Answer:
10.6 moles of CO₂ are produced in this combustion
Explanation:
The combustion reaction is:
2C₂H₆ (g) + 7O₂ (g) ⟶ 4CO₂ (g) + 6H₂O (g)
We assume the ethane as the limiting reactant because the excersise states that the O₂ is in excess.
We make a rule of three:
2 moles of ethane can produce 4 moles of CO₂
Therefore 5.30 moles of ethane will produce (5.3 . 4) /2 = 10.6 moles
As their name suggests, calcium ion channel blockers block calcium ion channels in the sarcolemma of pacemaker and contractile cardiac muscle cells and slow calcium ion entry into the cell during an action potential. What effects would these drugs have on the myocardial cells? View Available Hint(s)
Answer:
Effects of Calcium channel blockers on the myocardial cells include lower excitability of the cells due lower influx of calcium ions during an action potential. This effect helps to regulate abnormal rapid heart rhythms (arrhythmias).
Calcium channel blockers also reduce blood pressure by dilating the arteries thereby reducing pressure in the arteries. This in turn also reduces the workload on the heart and therefore allows for adequate oxygen supply to heart in the treatment of Angina.
Examples of calcium ion channel blockers are amlodipine, nimodipine, verapamil, diltiazem etc.
The molar mass of an unknown gas was measured by an effusion experiment. It was found that it took 60 s for the gas to effuse, whereas nitrogen gas required 48 s. The molar mass of the gas is ________
The molar mass of the unknown gas can be determined by comparing its effusion rate to the effusion rate of a known gas, such as nitrogen gas. According to Graham's law of effusion, the rate of effusion is inversely proportional to the square root of the molar mass of the gas. Let's set up a proportion to find the molar mass of the unknown gas.
Explanation:The molar mass of the unknown gas can be determined by comparing its effusion rate to the effusion rate of a known gas, such as nitrogen gas. In this case, it took 60 seconds for the unknown gas to effuse, whereas nitrogen gas required 48 seconds. According to Graham's law of effusion, the rate of effusion is inversely proportional to the square root of the molar mass of the gas. Using this information, we can set up a proportion to find the molar mass of the unknown gas.
Let's assume the molar mass of the unknown gas is represented by 'X'. The proportion can be set up as follows:
(rate of unknown gas)/(rate of nitrogen gas) = sqrt(molar mass of nitrogen gas)/sqrt(molar mass of unknown gas)
Substituting the known values, we have:
(60 s)/(48 s) = sqrt(28.01 g/mol)/sqrt(X)
Solving for X, the molar mass of the unknown gas is approximately 37.07 g/mol.
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A calorimeter contains 32.0 mLmL of water at 15.0 ∘C∘C . When 1.20 gg of XX (a substance with a molar mass of 54.0 g/molg/mol ) is added, it dissolves via the reaction X(s)+H2O(l)→X(aq)X(s)+H2O(l)→X(aq) and the temperature of the solution increases to 29.0 ∘C∘C . Calculate the enthalpy change, ΔHΔHDelta H, for this reaction per mole of XX. Assume that the specific heat of the resulting solution is equal to that of water [4.18 J/(g⋅∘C)J/(g⋅∘C)], that density of water is 1.00 g/mLg/mL, and that no heat is lost to the calorimeter itself, nor to the surroundings.
The enthalpy change [tex](\( \Delta H \))[/tex] for this reaction per mole of [tex]\( X \)[/tex] is approximately 88.27 kj/mol.
To calculate the enthalpy change[tex](\( \Delta H \))[/tex] for the reaction per mole of X , we can use the formula:
[tex]\[ \Delta H = \frac{q}{n} \][/tex]
where:
- [tex]\( q \)[/tex] is the heat absorbed or released by the reaction,
- [tex]\( n \)[/tex] is the moles of [tex]\( X \).[/tex]
First, let's find the moles of \( X \):
[tex]\[ \text{moles of } X = \frac{\text{mass}}{\text{molar mass}} \][/tex]
[tex]\[ \text{moles of } X = \frac{1.20 \, \text{g}}{54.0 \, \text{g/mol}} \][/tex]
[tex]\[ \text{moles of } X \approx 0.0222 \, \text{mol} \][/tex]
Now, we need to find the heat [tex](\( q \))[/tex] absorbed or released by the reaction. The heat can be calculated using the formula:
[tex]\[ q = mc\Delta T \][/tex]
where:
- m is the mass of the solution [tex](water + \( X \)),[/tex]
- c is the specific heat of the solution,
- [tex]\( \Delta T \)[/tex]is the change in temperature.
First, calculate the mass of the solution:
[tex]\[ \text{mass of solution} = \text{mass of water} + \text{mass of } X \][/tex]
[tex]\[ \text{mass of solution} = 32.0 \, \text{g} + 1.20 \, \text{g} \][/tex]
[tex]\[ \text{mass of solution} = 33.20 \, \text{g} \][/tex]
Now, calculate [tex]\( \Delta T \)[/tex]:
[tex]\[ \Delta T = T_f - T_i \][/tex]
[tex]\[ \Delta T = 29.0^\circ \text{C} - 15.0^\circ \text{C} \][/tex]
[tex]\[ \Delta T = 14.0^\circ \text{C} \][/tex]
Now, plug in the values into the heat formula:
[tex]\[ q = (33.20 \, \text{g}) \times (4.18 \, \text{J/(g} \cdot ^\circ \text{C})) \times (14.0 \, ^\circ \text{C}) \][/tex]
[tex]\[ q \approx 1961.12 \, \text{J} \][/tex]
Now, use the first formula to find [tex]\( \Delta H \):[/tex]
[tex]\[ \Delta H = \frac{1961.12 \, \text{J}}{0.0222 \, \text{mol}} \][/tex]
[tex]\[ \Delta H \approx 88272 \, \text{J/mol} \][/tex]
Therefore, the enthalpy change[tex](\( \Delta H \))[/tex] for this reaction per mole of X is approximately[tex]\( 88.27 \, \text{kJ/mol} \).[/tex]
Calculate the heat absorbed using q = mcΔT and convert to kJ. Determine the moles of substance X, then compute the enthalpy change per mole. The enthalpy change (ΔH) for the reaction per mole of X is 84.34 kJ/mol.
To calculate the enthalpy change (ΔH) for the reaction per mole of substance X, follow these steps:
Determine the heat absorbed by the solution using the formula q = mcΔT, where m is mass, c is specific heat capacity, and ΔT is the change in temperature.The mass of the solution is the volume of water (32.0 mL) converted to grams (32.0 g) and the specific heat capacity is 4.18 J/g°C. The temperature change (ΔT) is 29.0°C - 15.0°C = 14.0°C.Calculate q: q = 32.0 g * 4.18 J/g°C * 14.0°C = 1874.56 J.Convert q to kJ: 1874.56 J * (1 kJ / 1000 J) = 1.87456 kJ.Calculate the moles of substance X: Moles = 1.20 g / 54.0 g/mol = 0.02222 mol.Finally, calculate the ΔH per mole of X: ΔH = 1.87456 kJ / 0.02222 mol = 84.34 kJ/mol.A 100.0g sample of tin is heated to 100.0 oC (Celsius) and is placed in a coffee cup calorimeter containing 150. g of water at 25.0 oC. After the metal cools, the final temperature of the metal and the water is 27.4 oC. Calculate the specific heat capacity of tin from these experimental data, assuming that no heat escapes to the surroundings or is transferred to the calorimeter. Specific heat of water
Explanation:
It is known that specific heat of water is 4.184 [tex]J/g^{o}C[/tex] and atomic mass of tin is 118.7 g/mol. For the given situation,
[tex]Q_{lost} = Q_{gained}[/tex]
Let us assume that,
[tex]m_{1}[/tex] = mass of Sn
[tex]m_{2}[/tex] = mass of [tex]H_{2}O[/tex]
Therefore, heat energy expression for heat lost and gained is as follows.
[tex]Q_{lost} = Q_{gained}[/tex]
[tex]m_{1}C_{1}(T_{2} - T_{1}) = m_{2}C_{2}(T_{1} - T_{2})[/tex]
[tex]100 g \times C_{1} \times (100^{o}C - 27.4^{o}C) = 150 g \times 4.184 /g^{o}C \times (27.4^{o}C - 25^{o}C)[/tex]
[tex]7260C_{1} = 150 \times 4.184 \times 2.4[/tex]
[tex]C_{1} = \frac{1506.24}{7260}[/tex]
= 0.207 [tex]J/g^{o}C[/tex]
For, 118.7 g the specific heat of tin will be calculated as follows.
[tex]C_{1} = 0.207 J/g^{o}C \times 118.7 g[/tex]
= 24.5 [tex]J/mol^{o}C[/tex]
Thus, we can conclude that specific heat of tin is 24.5 [tex]J/mol^{o}C[/tex].
To calculate the specific heat capacity of the metal, we can use the equation q = m * c * ΔT, where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. By using the known temperatures and masses of the metal and water, we can solve for the specific heat capacity of the metal.
Explanation:To calculate the specific heat capacity of the metal, we can use the equation q = m * c * ΔT, where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. In this case, we know the initial and final temperatures of the metal and water, as well as the masses of the metal and water. We can rearrange the equation to solve for the specific heat capacity of the metal, c.
We can start by calculating the heat transferred from the metal to the water. The heat transferred to the water can be calculated using the equation q_water = m_water * c_water * ΔT_water, where m_water is the mass of the water, c_water is the specific heat capacity of water, and ΔT_water is the change in temperature of the water.
Next, we can calculate the heat transferred from the metal to the water using the equation q_metal = m_metal * c_metal * ΔT_metal, where m_metal is the mass of the metal, c_metal is the specific heat capacity of the metal, and ΔT_metal is the change in temperature of the metal.
Since the metal and water reach the same final temperature, we can set q_water equal to q_metal and solve for the specific heat capacity of the metal, c_metal.
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PLEASE HELP WILL GIVE BRAINLIEST
C8H18 + O2à CO2 + H2O is the unbalanced combustion reaction between octane and oxygen. What is the coefficient for carbon dioxide in the balanced equation?
A.2
B.8
C.16
D.18
Answer:
16. Option c.
Explanation:
The reaction is:
C₈H₁₈ + O₂ → CO₂ + H₂O
Let's count, we have 8 C, 18 H and 2 O in reactant side
We have 1 C, 2 H and 3 O in product side
In order to balance the C, we add 8 to CO₂ and to balance the H, we add a 9 to H₂O
Now we have 8 C on both sides and 18 H On both sides. Finally we have 25 O on product side.
C₈H₁₈ + O₂ → 8CO₂ + 9H₂O
To balance the O in product side we must add 25/2, but as it is a rational number, we must multiply x2 to get an integer number (x2 in all the stoichiometry). The balanced reaction is :
2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O
with 16 C, 36 H and 50 O, on both sides
Carbon, nitrogen, oxygen and hydrogen are considered _______ because they are required in relatively large quantities, whereas manganese, zinc, and copper are considered _____ because they are required in very small quantities.
Answer:
macro-nutrients , micro-nutrients
Explanation:
For plant growth soil needs some basic elements which is basic need for plant growth and metabolic synthesis.
plant need approx 17 essential nutrients for growth and for completing its life cycle among which some are found in water and air. Nitrogen, carbon, hydrogen and oxygen needed the most and in large quantity as nitrogen helps in rapid growth and green color and is mobile and its deficiency lead to reduced growth and yellow foliage.
Write a statement that reads a floating point (real) value from standard input into temperature. Assume that temperature. has already been declared as an double variable.
Answer:
temperature = stdin.nextDouble();
Explanation: since the temperature has already been declared as a double variable, the above statement will read so.
As temperature equal standard input dot next double variable.
To read a floating point value from standard input into a double variable, use the scanf function with %lf format specifier.
Explanation:To read a floating point (real) value from standard input into a double variable named temperature, you can use the scanf function in the C programming language. The scanf function allows you to read formatted input, and the %lf format specifier is used to read a double value. Here's an example:
scanf("%lf", &temperature);This line of code will read a floating point value from the standard input and store it in the temperature variable declared as a double.
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Describe a simple chemical test that could be used to distinguish between members of each of the following pairs of compounds (a) 4-Chlorophenol and 4-chloro-1-methylbenzene (c) 4-Methylphenol and 2,4,6-trinitrophenol (b) 4-Methylphenol and 4-methylbenzoic acid (d) Ethyl phenyl ether and 4-methylphenol
Answer:
Explanation:
(A) Phenol gives violet color by complexation with Fe3+ solution . It is best identification...
(B) For b , same phenolic test can be done ...But other esterification is also possibility...
(C) Picric acid has a particular identification test ,
(D) Here , like the first one , phenolic test with FeCl3 gives violet color for 4-ethylphenol and no color for ethyl phenyl ether...
Choose the correct statement about melting points. * Melting point can tell us if we have a mixture of compounds (for example, compound of interest impurity). Melting point can tell us the identity of the components of a mixture. If we have a sample with a lowered melting range as compared to our standards, we can do mixed melting point determinations, combining our sample with each standard, to determine the identity of our sample. All of these are true. None of these are true.
Answer: All of these statements are true
Explanation:
Melting point help us to determine if a mixture is pure or has impurities by the virtues of it melting range..
When a nonvolatile solute is added to a solvent, it tends to ___________________ the freezing point and ___________________ the boiling point, compared to the pure solvent.
Answer:
decreases and increases
Explanation:
the addition of solutes can alter the freezing point and the boiling point .the changes in solute content where there is an addition of solutes results in the dropping of freezing point and increase in boiling point.this is known as freezing point depression and boiling point depression.
What is true about the chemiosmotic synthesis of atpt?
Explanation:
The chemiosmotic theory
The theory suggests that the actual production of adenosine triphosphate ATP in cellular respiration of respiring cells takes place through the process of chemiosmosis that makes the electrochemical gradient across the inner and outer membranes of mitochondria.Breaking down of energy-rich molecules such as glucose the energy is released in the form of NADH and FADH2.It is found that for each glucose molecule 36 molecules of ATP can be produced during cellular respiration. These molecules are generated by the Krebs cycle, the electron transport system, and chemiosmosis.Two ATP molecules are produced through glycolysis, so the total number of molecules formed is 38 molecules of ATP.Consider the reaction of hydrogen peroxide with iodine: H2O2(aq)+I2(aq)⇌OH−(aq)+HIO(aq) The reaction is first order in I2 and second order overall. What is the rate law?
Answer:
Consider the reaction of hydrogen peroxide with iodine: H2O2(aq)+I2(aq)⇌OH−(aq)+HIO(aq) The reaction is first order in I2 and second order overall.
What is the rate law?
If the concentration of H2O2 is increased by half and the concentration of I2 is quadrupled, by what factor does the reaction rate increase?
Rate law = k[H₂O₂] [I₂]
The new rate is six times that of the old rate
Explanation:
Rate law = k[H₂O₂] [I₂]
lets input
[H₂O₂] = x
[I₂] = y
Rate = kxy
[H₂O₂] = x + x/2
= 3x/2
[I₂] = 4y
New rate = [tex]k * \frac{3x}{2} *4y[/tex]
New rate = 6
The new rate is six times that of the old rate
The initial concentration of C2H6N2 in the first-order reaction C2H6N2→N2+C2H6 is 0.427 M. After 30. seconds, the concentration of C2H6N2 is 0.395 M. What is the rate constant k for the reaction?
Answer:
2.6×10^-3 mols-1
Explanation:
From
k= 2.303/t × log[A]o/[A]t
Time taken t= 30 seconds
Initial concentration [A]o= 0.427M
Concentration at time (t) [A]t= 0.395M
k= 2.303/30 ×log[0.427]/[0.395]=2.6×10^-3 mols-1
Determine the volume of SO2 (at STP) formed from the reaction of 96.7 g of FeS2 and 55.0 L of O2 (at 398 K and 1.20 atm). The molar mass of FeS2 is 119.99 g/mol. 4 FeS2(s) + 11 O2(g) → 2 Fe2O3(s) + 8 SO2(g)
Answer: 32.9 Liters
Explanation:
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}[/tex]
1. moles of [tex]FeS_2=\frac{96.7g}{119.99g/mol}=0.806mol[/tex]
2. moles of [tex]O_2[/tex]
[tex]PV=nRT[/tex]
P = pressure of the gas = 1.20 atm
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature of the gas = [tex]398K[/tex]
[tex]1.20\times 55.0=n\times 0.0821\times 398[/tex]
[tex]n=2.02[/tex]
[tex]4FeS_2(s)+11O_2(g)\rightarrow 2Fe_2O_3(s)+8SO_2(g)[/tex]
According to stoichiometry:
11 moles of oxygen reacts with 4 moles of [tex]FeS_2[/tex]
Thus 2.02 moles of oxygen reacts with =[tex]\frac{4}{11}\times 2.02=0.73[/tex] moles of [tex]FeS_2[/tex]
Thus oxygen acts as limiting reagent and [tex]FeS_2[/tex] is excess reagent.
As 11 moles of oxygen gives = 8 moles of [tex]SO_2[/tex]
2.02 moles of oxygen gives =[tex]\frac{8}{11}\times 2.02=1.47[/tex] moles of [tex]SO_2[/tex]
[tex]PV=nRT[/tex]
P = pressure of the gas = 1 atm (at STP)
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature of the gas = [tex]273K[/tex] (at STP)
[tex]1\times V=1.47\times 0.0821\times 273[/tex]
[tex]V=32.9L[/tex]
Thus volume of [tex]SO_2[/tex] (at STP) formed from the reaction is 32.9 L
The volume of SO₂ formed from the reaction at STP is 32.92 L
From the question,
We are to determine the volume of SO₂ formed from the reaction
The given balanced chemical equation for the reaction is
4FeS₂(s) + 11O₂(g) → 2Fe₂O₃(s) + 8SO₂(g)
This means,
4 moles of FeS₂ reacts with 11 moles of oxygen to produce 2 moles of Fe₂O₃ and 8 moles of SO₂
First, we will determine the number of moles of each reactant present
For FeS₂Mass = 96.7 g
From the formula
[tex]Number\ of\ moles =\frac{Mass}{Molar\ mass}[/tex]
Molar mass of FeS₂ = 119.99 g/mol
∴ Number of moles of FeS₂ present =[tex]\frac{96.7}{119.99}[/tex]
Number of moles of FeS₂ present = 0.8059 mole
For O₂
Using the formula
PV = nRT
[tex]n =\frac{PV}{RT}[/tex]
Putting the given parameters into the formula, we get
[tex]n = \frac{1.2 \times 55.0}{0.08206 \times 398}[/tex]
[tex]n = \frac{66}{32.65988}[/tex]
n = 2.0208 moles
Since,
4 moles of FeS₂ reacts with 11 moles of oxygen to produce 8 moles of SO₂
Then,
[tex]\frac{2.0208 \times 4}{11}[/tex] moles of FeS₂ will react with 2.0208 moles of oxygen to produce [tex]\frac{2.0208 \times 4}{11} \times 2[/tex] moles of SO₂
That is,
0.7348 moles of FeS₂ will react with 2.0208 moles of oxygen to produce 1.4697 moles of SO₂
∴ Number of moles of SO₂ formed is 1.4697 moles
Now for the volume of SO₂ formed at STP
Since
1 mole of a gas has a volume of 22.4 L at STP
Then
1.4697 moles of the SO₂ will have a volume of 1.4697 × 22.4 L
1.4697 × 22.4 L = 32.92L
Hence, the volume of SO₂ formed from the reaction at STP is 32.92 L
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"In carrying out a titration of a hydrochloric acid solution with a standard sodium hydroxide solution, a student went beyond the end point before reading the volume on the burette. That is, the volume used was larger than the volume required to reach the end point. How will this error affect the calculated concentration of the hydrochloric acid?"
Answer:
The calculated concentration of HCl will be less than actual.
Explanation:
Suppose during titration, the HCl was taken in burette and the NaOH in the volumetric flask.
Now we will use equivalence formula for the calculation of concentration of HCl.
[tex]N_{1} V_{1} = N_{2} V_{2}[/tex]
Where L.H.S is for hydrochloric acid and R.H.S is for sodium hydroxide. The terms N and V represent normality and volume respectively.
If we calculate for
[tex]N_{1} = \frac{N_{2}V_{2} }{V_{1} }[/tex]
We see that if the volume of the HCl is greater then the concentration of the HCl will be reduced.
The error will cause the concentration of the hydrochloric acid to be underestimated.
The concentration of a solution is calculated from the ratio of the number of moles of the solutes and that of the volume of the solution.
Mathematically; concentration = mole/volume
Thus, with the number of moles being constant, the higher the volume of the solution, the lower the concentration that would be derived and vice versa.
This means that any volume that exceeds that of the accurate endpoint will cause the concentration to be underestimated and below the endpoint, the concentration would be overestimated.
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On the periodic table, calcium and bromine are located in period 4. What is one difference between the two elements?
A. Calcium atoms are more reactive than bromine atoms.
B. Bromine and calcium atoms will each combine with the same elements to form compounds.
C. Calcium atoms combine more easily with other atoms in Group 2.
D. Bromine atoms have more electrons in their outer shell than calcium atoms.
Explanation:
it's is d cause bromine can even react with water so
Bromine atoms have more electrons in their outer shell than calcium atoms. So the correct option is D.
What is valency?In chemistry, valence, usually spelled valency, is the characteristic of an element that establishes the maximum number of other atoms that one atom of the element may combine with. The phrase, which was first used in 1868, is used to indicate both the broad power of a combination of elements as well as its numerical value.
For 19th-century chemists, the explanation and systematization of valence posed significant difficulties. The majority of the work was focused on developing empirical guidelines for finding the valences of the elements because there was no good hypothesis for its origin.
The amount of hydrogen atoms that an element's atom may mix with or replace in a compound is used to quantify the characteristic valences of the various elements. But it soon became clear that various compounds had varied valences for numerous elements.
The identification of the chemical bond of organic compounds with a pair of electrons held jointly by two atoms and acting to hold them together in 1916 by the American chemist G.N. Lewis represented the first significant step in the development of a satisfactory explanation of valence and chemical combination. The German scientist W. Kossel studied the nature of the chemical connection between electrically charged atoms (ions) in the same year.
The theory of valence was reconstructed in terms of electronic structures and interatomic forces following the advent of the precise electronic theory of the periodic system of the elements. In response to the many ways that atoms interact, new notions such as ionic valence, covalence, oxidation number, coordination number, and metallic valence were developed.
Therefore the correct option is D.
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What mass of silver chloride can be produced from 1.30 L of a 0.245 M solution of silver nitrate? The reaction described in Part A required 3.36 L of calcium chloride. What is the concentration of this calcium chloride solution?
Please explain, I don't know where to begin.
Answer:
Explanation:
Given parameters:
Volume of AgNO₃ = 1.3L
Molarity or concentration AgNO₃ = 0.245M
Unknown:
Mass of AgCl produced = ?
Solution:
To solve this problem, we have to work from the known compound to the unknown one using the mole concept.
The known compound is the one that we can obtain the value of the number of moles from. Here it is the given AgNO₃ that can furnish us with this piece of information;
Now let us establish the balanced reaction equation;
2AgNO₃ + CaCl₂ → 2AgCl + Ca(NO₃)₂
Now;
Find the number of moles of AgNO₃;
Number of moles of AgNO₃ = concentration x volume
= 1.3 x 0.245
= 0.32moles
From the balanced reaction equation;
2 mole of AgNO₃ produced 2 moles of AgCl
0.32 moles of AgNO₃ will produce 0.32moles of AgCl
Now that we know the number of moles of the AgCl, we can find the mass;
Mass of AgCl = number of moles x molar mass
Molar mass of AgCl = 107.9 + 35.5 = 143.4g/mol
Mass of AgCl = 0.32 x 143.4 = 45.89g
Copper(I) ions in aqueous solution react with NH 3 ( aq ) according to Cu + ( aq ) + 2 NH 3 ( aq ) ⟶ Cu ( NH 3 ) + 2 ( aq ) K f = 6.3 × 10 10 Calculate the solubility (in g·L−1) of CuBr ( s ) ( K sp = 6.3 × 10 − 9 ) in 0.76 M NH 3 ( aq ) .
Answer:
53.18 gL⁻¹
Explanation:
Given that:
[tex]Cu^{2+}_{(aq)}[/tex] [tex]+[/tex] [tex]2NH_{3(aq)}[/tex] [tex]------>[/tex] [tex][Cu(NH_3)_2]^+_{(aq)}[/tex] ------equation (1)
where;
Formation Constant [tex](k_f) =[/tex] [tex]6.3*10^{10}[/tex]
However, the Dissociation of [tex]CuBr_{(s)[/tex] yields:
[tex]CuBr_{(s)}[/tex] ⇄ [tex]Cu^{+}_{(aq)}[/tex] [tex]+[/tex] [tex]Br^-_{(aq)}[/tex] -------------- equation (2)
where;
the Solubility Constant [tex](k_{sp})[/tex] [tex]= 6.3 *10^{-9[/tex]
From equation (1);
[tex](k_f) =[/tex] [tex]\frac{[[Cu(NH_3)_2]^+]}{[Cu^{2+}][NH_3]^{2}}[/tex] --------- equation (3)
From equation (2)
[tex](k_{sp})[/tex] [tex]= [Cu^+][Br^-][/tex] --------- equation (4)
In [tex]NH_3[/tex], the net reaction for [tex]CuBr_{(s)[/tex] can be illustrated as:
[tex]CuBr_{(s)[/tex] [tex]+[/tex] [tex]2NH_{3(aq)}[/tex] ⇄ [tex][Cu(NH_3)_2]^+_{(aq)}[/tex] [tex]+ Br^-_{(aq)}[/tex]
The equilibrium constant (K) can be written as :
[tex]K=[/tex][tex]\frac{[[Cu(NH_3)_2]^+][Br^-]}{[NH_3]^2}[/tex]
If we multiply both the numerator and the denominator with [tex][Cu^+][/tex] ; we have:
[tex]K=[/tex][tex]\frac{[[Cu(NH_3)_2]^+][Br^-]}{[NH_3]^2}*\frac{[Cu^+]}{[Cu^+]}[/tex]
[tex]K=[/tex][tex]\frac{[[Cu(NH_3)_2]^+}{[NH_3]^2[Cu^+]}*{[Cu^+][Br^-]}[/tex]
[tex]K = k_f *k_{sp}[/tex]
[tex]K= (6.3*10^{10})*(6.3*10^{-9})[/tex]
[tex]K= 3.97*10^2[/tex]
[tex]K[/tex] ≅ [tex]4.0*10^2[/tex]
Now; we can re-write our equilibrium constant again as:
[tex]K=[/tex][tex]\frac{[[Cu(NH_3)_2]^+][Br^-]}{[NH_3]^2}[/tex]
[tex]4.0*10^2 = \frac{(x)(x)}{(0.76-2x)^2}[/tex]
[tex]4.0*10^2 = \frac{(x)^2}{(0.76-2x)^2}[/tex]
[tex]4.0*10^2 = (\frac{(x)}{(0.76-2x)})^2[/tex]
By finding the square of both sides, we have
[tex]\sqrt {4.0*10^2} = \sqrt {(\frac{(x)}{(0.76-2x)})^2[/tex]
[tex]2.0*10 = \frac{x}{(0.76-2x)}[/tex]
[tex]20(0.76-2x) =x[/tex]
[tex]15.2 -40x=x[/tex]
[tex]15.2 = 40x +x[/tex]
[tex]15.2 = 41x[/tex]
[tex]x = \frac{15.2}{41}[/tex]
[tex]x = 0.3707 M[/tex]
In gL⁻¹; the solubility of [tex]CuBr_{(s)[/tex] in 0.76 M [tex]NH_3[/tex] solution will be:
[tex]= \frac{0.3707 mole of CuBr}{1L}*\frac{143.45 g}{mole of CuBr}[/tex]
= 53.18 gL⁻¹
The problem involves finding the solubility of copper(I) bromide in a solution of ammonia. The solution entails expressing the formation of complex ions and dissolution of CuBr in terms of equilibrium expressions. Solving the system of expressions for solubility 's' can eventually give the solubility in g·L-1.
Explanation:In this problem, we are looking for the solubility of copper(I) bromide (CuBr) in a solution with 0.76 M NH3. Since CuBr does not have high solubility in water, we are using ammonia which forms complex ions with copper, enhancing its solubility. We will utilize the given equilibrium constant (Kf) which helps us to understand how far the forward reaction (complex ion formation) proceeds.
To start, we'll write the reaction of Cu+ ion with NH3 and the corresponding Kf expression:
Cu+ (aq) + 2NH3 (aq) <=> Cu(NH3)2+ (aq); Kf = [Cu(NH3)2+] / [Cu+] [NH3]^2
Next, we need to find an expression for [Cu+] in terms of solubility and solve for solubility. Let's denote the solubility of CuBr as 's'. The dissolution of CuBr can be represented as:
CuBr (s) <=> Cu+ (aq) + Br- (aq)
We can see that for every mole of CuBr that dissolves, one mole of Cu+ ion is produced. Thus, [Cu+] = s.
Now using the Kf expression, we can solve the system of equations to find s, which effectively gives us the solubility of CuBr in 0.76 M NH3. After that, by taking the molecular weight of CuBr into account, we can finally express the solubility in g·L−1.
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The first-order rate constant for the reaction of methyl chloride (CH3Cl) with water to produce methanol (CH3OH) and hydrochloric acid (HCl) is 3.32 × 10−10 s−1 at 25°C. Calculate the rate constant at 48.5°C if the activation energy is 116 kJ/mol.
Answer:
K(48.5°C) = 1.017 E-8 s-1
Explanation:
CH3Cl + H2O → CH3OH + HClat T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1
at T2 = 48.5°C (321.5 K) ⇒ K2 = ?
Arrhenius eq:
K(T) = A e∧(-Ea/RT)Ln K = Ln(A) - [(Ea/R)(1/T)]∴ A: frecuency factor
∴ R = 8.314 E-3 KJ/K.mol
⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)
⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)
(1)/(2):
⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)
⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)
⇒ Ln (K1/K2) = (13952.37 K)*(- 2.453 E-4 K-1)
⇒ Ln (K1/K2) = - 3.422
⇒ K1/K2 = e∧(-3.422)
⇒ (3.32 E-10 s-1)/K2 = 0.0326
⇒ K2 = (3.32 E-10 s-1)/0.0326
⇒ K2 = 1.017 E-8 s-1
The rate constant at 48.5°C if the activation energy is 116 kJ/mol is [tex]1.017 E-8 s^{-1}[/tex]
Chemical reaction:CH₃Cl + H₂O → CH₃OH + HCl
At T₁ = 25°C (298 K) ⇒ K₁ = 3.32 E-10 s-1
At T₂ = 48.5°C (321.5 K) ⇒ K₂ = ?
According to Arrhenius equation:
[tex]K(T) = A*e^{(-Ea/RT)}\\\\ln K = ln(A) - [(Ea/R)(1/T)][/tex]
where,
A = frequency factor
R = 8.314 E-3 KJ/K.mol
[tex]ln K_1 = ln(A) - [Ea/R)*(1/T_1)][/tex]..........(1)
[tex]ln K_2 = ln(A) - [(Ea/R)*(1/T_2)][/tex]...........(2)
On dividing equation 1 by 2:
ln (K₁/K₂) = (Ea/R)* (1/T₂ - 1/T₁)
ln (K₁/K₂) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)
ln (K₁/K₂) = (13952.37 K)*(- 2.453 E-4 K-1)
ln (K₁/K₂) = - 3.422
(K₁/K₂)= [tex]e^{(-3.422)}[/tex]
[tex](3.32 E-10 s^{-1})[/tex] / K₂ = 0.0326
K₂ = [tex](3.32 E-10 s^{-1})[/tex] /0.0326
K₂ = [tex]1.017 E-8 s^{-1}[/tex]
Thus, the rate constant at 48.5°C if the activation energy is 116 kJ/mol is 1[tex]0.017 E-8 s^{-1}[/tex]
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An aqueous solution of calcium hydroxide is standardized by titration with a 0.164 M solution of hydroiodic acid. If 17.2 mL of base are required to neutralize 14.7 mL of the acid, what is the molarity of the calcium hydroxide solution?
Answer:
0.07 M
Explanation:
We have to start with the reaction between the Calcium hydroxide and the Hydroiodic acid:
[tex]Ca(OH)_2~+~HF~->~CaF_2~+~H_2O[/tex]
Then we have to balance the reaction:
[tex]Ca(OH)_2~+~2HF~->~CaF_2~+~2H_2O[/tex]
We have to keep in mind that we have to do use the volume in "L" units (14.7 mL= 0.0147L). When we plug the values into the equation we will get:
[tex]M=\frac{#mol}{L}[/tex]
[tex]0.164~M=\frac{#mol}{0.0147}[/tex]
[tex]#mol=~0.0024[/tex]
Now, we have to use the information from the balanced reaction to finding the moles of Calcium hydroxide. When we check the reaction we found a molar ratio of 1:2 ( 1 mol of [tex]Ca(OH)_2[/tex]: 2 mol of [tex]HF[/tex]). With this molar ratio, we can find the moles of [tex]HF[/tex].
[tex]0.0024~mol~HF\frac{1~mol~Ca(OH)_2}{2~mol~HF}[/tex]
[tex]0.0012~mol~Ca(OH)_2[/tex]
Finally, to find the molarity we have to divide the moles of [tex]Ca(OH)_2[/tex] and the volume of [tex]Ca(OH)_2[/tex] in liters (17.2 mL=0.0172L) so:
[tex]M=\frac{0.012}{0.0172}=0.07[/tex]
The molarity of [tex]Ca(OH)_2[/tex] is 0.07M.
How many moles of sodium acetate must be added to 2.0 L of 0.10 M acetic acid to give a solution that has a pH equal to 5.00? Ignore the volume change due to the addition of sodium acetate.
Here is the full question
How many moles of sodium acetate must be added to 2.0 L of 0.10 M acetic acid to give a solution that has a pH equal to 5.00? Ignore the volume change due to the addition of sodium acetate. Ka of acetic acid is 1.7×10-5.
Answer:
0.3396 moles
Explanation:
pH = pKA + log[tex]\frac{[acetate]}{[aceticacid]}[/tex]
[acetic acid] = 0.1 M
pH = 5.0
[tex]K_a[/tex] = [tex]1.7*10^{-5[/tex]
5.0 = -log ([tex]1.7*10^{-5[/tex]) + log [tex]\frac{[acetate]}{0.1}[/tex]
5.0 = 4.77 + log [acetate] + 1
5.0 = 5.77 + log [acetate]
- log [acetate] = 5.77 - 5.0
- log [acetate] = 0.77
log [acetate] = -0.77
[acetate] = log⁻¹ (-0.77)
[acetate] = 0.1698
∴ for 2L = 2 × 0.1698
= 0.3396 moles
The sodium acetate must be added to 2.0 L of 0.10 M acetic acid to give a solution that has a pH equal to 5.00 - 0.3396 moles
Given:
The concentration of acetic acid = 0.10 M
pH = 5.0
[tex]K_a[/tex] = [tex]1.7\times10^{-5[/tex]
Solution:
According to Hasselbalch equation:
[tex]pH=pKa + log\frac{[A^-]}{[HA]}\\\\or\\\\pH = pKa + log\frac{(acetate)}{(acetic acid)}[/tex]
5.0 = -log [tex](1.7\times10^{-5})[/tex] + log [tex]\frac{(acetate)}{(0.1)}[/tex]
5.0 = 4.77 + log [acetate] + 1
5.0 = 5.77 + log [acetate]
- log [acetate] = 5.77 - 5.0
- log [acetate] = 0.77
log [acetate] = -0.77
[acetate] = log⁻¹ (-0.77)
[acetate] = 0.1698
∴ for 2L = 2 × 0.1698
= 0.3396 moles
Thus, the sodium acetate must be added to 2.0 L of 0.10 M acetic acid to give a solution that has a pH equal to 5.00 - 0.3396 moles
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Argon, which comprises almost 1 percent of the atmosphere, is approximately 27 times more abundant than CO2, but does not contribute to global warming. Which explanation accounts for this fact
Explanation:
Argon, an element comprises of almost 1 percent in the Earth’s atmosphere, is approximately 27 times more abundant than CO2, is not a greenhouse gas. The reason that it is not a greenhouse gas or not contribute in global warming is that it is largely transparent to the wavelengths of light that is responsible for trapping heat that is similar to that of oxygen, nitrogen and other gasesArgon is not able to form large and complex molecules that are enough to block infrared light, which is generally done by greenhouse gases such as carbon dioxide and methane also.N2O5 decomposes to form NO2 and O2 with first-order kinetics. How long does it take for the N2O5 concentration to decrease from its initial value of 2.75 M to its final value of 1.85 M, if the rate constant, k, equals 5.89 × 10−3?
Answer:
67.3 s
Explanation:
The equation for the reaction can be represented as:
N₂O₅ ⇄ NO₂ + O₂
Rate (k) = [tex]5.89 * 10^{-3[/tex]
Rate law for first order is expressed as:
In [A] = -kt + In [A]₀
Given that:
[A] = Final Concentration = 1.85 M
[A]₀ = Initial Concentration = 2.75 M
time-taken = ???
substituting our given data; we have:
In[1.85] = -[5.89 × 10⁻³](t) + In [2.75]
t = [tex]\frac{In(2.75)-In(1.85)}{(5.89*10^{-3}}[/tex]
t = [tex]\frac{0.3964}{5.89*10^{-3}}[/tex]
t = 67.3 s
Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.
Here is the full question:
Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.
What is the concentration at 10 minutes? (Round your answer to three decimal places.
Answer:
0.046 %
Explanation:
The rate-in;
[tex]R_{in}[/tex] [tex]= \frac{0.04}{100}*2000[/tex]
[tex]R_{in}[/tex] = 0.8
The rate-out
[tex]R_{out}[/tex] = [tex]\frac{A}{6000}*2000[/tex]
[tex]R_{out}[/tex] = [tex]\frac{A}{3}[/tex]
We can say that:
[tex]\frac{dA}{dt}=[/tex] [tex]0.8-\frac{A}{3}[/tex]
where;
A(0)= 0.2% × 6000
A(0)= 0.002 × 6000
A(0)= 12
[tex]\frac{dA}{dt} +\frac{A}{3} =0.8[/tex]
Integration of the above linear equation =
[tex]e^{\int\limits \frac {1}{3}dt } =[/tex] [tex]e^{\frac{1}{3}t[/tex]
so we have:
[tex]e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A[/tex] [tex]= 0.8e^{\frac{1}{3}t[/tex]
[tex]\frac{d}{dt}[e^{\frac{1}{3}t}A][/tex] [tex]= 0.8e^{\frac{1}{3}t[/tex]
[tex]Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C[/tex]
∴ [tex]A(t) = 2.4 +Ce^{-\frac{1}{3}t[/tex]
Since A(0) = 12
Then;
[tex]12 =2.4 + Ce^{-\frac{1}{3}}(0)[/tex]
[tex]C= 12-2.4[/tex]
[tex]C =9.6[/tex]
Hence;
[tex]A(t) = 2.4 +9.6e^{-\frac{t}{3}}[/tex]
[tex]A(0) = 2.4 +9.6e^{-\frac{10}{3}}[/tex]
[tex]A(t) = 2.74[/tex]
∴ the concentration at 10 minutes is ;
= [tex]\frac{2.74}{6000}*100[/tex]%
= 0.0456667 %
= 0.046% to three decimal places
A total of 2.00 molmol of a compound is allowed to react with water in a foam coffee cup and the reaction produces 137 gg of solution. The reaction caused the temperature of the solution to rise from 21.00 to 24.70 ∘C∘C. What is the enthalpy of this reaction? Assume that no heat is lost to the surroundings or to the coffee cup itself and that the specific heat of the solution is the same as that of pure water.
Answer : The enthalpy of this reaction is, 1.06 kJ/mol
Explanation :
First we have to calculate the heat produced.
[tex]q=m\times c\times (T_2-T_1)[/tex]
where,
q = heat produced = ?
m = mass of solution = 137 g
c = specific heat capacity of water = [tex]4.18J/g^oC[/tex]
[tex]T_1[/tex] = initial temperature = [tex]21.00^oC[/tex]
[tex]T_2[/tex] = final temperature = [tex]24.70^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]q=137g\times 4.18J/g^oC\times (24.70-21.00)^oC[/tex]
[tex]q=2118.842J=2.12kJ[/tex]
Now we have to calculate the enthalpy of this reaction.
[tex]\Delta H=\frac{q}{n}[/tex]
where,
[tex]\Delta H[/tex] = enthalpy change = ?
q = heat released = 2.12 kJ
n = moles of compound = 2.00 mol
Now put all the given values in the above formula, we get:
[tex]\Delta H=\frac{2.12kJ}{2.00mole}[/tex]
[tex]\Delta H=1.06kJ/mol[/tex]
Thus, the enthalpy of this reaction is, 1.06 kJ/mol
In order to get high-quality energy, you must use high-quality energy. This rule is the result of which of the following? a.the law of gravityb.the universal law of energyc.the second law of thermodynamicsd.the law of diminishing returnse.the first law of thermodynamics
Answer: C. The Second Law of Thermodynamics.
Explanation:
The law of gravity is related to the force interaction due to gravitation between two bodies (i.e. a person and planet Earth, a star and a planet).
The law of diminishing return is not used in Physics, but in Economics and describes the diminishing of marginal returns in time.
The first law of Thermodynamics analyses the energy interactions of a system in a quantitative manner and it is a generalized form of the Principle of Energy Conservation. It is oriented to the analyses of energy inflows and outflows.
The second law of Thermodynamics analyses the energy interactions of a system in a qualitative manner and it is based on thermodynamic property named Entropy, which may helpful to measure irreversibilities associated with system and irreversibility generation as well, provided that a comparison with another equivalent system exists.
Irreversibilities depends on the characteristics of system and nature of energy sources. Empirically, it is known that heat offers a lower quality than electricity in order to get a available work. That is to say, there is a lower of obtaining available work from heat than from electricity.
Hence, the statement is a consequence of using the second law of Thermodynamics and, therefore, the correct answer is C.
A sample of ethane (C2H6)was combusted completely and the water that formed has a mass of 1.61 grams.How much ethane, in grams, was in the sample?Put your answer in thespace provided belo socratic.org
Answer:
0.89 g of ethane
Explanation:
The balanced reaction equation is
C2H6(g) + 7/2 O2(g) -----------> 2CO2(g) + 3H2O(g)
From this balanced reaction
30g of ethane yields 54g of water
Therefore mass of ethane necessary to obtain 1.61g of water we have:
30 × 1.61/54 = 0.89 g of ethane
Calculate the molecular mass or formula mass (in amu) of each of the following substances: (a) BrN3 amu (b) C2H6 amu (c) NF2 amu (d) Al2S3 amu (e) Fe(NO3)3 amu (f) Mg3N2 amu (g) (NH4)2CO3 amu
Answer:
Explanation:
The atomic unit of mass (a.m.u) is the unit used to compare the relative masses of atoms. An atomic mass unit is a twelfth of the mass of a carbon-12 atom.
The calculation of the molecular mass of a compound is done by adding the relative atomic masses of the atoms of the formula of said substance, taking into account its abundance in the compound. The atomic mass of the atoms that make up the molecule are obtained in the Periodic Table.
So, in these cases you have:
BrN₃:You know the atomic masses of the elements that make up the molecule, obtained from the periodic table:
Br: 79.9 amu
N: 14 amu
I take into account the abundance of each element in the compound you get:
BrN₃=79.9 amu + 3* 14 amu= 121.9 amu
C₂H₆:You know the atomic masses of the elements that make up the molecule, obtained from the periodic table:
C: 12 amu
H: 1 amu
I take into account the abundance of each element in the compound you get:
C₂H₆= 2*12 amu + 6*1 amu= 30 amu
NF₂:You know the atomic masses of the elements that make up the molecule, obtained from the periodic table:
N: 14 amu
F: 19 amu
I take into account the abundance of each element in the compound you get:
NF₂= 14 amu + 2*19 amu= 52 amu
Al₂S₃:You know the atomic masses of the elements that make up the molecule, obtained from the periodic table:
Al: 27 amu
S: 32 amu
I take into account the abundance of each element in the compound you get:
Al₂S₃= 2*27 amu + 3*32 amu= 150 amu
Fe(NO₃)₃:You know the atomic masses of the elements that make up the molecule, obtained from the periodic table:
Fe: 55.85 amu
N: 14 amu
O: 16 amu
I take into account the abundance of each element in the compound you get:
Fe(NO₃)₃= 55.85 amu + 3*(14 amu + 3*16 amu)= 241.85 amu
Mg₃N₂:You know the atomic masses of the elements that make up the molecule, obtained from the periodic table:
N: 14 amu
Mg: 24 amu
I take into account the abundance of each element in the compound you get:
Mg₃N₂= 3*24 amu + 2*14 amu= 100 amu
(NH₄)₂CO₃:You know the atomic masses of the elements that make up the molecule, obtained from the periodic table:
N: 14 amu
H: 1 amu
C: 12 amu
O: 16 amu
I take into account the abundance of each element in the compound you get:
(NH₄)₂CO₃: 2*(14 amu + 4*1 amu) + 12 amu+3*16 amu= 96 amu
The molecular mass or formula mass of the listed substances have been calculated using atomic masses from the periodic table.
Explanation:The molecular mass or formula mass of a substance can be calculated by summing up the atomic masses of all atoms in the substance. Using the atomic masses from the periodic table, we have:
BrN3 = 1(Br) + 3(N) = 79.9 + 3(14.0) = 122.9 amu C2H6 = 2(C) + 6(H) = 2(12.0) + 6(1.0) = 30.0 amu NF2 = 1(N) + 2(F) = 14.0 + 2(19.0) = 52.0 amu Al2S3 = 2(Al) + 3(S) = 2(26.98) + 3(32.06) = 150.16 amu Fe(NO3)3 = 1(Fe) + 3(1(N) + 3(O)) = 55.85 + 3(1(14.0) + 3(16.0)) = 241.85 amu Mg3N2 = 3(Mg) + 2(N) = 3(24.3) + 2(14.0) = 100.9 amu (NH4)2CO3 = 2(1(N) + 4(H)) + 1(C) + 3(O) = 2(1(14.0) + 4(1.0)) + 12.0 + 3(16.0) = 96.0 amu Learn more about Molecular Mass Calculation here:https://brainly.com/question/33791791
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If the detergent requires using 0.61 kg detergent per load of laundry, determine what percentage (by mass) of the detergent should be sodium carbonate in order to completely precipitate all of the calcium and magnesium ions in an average load of laundry water.
Here is the full question.
Sodium carbonate is often added to laundry detergents to soften hard water and make the detergent more effective. Suppose that a particular detergent mixture is designed to soften hard water that is 3.8×10−3M in Ca2+ and 1.1×10−3M in Mg2+ and that the average capacity of a washing machine is 24.5 gallons of water. 1gallon=3.785L
If the detergent requires using 0.61 kg detergent per load of laundry, determine what percentage (by mass) of the detergent should be sodium carbonate in order to completely precipitate all of the calcium and magnesium ions in an average load of laundry water.
Answer:
7.90%
Explanation:
The equation for the reaction can be written as:
[tex]Na_2CO_3+Ca^{2+} ---------> CaCO_3(s) +2Na^+[/tex]
[tex]Na_2CO_3+Mg^{2+} ---------> MgCO_3(s) +2Na^+[/tex]
Molar mass of [tex]Na_2CO_3[/tex] = 106 g/mol
1.00 gallon of water = 3.785 L
∴ 24.5 gallons of water = 24.5 × 3.785
= 92.7325 L
mass precipitate = [tex][(3.8*10^{-3})+(1.1*10^{-3})]\frac{mol}{L} *92.7325*\frac{106g}{mol}[/tex]
mass precipitate = 48.162605
mass precipitate = 48.2 g
mass % = [tex]\frac{mass ofprecipitate}{mass of detergent} *100%[/tex]%
mass % = [tex]\frac{48.2g}{0.61kg}*\frac{1kg}{1000g}*100[/tex]%
mass % = 7.901639344
mass % = 7.90 %
The percentage (by mass) of the detergent should be 7.90%.
Calculation of the percentage:Since 3.8×10−3M in Ca2+ and 1.1×10−3M in Mg2+ and that the average capacity of a washing machine is 24.5 gallons of water. 1gallon=3.785L
The molar mass of Na_2CO_3 is 106g/mol
Since 1.00 gallon of water = 3.785 L
so, for 24.5 gallons of water, it is
= 24.5 × 3.785
= 92.7325 L
Now mass precipitate is
= ((3.8*10^-3) * (1.1*10^-3) * 92.7325 * 106g/mol
= 48.162605
= 48.2g
Now the mass percentage is
= mass of precipitate / mass of detergent
= 48.3g / 0.61 * 1kg / 1000g
= 7.90%
This is an incomplete question. Please find the full question below.
Sodium carbonate is often added to laundry detergents to soften hard water and make the detergent more effective. Suppose that a particular detergent mixture is designed to soften hard water that is 3.8×10−3M in Ca2+ and 1.1×10−3M in Mg2+ and that the average capacity of a washing machine is 24.5 gallons of water. 1gallon=3.785L
If the detergent requires using 0.61 kg detergent per load of laundry, determine what percentage (by mass) of the detergent should be sodium carbonate in order to completely precipitate all of the calcium and magnesium ions in an average load of laundry water.
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For the reaction 2 H2O(g) equilibrium reaction arrow 2 H2(g) + O2(g), K = 2.4 ✕ 10−3 at a given temperature. At equilibrium in a 2.7 L container, it is found that [H2O(g)] = 2.0 ✕ 10−1 M and [H2(g)] = 2.5 ✕ 10−2 M. Calculate the moles of O2(g) present under these conditions.
To calculate the moles of O2(g) present at equilibrium, use the equilibrium constant expression and the given concentrations of H2O(g) and H2(g). The moles of O2(g) present at equilibrium is approximately 7.23 × 10-3 mol.
Explanation:To calculate the moles of O2(g) present at equilibrium, we can use the equilibrium constant expression and the given concentrations of H2O(g) and H2(g). The equilibrium constant expression for the given reaction is K = [H2]^2/[H2O]. Since we have the values for [H2] and [H2O], we can rearrange the expression to solve for [O2], which is the unknown.
By substituting the known values into the expression and solving for [O2], we get [O2] ≈ 2.67 × 10-3 M. This represents the concentration of O2(g), which we can then convert to moles of O2 using the volume of the container.
Given that the container has a volume of 2.7 L, we can multiply the concentration of O2 by the volume to get the moles of O2(g). In this case, the moles of O2(g) present at equilibrium would be approximately 7.23 × 10-3 mol.
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To calculate the moles of O2(g) present under the given conditions, divide the concentration of H2(g) by 2 and use the ideal gas law to find the number of moles in the 2.7 L container. The number of moles of O2(g) in the 2.7 L container is equal to the number of moles in any other volume multiplied by the ratio of the volumes.
Explanation:To calculate the moles of O2(g) present under the given conditions, we can first start by calculating the concentration of O2(g) using the given equilibrium concentrations of H2O(g) and H2(g). Since the reaction is 2 H2O(g) <=> 2 H2(g) + O2(g), the concentration of O2(g) is equal to half the concentration of H2(g). Therefore, the concentration of O2(g) is 2.5 x 10^-2 M / 2 = 1.25 x 10^-2 M.
Next, we can use the ideal gas law to calculate the number of moles of O2(g) in the 2.7 L container. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin. In this case, we don't have the pressure or the temperature, so we can't directly calculate the number of moles. However, since the pressure and temperature are not changing, we can assume that the pressure and temperature are constant, and therefore the number of moles is directly proportional to the volume. So, if we know the number of moles of O2(g) in the 2.7 L container, we can calculate the number of moles in any other volume using the equation: n1V1 = n2V2.
Therefore, if we let n1 be the number of moles of O2(g) in the 2.7 L container and n2 be the number of moles of O2(g) in another volume, we can set up the equation: n1 x 2.7 = n2 x V2. Substituting the values, we get: n1 x 2.7 = 1.25 x 10^-2 x V2. Solving for n2, we get: n2 = n1 x 2.7 / (1.25 x 10^-2). Plugging in the value of n1 as 1.25 x 10^-2 M x 2.7 L, we can calculate n2.
Therefore, the moles of O2(g) present under the given conditions can be calculated as follows: n2 = (1.25 x 10^-2 M) x (2.7 L) / (1.25 x 10^-2) = 2.7 moles.
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