 0.04807 K+ moles from KBr = 0.539054 K+ moles from K2Cr2O7
If we go back and take a look a the chemical reactions, we can see that 1 mole of K2Cr2O7 dissociates into 2 moles of K+ ions, so:
2 K+ moles ----- 1 K2Cr2O7 mole
0.539054 K+ moles ---- x = 0.269527 K2Cr2O7 moles
Now this quantity of potassium dichromate moles came from the respective solution, that is 441 mL, so we calculate the amount of them that would be present in 1000 mL to determine de molar concentration:
441 mL ----- 0.269527 K2Cr2O7 moles
1000 mL ----- x = 0.6112 K2Cr2O7 moles = 0.6112 M
The concentration of the original potassium dichromate solution used in the mixture is 1.222 M. This was calculated by analyzing the total potassium concentration in the final solution, subtracting the contribution of potassium bromide, and then considering the volume of the potassium dichromate solution.
Explanation:This is a problem related to the principle of conservation of matter focusing on concentration and volume in a chemical solution. As per the question, the potassium ion concentration ([K+]) in the final solution is 0.846 M. We know that the total amount of potassium (in moles) comes from both the potassium bromide and the potassium dichromate.
First, we calculate the moles of potassium from the potassium bromide: volume (L) x concentration (M) = 0.253 L x 0.19 M = 0.04807 moles. Now, consider that the total volume of the solution is 253 mL + 441 mL = 694 mL or 0.694 L. Since the given final concentration of the mixed solution is 0.846 M, the total moles of potassium in the solution would be: 0.694 L x 0.846 M = 0.587124 moles. We subtract the moles of potassium from the potassium bromide to find the moles contributed by potassium dichromate: 0.587124 moles - 0.04807 moles = 0.539054 moles.
This is the amount of potassium in the potassium dichromate solution. To find concentration, we divide this by the volume of the potassium dichromate solution: 0.539054 moles / 0.441 L = 1.222 M. So, the concentration of the original potassium dichromate solution is 1.222 M.
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One solution has a formula C (n) H (2n) O (n) If this material weighs 288 grams, dissolves in weight 90 grams, the solution will have a boiling point of 101.24 ° C. Find the formula
The molecules of this substance When determining the Kb value of water = 0.512 ° C / m and the atomic weight H = 1, C = 12 and O = 16.
Explanation:
The given data is as follows.
Boiling point of water ([tex]T^{o}_{b}) = 100^{o}C[/tex] = (100 + 273) K = 323 K,
Boiling point of solution ([tex]T_{b}) = 101.24^{o}C[/tex] = (101.24 + 273) K = 374.24 K
Hence, change in temperature will be calculated as follows.
[tex]\Delta T_{b} = (T_{b} - T^{o}_{b})[/tex]
= 374.24 K - 323 K
= 1.24 K
As molality is defined as the moles of solute present in kg of solvent.
Molality = [tex]\frac{\text{weight of solute \times 1000}}{\text{molar mass of solute \times mass f solvent(g)}}[/tex]
Let molar mass of the solute is x grams.
Therefore, Molality = [tex]\frac{\text{weight of solute \times 1000}}{\text{molar mass of solute \times mass f solvent(g)}}[/tex]
m = [tex]\frac{288 g \times 1000}{x g \times 90}[/tex]
= [tex]\frac{3200}{x}[/tex]
As, [tex]\Delta T_{b} = k_{b} \times molality[/tex]
[tex]1.24 = 0.512 ^{o}C/m \times \frac{3200}{x}[/tex]
x = [tex]\frac{0.512 ^{o}C/m \times 3200}{1.24}[/tex]
= 1321.29 g
This means that the molar mass of the given compound is 1321.29 g.
It is given that molecular formula is [tex]C_{n}H_{2n}O_{n}[/tex].
As, its empirical formula is [tex]CH_{2}O[/tex] and mass is 30 g/mol. Hence, calculate the value of n as follows.
n = [tex]\frac{\text{Molecular mass}}{\text{Empirical mass}}[/tex]
= [tex]\frac{1321.29 g}{30 g/mol}[/tex]
= 44 mol
Thus, we can conclude that the formula of given material is [tex]C_{44}H_{88}O_{44}[/tex].
Our subjective feeling of hot and cold depends on the rate at which heat is lost through our skin. In this problem, we will model the skin as a flat layer of fatty tissue (k 0.3 W/m °K) with a thickness of 4 mm, separating an environment with a constant temperature of 37°C (normal body temperature) and the outside air. On a calm winter day with an ambient temperature of 0'°C, the convective heat transfer coefficient between the skin and the ambient air is approximately equal to 20 W/m2 °K. Find the heat flux lost through the skin.
Answer:
q = 2.343 W/m^2
Explanation:
Given data:
Ambient temperature = 0°C
Normal Body temperature = 37 °C
Thermal conductivity of tissue is K 0.3W/m °K
Heat transfer coefficient 20 W/m2 °K
Heat flux can be determined by using following formula
[tex]q = \frac{\Delta T}{\frac{c}{K} + \frac{1}{h}}[/tex]
putting all value to get flux value
[tex]q = \frac{37 -0}{\frac{4\times 10^{-3}}{0.3} + \frac{1}{20}}[/tex]
q = 2.343 W/m^2
In a particular experiment at 300 ∘C, [NO2] drops from 0.0100 to 0.00800 M in 100 s. The rate of appearance of O2 for this period is ________ M/s. In a particular experiment at 300 , drops from 0.0100 to 0.00800 in 100 . The rate of appearance of for this period is ________ . 4.0×10−3 2.0×10−3 2.0×10−5 4.0×10−5 1.0×10−5
Answer:
[tex]1\times{10}^{-5}\frac{M}{s}[/tex]
Explanation:
The stoichiometry for this reaction is
[tex]2NO_2\rightarrow2NO+O_2[/tex]
The rate for this reaction can be written as
[tex]-r_{NO_2}=-\frac{d\left[NO_2\right]}{dt}=\frac{(0.01-0.008)M}{100s}=2\times{10}^{-5}\frac{M}{s}[/tex]
This rate of disappearence of [tex]NO_2[/tex] can be realated to the rate of appearence of [tex]O_2[/tex] as follows (the coefficients of each compound are defined by the stoichiometry of the reaction)
[tex]-r_{O_2}=-r_{NO_2}\times\frac{coefficient\ O_2\ }{coefficient\ NO_2}=2\times{10}^{-5}\frac{M}{s}\times\frac{1\ mole\ O_2\ }{2\ mole\ NO_2}=1\times{10}^{-5}\frac{M}{s}[/tex]
Aluminium does not corrode in a normal atmosphere because: a) Aluminum is unreactive with oxygen b) The reaction with oxygen has too high and activation energy. c) The formation of an oxide layer inhibits further diffusion and corrosion d) None of the above
Answer:
The correct option is: c) The formation of an oxide layer inhibits further diffusion and corrosion
Explanation:
Aluminium is a chemical element which belongs to the group 13 of the periodic table and has atomic number 13. It is a soft metal and a member of the p-block.
When aluminium is exposed to the normal atmosphere, the top layer of the metal gets oxidized to form a thin protective layer of aluminium oxide. The thin and hard protective aluminium oxide layer then protects the metal from getting further corroded. This process is called passivation.
Predict the direction and the relative value of the dipole moment of the following bond:
(A) C←Cl, electronegativity difference > 0.5
(B) C←Cl, electronegativity difference < 0.5
(C) C→Cl, electronegativity difference > 0.5
(D) C→Cl, electronegativity difference < 0.5
Final answer:
The dipole moment points toward the more electronegative atom, Cl, in a C-Cl bond. The greater the electronegativity difference, the larger the dipole moment. Thus, (A) C←Cl with an electronegativity difference > 0.5 correctly predicts the dipole direction and suggests a significant dipole moment.
Explanation:
The direction and relative value of the dipole moment in a bond between two atoms depends on the difference in their electronegativities. Electronegativity is a measure of an atom's ability to attract electrons in a chemical bond. When two atoms form a bond, the more electronegative atom will attract the bonding electrons more strongly and will acquire a partial negative charge, while the less electronegative atom will have a partial positive charge.
Regarding the options given:
(A) C←Cl, with an electronegativity difference > 0.5 would have a dipole moment pointing towards the Cl, because Cl is more electronegative than C.
(B) C←Cl, with an electronegativity difference < 0.5 might suggest a relatively nonpolar bond, which is uncommon for C-Cl and might not be a realistic scenario.
(C) C→Cl, with an electronegativity difference > 0.5 would incorrectly suggest that C is more electronegative than Cl, which is not the case; thus this depiction of the dipole direction is incorrect.
(D) C→Cl, with an electronegativity difference < 0.5 would also be incorrect as it suggests the incorrect direction of the dipole moment.
The relative value of the dipole moment will be greater when the electronegativity difference is greater, leading to a stronger separation of charges, and hence a larger dipole moment.
A lead atom has a mass of 3.14 x 10 to the negative 22nd g.How
many lead atoms are in a cube of lead that has a volume of 2.00cm
cubed, if the density of lead is 11.3 g/cm cubed?
Explanation:
The given data is as follows.
Mass of a lead atom = [tex]3.14 \times 10^{-22}[/tex]
Volume = 2.00 [tex]cm^{3}[/tex]
Density = 11.3 [tex]g/cm^{3}[/tex]
As it is mentioned that 1 cubic centimeter contains 11.3 grams of lead.
So, in 2 cubic centimeter there will be [tex]2 \times 11.3 g = 22.6 g[/tex] of lead atoms.
One lead atom has a mass of [tex]3.14 \times 10^{-22}[/tex]. Therefore, number of atoms present in 22.6 g of lead will be as follows.
[tex]\frac{22.6 g}{3.14 \times 10^{-22}}[/tex]
= [tex]7.197 \times 10^{22} atoms[/tex]
Thus, we can conclude that there are [tex]7.197 \times 10^{22} atoms[/tex] of lead are present.
Define stereochemistry
Answer:
Stereochemistry is a branch of chemistry that studies the spatial arrangement of atoms or groups in a molecule.
The molecules with the same molecular formula, bond connectivity, and reactivity but a different arrangement of atoms in the space are known as stereoisomers. These molecules interact differently in a chiral environment or optical light.
The speed of light in a vacuum is 2.998 x 108 m/s. What is its speed in kilometers per hour (km/h)? העתK speed = What is its speed in miles per minute (mi/min)? speed = mi/min
Final answer:
The speed of light in a vacuum converted to kilometers per hour (km/h) is 1.079 x 10^12 km/hr, and in miles per minute (mi/min) is 11184.71 mi/min.
Explanation:
The speed of light in a vacuum is 2.998 x 108 m/s. To convert this speed into kilometers per hour (km/h), you multiply by the number of meters in a kilometer (1000) and the number of seconds in an hour (3600). This calculation gives us:
2.998 x 108 m/s x 1000 m/km x 3600 s/hr = 1.079 x 1012 km/hr.
Similarly, to find the speed in miles per minute (mi/min), you must use the conversion factor that 1 meter is approximately equal to 0.000621371 miles, and there are 60 seconds in a minute:
2.998 x 108 m/s x 0.000621371 mi/m x 60 s/min = 11184.71 mi/min.
In a five-fold serial dilution of a 20 pg/ml solution, all tube dilutions are 1/5. What is the substance concentration in the third tube of this series?
Answer:
0.8 pg/ml
Explanation:
To make the dilutions, you will take 1 ml of the original solution (tube 1) and add 4 ml of solvent. You will now have 20 pg per 5 ml of solution, so your new concentration will be 4 pg/ml (tube 2). Then you will repeat the process, so you will have 4 pg per 5 ml of solution, resulting in a concentration of 0.8 pg/ml (tube 3). The same process will be repeated for tubes 4 and 5.
A piece of an unknown metal has a volume of 16.6 mL and a mass of 190.1 grams. The density of the metal is g/mL A piece of the same metal with a mass of 94.6 grams would have a volume of ml. Submit Answer
Answer: The density of the metal is 11.45 g/mL and the volume occupied by 94.6 grams is 8.26 mL
Explanation:
To calculate the density of unknown metal, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex] ......(1)
Volume of unknown metal = 16.6 mL
Mass of unknown metal = 190.1 g
Putting values in equation 1, we get:
[tex]\text{Density of unknown metal}=\frac{190.1g}{16.6mL}\\\\\text{Density of unknown metal}=11.45g/mL[/tex]
The density of the metal remains the same.
Now, calculating the volume of unknown metal, using equation 1, we get:
Density of unknown metal = 11.45 /mL
Mass of unknown metal = 94.6 g
Putting values in above equation, we get:
[tex]11.45g/mL=\frac{94.6g}{\text{Volume of unknown metal}}\\\\\text{Volume of unknown metal}=8.26mL[/tex]
Hence, the density of the metal is 11.45 g/mL and the volume occupied by 94.6 grams is 8.26 mL
write and the integrated rate laws hor zeroth-first- second-order rate laws.
Explanation:
The integrated rate law for the zeroth order reaction is:
[tex][A]=-kt+[A]_0[/tex]
The integrated rate law for the first order reaction is:
[tex][A]=[A]_0e^{-kt}[/tex]
The integrated rate law for the second order reaction is:
[tex]\frac{1}{[A]}=kt+\frac{1}{[A]_0}[/tex]
Where,
[tex][A][/tex] is the active concentration of A at time t
[tex][A]_0[/tex] is the active initial concentration of A
t is the time
k is the rate constant
Answer:
- 0th: [tex]C_A=C_{A0}-kt[/tex]
- 1st: [tex]C_A=C_{A0}exp(-kt)[/tex]
- 2nd: [tex]\frac{1}{C_A}=kt+\frac{1}{C_{A0}}[/tex]
Explanation:
Hello,
For the ideal reaction A→B:
- Zeroth order rate law: in this case, we assume that the concentration of the reactants is not included in the rate law, therefore the integrated rate law is:
[tex]\frac{dC_A}{dt}=-k\\ \int\limits^{C_A}_{C_{A0}} {} \ dC_A= \int\limits^{t}_{0} {-k} \ dt\\C_A-C_{A0}=-kt\\C_A=C_{A0}-kt[/tex]
- First order rate law: in this case, we assume that the concentration of the reactant is included lineally in the rate law, therefore the integrated rate law is:
[tex]\frac{dC_A}{dt}=-kC_A\\ \int\limits^{C_A}_{C_{A0}} {\frac{1}{C_A} } \ dC_A= \int\limits^{t}_{0} {-k} \ dt\\ln(\frac{C_{A}}{C_{A0}} )=-kt\\C_A=C_{A0}exp(-kt)[/tex]
- Second order rate law: in this case, we assume that the concentration of the reactant is squared in the rate law, therefore the integrated rate law is
[tex]\frac{dC_A}{dt}=-kC_A^{2} \\ \int\limits^{C_A}_{C_{A0}} {\frac{1}{C_A^{2} } } \ dC_A= \int\limits^{t}_{0} {-k} \ dt\\-\frac{1}{C_A}+\frac{1}{C_{A0}}=-kt\\\frac{1}{C_A}=kt+\frac{1}{C_{A0}}[/tex]
Best regards.
Name two "Storage Polysaccharides" and two "Structural Polysaccharides"
Answer:
Examples of storage polysaccharides - starch and glycogen and structural polysaccharides - cellulose and chitin
Explanation:
Polysaccharides are the complex carbohydrate polymers, composed of monosaccharide units that are joined together by glycosidic bond.
In other words, polysaccharides are the carbohydrate molecules that give monosaccharides or oligosaccharides on hydrolysis.
The examples of storage polysaccharides are starch and glycogen. The examples of structural polysaccharides are cellulose and chitin.
Konvert the following temperatures from °F to PC: Ta 86°F, (b) -22°F, (C) 50°F, (d) -40°F, (e) 32°F, (f) -459.67°F. Convert each temperature to K.
Answer:
The answers are:
a) 30°C; 303.15K
b) -30°C; 243.15K
c) 50°C; 323.15K
d) -40°C; 233.15K
e) 0°C; 273.15K
f) -273.15 °C ; 0K
Explanation:
To convert the temperature from ° F to ° C we use the following expression:
[tex]C=(F-32)\frac{5}{9}[/tex]
where C es temperature en °C and F is temperature in °F
To obtain the temperature in K we need to add 273.15 to each Celcius temperature
[tex]K=C+273.15[/tex]
Two solutions of sodium acetate are prepared, one having a concentration of 0.1 M and the other having a concentration of 0.01 M. Calculate the pH values when the following concentrations of HCl have been added to each of these solutions: 0.0025 M, 0.005 M, 0.01 M, and 0.05 M.
Answer:
For 0.1 M sodium acetate solution
if concentration of acid is 0.0025 then pH will 6.075
if concentration of acid is 0.005 then pH will 5.775
if concentration of acid is 0.01 then pH will 5.475
if concentration of acid is 0.05 then pH will 4.775
For 0.01 M sodium acetate solution
if concentration of acid is 0.0025 then pH will 5.075
if concentration of acid is 0.005 then pH will 4.775
if concentration of acid is 0.01 then pH will 4.475
if concentration of acid is 0.05 then pH will 3.775
Explanation:
to calculate the pH of a buffer solution we use the following formula
pH = pKa + log [B]/[A] ------------- eq (1)
[B] = concentration of base
[A] = concentration of acid
Given data
[B] = 0.1 M , 0.01M
[A] = 0.0025 M , 0.005 M, 0.01 M, 0.05 M
pKa value for sodium acetate is 4.75
1. First we will calculate the pH values for 0.1 M acetate solution.
If the concentration of acid is 0.0025, then:
[B] = 0.1 M
[A] = 0.0025 M
put these values in eq 1. which is:
pH = pKa + log [B]/[A]
pH = 4.75 + log [0.1]/[0.0025]
pH = 4.75 + log [40]
pH = 4.475 + 1.6
pH = 6.075
If the concentration of acid is 0.005 M, then:
[B] = 0.1 M
[A] = 0.005 M
put these values in eq 1. which is:
pH = pKa + log [B]/[A]
pH = 4.75 + log [0.1]/[0.005]
pH = 4.75 + log [20]
pH = 4.475 + 1.3
pH = 5.775
If the concentration of acid is 0.01, then:
[B] = 0.1 M
[A] = 0.01 M
put these values in eq 1. which is:
pH = pKa + log [B]/[A]
pH = 4.75 + log [0.1]/[0.01]
pH = 4.75 + log [10]
pH = 4.475 + 1
pH = 5.475
If the concentration of acid is 0.05, then:
[B] = 0.1 M
[A] = 0.05 M
put these values in eq 1. which is:
pH = pKa + log [B]/[A]
pH = 4.75 + log [0.1]/[0.05]
pH = 4.75 + log [2]
pH = 4.475 + 0.3
pH = 4.775
2. Now we will calculate the pH values for 0.01 M acetate solution.
If the concentration of acid is 0.0025, then:
[B] = 0.01 M
[A] = 0.0025 M
put these values in eq 1. which is:
pH = pKa + log [B]/[A]
pH = 4.75 + log [0.01]/[0.0025]
pH = 4.75 + log [4]
pH = 4.475 + 0.6
pH = 5.075
If the concentration of acid is 0.005 M, then:
[B] = 0.01 M
[A] = 0.005 M
put these values in eq 1. which is:
pH = pKa + log [B]/[A]
pH = 4.75 + log [0.01]/[0.005]
pH = 4.75 + log [2]
pH = 4.475 + 0.3
pH = 4.775
If the concentration of acid is 0.01 M, then:
[B] = 0.01 M
[A] = 0.01 M
put these values in eq 1. which is:
pH = pKa + log [B]/[A]
pH = 4.75 + log [0.01]/[0.01]
pH = 4.75 + log [1]
pH = 4.475 + 0
pH = 4.475
If the concentration of acid is 0.05 M, then:
[B] = 0.01 M
[A] = 0.05 M
put these values in eq 1. which is:
pH = pKa + log [B]/[A]
pH = 4.75 + log [0.01]/[0.05]
pH = 4.75 + log [0.2]
pH = 4.475 + (-0.7)
pH = 4.475 - 0.7
pH = 3.775
Answer:
a) pH = 4.71
b) pH = 4.704
c) pH = 4.57
d) No buffer here, the pH will be between 2-3
Explanation:
Applying Henderson Hasselbach equation:
pH = pKa + log([A]/[HA])
a) For 0.0025 M:
[A] = 0.1/2 = 0.05 M
[HA] = 0.05 M
After add 0.0025 M of acid:
[A] = 0.05 - 0.0025 = 0.0475 M
[HA] = 0.05 + 0.0025 = 0.0525 M
[tex]pH=4.75+log(\frac{0.0475}{0.0525} )=4.71[/tex]
b) For 0.005 M:
[A] = 0.1/2 = 0.05 M
[HA] = 0.05 M
After add 0.005 M of acid:
[A] = 0.05 - 0.005 = 0.0495 M
[HA] = 0.05 + 0.005 = 0.055 M
[tex]pH=4.75+log(\frac{0.0495}{0.055} )=4.704[/tex]
c) For 0.01 M:
[A] = 0.1/2 = 0.05 M
[HA] = 0.05 M
After add 0.01 M of acid:
[A] = 0.05 - 0.01 = 0.04 M
[HA] = 0.05 + 0.01 = 0.06 M
[tex]pH=4.75+log(\frac{0.04}{0.06} )=4.57[/tex]
d) For 0.05 M:
[A] = 0.1/2 = 0.05 M
[HA] = 0.05 M
After add 0.05 M of acid:
[A] = 0.05 - 0.05 = 0
[HA] = 0.05 + 0.05 = 0.1 M
No buffer here, the pH will be between 2-3