Explanation:
A solution comprises of a homogeneous mixture that means it is composed of one phase (e.g., solid, liquid, gas).Particles do not settle out of in the mixture of a solution and are not visible to our eyes.Components of a solution cannot be separated by using any simple processes like mechanical filtration. Therefore, the given statement is False.0.055 grams of PbSO4 is dissolved in 200.00 grams of H2O.
1. What is the mass of the total solution?
2. What is the concentration of the PbSO4 in ppm?
Answer:
Solution's mass = 200.055 g
[PbSO₄] = 275 ppm
Explanation:
Solute mass = 0.055 g of lead(II) sulfate
Solvent mass = 200 g of water
Solution mass = Solvent mass + Solution mass
0.055 g + 200 g = 200.055 g
ppm = μg of solute / g of solution
We convert the mass of solute from g to μg
0.055 g . 1×10⁶ μg/ 1g = 5.5×10⁴μg
5.5×10⁴μg / 200.055 g = 275 ppm
ppm can also be determined as mg of solute / kg of solution
It is important that the relation is 1×10⁻⁶
Let's verify: 0.055 g = 55 mg
200.055 g = 0.200055 kg
55 mg / 0.200055 kg = 275 ppm
A decrease of one unit in the pH scale above represents a tenfold increase in the hydrogen ion concentration of a solution. For example, a solution having a pH of 4 is 10 times more acidic than a solution with a pH of 5. If acid precipitation rain changes the pH of a pond from 7.5 to 6.5, the level of hydrogen ion has changed by a factor of ________
A) 2.0
B) 10
C) 13.0
D) 100
E) 0.01
Answer:
The level of hydrogen ion has increased by a factor of 10.
Explanation:
We know, [tex]pH=-log[H^{+}][/tex]
where [tex][H^{+}][/tex] represents concentration of [tex]H^{+}[/tex] in molarity
Here [tex](pH)_{final}=6.5[/tex]
So, [tex][H^{+}]_{final}=10^{-6.5}M[/tex]
Here [tex](pH)_{initial}=7.5[/tex]
So, [tex][H^{+}]_{initial}=10^{-7.5}M[/tex]
Hence, change in pH = [tex]\frac{[H^{+}]_{final}}{[H^{+}]_{initial}}[/tex] = [tex]\frac{10^{-6.5}M}{10^{-7.5}M}=10[/tex]
So, the level of hydrogen ion has increased by a factor of 10.
Option (B) is correct
0.70 moles of an unknown solid is placed into water to make 150.0 mL of solution. The solution's temperature decreases by 8.4°C. Calculate ∆H for the dissolution of the unknown solid. (The specific heat of the solution is 4.18 J/g･°C and the density of the solution is 1.02 g/mL).
Answer:
[tex]\large \boxed{\text{7.7 kJ}\cdot\text{mol}^{-1} }[/tex]
Explanation:
1. Mass of solution
[tex]\text{Mass of solution} = \text{150 mL} \times \dfrac{\text{1.02 g}}{\text{1 mL}} = \text{153 g}[/tex]
2. Calorimetry
There are two energy flows in this reaction.
q₁ = heat from reaction
q₂ = heat to warm the water
[tex]\begin{array}{ccccl}n\Delta H & + & mC\Delta T &=&0\\\text{0.70 mol}\times \Delta H& + & \text{153 g} \times 4.18 \text{ J}\cdot^{\circ}\text{C}^{-1}\cdot \text{g⁻1} \times (-8.4\, ^{\circ}\text{C}) & = & 0\\0.70\Delta H \text{ mol} & - & \text{5372 J} & = & 0\\& & 0.70\Delta H \text{ mol} & = & \text{5372 J} \\\end{array}[/tex]
[tex]\begin{array}{ccrcl} & & \Delta H & = & \dfrac{\text{ 5372 J}}{\text{0.70 mol}}\\\\ & & & = & \text{7670 J/mol}\\ & & & = & \textbf{7.7 kJ}\cdot\textbf{mol}^{ -\mathbf{-1}} \\\end{array}\\\text{The heat of solution of the compound is $\large \boxed{\textbf{7.7 kJ}\cdot\textbf{mol}^{\mathbf{-1}}}$}[/tex]
The heat of solution is positive, because the water cooled down when the substance dissolved,
To calculate ∆H for the dissolution process, we first calculate the total heat released using the equation q = m × c × ∆T and the provided properties of the solution. We then find ∆H by dividing this total heat by the number of moles of solute, yielding an enthalpy change of -7.57 kJ/mol.
Explanation:The subject of this question relates to chemistry, specifically the concept of enthalpy change (∆H) in the dissolution of a solid in a solution. The first step to calculating ∆H is determining the total heat absorbed or released during the dissolution, which is calculated using the equation q = m × c × ∆T, where m is the mass of the solution, c is the specific heat capacity, and ∆T is the change in temperature.
In this case, we first convert the volume of the solution to mass by using the provided density of the solution (1.02 g/mL). So, the total mass of the solution is 150 mL × 1.02 g/mL = 153.0 g. Next, using the provided temperature and specific heat, we calculate q = (153.0 g) × (4.18 J/g･°C) × (-8.4°C) = -5300.472 J. We use a negative sign because the temperature decreases, meaning the reaction is exothermic and heat is released.
Finally, we calculate ∆H, the enthalpy change per mole of solute, by dividing q by the number of moles of the solute, 0.70 moles. ∆H = -5300.472 J / 0.70 mol = -7572 J/mol, or -7.57 kJ/mol.
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Part 1: Ionic Bonding 1. Choose Sodium (Na). a. What type of element is it? b. How many valence electrons does it have? 2. Choose Fluorine (F). a. What type of element is it? b. How many valence electrons does it have? 3. Answer the question on the screen, "What type of bond is this combination likely to form?" a. Circle: Ionic or Covalent? b. Choose the appropriate number of atoms to make the bond. Record the number of each atom below: 4. Watch the final animation closely (it will play continuously). a. Describe the change in the number of valence electrons in the atoms as the bond is successfully formed: b. What does the positive (+) charge indicate (mention specific subatomic particles in your answer)? c. What does the negative (-) charge indicate (mention specific subatomic particles in your answer)? d. Record the name and molecular formula for the compound below:
Answer:
Explanation:
Sodium is a group 1 element with atomic number 1. It has 11 electrons. It is soft reactive metal. It has 1 valence electron.
Fluorine is a group 7 element, a hologen with 7 valence electron. It is a most reactive non metal.
When sodium react with fluorine, ionic bond is formed in the resulting compound sodium fluoride.
One sodium and fluorine each totaling 2 atoms are enough to make the bond.
As the bond is formed, both atoms have octet structure. That is they each have 8 electrons on their outermost shells.
The positive charge on sodium indicates that sodium had lost 1 electron to fluorine atom.
The negative charge on fluorine ion indicates that fluorine atom had gained 1 electron from sodium atom to form negative ion.
The name of the compound is sodium fluoride with formula NaF.
Answer:
The reaction of 1st group element Na and 7th group element F, results in formation of Sodium fluoride (NaF).
Explanation:
1. (a) Sodium (Na) is element belonging to the first group of the periodic table.
(b) The atomic number of Na is 11 and after distribution it has 1 valence electron.
2. (a) Fluorine (F) is the halogen element and belongs to group 7 of the periodic table.
(b) The atomic number of F is 19 and it has 7 valence electrons.
3. The reaction of Na with F results in an ionic interaction with the formation of NaF (Sodium fluoride).
(a) The reaction and combination of Na and F is likely to be ionic in nature.
(b) The one valence electron of Na and F are involved in formation of bond. Total 2 electrons are involved.
4. (a) The successful formation of bond result in octet completion of NaF by sharing of electrons. Both the elements share 8 electrons, i.e. 7 electrons of F and 1 electron of Na.
(b) The +ve charge is imparted for the loss of electron . Na lost its 1 valence electron to F for bond formation and thus imparts the +ve charge.
(c) The -ve charge is imparted with the gain of electron. In reaction of Na with F, fluorine being more electronegative gains its electrons from Na and imparts a -ve charge.
(d) The reaction of Na with F results in the formation of compound known as sodium fluoride with the molecular formula NaF.
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A sample of 8.4 grams of NaOH is dissolved into 620 mL of aqueous 0.250 M NaOH (assume no volume change). This solution is then poured into 1.65 gallons of water. (You may assume that the two volumes can be added.) What is the concentration of NaOH in the final solution
Answer:
The concentration of NaOH in the final solution 0.0584 mole/lit
Explanation:
[tex]moles of NaOH =[/tex] [tex]\frac{Weight}{Molecular weight}[/tex] [tex]=\frac{8.4}{40}=0.21 mole[/tex]620 ml of aqueous 0.25 M NaOH [tex]=0.25X 620 = 0.155mole[/tex]Total moles = 0.21 + 0.155 = 0.365 moles1.65 gallons = 3.785 lit x 1.65 = 6.245929 lit (∵ 1 gallon = 3.785 lit)[tex]Molarity =\frac{No. of moles}{Volume(lit)}[/tex]So Molarity of the solution after mixing [tex]=\frac{0.365}{6.245929} = 0.0584 (M)[/tex]
∴ The concentration of NaOH in the final solution 0.0584 mole/lit
The concentration of NaOH in the final solution is thus 0.0531 M.
To determine the concentration of NaOH in the final solution after 8.4 grams of NaOH are dissolved into 620 mL of 0.250 M NaOH solution, which is then mixed into 1.65 gallons of water, follow these steps:
First, calculate the moles of NaOH initially dissolved. Knowing NaOH has a molar mass of 40.0 g/mol, 8.4 g of NaOH translates to 0.21 moles of NaOH.
Add the moles from the initial 620 mL of 0.250 M solution. That will be 0.250 moles/L * 0.62 L = 0.155 moles.
Combine the moles from both sources for a total of 0.365 moles of NaOH.
Convert 1.65 gallons of water to liters (1 gallon = 3.78541 L), making 1.65 gallons equal to 6.245 liters. Adding this to the initial 0.62 L yields a total final volume of 6.865 liters.
Finally, calculate the final concentration by dividing the total moles of NaOH by the final volume of solution in liters to find the molarity of the final solution.
The concentration of NaOH in the final solution is thus 0.0531 M.
NH4CO2NH2(s) equilibrium reaction arrow 2 NH3(g) + CO2(g) Ammonium carbamate decomposes according to the equation above. At 40°C, an equilibrium mixture for this reaction contains 0.0640 atm CO2 and 0.370 atm NH3. What is the value of the equilibrium constant Kp for the reaction?
Answer:
Kp = 8.76×10⁻³
Explanation:
We determine the carbamate decomposition in equilibrium:
NH₄CO₂NH₂ (s) ⇄ 2NH₃(g) + CO₂(g)
Let's build the expression for Kp
Kp = (Partial pressure NH₃)² . Partial pressure CO₂
We do not consider, the carbamate because it is solid and we only need the partial pressure from gases
Kp = (0.370atm)² . 0.0640 atm
Kp = 8.76×10⁻³
Remember Kp does not carry units
The value of the equilibrium constant Kp for the decomposition of ammonium carbamate at 40°C, given the partial pressures of 0.370 atm for NH3 and 0.0640 atm for CO2, is 0.0087616.
The decomposition of ammonium carbamate to ammonia and carbon dioxide at equilibrium can be represented by the following equation:
NH4CO2NH2(s) ⇌ 2 NH3(g) + CO2(g)
The equilibrium constant for a reaction involving gases, Kp, is based on the partial pressures of the gaseous products and reactants. The equilibrium constant expression for this reaction is:
Kp = (PNH3)^2 × (PCO2)
Given the equilibrium partial pressures, 0.370 atm of NH3 and 0.0640 atm of CO2, we can calculate the Kp value:
Kp = (0.370)^2 × (0.0640) = 0.1369 × 0.0640 = 0.0087616
So, the value of the equilibrium constant Kp for the reaction at 40°C is 0.0087616.
Consider the following reaction: 2 H2(g) + 2 NO(g) → 2 H2O(g) + N2(g) If the concentration of NO changed from 0.100 M to 0.025 M in the first 15 minutes of the reaction, what is the average rate of the reaction during this time interval?
Final answer:
The average rate of the reaction is -0.005 M/min.
Explanation:
The average rate of a reaction can be calculated by using the change in concentration of a reactant or product divided by the time interval. In this case, the concentration of NO changed from 0.100 M to 0.025 M in 15 minutes. You can calculate the average rate of the reaction by dividing the change in concentration by the time interval:
Average rate = (Final concentration - Initial concentration) / Time
Average rate = (0.025 M - 0.100 M) / 15 minutes
Average rate = -0.075 M / 15 minutes = -0.005 M/min
Briefly explain the observed effect of the acetylcholine concentration on the rate of the enzyme-catalyzed reaction
Explanation:
It is observed that when the concentration of acetylcholine remains constant in the reaction of an aqueous solution, the speed of the enzyme-catalyzed reaction or the formation of the product increases with increasing concentrations of substrate. The reaction rate is directly proportional to the concentration of acetylcholine. At very low concentrations of acetylcholine, there is a small increase in the concentration of the substrate which results in a large increase of the rate in reaction.What is the molar mass of 37.96 g of gas exerting a pressure of 3.29 on the walls of a 4.60 L container at 375 K?
The molar mass of the gas is 77.20 gm/mole.
Explanation:
The data given is:
P = 3.29 atm, V= 4.60 L T= 375 K mass of the gas = 37.96 grams
Using the ideal Gas Law will give the number of moles of the gas. The formula is
PV= nRT (where R = Universal Gas Constant 0.08206 L.atm/ K mole
Also number of moles is not given so applying the formula
n= mass ÷ molar mass of one mole of the gas.
n = m ÷ x ( x molar mass) ( m mass given)
Now putting the values in Ideal Gas Law equation
PV = m ÷ x RT
3.29 × 4.60 = 37.96/x × 0.08206 × 375
15.134 = 1168.1241 ÷ x
15.134x = 1168.1241
x = 1168.1241 ÷ 15.13
x = 77.20 gm/mol
If all the units in the formula are put will get cancel only grams/mole will be there. Molecular weight is given by gm/mole.
Answer:
77.2
Explanation:
In each case of a chemical reaction shown below, state the indicator of the chemical reaction. Identifying Indicators of Chemical Reactions Bread dough rising
Bread dough rises due to the formation of gas.
Explanation:
If we observe the outer part of the bread, which contains tiny hole like structure, due to the gas formation inside the bread dough.In the bread dough rising process, the industries are using the yeast, which makes the gas that separates the protein particles in the bread move apart and makes the dough of the bread rise.In this reaction, the yeast utilizing the carbohydrate to make a gas namely carbon-di-oxide gas which makes the dough to riseThe collapse point is an indicator of sufficient Bread dough rise.
Discussion;
When yeast is added to water and flour to create dough, it eats up the sugars in the flour and in turn produces carbon dioxide gas and ethanol in a process termed fermentation.
The elastic protein, gluten in the dough traps the carbon dioxide gas, preventing it from escaping.
In essence, the bread volume increases as the dough rises.
The collapse point is an indicator of sufficient dough rise.
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Rank the following solutes in order of increasing entropy when 0.0100 moles of each dissolve in 1.00 liter of water.
Answer:
CH3OH, NaBr, CaCl2, Cr(NO3)3
Based on disorderliness, the ionization strength gives the entropy.
CH3OH - Does not ionize, it's a solvent
NaBr- Ionizes intwo two ions
CaCl2- Ionizes to three ions 1Ca and 2 Cl ions
Cr(NO3)3- Ionizes to four ions
Explanation:
Here is the complete question:.
Rank the following solutes in order of increasing entropy when 0.0100 moles of each dissolve in 1.00 liter of water.
CaCl2, CH3OH, Cr(NO3)3, NaBr
A cylinder with a moveable piston contains 92g of Nitrogen. The external pressure is constant at 1.00 atm. The initial temperature is 200K. When the temperature is decreased by 85 K, by putting it in a lower temperature freezer, the volume will decrease, according to the Ideal Gas Law. Calculate the work for this process. Express your answer in J. The conversion factor between liter atmospheres and joules is 101.3 J
Answer:
Work done in this process = 4053 J
Explanation:
Mass of the gas = 0.092 kg
Pressure is constant = 1 atm = 101325 pa
Initial temperature [tex]T_{1}[/tex] = 200 K
Final temperature [tex]T_{2}[/tex] = 200 - 85 = 115 K
Gas constant for nitrogen = 297 [tex]\frac{J}{kg k}[/tex]
When pressure of a gas is constant, volume of the gas is directly proportional to its temperature.
⇒ V ∝ T
⇒ [tex]\frac{V_{2} }{V_{1} }[/tex] = [tex]\frac{T_{2} }{T_{1} }[/tex] ------------ ( 1 )
From ideal gas equation [tex]P_{1}[/tex] [tex]V_{1}[/tex] = m R [tex]T_{1}[/tex] ------ (2)
⇒ 101325 × [tex]V_{1}[/tex] = 0.092 × 297 × 200
⇒ [tex]V_{1}[/tex] = 0.054 [tex]m^{3}[/tex]
This is the volume at initial condition.
From equation 1
⇒ [tex]\frac{V_{2} }{0.054}[/tex] = [tex]\frac{200}{115}[/tex]
⇒ [tex]V_{2}[/tex] = 0.094 [tex]m^{3}[/tex]
This is the volume at final condition.
Thus the work done is given by W = P [[tex]V_{2}[/tex] - [tex]V_{1}[/tex] ]
⇒ W = 101325 × [ 0.094 - 0.054]
⇒ W = 4053 J
This is the work done in that process.
Which will evaporate more quickly: 55 mLmL of water in a beaker with a diameter of 4.5 cmcm, or 55 mLmL of water in a dish with a diameter of 12 cmcm ?
Explanation:
More is the surface area of water comes in contact with the atmosphere or its surrounding more readily it will evaporate.
So, when 55 mL of water in a beaker with a diameter of 4.5 cm is placed then due to the small diameter of the beaker less surface area is in contact with the atmosphere.
But when 55 mL of water in a dish with a diameter of 12 cm is placed then it has larger diameter as compared to 4.5 cm diameter. Therefore, water in a beaker of 12 cm diameter will evaporate more quickly.
Thus, we can conclude that water in 55 mL of water in a dish with a diameter of 12 cm will evaporate more quickly.
The larger the surface area exposed to air, the faster the rate of evaporation. Therefore, a dish with a larger diameter will allow water to evaporate more quickly than a beaker with a smaller diameter while holding the same volume of liquid.
Explanation:The 55 mL of water will evaporate more quickly in the dish with a diameter of 12 cm. This is because evaporation is a surface phenomenon - it depends on the surface area exposed to air. In this case, the dish with a more significant diameter will have a larger surface area exposed to air, leading to faster evaporation.
An analogy might be to consider two scenarios: the first is pouring a cup of coffee into a narrow, high cylinder and the second is into a broad, low dish. Both containers might have the same volume, but the surface area is significantly different. The coffee in the broad dish is more exposed and will cool faster due to a higher degree of evaporation.
Just as the coffee cools in the broad dish, the water in the dish with the larger diameter will evaporate more quickly. While volume might be constant in both the beaker and the dish, the shape and structure of the container significantly impact the rate of evaporation.
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a solution of hydrochloric acid of unknown concentration was titrated with .21 M NaOH. if a 75 ml sample of the HCl solution required exactly 13ml of the NaOH solution to reach the equivalence point, what was the ph of the HCl solution
Answer:
pH of HCl solution is 1.44
Explanation:
NaOH is a monoprotic base and HCl is a monobasic acid.
Neutralization reaction: [tex]NaOH+HCl\rightarrow NaCl+H_{2}O[/tex]
According to balanced reaction, 1 mol of NaOH neutralizes 1 mol of HCl.
Number of moles of NaOH in 13 mL of 0.21 M NaOH
= [tex]\frac{0.21}{1000}\times 13mol=0.00273mol[/tex]
let's assume concentration of HCl is C (M)
Then, number of moles of HCl in 75 mL of C (M) HCl solution
= [tex]\frac{75}{1000}\times Cmol=\frac{75C}{1000}mol[/tex]
So, we can write, [tex]\frac{75C}{1000}=0.00273[/tex]
or, [tex]C=0.0364[/tex]
1 mol of HCl contains 1 mol of [tex]H^{+}[/tex]
So, concentration of [tex]H^{+}[/tex] in 0.0364 M HCl, [tex][H^{+}]=0.0364M[/tex]
Hence, [tex]pH=-log[H^{+}]=-log(0.0364)=1.44[/tex]
Balance the following equation with the smallest whole number coefficients. Choose the answer that is the sum of the coefficients in the balanced equation. Do not forget coefficients of "one". RbOH + H3PO4 → Rb3PO4 + H2O
Final answer:
To balance the equation RbOH + H3PO4 → Rb3PO4 + H2O, you use the coefficients 3, 1, 1, and 3, resulting in a sum of 8.
Explanation:
The question involves balancing a chemical equation to ensure that the same number of atoms of each element is present on both sides, following the law of conservation of mass. The balanced equation for RbOH + H3PO4 → Rb3PO4 + H2O is 3RbOH + H3PO4 → Rb3PO4 + 3H2O, giving us coefficients 3, 1, 1, 3, and the sum of these coefficients is 8.
It's important to never change the subscripts in a chemical formula while balancing an equation. Instead, coefficients are placed in front of the chemical formulae to indicate the relative amounts of reactants and products. The lowest whole number ratio that balances the equation should be used.
An element has two naturally-occurring isotopes. The mass numbers of these isotopes are 123 amu and 125 amu, with natural abundances of 95% and 5%, respectively. Calculate its average atomic mass. Report your answer to 1 decimal place.
Answer:
123.1
Explanation:
The average atomic mass of an element which exhibits isotopy is the average mass of its various isotopes as they occur naturally in any quantity of the element.
Average atomic mass = (95% of 123 amu)+(5% 0f 125 amu)
= (0.95 × 123) + (0.05 × 125)
= 116.85 + 6.25
= 123.1 (to 1 decimal place)
A student performs an experiment to determine the density of a sugar solution. She obtains the following results: 1.11 g/mL, 1.81 g/mL, 1.95 g/mL, 1.75 g/mL. If the actual value for the density of the sugar solution is 1.75 g/mL, which statement below best describes her results?
A) Her results are precise, but not accurate.
B) Her results are accurate, but not precise.
C) Her results are both precise and accurate
D) Her results are neither precise nor accurate.
E) It isn't possible to determine with the information given.
The answer is D, I don't know why they choose D?
Answer:
D. her results are neither precise, nor accurate
Explanation:
Reason being that: First, she obtained four different result for an experiment ( obviously it's supposed to be one value for the different procedure or atleast very close range) sugar solution and all four values were far apart.
Final answer:
The student's density measurements of a sugar solution were neither precise nor accurate because they varied widely from each other and, except one value, they also differed significantly from the known actual density value.
Explanation:
When a student performs an experiment to determine the density of a sugar solution and obtains results that vary greatly from the actual value, we must assess the precision and accuracy of the measurements. Accuracy refers to how close each measurement is to the true value, in this case, the actual density of 1.75 g/mL. Precision indicates how close the measurements are to each other, regardless of how close they are to the actual value.
In this scenario, the student's results show one measurement that is the same as the known actual value (1.75 g/mL), suggesting a single instance of accuracy. Therefore, the statement that best describes the student's results is that they are neither precise nor accurate, which corresponds to option D.
Woody plants evolve in the mid-Paleozoic Era. A. atmospheric CO2 increases B. atmospheric CO2 decreases C. There is no change in atmospheric CO2 levels.
Answer:
A. atmospheric CO2 increases
Explanation:
The concentration of CO2 during the Paleozoic era was much higher than the current one, due to geographical factors, but not the chemical composition of the air, which should have been more important in increasing this gas.
The highest concentrations of CO2 during the Paleozoic era occurred during the Cambrian, approximately 7000 ppm (18 times higher than current measurements). During the Late Ordovic Period the CO2 concentration was approximately 12 times higher than the current ones, with a value of 4400 ppm. These high concentrations of CO2 favored the plants colonizing the continents and thus affecting the world climate.
At their centers, all the jovian planets have cores made of:
new elements produced by the high pressure; elements which we do not have on Earth
hydrogen and helium in the form of gas
a solid mixture of rocky and icy materials under great pressure
methane, ammonia, and sulfur compounds
hydrogen and helium in the form of liquids
Final answer:
The cores of the Jovian planets are composed of a solid mixture of rocky and icy materials under great pressure, and their compositions are dominated by elements like carbon, nitrogen, and oxygen.
Explanation:
The cores of the jovian planets, such as Jupiter, Saturn, Uranus, and Neptune, are composed of a solid mixture of rocky and icy materials under great pressure.
This information is supported by studies of the gravitational fields of these planets and the fact that their compositions are dominated by elements like carbon, nitrogen, and oxygen, along with rock and ice.
Although the cores exist at pressures of tens of millions of bars, the materials referred to as 'rock' and 'ice' do not assume familiar forms at such extreme pressures and temperatures.
A compound consists of 65.45% C, 5.492% H, and 29.06% O on a mass basis and has a molar mass of 110 g/mol. Determine the molecular formula of the compound.
Answer:
The molecular formula is C6H6O2
Explanation:
Step 1: Data given
Suppose the mass of the compound = 100 grams
The compound has:
65.45 % C = 65.45 grams
5.492 % H = 5.49 grams
29.06 % O = 29.06 grams
Molar mass C = 12.01 g/mol
Molar mass H = 1.01 g/mol
Molar mass O = 16.00 g/mol
Molar mass = 110 g/mol
Step 2: Calculate moles
Moles = mass / molar mass
Moles C = 65.45 grams / 12.01 g/mol
Moles C = 5.450 moles
Moles H = 5.49 grams / 1.01 g/mol
Moles H = 5.44 moles
Moles O = 29.06 grams / 16.00 g/mol
Moles O = 1.816 moles
Step 3: Calculate mol ratio
We divide by the smallest amount of moles
Moles C = 5.450 moles / 1.816 = 3
Moles H = 5.44 moles / 1.816 = 3
Moles O = 1.816/1.816 = 1
The empirical formula is C3H3O
The molar mass of this formula is 55 g/mol
We have to multiply the empirical formula by n
n = 110/55 = 2
2*(C3H3O) = C6H6O2
The molecular formula is C6H6O2
The empirical formula for the compound, given the percentage composition and molecular mass, is C3H3O. The molecular formula, given that the molar mass of the compound is about 2.66 times the empirical formula's molar mass, is C9H9O3.
Explanation:To determine the molecular formula of the compound, we first assume that we have 100g of the compound. This means we have 65.45g of Carbon (C), 5.492g of Hydrogen (H), and 29.06g of Oxygen (O). We must then convert these masses to moles by dividing by the atomic mass (12.01 for C, 1.008 for H, and 16 for O).
Therefore, we have:
For Carbon: 65.45g ÷ 12.01g/mol = 5.45 molFor Hydrogen: 5.492g ÷ 1.008g/mol = 5.45 molFor Oxygen: 29.06g ÷ 16g/mol = 1.816 molWe then divide these mole quantities by the smallest obtained mole number to try to get whole numbers. This should give an atomic ratio of 3:3:1 for C, H, and O respectively.
By combining these ratios into a chemical formula, we get C3H3O. This is the empirical formula. The molar mass of this formula (41 g/mol) goes into the given molar mass of the compound (110 g/mol) about 2.66 times, indicating that the true molecular formula of this compound is C9H9O3.
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Hydrochloric acid is purchased in 12 molar concentrations. If I needed 200 ml of 2.0 molar HCl, how many milliliters of the concentrated solution would I need
Answer:
The answer to your question is 33.3 ml
Explanation:
Data
Stock solution 12M
Volume 2 = 200 ml
Concentration 2 = 2 M
Volume stock = ?
Process
1.- Use the formula for dilutions
C1V1 = C2 V2
- Solve for V1
V1 = C2V2 / C1
2.- Substitution
V1 = (2)(200) / 12
- Simplification
V1 = 400 / 12
3.- Result
V1 = 33.3 ml
4.- Conclusion
We need 33.3 ml of the concentrated solution.
Classify each of the following possible reactions according to whether a precipitate will form or will not form.
1.) KI+NaCl
2.) Na3PO4+CoCl2
3.) Na2CO3+CuCl2
4.) LiNO3+Na2SO4
5.) CrCl2+Li2CO3
The reactions Na3PO4+CoCl2, Na2CO3+CuCl2 and CrCl2+Li2CO3 produce precipitates.
A precipitate is said to be formed when reaction of two aqueous solutions produces an insoluble product. The insoluble solid product separates from the solution and is called a precipitate.
The full reactions are written below;
KI(aq) + NaCl(aq) ----> KCl(aq) + NaI(aq) 2Na3PO4(aq) + 3CoCl2(aq) ------> Co3(PO4)2(s) + 6NaCl(aq) Na2CO3(aq) + CuCl2(aq) -------> CuCO3(s) + 2NaCl(aq)2LiNO3(aq) + Na2SO4(aq) -------> Li2(SO4)(aq) + 2NaNO3(aq)CrCl2(aq) + Li2CO3(aq) -------> CrCO3(s) + 2LiCl(aq)Hence, the reactions;
2Na3PO4(aq) + 3CoCl2(aq) ------> Co3(PO4)2(s) + 6NaCl(aq) Na2CO3(aq) + CuCl2(aq) -------> CuCO3(s) + 2NaCl(aq)CrCl2(aq) + Li2CO3(aq) -------> CrCO3(s) + 2LiCl(aq)Produce precipitates.
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To determine if a precipitate will form or not, we need to consult solubility rules. By analyzing the compounds in each reaction, we can determine whether they contain soluble or insoluble ions. Based on this analysis, we can classify each reaction as either forming a precipitate or not.
Explanation:In order to determine if a precipitate will form or not, we need to consult solubility rules. According to the solubility rules, compounds containing Li+, Na+, K+, Rb+, Cs+, and NH4+ cations, as well as compounds containing NO3- and C2H3O2- anions, are generally soluble. On the other hand, compounds containing Cl-, Br-, I-, CO32-, and most O2- anions are insoluble, except when paired with the mentioned cations. By applying these rules, we can analyze each reaction:
KI + NaCl: Both KI and NaCl contain soluble ions, so no precipitate will form.Na3PO4 + CoCl2: Since both compounds contain soluble ions, no precipitate will form.Na2CO3 + CuCl2: Na2CO3 contains a soluble ion (Na+), but CuCl2 contains insoluble ions. Thus, a precipitate of CuCO3 will form.LiNO3 + Na2SO4: Both LiNO3 and Na2SO4 contain soluble ions, so no precipitate will form.CrCl2 + Li2CO3: CrCl2 contains soluble ions, but Li2CO3 contains insoluble ions. Therefore, a precipitate of LiCl will form.Learn more about Classifying Reactions here:https://brainly.com/question/21392527
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In a particular experiment, the partial pressures of H 2 and I 2 at equilibrium are 0.710 and 0.888 atm, respectively. The partial pressure of HI is __________ atm.
This is an incomplete question, here is a complete question.
Kp = 0.0198 at 721 K for the reaction:
[tex]2HI(g)\rightleftharpoons H_2(g)+I_2(g)[/tex]
In a particular experiment, the partial pressures of H₂ and I₂ at equilibrium are 0.710 and 0.888 atm, respectively. The partial pressure of HI is __________ atm.
Answer : The partial pressure of HI is, 5.64 atm
Explanation :
For the given chemical reaction:
[tex]2HI(g)\rightleftharpoons H_2(g)+I_2(g)[/tex]
The expression of [tex]K_p[/tex] for above reaction follows:
[tex]K_p=\frac{P_{H_2}\times P_{I_2}}{(P_{HI})^2}[/tex]
We are given:
[tex]P_{H_2}=0.710atm[/tex]
[tex]P_{H_2}=0.888atm[/tex]
[tex]k_p=0.0198[/tex]
Now put all the given values in above equation, we get:
[tex]0.0198=\frac{0.710\times 0.888}{(P_{HI})^2}[/tex]
[tex]P_{HI}=5.64atm[/tex]
Thus, the partial pressure of HI is, 5.64 atm
The correct partial pressure of HI cannot be determined without the accurate initial conditions and changes at equilibrium, since the provided equilibrium partial pressures of H2 and I2 do not match the provided equilibrium constant.
The question states that the partial pressures of H2 and I2 at equilibrium are 0.710 atm and 0.888 atm, respectively, in a reaction H2(g) + I2(g) = 2 HI(g). Given the partial pressure at the start for HI was 1 atm and we are looking to find the equilibrium partial pressure of HI, we can apply the equilibrium constant Kp. However, the values given are part of a typo or irrelevant to the scenario as they do not match with the equilibrium constant Kp provided (1.82 imes 10-2 at 698K), and since such calculations require an accurate accounting of initial conditions and changes at equilibrium, providing an exact value for the partial pressure of HI would be speculative.
A solution of methanol and water has a mole fraction of water of 0.312 and a total vapor pressure of 211 torr at 39.9 ∘C. The vapor pressures of pure methan ol and pure water at this temperature are 256 torr and 55.3 torr, respectively. ls the solution ideal? If not, what can you say about the relative strengths of the solute-solvent interactions compared to the solute-solute and solvent-solvent interactions?
Answer:
Weaker
Explanation:
The strategy here is to use Raoult´s law to calculate the theoretical vapor pressure for the concentrations given and compare it with the experimental value of 211 torr.
Raoult´s law tell us that for a binary solution
P total = partial pressure A + partial pressure B = Xa PºA + Xb PºB
where Xa and Xb are the mol fractions, and PºA and PºB are the vapor pressures of pure A and pure B, respectively
For the solution in question we have
Ptotal = 0.312 x 55.3 torr + ( 1- 0.312 ) x 256 torr ( XA + XB = 1 )
Ptotal = 193 torr
Since experimentally, the total vapor pressure is 211 and our theoretical value is smaller ( 193 torr ), we can conclude the interactions solute-solvent are weaker compared to the solute-solute and solvent-solvent interactions.
Using Raoult's law, we can conclude that if the actual total vapor pressure of the solution does not match the calculated value, the solution is not ideal. Deviations from the ideal behavior indicate that the solute-solvent interactions differ in strength from solute-solute and solvent-solvent interactions.
To determine if the solution is ideal and to discuss the strengths of solute-solvent interactions compared to solute-solute and solvent-solvent interactions, we can use Raoult's law. Raoult's law states that the partial vapor pressure of each component in an ideal solution is equal to the product of the mole fraction of the component in the liquid phase and the vapor pressure of the pure component.
The expected total vapor pressure for an ideal solution can be calculated by summing the partial pressures of each component. For methanol (CH3OH), with a mole fraction of 1 - 0.312 = 0.688 and a vapor pressure of 256 torr, the expected partial pressure is partial pressure of methanol = 0.688 × 256 torr. For water (H2O), with a mole fraction of 0.312 and a vapor pressure of 55.3 torr, the expected partial pressure is partial pressure of water = 0.312 × 55.3 torr. By adding these two values together, we would get the expected total vapor pressure of an ideal solution.
If the calculated total vapor pressure using Raoult's law does not match the actual total vapor pressure of 211 torr, the solution is not ideal. In that case, the deviation indicates that the solute-solvent interactions differ in strength from the solute-solute and solvent-solvent interactions. If the actual vapor pressure is lower than expected, the solute-solvent interactions are stronger, suggesting a negative deviation. Conversely, a higher actual vapor pressure indicates weaker solute-solvent interactions relative to pure components, leading to a positive deviation.
In a constant‑pressure calorimeter, 70.0 mL of 0.310 M Ba ( OH ) 2 was added to 70.0 mL of 0.620 M HCl . The reaction caused the temperature of the solution to rise from 21.91 ∘ C to 26.13 ∘ C. If the solution has the same density and specific heat as water, what is heat absorbed by the solution?
Explanation:
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The mass of a gold atom is 3.27 × 10−25 kg . If 3.6 kg of gold is deposited on the negative electrode of an electrolytic cell in a period of 14.88 h , what is the current in the cell in this period? Assume that each gold ion carries one fundamental unit (1.602 × 10−19 C) of positive charge. Answer in units of A.
Answer:
Current is 32.8 A
Explanation:
1 mass of Au atom = 3.27×10⁻²⁵kg
Mass of gold = 3.6 kg
We determine the amount of atoms, in that mass of gold therefore we need a rule of three:
3.27×10⁻²⁵kg is the mass for 1 atom
3.6 kg will be the mass for (3.6 . 1)/3.27×10⁻²⁵ = 1.10×10²⁵ atoms
Now we have to find out the total charge of the 3.6 kg of gold so we make another rule of three:
1 atom of Au has a charge of 1.602 × 10⁻¹⁹ C
1.10×10²⁵ atoms of Au will have a charge of (1.10×10²⁵.1.602 × 10⁻¹⁹) / 1 = 1.76×10⁶ C
Formula for the current is: q = i . t
where q is the C, i the value of current and t, time (s)- We make time conversion from h to s
14.88 h . 3600s / 1h = 53568 s → We replace values:
1.76×10⁶ C = i . 53568s
1.76×10⁶ C / 53568s = i → 32.8 A
In Keynes's liquidity preference framework, as the expected return on bonds increases (holding everything else unchanged), the expected return on money ________, causing the demand for ________ to fall.
Answer:
the correct words to fill up the blank spaces are falls and money
Explanation:
.1. the expected return on money falls
.2. causing the demand for money to fall.
Fatty acids may differ from one another in: a. chain length and degree of saturation b. the number of calories and in whether they contain carbon atoms c. degree of saturation and in the number of sulfur atoms d. chain length and in the number of nitrogen atoms e.
Answer:
a. chain length and degree of saturation.
Explanation:
Fatty acids can be classified according to the length of their chain, for example, in short (if it has less than 8 carbons), medium (between 8-12 carbons), long (between 12-18 carbons) and very long (if it has less than 18 carbons); they are also classified according to their degree of unsaturation, in saturated, monounsaturated and polyunsaturated; and according to the isomerism in cis and trans fatty acids.
Fatty acids may differ from one another in chain length and degree of saturation. Therefore, the correct option is option A.
Fatty acid is a crucial part of lipids, which are the parts of living things including plants, animals, and microbes that may be dissolved in fat. A fatty acid normally consists of a chain that is straight with a even amount of carbon atoms, hydrogen atoms running along the entire length of the chain, a carboxyl group (COOH) at one end, and hydrogen atoms running the duration of the chain. It forms an acid (carboxylic acidic solution) because of the carboxyl group. The acid is saturated if there are only single carbon-to-carbon bonds. Fatty acids may differ from one another in chain length and degree of saturation.
Therefore, the correct option is option A.
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The _______________ system collects solid wastes—such as undigested food, intestinal bacteria, and old cells—and removes them from the body, while the _____________ system filters the blood and collects wastes to remove from the body as urine.
Answer:
Excretory system, and Kidney.
Explanation:
The excretory system helps to regulate the chemical composition of the body fluids by removing metabolic waste such as undigested food, intestinal bacteria, and old cells and helping to retain the proper amount of water, nutrients, and salts in the body.
The main function of the kidney is to filter the blood, kidney help to remove waste from the body as urine, control fluid balance in the body, and keep the right level of electrolytes. Each kidney is bean-shaped in structure and 4-5 inches long in size.
Calculate the energy required to change the temperature of 1.00 kg of ethane (C2H6) from 25.0"C to 73.4"C in a rigid vessel. (Cv for C2H6 is 44.60 J K%1 mol%1.) Calculate the energy required for this same temperature
To find the energy required to heat 1.00 kg of ethane from 25.0°C to 73.4°C, one must calculate the number of moles of ethane, determine the temperature change, and apply the formula Q = n • Cv • ΔT using the given specific heat capacity at constant volume (Cv).
Explanation:The question pertains to the calculation of energy required for a temperature change in ethane (C2H6) and is based on thermodynamic principles involving heat, enthalpy, and specific heat capacities. To calculate the energy required to change the temperature of 1.00 kg of ethane from 25.0°C to 73.4°C in a rigid vessel, we need the specific heat capacity at constant volume (Cv) and the molar mass of ethane to convert the mass to moles. The energy, Q, needed for the temperature change is given by Q = n • Cv • ΔT, where n is the number of moles, and ΔT is the change in temperature.
First, calculate the number of moles (n) of ethane: its molar mass is 30.07 g/mol (add the atomic masses of 2 carbon atoms and 6 hydrogen atoms). Given that we have 1.00 kg (or 1000 g) of ethane, divide 1000 g by 30.07 g/mol to find n. Then, use ΔT = 73.4°C - 25.0°C to find the temperature change. Finally, plug these values into the equation Q = n • Cv • ΔT with the given Cv of 44.60 J/K·mol to calculate the required energy in joules.
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The energy required to change the temperature of 1.00 kg of ethane (C₂H₆) from 25.0°C to 73.4°C in a rigid vessel is 71622 J.
To calculate the energy required to change the temperature of 1.00 kg of ethane (C₂H₆) from 25.0°C to 73.4°C in a rigid vessel, we need to use the specific heat capacity at constant volume ([tex]C_v[/tex]) and the number of moles of ethane.
Given:
Mass of ethane (m) = 1.00 kgInitial temperature (T₁) = 25.0°CFinal temperature (T₂) = 73.4°CMolar specific heat capacity at constant volume ([tex]C_v[/tex]) = 44.60 J/K/molMolar mass of ethane [tex](\text{C}_2\text{H}_6) = 30.07 g/mol[/tex]Step-by-Step Calculation
1. Convert the mass of ethane to moles:
[tex]n = \frac{m}{M}[/tex]
where
m is the mass of ethane in grams,M is the molar mass of ethane.[tex]n = \frac{1000 \, \text{g}}{30.07 \, \text{g/mol}} \\\\n \approx 33.25 \, \text{mol}[/tex]
2. Calculate the temperature change:
[tex]\Delta T = T_2 - T_1 \\\\\Delta T = 73.4^\circ\text{C} - 25.0^\circ\text{C}\\\\\Delta T = 48.4 \, \text{K}[/tex]
3. Calculate the energy required using the formula:
[tex]Q = n \cdot C_v \cdot \Delta T[/tex]
where
Q is the energy required,n is the number of moles,[tex]C_v[/tex] is the molar specific heat capacity at constant volume,ΔT is the temperature change.[tex]Q = 33.25 \, \text{mol} \cdot 44.60 \, \text{J/K/mol} \cdot 48.4 \, \text{K}\\\\Q \approx 33.25 \cdot 44.60 \cdot 48.4\\\\Q \approx 71621.9 \, \text{J}[/tex]
So, the energy required to change the temperature of 1.00 kg of ethane from 25.0°C to 73.4°C in a rigid vessel is approximately:
[tex]Q \approx 71622 \, \text{J}[/tex]
The energy required is 71622 J.