Answer:
At the start of the process, the volume not occupied by the water is 2 m3
Explanation:
At the start of the process you have a half full tank. It means that also a half is empty (not occupied by water).
Since the volume is 4 m3, 2 m3 are full (occupied by water) and 2 m3 (not occupied by water).
The volume in time will be
[tex]V(t)=V_0+(f_i-f_o)*t\\\\V(t) = 2 +(6.33/1000-3.25/1000)*t=2+0.00308*t \, \, [m3][/tex]
Which would usually influence a reaction rate more? OH os O H&S O Neither H nors
Answer:
As the average kinetic energy increases, the particles move faster and collide more frequently per unit time and possess greater energy when they collide. Both of these factors increase the reaction rate. Hence the reaction rate of virtually all reactions increases with increasing temperature
HOPE THIS HELPED!!!!!!!!!!!!!! XDDDDD
The total volume required to reach the endpoint of a titration required more than the 50 mL50 mL total volume of the buret. An initial volume of 49.37±0.06 mL49.37±0.06 mL was delivered, the buret was refilled, and an additional 1.34±0.05 mL1.34±0.05 mL was delivered before the endpoint was reached. The titration of a blank solution without the analyte required 0.51±0.04 mL0.51±0.04 mL . Calculate the endpoint volume corrected for the blank and its absolute uncertainty.
The corrected endpoint volume is approximately 50.20 mL, and its absolute uncertainty is 0.11 mL.
We have,
To calculate the corrected endpoint volume and its absolute uncertainty, we need to subtract the volume of the blank from the total volume used to reach the endpoint.
Given data:
Initial volume delivered = 49.37 ± 0.06 mL
Additional volume delivered = 1.34 ± 0.05 mL
Volume of blank = 0.51 ± 0.04 mL
Total volume without blank = Initial volume + Additional volume
Total volume without blank = (49.37 ± 0.06) mL + (1.34 ± 0.05) mL
Corrected endpoint volume = Total volume without blank - Volume of blank
Calculate the corrected endpoint volume:
Total volume without blank = 49.37 mL + 1.34 mL = 50.71 mL
Corrected endpoint volume = 50.71 mL - 0.51 mL = 50.20 mL
Next, let's calculate the absolute uncertainty in the corrected endpoint volume. Since the absolute uncertainties are given in the problem, we'll simply add them:
Absolute uncertainty in corrected endpoint volume
= Absolute uncertainty in total volume without blank + Absolute uncertainty in the volume of blank
= 0.06 mL + 0.05 mL = 0.11 mL
Therefore,
The corrected endpoint volume is approximately 50.20 mL, and its absolute uncertainty is 0.11 mL.
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The corrected endpoint volume of the titration is 50.20 ± 0.09 mL, which is found by subtracting the blank's volume from the total volume used and calculating the combined uncertainty.
Explanation:To calculate the endpoint volume corrected for the blank and its absolute uncertainty, we need to subtract the volume used for the blank from the total volume used in the titration and then calculate the combined uncertainty of these measurements. Here is how you can do it:
The total volume delivered in the titration is the sum of the initial volume and the additional volume, which equals 49.37 mL + 1.34 mL = 50.71 mL. To find the volume corrected for the blank, we subtract the volume for the blank, which gives us 50.71 mL - 0.51 mL = 50.20 mL.
The absolute uncertainty of the titration volume is found by combining the uncertainties of the individual measurements using the square root of the sum of the squares of the uncertainties (assuming they are independent):
[tex]\(\sqrt{0.06^2 + 0.05^2 + 0.04^2}[/tex] =[tex]\sqrt{0.0036 + 0.0025 + 0.0016}[/tex] = [tex]\sqrt{0.0077} \approx 0.088\ mL\)[/tex]
Therefore, the corrected endpoint volume of the titration is 50.20 ± 0.09 mL.
The solubility constant of AgBr is 5.0 x 10-13. The formation constant of the silver(I) ammonia complex, Ag(NH3)2, is 1.6 x 107 Calculate the solubility, in moles/L, of AgBr in 1.00 Mammonia solution.
Answer:
S AgBr = 2.82 E-3 mol/L
Explanation:
AgBr ↔ Ag+ + Br- .....(1)∴ Ksp = [ Ag+ ] * [ Br- ] = 5.0 E-13
Ag+ + 2NH3 ↔ [ Ag(NH3)2 ]+ ..........(2)∴ Kf = 1.6 E7 = α[Ag(NH3)2]+ / ( αAg+ )*( αNH3 )²
∴ C NH3(sln) = 1.00 M
from (1) + (2):
AgBr(s) + 2NH3(aq) ↔ Ag(NH3)2+(aq) + Br-(aq)1 M 0 0
1 - 2x x x
∴ K = Ksp*Kf = ( 5.0 E-13 )*( 1.6 E7 ) = 8.0 E-6
⇒ K = ( [ Br- ] * [ Ag(NH3)2+] ) / [ NH3 ]² = 8.0 E-6
⇒ K = (( x )*( x )) / ( 1 - 2x ) = 8.0 E-6
⇒ x² = 8.0 E-6*( 1- 2x )
⇒ x² + 1.6 E-5x - 8.0 E-6 = 0
⇒ x = 2.82 E-3 M
⇒ [ Br- ] = [ AgBr ] = 2.82 E-3 M
To calculate the solubility of AgBr in a 1.00 M ammonia solution, we can use the equation for the formation constant of the silver(I) ammonia complex.
Explanation:The solubility constant of AgBr is 5.0 x 10-13. The formation constant of the silver(I) ammonia complex, Ag(NH₃)₂, is 1.6 x 107. To calculate the solubility of AgBr in a 1.00 M ammonia solution, we can use the equation:
AgBr(s) → Ag+(aq) + Br-(aq)
Let x be the concentration of AgBr(s). Then, the equilibrium concentrations of Ag+ and Br- would be x and x, respectively.
Using the equation for the formation constant:
Kf = [Ag(NH₃)₂]+ / [Ag+][NH₃]₂
we can substitute the given values and solve for [Ag+] and [NH₃].
Classify these properties of the metal lithium as physical or chemical. Physical Chemical light enough to float on water silvery gray in color changes from silvery gray to black when placed in moist air can be cut with a sharp knife in the liquid state, it boils at 1317 °C reacts violently with chlorine to form a white solid in the liquid state, it reacts spontaneously with its glass container, producing a hole in the container burns in oxygen with a bright red flame
The psychical properties do no change the composition of the lithium metal.
The chemical properties will change the composition of the lithium producing other substances.
light enough to float on water - physical property
silvery gray in color - physical property
changes from silvery gray to black when placed in moist air - chemical property
(in air the lithium will react with the oxygen forming the oxide which have a black color)
can be cut with a sharp knife - physical property
in the liquid state, it boils at 1317 °C - physical property
reacts violently with chlorine to form a white solid - chemical property
in the liquid state, it reacts spontaneously with its glass container, producing a hole in the container - chemical property
burns in oxygen with a bright red flame - chemical property
Explanation:
A property that does not bring any change in chemical composition of a substance are known as physical properties.
For example, shape, size, mass, volume, density, hardness etc of a substance are all physical properties.
On the other hand, a property that changes chemical composition of a substance is known as chemical property.
For example, precipitation, reactivity, toxicity etc are chemical property.
Hence, the given properties of lithium are classified as follows.
Light enough to float on water - Physicalsilvery gray in color - Physicalchanges from silvery gray to black when placed in moist air - Chemicalcan be cut with a sharp knife - Physicalin the liquid state, it boils at [tex]1317^{o}C[/tex] - Physicalreacts violently with chlorine to form a white solid in the liquid state - Chemicalit reacts spontaneously with its glass container, producing a hole in the container - Chemicalburns in oxygen with a bright red flame - ChemicalDuring the phosphatase experiment you will use a 1% w/v of phenolphthalein di-phosphate (PPP). How much PPP do you need to make 500ml of the solution if you only have 25g and you know that the MW of PPP is 478 g/mole?
Answer:
To make 500 mL of a solution of PPP 1% w/v you need 5g
Explanation:
A percent w/v solution is calculated with the following formula using the gram as the base measure of weight (w):
% w/v = g of solute/ mL of solution × 100
A 1% w/v means that you have 1 g of PPP per 100 mL of solution
Thus, if you need to make 500 mL of solution you will need 5 g of PPP.
I hope it helps!
To prepare a 1% w/v solution of phenolphthalein di-phosphate (PPP) for 500ml, you need to weigh out 5 grams of PPP and dissolve it in enough water to make the total solution volume up to 500 ml.
Explanation:To prepare a 1% w/v solution of phenolphthalein di-phosphate (PPP) for 500ml, you first need to understand that 1% w/v means you have 1 gram of solute for every 100 milliliters of solution. Then, for 500 milliliters, you would need 5 grams of PPP. The molar mass of PPP (478 g/mol) is only used to calculate moles from grams or vice versa and is not needed to prepare a percent weight/volume solution. Since you have enough PPP (25g), you can weigh out 5 grams of PPP and dissolve it in water to make a total volume of 500ml to achieve a 1% w/v concentration.
Steps to prepare the solution:
Calculate the total amount of PPP needed: 500 ml * 1%/100ml = 5 grams.Weigh out 5 grams of PPP.Dissolve the PPP in a volume of water less than 500 ml.After PPP has dissolved, adjust the solution volume to 500 ml with water.Consider these compounds:
A. AgOH
B. FeCO3
C. Pb3(PO4)2
D. Zn3(PO4)2
Complete the following statements by entering the letter(s) corresponding to the correct compound(s). (If more than one compound fits the description, include all the relevant compounds by writing your answer as a string of characters without punctuation, e.g, ABC.)
Without doing any calculations it is possible to determine that silver chloride is more soluble than ___, and silver chloride is less soluble than___ .
It is not possible to determine whether silver chloride is more or less soluble than ___ by simply comparing Kspvalues.
Answer:
Without doing any calculations it is possible to determine that silver chloride is more soluble than ACD, and silver chloride is less soluble than ___.
It is not possible to determine whether silver chloride is more or less soluble than B by simply comparing Ksp values.
Explanation:
The Lead (Pb) is a heavy metal and Zinc (Zn) is a transition metal, both combine with the anion phosphate produces heavy complexes very insoluble.
The silver cation (Ag+) is an acid cation which in basic conditions rapidly precipitates as AgOH which is highly insoluble and unstable.
The FeCO3 is a complex with a solubility relatively close between AgCl. The Kps of these two are [tex]3.2x10^{-11}[/tex] and [tex]1.8x10^{-10}[/tex] respectively at room temperature.
A poundal is the force required to accelerate a mass of 1 lb,m (pound mass) at a rate of 1 ft/s^2. 1 poundal 1 lb,m*ft/(s^2) Calculate the weight in poundals of an 148 lb,m object on Earth
Explanation:
According to Newtons second law,
F = m × a
where, F = force acting on object
m = mass of object
a = acceleration
On Earth, a = g
where, g = gravitational acceleration constant = 9.8 [tex]m/sec^{2}[/tex]
Now, convert g into [tex]ft/sec^{2}[/tex] as follows.
1 m = 3.280 ft
g = [tex]9.8 m/s^{2} \times 3.280 ft/m[/tex]
= 32.144 [tex]ft/s^{2}[/tex]
It is given that m = 148 lbm. Hence, calculate force as follows.
F = [tex]148 lbm \times 32.144 ft/s^{2}[/tex]
= 4757.312 [tex]lbm.ft/s^{2}[/tex]
= 4757.312 poundal
Thus, we can conclude that the weight in poundals of an 148 lbm object on Earth is 4757.312 poundal.
Solve for x: ln x = 0.863. Report your answer to 3 significant figures.
By converting the given natural logarithm equation to its equivalent exponential form, we find that e^0.863 equals x. Therefore, calculating this gives us x as approximately 2.37 when rounded to three significant figures.
Explanation:The equation given is in the form of a natural logarithm (ln x), where 'ln' stands for the natural logarithm and 'x' is the variable we need to solve for. The natural logarithm of a number is the power to which the constant 'e' (approximately equal to 2.7182818) must be raised to equal the number. Therefore, to solve the equation, we need to convert the equation back to the exponential form using the property of logarithms that states a^b=c is equivalent to log_a c = b. As such, ln x = 0.863 becomes e^0.863 = x.
When you calculate e^0.863, you will get x to be approximately 2.37 when rounded to three significant figures (the zero after the decimal point is a significant figure because it's located to the right).
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Final answer:
To solve the equation 'ln x = 0.863', rewrite it as 'e0.863 = x' and calculate the value, which is approximately 2.37 to three significant figures.
Explanation:
To solve for x when given the equation ln x = 0.863, we need to use the property of logarithms that allows us to rewrite the equation in exponential form. The base of the natural logarithm (ln) is e, so we can rewrite the equation as e0.863 = x. Using a calculator, we find that e0.863 is approximately 2.37. Therefore, x ≈ 2.37 to three significant figures.
The activation energy for the reaction NO2 (g )+ CO (g) ⟶ NO (g) + CO2 (g) is Ea = 218 kJ/mol and the change in enthalpy for the reaction is ΔH = -252 kJ/mol . What is the activation energy for the reverse reaction? Enter your answer numerically and in terms of kJ/mol.
The activation energy for the reverse reaction of NO₂ (g) + CO (g) ⇒ NO (g) + CO₂ (g) is calculated using the given activation energy for the forward reaction and the enthalpy change. It is found to be 470 kJ/mol.
To determine the activation energy for the reverse reaction of NO₂ (g) + CO (g) ⇒ NO (g) + CO₂ (g), we must first understand the relationship between activation energy, enthalpy change, and the reverse reaction's activation energy.
The activation energy (Ea) for the forward reaction is given to be 218 kJ/mol. The enthalpy change for the forward reaction is -252 kJ/mol. This means that the products have 252 kJ/mol less energy than the reactants. Knowing this, we can use the equation:
Ea_reverse = Ea_forward +
Substituting the given values:
Ea_reverse = 218 kJ/mol - (-252 kJ/mol)
Ea_reverse = 218 kJ/mol + 252 kJ/mol
Ea_reverse = 470 kJ/mol
Therefore, the activation energy for the reverse reaction is 470 kJ/mol.
Consider the reaction between hydrogen gas and chlorine gas to form hydrogen chloride gas in a closed cylinder with a movable piston. H2(g) + Cl2(g) → 2 HCl(g) Predict what would happen to the reaction rate for a mixture containing both reactants if the following occur. (a) The pressure on the reaction mixture is increased at constant temperature (while keeping the reaction components in the gaseous phase).
Answer:
In this case, the system doesn't be affected by the pressure change. This means that nothing will happen
Explanation:
We can answer this question applying the Le Chatelier's Principle. It says that changes on pressure, volume or temperature of an equilibrium reaction will change the reaction direction until it returns to the equilibrium condition again.
The results of these changes can define as:
Changes on pressure: the reaction will move depending the quantity of moles on each side of the reaction
Changes on temperature: The reaction will move depending on if it's endothermic or exothermic
Changes on volume: The reaction will move depending the limit reagent and the quantity of moles on each side of the reaction
In the exercise, they mention a change on pressure of the system at constant temperature (that means the temperature doesn't change). As Le Chatelier Principle's says, we must analyze what happens if the pressure increase or decrease. If pressure increase the reaction will move on the side that have less quantity of moles, otherwise, if the pressure decreases the reaction will move to the side that have more quantity of moles. In this case, we can see that both sides of the equation have the same number of moles (2 for the reactants and 2 for the products). So, in this case, we can conclude that, despite the change on pressure (increase or decrease), nothing will happen.
A certain liquid X has a normal freezing point of 7.60 °C and a freezing point depression constant K= 6.90 °C-kg-mol. Calculate the freezing point of a solution made of 7.57g of sodium chloride (NaCl) dissolved in 350. g of X Round your answer to 3 significant digits. lºc X 5
The freezing point of the solution made by dissolving the sodium chloride in the liquid is 2.48 °C, calculated using the freezing point depression equation and given constants.
Explanation:This problem involves the Physical Chemistry concept of freezing point depression.
First, we need to calculate the molality of the solution. Molality (m) is defined as moles of solute (mol) divided by mass of solvent (kg). Given 7.57g of NaCl, and we know the molar mass of NaCl is approximately 58.44 g/mol, we therefore have 7.57g / 58.44 g/mol = 0.1297 mol. So, the molality (m) is 0.1297 mol / 0.350 kg = 0.370 mol/kg.
Next, we use the freezing point depression equation: ΔTF = i * K * m, where ΔTF is the change in freezing point, i is the number of ions produced per formula unit (for NaCl, i=2 because NaCl dissociates into 2 ions), K is the cryoscopic constant (given as 6.9 ºC-kg/mol) and m is the molality. Substituting the values we get ΔTF = 2 * 6.9 °C-kg/mol * 0.370 mol/kg = 5.118 °C.
Finally, we subtract this value from the original freezing point of the liquid X (7.60 °C) to get the new freezing point: FP_new =FP_old - ΔTF = 7.60°C - 5.118°C = 2.482°C. However, rounding to three significant figures, it should be 2.48 °C.
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The freezing point of the solution made of 7.57 g of NaCl dissolved in 350 g of liquid X is approximately [tex]\( 5.04 \, \text{degree\ C} \).[/tex]
The freezing point of the solution, we'll use the formula for freezing point depression.
[tex]\[ \Delta T_f = K_f \cdot m \][/tex]
where.
[tex]- \( \Delta T_f \)[/tex] is the freezing point depression,
[tex]- \( K_f \)[/tex] is the freezing point depression constant of the solvent liquid X.
[tex]- \( m \)[/tex] is the molality of the solution.
Let's find the molality [tex](\( m \))[/tex] of the solution.
1. Calculate moles of NaCl.
Molar mass of NaCl = [tex]\( 22.99 \, \text{g/mol} + 35.45 \, \text{g/mol} = 58.44 \, \text{g/mol} \)[/tex].
Moles of NaCl [tex]\( = \frac{7.57 \, \text{g}}{58.44 \, \text{g/mol}} = 0.1295 \, \text{mol} \)[/tex].
2. Calculate molality [tex](\( m \)).[/tex]
[tex]\( m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \)[/tex]
Mass of solvent [tex](liquid X) \( = 350 \, \text{g} = 0.350 \, \text{kg} \)[/tex].
3. Calculate the freezing point depression[tex](\( \Delta T_f \)).[/tex]
Given [tex]\( K_f = 6.90 \, \text{degree\ C-kg/mol} \),[/tex]
[tex]\( \Delta T_f = K_f \cdot m = 6.90 \, \text{°C-kg/mol} \cdot 0.370 \, \text{mol/kg} \)[/tex]
[tex]\( \Delta T_f = 2.557 \, \text{degree\ C} \)[/tex]
4. Calculate the freezing point of the solution.
Freezing point of solution = Normal freezing point of solvent -[tex]\( \Delta T_f \)[/tex]
Normal freezing point of X = 7.60 °C
Freezing point of solution [tex]\( = 7.60 \, \text{degree\ C} - 2.557 \, \text{degree\ C} \)[/tex]
Freezing point of solution [tex]\( = 5.043 \, \text{degree\ C} \)[/tex]
Round your answer to 3 significant digits [tex]\[ \\5.04 \, \text{degree\ C}} \][/tex]
Draw the Lewis Structure for H2O.
Answer:
Structure has been shown below
Explanation:
In a Lewis structure, least electronegative atom is taken as central atom (except H)Total valence electrons of constituting atoms are used to construct Lewis structureA bond between two atoms are represented by a horizontal line or a dotted line (two dots).Spatial rearrangement of atoms are neglected in Lewis structure.H has one valence electron and O has six valence electrons. So, total valence electrons in [tex]H_{2}O[/tex] are 8Lewis structure of [tex]H_{2}O[/tex] has been given below.To draw the Lewis structure for H2O, count the valence electrons, place the atoms, represent the bonding pairs and lone pairs, and check for octet completeness.
The Lewis structure for H2O can be drawn using the following steps:
Count the total number of valence electrons for each atom. Hydrogen has 1 valence electron and oxygen has 6 valence electrons.
Place the atoms in a way that the central atom (in this case, oxygen) is surrounded by the other atoms (hydrogen).
Use single lines to represent the bonding pairs between the central atom and the surrounding atoms. In this case, draw two lines between the oxygen and hydrogen.
Distribute the remaining valence electrons around the atoms to complete their octets. Oxygen will have two lone pairs of electrons and hydrogen will have no lone pairs.
Check if every atom has a complete octet. In this case, oxygen has 8 electrons and hydrogen has 2.
Write the Lewis structure, making sure to correctly represent the bonding pairs and lone pairs. In this case, it would be H—O—H.
The osmotic pressure of an aqueoussolution of urea at 300 K is
120 kPa. Calculate the freezing pointof the same solution.
The freezing point of the aqueous solution of urea at 300 K with an osmotic pressure of 120 kPa is -0.09°C
To calculate the freezing point of the solution, we can use the following equation:
ΔTf = Kf * m
where:
ΔTf is the freezing point depression in C
Kf is the molal freezing point constant of water in C·kg/mol (1.86 C·kg/mol)
m is the molality of the solution in mol/kg
The molality of the solution can be calculated from the osmotic pressure using the following equation:
Π = MRT
where:
Π is the osmotic pressure in Pa
M is the molarity of the solution in mol/m³
R is the ideal gas constant in J/mol·K (8.314 J/mol·K)
T is the temperature in K
Substituting the given values into the equation, we get:
M = Π / (RT) = 120 kPa / (8.314 J/mol·K * 300 K) = 0.4918 mol/m³
The molality of the solution is equal to the molarity divided by the density of the solvent. The density of water is 1 kg/L, so the molality of the solution is:
m = M / d = 0.4918 mol/m³ / 1 kg/L = 0.4918 mol/kg
Now that we know the molality of the solution, we can calculate the freezing point depression using the first equation:
ΔTf = Kf * m = 1.86 C·kg/mol * 0.4918 mol/kg = 0.09 °C
Finally, we can calculate the freezing point of the solution by subtracting the freezing point depression from the freezing point of pure water:
Freezing point of solution = freezing point of pure water - freezing point depression
Freezing point of solution = 0 °C- 0.09 °C = -0.09°C
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The freezing point of an aqueous solution of urea at 300 K with an osmotic pressure of 120 kPa is -0.08928 °C.
1. To find the freezing point depression (ΔTf) of the solution, we can use the formula: ΔTf = i x k x m, where:
i is the van 't Hoff factor (for urea, i = 1 because it does not dissociate in solution). Kf is the freezing point depression constant for water (1.86 °C kg/mol).m is the molality of the solution.2. We need to calculate the molality (m) using osmotic pressure (π), which we can find from the formula: π / (i x R x T), given:
Osmotic pressure (π) = 120 kPa (convert to atm: 120 kPa / 101.325 = 1.184 atm).R is the universal gas constant = 0.0821 L atm/mol K.T is the temperature = 300 K.M is the molarity.Rearranging for M (molarity): M = π / (i x R x T)Substitute values: M = 1.184 atm / (1 x 0.0821 L atm/mol K x 300 K) = 0.048 mol/L3. Since the density of water is approximately 1kg/L, the molality (m) ≈ molarity (M) in this case: m ≈ 0.048 mol/kg
Now, we can find the freezing point depression (ΔTf):ΔTf = 1 x 1.86 °C kg/mol x 0.048 mol/kg = 0.08928 °C
The molar mass of monopotassium phosphate is 136.09 g/mol. How many moles of monopotassium phosphate are needed to make 250.0 ml of 0.15 M monopotassium phosphate?
Answer:
0.0375 moles
Explanation:
Given that:
Molar mass of monopotassium phosphate = 136.09 g/mol
Given that volume = 250.0 mL
Also,
[tex]1\ mL=10^{-3}\ L[/tex]
So, Volume = 250 / 1000 L = 0.25 L
Molarity = 0.15 M
Considering:
[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]
[tex]0.15=\frac{Moles\ of\ solute}{0.25}[/tex]
Thus, moles of monopotassium phosphate needed = 0.0375 moles
Provide the Bronstead-Lowry de finition for an "Acid" and a "Base"
Answer:
According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.
For example:
[tex]H_3BO_3(aq.)+HS^-(aq.)\rightarrow H_2BO_3^-(aq.)+H_2S(aq.)[/tex]
Here, [tex]H_3BO_3[/tex] is loosing a proton, thus it is considered as a bronsted acid and after losing a proton, it forms [tex]H_2BO_3^-[/tex] which is a conjugate base.
And, [tex]HS^-[/tex] is gaining a proton, thus it is considered as a bronsted base and after gaining a proton, it forms [tex]H_2S[/tex] which is a conjugate acid.
The sequence of the acids and bases in the above equation are:
[tex]Acid+Base\rightarrow Base+Acid[/tex]
Please select statements that satisfy the Pauli exe a) Electron state can hold no more than two electrons. b) Electrons with the same state must have opposite spins. c) Electron state can hold no more than four electrons. d) Electrons with the same state must have the same spins. its that satisfy the Pauli exclusion principle.
Answer:
b) Electrons with the same state must have opposite spins.
Explanation:
Pauli Exclusion Principle-
The principle states that in an atom or in a molecule, no two electrons present can have same set of the four quantum numbers.
Even if the two electrons are present in same state which means that if it is present in the same orbital , n, l and m are same but still the forth quantum number which is spin quantum number is different for them. Both the two electrons have opposing spins having value either [tex]\frac {1}{2}[/tex] or [tex]\frac {-1}{2}[/tex].
Hence, option B is correct.
Answer:
Options A and B
Explanation:
The Pauli Exclusion Principle states that, in an atom or molecule, no two electrons can have the same four electronic quantum numbers. As an orbital can contain a maximum of only two electrons, the two electrons must have opposing spins.
Going through the options;
Option A - An electron shell can have more than two electrons. An orbital however can hold a maximum of two electrons.
Option B - This is correct because No two electrons in a atom can have the same four quantum numbers . This means that only two electrons are allowed in the same orbital, and then they have opposite spin, +1/2 and -1/2.
Option C - Each electron orbital (state) can only contain a maximum of two lectrons.
Option D - This is wrong because no two electrons in a atom can have the same four quantum numbers . This means that only two electrons are allowed in the same orbital, and then they must have opposite spin, +1/2 and -1/2.
Which of the following solvents are polar? (eg Have an overall molecular dipole) (Hexanes, Ethyl Acetate, Ethanol) Ethyl acetate Ethyl acetate and ethanol Ethanol Hexanes Ethyl acetate and hexanes
Answer:
Ethanol
Explanation:
Hello,
Since the hexanes and the ethyl acetate are mostly composed by C-C, C-H and for the ester C-O bonds which are apolar, just the ethanol which has one O-H bond could be classified as polar, allowing H-bridges to be present.
Best regards.
Final answer:
Ethyl Acetate and Ethanol are the polar solvents among the given options. Hexanes are nonpolar and would not dissolve readily in water. Ethanol, being highly polar, and Ethyl Acetate, with moderate polarity, would indeed have an overall molecular dipole.
Explanation:
The question you've asked is related to the polarity of solvents and their solubility in water, which is a polar solvent. Among the choices given (Hexanes, Ethyl Acetate, and Ethanol), we need to determine which ones are polar and would have an overall molecular dipole.
Hexanes are hydrocarbons with a nonpolar nature due to the C-H bonds that possess minimal electronegativity difference. Therefore, hexanes are not polar and are insoluble in water.
Ethyl Acetate has a polar carbonyl group (C=O) and is overall a polar molecule, although it has a slight nonpolar character due to its ethyl groups. Because of this, it has moderate solubility in water.
Ethanol is highly polar due to the presence of an -OH group allowing hydrogen bonding. This makes ethanol soluble in water.
Therefore, the solvents that are polar and have an overall molecular dipole from the given options are Ethyl Acetate and Ethanol.
How many moles of solute particles are present in 4.98 mL of 0.72 M (NH4)2SO4? When you have your answer, take the LOG (base 10) of your answer and enter it with 2 decimal places.
Answer: The number of moles of ammonium sulfate is 0.0036 moles and its logarithmic value is -2.44
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]
Molarity of ammonium sulfate solution = 0.72 M
Volume of solution = 4.98 mL
Putting values in above equation, we get:
[tex]0.72M=\frac{\text{Moles of ammonium sulfate}\times 1000}{4.98mL}\\\\\text{Moles of ammonium sulfate}=0.0036mol[/tex]
Taking the log (base 10) of the calculated moles of ammonium sulfate we get:
[tex]\log_{10}(0.0036)=-2.44[/tex]
Hence, the number of moles of ammonium sulfate is 0.0036 moles and its logarithmic value is -2.44
Final answer:
First, calculate the moles of (NH4)2SO4 using the molarity and volume, then find the base 10 logarithm of the result. The moles of (NH4)2SO4 are approximately 0.003586, and the logarithm is roughly -2.45.
Explanation:
To compute the moles of solute in a given volume of solution, one can use the formula moles of solute = Molarity (mol/L) × Volume of solution (L). For a 0.72 M solution of (NH4)2SO4, when given a volume of 4.98 mL (which is 0.00498 L), we calculate:
moles of solute = 0.72 mol/L × 0.00498 L = 0.0035856 mol
Now, to find the base 10 logarithm, we perform the following calculation:
LOG(moles of solute) = LOG(0.0035856) ≈ -2.45
List emergency equipment, those are available in the laboratory?
Answer:
Hands-free eyewash stations, sand bucket, fire blankets, fire extinguishers, fire alarm system and first aid kit.
Explanation:
Working in laboratories has many risks, therefore, preventive measures that should be incorporated to avoid the occurrence of any laboratory accidents.
Some of the important emergency equipment that should be available in laboratories are: hands-free eyewash stations, sand bucket, fire blankets, fire extinguishers, fire alarm system, chemical storage cabinet, first aid kits and fume hood.
Some of the personal protective equipment include lab coats, goggles, safety gloves and face shield.
The thallium Subscript 81 Superscript 208 Baseline Tl nucleus is radioactive, with a half-life of 3.053 min. At a given instant, the activity of a certain sample of thallium is 2400 Bq. Using the concept of a half-life, and without doing any written calculations, determine what the activity 9 minutes later is.
(A) 150 bq
(B) 2400 bq
(C) 1200 bq
(D) 600 bq
(E) 300 bq
Answer: (E) 300 bq
Explanation:
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
Radioactive decay process is a type of process in which a less stable nuclei decomposes to a stable nuclei by releasing some radiations or particles like alpha, beta particles or gamma-radiations. The radioactive decay follows first order kinetics.
Half life is represented by [tex]t_{\frac{1}{2}[/tex]
Half life of Thallium-208 = 3.053 min
Thus after 9 minutes , three half lives will be passed, after ist half life, the activity would be reduced to half of original i.e. [tex]\frac{2400}{2}=1200[/tex], after second half life, the activity would be reduced to half of 1200 i.e. [tex]\frac{1200}{2}=600[/tex], and after third half life, the activity would be reduced to half of 600 i.e. [tex]\frac{600}{2}=300[/tex],
Thus the activity 9 minutes later is 300 bq.
Ethyl chloride vapor decomposes by the first-order reaction C2H5Cl → C2H4 + HCl The activation energy is 249 kJ/mol and the frequency factor is 1.60 × 1014 s−1. Find the value of the specific rate constant at 896 K . Enter your answer numerically (to 4 decimal places) and in terms of the appropriate units for a first order reaction.
Answer : The value of rate constant is, [tex]0.3607s^{-1}[/tex]
Explanation :
The Arrhenius equation is written as:
[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]
Taking logarithm on both the sides, we get:
[tex]\ln k=-\frac{Ea}{RT}+\ln A[/tex] ............(1)
where,
k = rate constant
Ea = activation energy = 249 kJ/mol = 249000 kJ/mol
T = temperature = 896 K
R = gas constant = 8.314 J/K.mole
A = pre-exponential factor or frequency factor = [tex]1.60\times 10^{14}s^{-1}[/tex]
Now we have to calculate the value of rate constant by putting the given values in equation 1, we get:
[tex]\ln k=-\frac{249000J/mol}{8.314J/K.mol\times 896K}+\ln (1.60\times 10^{14}s^{-1})[/tex]
[tex]\ln k=-1.0198[/tex]
[tex]k=0.3607s^{-1}[/tex]
Therefore, the value of rate constant is, [tex]0.3607s^{-1}[/tex]
Ethyl chloride vapor decomposes by the first-order reaction. Given the activation energy is 249 kJ/mol and the frequency factor is 1.60 × 10¹⁴ s⁻¹, the rate constant at 896 K is 0.4870 s⁻¹.
What is a first-order reaction?A first-order reaction is a chemical reaction in which the rate of reaction is directly proportional to the concentration of the reacting substance.
Let's consider the first-order reaction for the decomposition of ethyl chloride.
C₂H₅Cl → C₂H₄ + HCl
The activation energy is 249 kJ/mol and the frequency factor is 1.60 × 10¹⁴ s⁻¹. We can find the value of the rate constant at 896 K using the Arrhenius equation.
[tex]k = A \times e^{-Ea/R \times T} \\\\k = (1.60 \times 10^{14}s^{-1} ) \times e^{-(249 \times 10^{3} J/mol)/(8.314 J/mol.K) \times 896K} = 0.4870 s^{-1}[/tex]
where,
k is the rate constant.A is the frequency factor.Ea is the activation energy.R is the ideal gas constant.T is the absolute temperature.Ethyl chloride vapor decomposes by the first-order reaction. Given the activation energy is 249 kJ/mol and the frequency factor is 1.60 × 10¹⁴ s⁻¹, the rate constant at 896 K is 0.4870 s⁻¹.
Learn more about first-order reactions here: https://brainly.com/question/518682
Calculate the specific volume of Helium using the compressibility factor. 500 kPa, 60 degrees Celsius
Explanation:
Formula for compressibility factor is as follows.
z = [tex]\frac{P \times V_{m}}{R \times T}[/tex]
where, z = compressibility factor for helium = 1.0005
P = pressure
[tex]V_{m}[/tex] = molar volume
R = gas constant = 8.31 J/mol.K
T = temperature
So, calculate the molar volume as follows.
[tex]V_{m} = \frac{z \times R \times T}{P}[/tex]
= [tex]\frac{1.0005 \times 8.314 \times 10^{-3} m^{3}.kPa/mol K \times (60 + 273)K}{500 kPa}[/tex]
= 0.0056 [tex]m^{3}/mol[/tex]
As molar mass of helium is 4 g/mol. Hence, calculate specific volume of helium as follows.
[tex]V_{sp} = \frac{V_{m}}{M_{w}}[/tex]
= [tex]\frac{0.0056 m^{3}/mol}{4 g/mol}[/tex]
= 0.00139 [tex]m^{3}/g[/tex]
= 0.00139 [tex]m^{3}/g \times \frac{1 g}{10^{-3}kg}[/tex]
= 1.39 [tex]m^{3}/kg[/tex]
Thus, we can conclude that the specific volume of Helium in given conditions is 1.39 [tex]m^{3}/kg[/tex].
Show the conversion factor from cm2/stoft2/hris 3.875.
Explanation:
In 1 centimeter square there are 0.00107639 square feets.
[tex] 1 cm^2=0.00107639 ft^2[/tex]
In 1 hour there 3600 seconds.
1 seconds = 0.000277778 Hours
[tex]1cm^2/s=(x)\times ft^2/h[/tex]..(1)
[tex]1 cm^2/h=\frac{1\times 0.00107639 ft^2}{1\times 0.000277778 h}[/tex]
[tex]=3.975000899 ft^2/h\approx 3.875 ft^2/h[/tex]
On comparing with (1) we get:
[tex]1 cm^2/s=(3.875)\times ft^2/h[/tex]
3.875 is the conversion factor.
State the "Physical Interpretation" of pKa
Answer:
pka = -logKa
Explanation:
pKa is defined as negative logarithm of dissociation constant, Ka.
Or, pka = -logKa
pKa define strength of an acid.
Higher pKa indicates lower strength of the acid or low dissociation of the acid.
Whereas lower pKa indicates higher strength of the acid or high dissociation of the acid.
pKa is also related with pH of the solution.
Higher the pKa, higher is the pH of the solution. This can be understood as:
Higher pKa means lower dissociation dissociation or low concentration of H+. Low H+ means high pH.
Relation between pKa and pH is given by Henderson-Hasselbalch equation,
[tex]pH=p_{Ka} + \frac{[Salt]}{[Acid]}[/tex]
A Chemist measured 184 mL using a graduated cylinder. Based on her experience, the volume did not seem to be correct. This volume was again measured using a calibrated standard device, from which it was determined that the volume of the liquid was actually 230.0 mL. What is the % error in this measurement?
Answer:
Percent error = 20%
Explanation:
The percent error is calculated using the following equation:
Percent error = |(approximate value - exact value)| / (exact value) x 100%
In this problem, the approximate value was 184 mL and the exact value was 230.0 mL
Percent error = |(184 mL - 230.0 mL)| / (230.0 mL) x 100% = 20%
Final answer:
The percent error of the chemist's measurement is 20%, which reflects the difference between the measured value using a graduated cylinder and the actual value obtained using a calibrated device.
Explanation:
The percent error in measurement can be determined using the formula: percent error = (|actual value - measured value| / actual value) × 100%. In this case, the actual volume is 230.0 mL, while the measured volume is 184 mL.
To calculate the percent error: percent error = (|230.0 mL - 184 mL| / 230.0 mL) × 100% = (46 mL / 230.0 mL) × 100% = 20%.
So, the percent error of the chemist's measurement is 20%. This value indicates how far the initial measurement was from the actual value, reflecting the importance of using calibrated instruments for accurate measurement. The example of the quality control chemist shows the significance of both precision and accuracy in chemical measurements.
In an old trunk, you find a cube of metal that you think may be aluminum, silver, lead, cobalt or tin. In lab you find that it has a mass of 222 g and a length of 2.7 cm. What is the metal? The density of each metal is given below at room temperature. O Lead (11.3 g/cm) O Silver (10.5 g/cm) Cobalt (8.90 g/cm) O Tin (7.31g/cm) Aluminum (2.70 g/cm)
Answer:
The density of old trunk is [tex]11.3 g/cm^3[/tex] and it is equal to the density of metal lead.
Explanation:
Mass of the trunk , m= 222 g
Length of the cubic trunk = 2.7 cm
Volume of the cube = [tex]a^3[/tex]
Density of the truck = [tex]\frac{Mass}{Volume}[/tex]
[tex]D=\frac{m}{a^3}=\frac{222 g}{(2.7 cm)^3}=11.2787 g/cm^3\approx 11.3 g/cm^3[/tex]
The density of old trunk is [tex]11.3 g/cm^3[/tex] and it is equal to the density of metal lead.
Final answer:
The density of the metal cube is calculated to be approximately 11.28 g/cm³, which closely matches the density of lead (11.3 g/cm³), suggesting that the cube is made of lead.
Explanation:
To determine the type of metal a cube is made of, you need to calculate its density and compare it with the given densities of possible materials. The density is calculated by dividing the mass of the cube by its volume.
The mass of the metal cube is given as 222 g, and the length of each side is 2.7 cm. The volume V of a cube is found by cubing the length of one of its sides: V = 2.7 cm × 2.7 cm × 2.7 cm = 19.683 cm³. The density ρ is then found by dividing the mass m by the volume V: ρ = m/V = 222 g / 19.683 cm³ ≈ 11.28 g/cm³.
Comparing this calculated density with the given densities, the closest match is lead (11.3 g/cm³). Therefore, the metal cube is most likely made of lead.
The Prandtl number, Pr, is a dimensionless group important in heat transfer. It is defined as Pr - Cp*mu/k where Cp is the heat capacity of a fluid, mu is the fluid viscosity, and k is the fluid thermal conductivity. For a given fluid, Cp 0.66 J/(g * deg C), k 0.36 W/(m * deg C), and mu 1896 lbm (ft * h}. Determine the value of the Prandtl number for this fluid
Answer:
The Prandtl number for this example is 14,553.
Explanation:
The Prandlt number is defined as:
[tex]Pr=\frac{C_{p}*\mu}{k}[/tex]
To compute the Prandlt number for this case, is best if we use the same units in every term of the formula.
[tex]\mu=1896 \frac{lbm}{ft*h}*\frac{1000 g}{2.205 lbm}*\frac{3.281 ft}{1 m}*\frac{1h}{3600s} =7938 \frac{g}{m*s}[/tex]
Now that we have coherent units, we can calculate Pr
[tex]Pr=\frac{C_{p}*\mu}{k}=0.66*7938/0.36=14553[/tex]
Which of the following increases as the concentration of a colored solution increases? wavelenght transmittance absorbance b) A blue-green solution will absorb light most strongly in this part of the spectrum: 450nm 580nm 800nm
Answer:
a) Absorbance
b) The absorb light most strongly in 580nm
Explanation:
Beer-Lambert law relates concentration with light absorbance. The more concentrated solution are the more molecules and the most absorbance.
Wavelenght depends of composition of solution and doesn't change with different concentrations of the same solution.
Transmittance is inversely proportional to absorbance. Thus, the more concentrated solution the less transmittance.
Colored compounds are absorb energy of visible radiation. The colour that we see is a result of the absortion of complimentary colour (Colour wheel). Thus, a blue-green solution absorb energy of ≈600 nm. Thus, the absorb light most strongly in 580nm.
I hope it helps!
Two elements are more likely to be ionically bonded when ....
Select one:
a. the elements in the material have similar electronegativities
b. the elements in the material are well apart (in different sides) in the periodic table
c. the elements in the material belong to the same group
d. the elements in the material has the same valence
Answer:
The correct option is: b. the elements in the material are well apart (in different sides) in the periodic table
Explanation:
Ionic bond is a type of chemical bonding that is formed by the transfer of electrons from one atom to another. It is formed between atoms having large electronegativity difference.
The more electronegative atom accepts electrons and becomes a negatively charged anion. Whereas, the less electronegative atom loses electrons and becomes a positively charged cation.
Simply, an ionic bond is formed when the electrons are transferred from a metal to a non-metal, which are present on different sides of the periodic table.
Question 8: A 150 mL of buffer contains 0.045 M carbonic acid (H2CO3) and 0.025 M sodium hydrogen carbonate (HCO3). If 3.0 mL of 0.15M HCl is added to the buffer, what is the new pH? pKa = 6.37
Answer: The new pH of resulting solution is 6.03
Explanation:
We are adding hydrochloric acid to the solution, so it will react with salt (sodium hydrogen carbonate) only.
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]
For hydrochloric acid:Molarity of hydrochloric acid = 0.15 M
Volume of solution = 3 mL
Putting values in above equation, we get:
[tex]0.15M=\frac{\text{Moles of hydrochloric acid}\times 1000}{3mL}\\\\\text{Moles of hydrochloric acid}=0.00045mol[/tex]
For sodium hydrogen carbonate:Molarity of sodium hydrogen carbonate = 0.025 M
Volume of solution = 150 mL
Putting values in above equation, we get:
[tex]0.025M=\frac{\text{Moles of sodium hydrogen carbonate}\times 1000}{150mL}\\\\\text{Moles of sodium hydrogen carbonate}=0.00375mol[/tex]
For carbonic acid:Molarity of carbonic acid = 0.045 M
Volume of solution = 150 mL
Putting values in above equation, we get:
[tex]0.045M=\frac{\text{Moles of carbonic acid}\times 1000}{150mL}\\\\\text{Moles of carbonic acid}=0.00675mol[/tex]
The chemical reaction for sodium hydrogen carbonate and hydrochloric acid follows the equation:
[tex]NaHCO_3+HCl\rightarrow NaCl+H_2CO_3[/tex]
Initial: 0.00375 0.00045 0.00675
Final: 0.00330 - 0.00720
Volume of solution = 150 + 3 = 153 mL = 0.153 L (Conversion factor: 1 L = 1000 mL)
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]
[tex]pH=pK_a+\log(\frac{[NaHCO_3]}{[H_2CO_3]})[/tex]
We are given:
[tex]pK_a[/tex] = negative logarithm of acid dissociation constant of carbonic acid = 6.37
[tex][NaHCO_3]=\frac{0.0033}{0.153}[/tex]
[tex][H_2CO_3]=\frac{0.0072}{0.153}[/tex]
pH = ?
Putting values in above equation, we get:
[tex]pH=6.37+\log(\frac{0.0033/0.153}{0.0072/0.153})\\\\pH=6.03[/tex]
Hence, the new pH of the solution is 6.03