Answer:
156.4g K
Explanation:
I'm not sure if it is correct but I think it should be this
What do we know so far?: 2K + 1Cl2 -> 2KCl, 2 mol of Cl2
What are we looking for?: #g of K
What is the ratio of K to Cl2?: 2:1
Set up equation: 2molCl2 x \(\frac{2mol K}{1 mol Cl2}\)
Cancel unwanted units: 2 x \(\frac{2mol K}{1}\)
Answer we got: 2 x 2mol K = 4mol K
Converting moles to grams: 4 x 39.1 (molar mass of K) = 156.4g K
Concentration (mol dm-³) 0.5- 0.4- 0.3- 0.2- 0.1 2. 3 5 The following equilibrium reaction is given: 2HI(g) = H₂(g) + I₂(g) Time (s) H₂/ HI Cy A change in pressure will not affect equilibrium in this case as the number of moles of gas is the same on both sides of the equation. AH> 0 A graph plotting the concentrations of the substances present versus time is given in Figure 7.10. a) b) Explain the physical situation in the container from t=0 s to t = 5 s. Which external factor was altered in order to bring about a change in the shape of the graph at t = 5 s? Explain. Calculate Kat t = 3 s. 1 dm³ COCI, decomposes
Based on the information provided, we have a reaction between hydrogen iodide (HI) gas and hydrogen gas (H₂) to form iodine gas (I₂). The equilibrium is represented by the equation:
2HI(g) = H₂(g) + I₂(g)
The concentration values given in the table correspond to the concentrations of H₂ and HI at different times.
a) From t=0 s to t=5 s: Without the specific graph mentioned in Figure 7.10, it is difficult to provide a precise explanation of the physical situation in the container during this time period. However, based on the equilibrium reaction given, we can make some general observations. At the start (t=0 s), the concentrations of H₂ and HI may be high. As time progresses, the reaction proceeds, and the concentrations of H₂ and HI may decrease while the concentration of I₂ increases. The specific behavior will depend on the rate of the forward and reverse reactions.
b) External factor altered at t=5 s: To bring about a change in the shape of the graph at t=5 s, some external factor must have been altered. The most likely factor is the total pressure within the container. Since the reaction involves gases, changes in pressure can affect the equilibrium position. However, according to the information given, a change in pressure will not affect equilibrium in this case since the number of moles of gas is the same on both sides of the equation. Therefore, if the shape of the graph changes at t=5 s, some other external factor, such as temperature or the addition of a catalyst, must have been altered.
c) Calculation of K at t=3 s: The equilibrium constant (K) can be calculated at any given time using the concentrations of the reactants and products. However, the concentrations of H₂ and HI at t=3 s are not provided in the information given. Without the necessary data, it is not possible to calculate K at t=3 s.
Lastly, the statement "1 dm³ COCI, decomposes" seems incomplete. If you provide additional information or clarify the question, I'll be happy to assist you further.
Repeated trials and replication are the same thing.
Group of answer choices
True
False
Answer:
false
Explanation:
Plz help ASAP
What is the molarity of a 0.6 L solution that contains 50.0 grams of sodium hydroxide(NaOH)?
A) 2.1 M
B) 2.9 M
C) 3.3 M
Answer: A 2.1
Explanation:
benzen has a boiling point of 80.10 c we know the change in boiling point for a solution of c6h14 in benzen is 2.25 what is the new boiling point for the solution
bp= ? c
The pressure of the environment affects the liquid's boiling point. The boiling point of the liquid is higher when it is under high pressure than when it is under normal atmospheric pressure. For a given pressure, various liquids have different boiling points.
The temperature at which a liquid's vapour pressure equals the surrounding atmosphere is known as the boiling point of the liquid. This temperature causes the liquid to become a vapour.
The temperature of the liquid, the pressure of the atmosphere, and the pressure of the vapour all affect its boiling point.
We know that change in temperature of a system is given by the following formula:
Initial boiling point (T₁) = 80.1 °C
ΔT = T₂ - T₁
2.25 = T₂ - 80.1
T₂ = 82.35 °C
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Answer:
82.35
Explanation:
acellus
Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce ammonia.
N2(g)+3H2(g)⟶2NH3(g)
There are four molecules of nitrogen and nine molecules of hydrogen present in the diagram.
When the reaction is complete, how many molecules of NH3 are produced?
What is the limiting reactant?
How many molecules of each reactant are remain after the reaction is complete?
After the reaction is complete, no nitrogen and no hydrogen molecules remain, and 8.00 x 1014 molecules of NH3 are produced.
In the equation, nitrogen and hydrogen react at a high temperature, in the presence of a catalyst, to produce ammonia, according to the balanced chemical equation:N2(g)+3H2(g)⟶2NH3(g)The coefficients of each molecule suggest that one molecule of nitrogen reacts with three molecules of hydrogen to create two molecules of ammonia.
So, to determine how many molecules of ammonia are produced when four nitrogen and nine hydrogen molecules are present, we must first determine which of the two reactants is the limiting reactant.
To find the limiting reactant, the number of moles of each reactant present in the equation must be determined.
Calculations:
Nitrogen (N2) molecules = 4Hence, the number of moles of N2 = 4/6.02 x 1023 mol-1 = 6.64 x 10-24 mol
Hydrogen (H2) molecules = 9Hence, the number of moles of H2 = 9/6.02 x 1023 mol-1 = 1.50 x 10-23 mol
Now we have to calculate the number of moles of NH3 produced when the number of moles of nitrogen and hydrogen are known, i.e., mole ratio of N2 and H2 is 1:3.
The mole ratio of N2 to NH3 is 1:2; thus, for every 1 mole of N2 consumed, 2 moles of NH3 are produced.
The mole ratio of H2 to NH3 is 3:2; thus, for every 3 moles of H2 consumed, 2 moles of NH3 are produced.
From these mole ratios, it can be observed that the limiting reactant is nitrogen.
Calculation for NH3 production:
Nitrogen (N2) moles = 6.64 x 10-24 moles
The mole ratio of N2 to NH3 is 1:2; therefore, moles of NH3 produced is 2 × 6.64 × 10−24 = 1.33 × 10−23 moles.
Now, to determine how many molecules of NH3 are produced, we need to convert moles to molecules.
1 mole = 6.02 x 1023 molecules
Thus, 1.33 x 10-23 moles of NH3 = 8.00 x 1014 molecules of NH3 produced.
To find the amount of each reactant remaining after the reaction is complete, we must first determine how many moles of nitrogen are consumed, then how many moles of hydrogen are consumed, and then subtract these from the initial number of moles of each reactant.
The moles of nitrogen consumed = 4 moles × 1 mole/1 mole N2 × 2 mole NH3/1 mole N2 = 8 moles NH3
The moles of hydrogen consumed = 9 moles × 2 mole NH3/3 mole H2 × 2 mole NH3/1 mole N2 = 4 moles NH3
Thus, the moles of nitrogen remaining = 6.64 × 10−24 mol – 8 × 2/3 × 6.02 × 10^23 mol-1 = 5.06 × 10−24 mol
The moles of hydrogen remaining = 1.50 × 10−23 mol – 4 × 2/3 × 6.02 × 10^23 mol-1 = 8.77 × 10−24 mol
Finally, the number of molecules of each reactant remaining can be calculated as follows:
Number of N2 molecules remaining = 5.06 × 10−24 mol × 6.02 × 10^23 molecules/mol = 3.05 × 10−1 molecules ≈ 0 molecules
Number of H2 molecules remaining = 8.77 × 10−24 mol × 6.02 × 10^23 molecules/mol = 5.28 × 10−1 molecules ≈ 0 molecules.
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What property is not used to separate a mixture ?
Answer:
Conductivity
Explanation:
How many significant digits are in this number: 0.0030670?
Answer:
Answer:
Number of Significant Figures: 5
The Significant Figures are 3 0 6 7 0
Explanation:
hope this helps
which atom has 20protons and 18 electrons?!
Explanation:
The atomic number 20 Is the element calcium (Ca). The electron deficit of 2 indicates that it has a positive charge.
Hope this helps
Sherman suggests that reproduction always creates individuals with adaptive traits. Does this seem correct? Why or why not?
If Sherman suggests that reproduction always creates individuals with adaptive traits, then he/she is not correct because variation may lead to non-adaptive phenotypes.
What is the presence of genetic variation in individuals of a given population?The presence of genetic variation in individuals of a given population is not adaptive per se, but instead, it provides the raw material for natural selection that leads to the differential survival and reproduction of the most adaptive phenotypes.
The genetic variation in individuals of a given population does not crate adaptive traits and it may derive from sexual reproduction.
Therefore, with this data, we can see that the presence of genetic variation in individuals of a given population may or not be adaptive in the function of the environment in which the individual produced by means of sexual reproduction is developed and therefore it is an advantage for the population but the individual may be harmful.
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Consider the complete reaction of 63.2 g of HCl with excess oxygen gas. What mass of chlorine gas product forms? 4 HCl(g) + O2(g) → 2 H2O(l) + 2 Cl2(g)
The mass of chlorine gas product formed is 61.4 g.
What is the mass ?
First, we need to determine which reactant is limiting and which is in excess. To do this, we can use the given mass of HCl and convert it to moles:
63.2 g HCl × (1 mol HCl/36.46 g HCl) = 1.73 mol HCl
Next, we can use stoichiometry to determine how many moles of \(Cl_{2}\) will be produced from 1.73 mol of HCl:
1.73 mol HCl × (2 mol \(Cl_{2}\)/4 mol HCl) = 0.865 mol \(Cl_{2}\)
Finally, we can convert the moles of \(Cl_{2}\) to grams using its molar mass:
0.865 mol \(Cl_{2}\) × (70.91 g \(Cl_{2}\)/1 mol \(Cl_{2}\)) = 61.4 g \(Cl_{2}\)
Therefore, the mass of chlorine gas product formed is 61.4 g.
What is stoichiometry ?
Stoichiometry is a branch of chemistry that deals with the calculation of the quantities of reactants and products involved in a chemical reaction. It involves using balanced chemical equations to calculate the amounts of reactants required to produce a certain amount of product, or the amount of product that can be obtained from a given amount of reactant. Stoichiometry is an important tool for predicting the outcomes of chemical reactions and for designing chemical processes in industry.
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How does the thermochemical equation S (s) + O2 (g) → SO2 (g), ΔH = –296 kJ, need to be modified to be used as a portion of the data needed to find the change in enthalpy for the chemical reaction 2SO2 (g) + O2 (g) → 2SO3 (g)?
Answer:2C(s) + H2(g) ---> C2H2(g) ΔH° = ??? kJ
Given the following thermochemical equations:
C2H2(g) + 5⁄2O2(g) ---> 2CO2(g) + H2O(ℓ) ΔH° = −1299.5 kJ
C(s) + O2(g) ---> CO2(g) ΔH° = −393.5 kJ
H2(g) + 1⁄2O2(g) ---> H2O(ℓ) ΔH° = −285.8 kJ
Solution:
1) Determine what we must do to the three given equations to get our target equation:
a) first eq: flip it so as to put C2H2 on the product side
b) second eq: multiply it by two to get 2C
c) third eq: do nothing. We need one H2 on the reactant side and that's what we have.
2) Rewrite all three equations with changes applied:
2CO2(g) + H2O(ℓ) ---> C2H2(g) + 5⁄2O2(g) ΔH° = +1299.5 kJ
2C(s) + 2O2(g) ---> 2CO2(g) ΔH° = −787 kJ
H2(g) + 1⁄2O2(g) ---> H2O(ℓ) ΔH° = −285.8 kJ
Notice that the ΔH values changed as well.
3) Examine what cancels:
2CO2 ⇒ first & second equation
H2O ⇒ first & third equation
5⁄2O2 ⇒ first & sum of second and third equation
4) Add up ΔH values for our answer:
+1299.5 kJ + (−787 kJ) + (−285.8 kJ) = +226.7 kJ
Explanation:
What is the final temperature after 840 Joules is absorbed by 10.0g of water at 25.0
C?
The final temperature of the water is: T_final = 45.0°C
We can use the formula for the specific heat capacity of the water to solve this problem:
q = mcΔT
First, we can calculate the initial energy of the water:
q = mcΔT
q = (10.0 g) (4.184 J/g°C) (25.0°C)
q = 1,046 J
Next, we can calculate the final temperature after absorbing 840 J:
q = mcΔT
840 J = (10.0 g) (4.184 J/g°C) (ΔT)
ΔT = 20.0°C
Therefore, the final temperature of the water is:
T_final = T_initial + ΔT
T_final = 25.0°C + 20.0°C
T_final = 45.0°C
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Suppose that a substance in a beaker is heated over a burner in a science lab. Which observation would most likely indicate that a chemical change has occurred in the substance?
Answer:
not sure if it's right because there are no choices but most likely a chemical change
Glucose, C6H12O6, is used to prepare intravenous feeding solutions. What volume of 5.0 % W/V glucose solution can be prepared using 125 g of glucose? Show your working.
Please if the answer is correct, ill give brainliest
250 L of 5.0% w/v glucose solution can be prepared using 125 g of glucose.
We use the below formula to solve our problem,w/v = [ mass of solute (g) / volume of solution (mL) ] × 100
Substitute the values from our problem,5.0 % w/v = [ 125 g / volume of solution (mL) ] × 100
Rearranging the formula, we havevolume of solution (mL) = [ 125 g / 5.0 % w/v ] x 100
Substitute further for w/v,volume of solution (mL) = [ 125 g / (5.0 / 100) ] x 100
Simplify the expression,volume of solution (mL) = [ 125 g / 0.05 ] x 100
Hence, the volume of solution (mL) = 250,000 mL or 250 LGeology
Please help!
The first image shows cleavage, same as the second. The third image shows fracture
What is shown in the images1. Image 1:
As it has clean surfaces and appears to have been sliced with a knife, it exhibits CLEAVAGE.Visible cleavage planes: 3There is not a 90-degree cleavage.Rhombohedral is the three-dimensional structure it symbolizes.2. Image 2
It exhibits CLEAVAGE.Visible cleavage planes: 1Since there is only one cleavage plane, cleavage angle is irrelevant.Sheet is the three-dimensional structure it symbolizes.Image 3:
It exhibits FRACTURE because it is fractured unevenly. In addition, the surfaces are not smooth.Inapplicable if there are obvious cleavage planesAngle of cleavage: inapplicableIt does not apply to the three-dimensional structure it portrays.Read more on rocks here:https://brainly.com/question/26046551
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Which of the following aqueous solutions are good buffer systems?
(Select all that apply.)
a. 0.22 M acetic acid + 0.15 M potassium acetate
b. 0.27 M nitric acid + 0.16 M potassium nitrate
c. 0.29 M ammonium nitrate + 0.32 M ammonia
d. 0.20 M calcium hydroxide + 0.23 M calcium chloride
e. 0.31 M hydrofluoric acid + 0.22 M potassium fluoride
The following aqueous solutions represents good buffer systems:
0.22 M acetic acid + 0.15 M potassium acetate0.29 M ammonium nitrate + 0.32 M ammonia What is a buffer?A buffer is a solution used to stabilize the pH (acidity) of a liquid.
A good buffer system is generally known to contain close or equal concentrations of a weak acid and its conjugate base.
Based on the above explanation, the following represents a good buffer system as they are between their weak acid and conjugate base:
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A beaker contains a total of 500 ml of solution which is 0.00050 M Ag^+, 0.00050 M Pb^2+, and 0.00050 M in Mn^2+ ions. If 10.00 ml of 1.0*10^-6 M Na2CO3 is added to the beaker, what will precipitate?
Ksp Ag2CO3 = 8.1*10^-12
Ksp PbCO3 = 7.4*10^-14
Ksp MnCO3 = 8.8*10^-11
Only Ag2CO3 will precipitate from the solution.
Precipitation reactionWhen Na2CO3 is added to the solution, it will react with the Ag^+ and Pb^2+ ions to form precipitates of Ag2CO3 and PbCO3. The Mn^2+ ion concentration is not high enough to form a precipitate with Na2CO3.
First, let's calculate the initial concentration of Ag^+ and Pb^2+ ions in the solution:
Ag^+: 0.00050 M
Pb^2+: 0.00050 M
Next, we need to calculate the concentration of Na2CO3 after it is added to the solution. Since we added 10.00 ml of 1.0*10^-6 M Na2CO3 to a total volume of 500 ml, the final concentration of Na2CO3 is:
[Na2CO3] = (10.00 ml / 500 ml) * 1.010^-6 M
[Na2CO3] = 2.010^-8 M
Now we can use the Ksp values to determine which precipitates will form.
For Ag2CO3:
Ksp = [Ag^+]^2[CO3^2-]
8.110^-12 = (2x)^2 (2x)
8.110^-12 = 4x^3
x = 2.0*10^-4 M
Since the concentration of CO3^2- is higher than the solubility product, Ag2CO3 will precipitate.
For PbCO3:
Ksp = [Pb^2+][CO3^2-]
7.410^-14 = (0.00050 M)(2x)
x = 9.210^-11 M
Since the concentration of CO3^2- is lower than the solubility product, PbCO3 will not precipitate.
Therefore, the only precipitate that will form is Ag2CO3.
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which is the graph of the function g(x) = f(-x)
To graph the function g(x) = f(-x), you can start with the graph of f(x) and then reflect it about the y-axis.
What is a graph of the function g(x) = f(-x)?To find the graph of the function g(x) = f(-x), we can start with the graph of the function f(x) and then reflect it about the y-axis.
If the graph of f(x) is symmetric with respect to the y-axis, meaning it is unchanged when reflected, then g(x) = f(-x) will have the same graph as f(x).
However, if the graph of f(x) is not symmetric with respect to the y-axis, then g(x) = f(-x) will be a reflection of f(x) about the y-axis.
In either case, the resulting graph of g(x) = f(-x) will be symmetric with respect to the y-axis.
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What happens to the molecules of a compound when it changes state from liquid to solid
Answer:
they lose energy. because of that, they slow down and get closer together
An average pumpkin weighs between 10 and 20 pounds. Those that win giant pumpkin contests weigh up to 1,000 pounds. Growers state that soil, fertilizer N-P-K (nitrogen-phosphorus-potassium) ratio, watering schedule, and fertilizer type are the main factors that lead to success in growing giant pumpkins. The table below shows an experimental design that compares three methods of growing a giant pumpkin. Which statement describes a solution for the error in the experiment? A. Manure should not be used. B. Fertilizer should not be used. C.A single fertilizer type should be used. D. Different types of seeds could be used.
The error in the experiment is that different types of seeds could be used. Option D
What is an error?We know that an experiment can be used to establish the relationship between the dependent and the independent variables. Now note that an experiment can be able to establish a cause and effect relationship.
Now In this case, we are trying to find out the factors that account for the size of the pumpkin. We have to note that we have to look at different types of seed in order to establish the validity of the experiment.
Hence, the error in the experiment is that different types of seeds could be used. Option D
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Answer:
D
Explanation:
Which is true regarding a water molecule?
Answer:
Has many answers, but one is that it consists of small polar v shaped molecules with a molecular formula H20.
Explanation:
Water molecules consists of 2 hydrogen atoms bonded with on oxygen atom. Each molecule is electrically neutral but polar, with the center of positive and negative charges located in different places.
Each hydrogen atom has a nucleus consisting of a single positively-charged proton surrounded by a 'cloud' of a single negatively-charged electron and the oxygen atom has a nucleus consisting of eight positively-charged protons and eight uncharged neutrons surrounded by a 'cloud' of eight negatively-charged electrons.
Hoped this helped!
A mixture of short branched
and unbranched
undigested
oligosaccharides is known as
Answer:monosaccharides
Explanation:
A solution containing 113. g of KCl in 270. g of H2O at 50 ∘C is cooled to 20 ∘C . It goes to 34 degrees Celsius when cooled down.
The solution of KCl in water will undergo precipitation of KCl crystals as the temperature decreases from 50 °C to 20 °C and then to 34 °C due to the decreasing solubility of KCl in water.
What happens to a solution of KCl in water when cooled from 50 °C to 20 °C and then to 34 °C?
When the solution of KCl in water is cooled from 50 °C to 20 °C, the solubility of KCl in water decreases. As a result, some of the KCl will start to come out of the solution and form solid crystals.
Then, when the temperature of the solution is further decreased from 20 °C to 34 °C, the solubility of KCl in water decreases even further, causing more KCl to come out of the solution and form solid crystals. This is known as precipitation.
The amount of KCl that will come out of the solution depends on the solubility of KCl at the respective temperatures. The solubility of KCl in water is about 34 g/100 mL at 20 °C and about 42 g/100 mL at 34 °C. Therefore, when the solution is cooled from 50 °C to 20 °C, some KCl will come out of solution until the concentration in solution reaches about 34 g/100 mL. Further cooling to 34 °C will cause more KCl to come out of solution until the concentration in solution reaches about 42 g/100 mL.
The exact amount of KCl that will come out of solution can be calculated using thermodynamic models and experimental data on the solubility of KCl in water at different temperatures.
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Any molecule within a cell that has a uracil group will also contain
A deoxyribose
B glycerol
C glucose
D ribose
Answer:
A
Explanation:
pretty sure..
Any molecule within a cell that has a uracil group will also contain a ribose.
What is a RNA?RNA refers to ribonucleic acid.
It is a nucleic acid that contains ribose sugar along with purine and pyrimidine bases.
The purine bases in RNA include Adenine and Guanine while the pyrimidine bases include Uracil and Cytosine.
This is unlike deoxyribonucleic acid (DNA) which contains deoxyribose sugar and Thymine instead of Uracil.
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What is the empirical formula for this crystalline material
The empirical formula for the given crystalline material is Cu₁₆O₂₀YBa₂
What is the empirical formula of a compound?The empirical formula of a compound is the simplest formula of a compound that shows the simplest ratio in which the elements are present in a compound.
The simplest whole-number ratio of atoms in a chemical molecule is its empirical formula. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S₂O₂, are two straightforward examples of this idea. Both have the same empirical formula of SO.
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2. What is the percent sulfur in iron (III) sulfate?
Step 1
The chemical formula of iron (III) sulfate is the next one:
Fe2(SO4)3
As we can see, there are 3 x 1 atom of S, 3 atoms of S
----------------
Step 2
Information needed:
The atomic masses of:
Fe) 55.8 g/mol
S) 32.0 g/mol
O) 16.0 g/mol
(Please, the periodic table is useful here)
----------------
Step 3
The % of S in Fe2(SO4)3 is calculated as follows:
\(\frac{3\text{ x 32.0 g/mol of S}}{(2\text{ x 55.8 g/mol Fe + 3 x 32.0 g/mol S + 3 x 4 x 16.0 g/mol O\rparen}}x100\text{ = 24.0 \%}\)Answer: 24 % of S
Calculate the molality of a 5.73 M ethanol (C2H5OH) solution whose density is 0.9327 g/mL.
Answer:
Molality = 8.57 m
Explanation:
Given data:
Molarity of solution = 5.73 M
density = 0.9327 g/mL
Molality of solution = ?
Solution:
Molality = moles of solute / kg of solvent.
Kg of solvent:
Mass of 1 L solution = density× volume
Mass of 1 L solution = 0.9327 g/mL × 1000 mL
Mass of 1 L solution = 932.7 g
Mass of solute:
Mass of 1 L = number of moles × molar mass
Mass = 5.73 mol × 46.068 g/mol
Mass = 263.97 g
Mass of solvent:
Mass of solvent = mass of solution - mass of solute
Mass of solvent = 932.7 g - 263.97 g
Mass of solvent = 668.73 g
In Kg = 668.73 /1000 = 0.6687 Kg
Molality:
Molality = number of moles of solute / mass of solvent in Kg
Molality = 5.73 mol / 0.6687 Kg
Molality = 8.57 m
Considering the definition of molality , you obtain that the molality of a 5.73 M ethanol (C₂H₅OH) solution whose density is 0.9327 g/mL is 8.57 \(\frac{moles}{kg}\).
Molality is the ratio of the number of moles of any dissolved solute to kilograms of solvent.
The Molality of a solution is determined by the expression:
\(Molality=\frac{number of moles of solute}{kilogramof solvent}\)
You have a 5.73 M ethanol (C₂H₅OH) solution whose density is 0.9327 g/mL.
Molarity is the number of moles of solute that are dissolved in a certain volume.
In this case, taking into account that the volume considered is 1 L, the number of moles of solute is 5.73 moles.
On the other side, density is the ratio of the weight (mass) of a substance to the volume it occupies. So, being 1 mL= 0.001 L, 0.9327 g/mL means that you have 0.9327 grams per 1 mL or 932.7 g per 1 L.
So, being the mass of solution calculated as number of moles multiplied by the molar mass, and being the mass of the solution 932.7 grams in 1 L, the mass of water is:
Mass of solvent = mass of solution - mass of solute
Mass of solvent = mass of solution - number of moles× molar mass
Mass of solvent = 932.7 g - 5.73 mol× 46.068 g/mol
Mass of solvent = 932.7 g - 263.97 g
Mass of solvent = 668.73 g= 0.66873 kg
Then, the molality can be calculated as:
\(Molality=\frac{5.73 moles}{0.66873 kg}\)
Solving:
molality= 8.57 \(\frac{moles}{kg}\)
Finally, the molality of a 5.73 M ethanol (C₂H₅OH) solution whose density is 0.9327 g/mL is 8.57 \(\frac{moles}{kg}\).
Learn more about:
density: brainly.com/question/952755?referrer=searchResults brainly.com/question/1462554?referrer=searchResults molalitybrainly.com/question/20366625?referrer=searchResults brainly.com/question/4580605?referrer=searchResults molarity with this example: brainly.com/question/15406534?referrer=searchResultsWhat is the mass of a 2.5 mole sample of MgO2
Why KHPo4 ignore effective as a buffer but kh2po4 is not
KH2PO4 is a more suitable choice as a buffer because it has a greater buffering capacity due to the presence of the weak acid and its conjugate base.
KHPo4 is not considered an effective buffer compared to KH2PO4 due to its limited buffering capacity. The effectiveness of a buffer is determined by the concentration and dissociation properties of its conjugate acid-base pair.
KH2PO4 is a salt composed of the weak acid H2PO4- and its conjugate base HPO4^2-. In an aqueous solution, KH2PO4 can dissociate to release H+ ions from the H2PO4- component, which acts as a weak acid, and the HPO4^2- component can accept H+ ions, acting as a weak base. This allows KH2PO4 to effectively resist changes in pH when small amounts of acid or base are added to the solution.
On the other hand, KHPo4 consists of the strong acid H3PO4 and the weak base HPO4^2-. H3PO4 fully dissociates in water, providing a large concentration of H+ ions, making it difficult for the HPO4^2- to effectively act as a base and maintain pH stability.
Therefore, KH2PO4 is a more suitable choice as a buffer because it has a greater buffering capacity due to the presence of the weak acid and its conjugate base.
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The rate of deforestation to create new buildings and farmland has increased over the last 100 years. What effect could this have on the atmosphere in the future?
The increase in deforestation can have a significant impact on the atmosphere in the future, mainly by reducing the number of trees that absorb carbon dioxide during photosynthesis.
Trees take in carbon dioxide and release oxygen as a byproduct, reducing the amount of carbon dioxide in the atmosphere. However, when trees are cut down, the carbon they have stored is released back into the atmosphere, contributing to increased levels of greenhouse gases.
The increase in greenhouse gases can lead to global warming and climate change, which can have a range of negative effects on the environment and human health, such as rising sea levels, increased frequency and severity of natural disasters, and more intense heat waves. Additionally, deforestation can lead to soil erosion, loss of biodiversity, and disruption of local ecosystems.
It is important to find ways to reduce deforestation and promote sustainable land use practices to mitigate the potential impacts on the atmosphere and the environment as a whole.
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