Answers
Answer:
4
Explanation:
the one you ARE ON
Related Questions
The density of a sample of gasoline is 0.70 g/cm3. What is the mass of 1 liter of this gasoline?
Group of answer choices
0.7 g
70 g
700 g
1,429 g
Answers
Answer:
700g
Explanation:
Given parameters:
Density of gasoline = 0.7g/cm³
Volume of gasoline = 1L = 1000cm³
Unknown:
Mass of the gasoline = ?
Solution:
Density is the mass per unit volume of a substance. It can be expressed as;
Density = [tex]\frac{mass}{volume}[/tex]
So;
Mass = density x volume
Mass = 0.7 x 1000 = 700g
Classify each of the following compounds as a strong acid, weak acid, strong base, or weak base, and write the Ka expression for any weak acid or weak base:
1. [ Select ] ["strong base", "weak base", "strong acid", "weak acid"] LiOH
2. [ Select ] ["weak acid", "strong acid", "strong base", "weak base"] HF
3. [ Select ] ["strong acid", "weak acid", "strong base", "weak base"] HCl
4. [ Select ] ["weak base", "strong base", "weak acid", "strong acid"] NH3
Ka expression: [ Select ] ["[H+][F-] / [HF]", "[Li+][OH-]/ [LiOH]", "[H+][Cl-} / [HCl]", "[NH4+] / [NH3]", "[HF] / [H+][F-}", "[LiOH] / [Li+][OH-]", "[HCl] / [H+][Cl-}", "none"]
Calculate the concentration of OHLaTeX: -? in a solution that has a concentration of H+ = 7 x 10LaTeX: -?6 M at 25°C. Multiply the answer you get by 1010 and enter that into the field to 2 decimal places.
Answers
Answer:
See explanation below
Explanation:
There are several ways to know if an acid or base is strong. One method is calculating the pH. If the pH is really low, is a strong acid, and if it's really high is a strong base.
However we do not have a pH value here.
The other method is using bronsted - lowry theory. If an acid is strong, then his conjugate base is weak. Same thing with the bases.
Now, Looking at the 4 compounds, we can say that only two of them is weak and the other two are strong compounds. Let's see:
LiOH ---> Strong. If you try to dissociate :
LiOH ------> Li⁺ + OH⁻ The Li⁺ is a weak conjugate acid.
HF -----> Weak
HF --------> H⁺ + F⁻ The Fluorine is a relatively strong conjugate base.
HCl -----> Strong
This is actually one of the strongest acid.
NH₃ ------> Weak
Now writting the Ka and Kb expressions:
Ka = [H⁺] [F⁻] / [HF]
Kb = [NH₄⁺] [OH⁻] / [NH₃]
Finally, to calculate the [OH⁻] we need to use the following expression:
Kw = [H⁻] [OH⁻]
Solving for [OH⁻] we have:
[OH⁻] = Kw / [H⁺]
Remember that the value of Kw is 1x10⁻¹⁴. So replacing:
[OH⁻] = 1x10⁻¹⁴ / 7x10⁻⁶
[OH⁻] = 1.43x10⁻⁹ M
And now, multiplying by 10¹⁰ we have:
[OH⁻] = 1.429x10⁻⁹ * 1x10¹⁰
[OH⁻] = 14.29Hope this helps
Strong acids and bases are those which completely ionized in body fluid, and weak acids and bases are those who does not completely ionized in body fluid.
Ka expression is used to differentiate between strong and weak acids.
Which are strong acids and base and weak acids and bases?LiOH - strong baseHF - weak acidHCl - strong acidNH3 - weak baseWhat are the Ka expression of the following?Weak acid – HF[tex]\bold{\dfrac{[H+][F-]}{[HF]}}[/tex]
Weak base – NH3[tex]\bold{\dfrac{[NH_4^+] [OH^-]}{[NH_3]} }[/tex]
Calculate the concentration of OH?Given, [tex]\bold{ [H^+]=1\times10^-^6\; at \;25^oC}[/tex]
We know, [tex]\bold{ [H^+]\times[OH^-]=1\times10^-^6\; at \;25^oC}[/tex]
[tex]\bold{[OH^-]=\dfrac{1\times10^-^1^4}{6.2\times10^-^6} = 1.43\times10^-^9}[/tex]
Now, multiplying the value by [tex]10^1^0[/tex]
[tex]\bold{( 1.429\times10^-^9) \times 1\times10^1^0= 14.29}[/tex]
Thus, the value is 14.29.
Learn more about acid and base, here:
https://brainly.com/question/10468518
Calculate the number of oxygen atoms in a 50.0g sample of scheelite CaWO4
Answers
Answer:
0.696 atoms of oxygen
Explanation:
We'll begin by calculating the number of mole in 50 g of scheelite CaWO₄. This can be obtained as follow:
Mass of CaWO₄ = 50 g
Molar mass of CaWO₄ = 40 + 184 + (4×16)
= 40 + 184 + 64
= 288 g/mol
Mole of CaWO₄ =?
Mole = mass / Molar mass
Mole of CaWO₄ = 50 / 288
Mole of CaWO₄ = 0.174 mole
Finally, we shall determine the number of oxygen atom in 50 g (i.e 0.174 mole) of CaWO₄. This can be obtained as follow:
1 mole of CaWO₄ contains 4 atoms of oxygen.
Therefore, 0.174 mole of CaWO₄ will contain = 0.696 atoms of oxygen.
Thus, 50 g (i.e 0.174 mole) of CaWO₄ contains 0.696 atoms of oxygen.
Gravity pulls rain and snow down to Earth from the atmosphere through a paire
process called precipitation Water is pulled from elevated areas such as
mountains and hills into lakes, oceans, and water reserviors. What is this
describing?*
role of gravity in the water cycle
role of gravity in condensation
O
role of gravity in evaporation
role of gravity in precipitation
Answers
LaKeisha is measuring the density of a solid piece of metal using the graduated cylinder method. She initially measures a volume of water in the cylinder to be 3.28 mL. After placing the metal into the graduated cylinder, the new volume was 8.72 mL. The mass of the metal was 42.26 g on a top loading balance.
Required:
What is the density of the metal calculated to the correct number of significant figures?
Answers
Answer: 7.77 g/ml
Explanation:
Volume of cylinder with only water = 3.28 mL
Volume of cylinder with water and metal = 8.72 mL
Volume of metal = (Volume of cylinder with water and metal ) -(Volume of cylinder with only water)
=8.72-3.28
=5.44 ml
Mass of metal = 42.26 g
Formula of Density = [tex]\dfrac{\text{Mass}}{\text{Volume}}[/tex]
i.e. the density of the metal = [tex]\dfrac{42.26}{5.44}\approx7.77\text{ g/ml}[/tex]
Hence, the density of metal = 7.77 g/ml
You want to clean a 500-ml flask that has been used to store a 0.9M solution. Each time the flask is emptied, 1.00 ml of solution adheres to the walls, and thus remains in the flask. For each rinse cycle, you pour 9.00 ml of solvent into the flask (to make 10.00 ml total), swirl to mix uniformly, and then empty it. What is the minimum number of such rinses necessary to reduce the residual concentration of 0.00001 M or below
Answers
Answer:
In the 5th cycle rinse, the residual concentration of the solution is < 0.00001M
Explanation:
In each rinse cycle, the dilution that you are doing of the solution is from 1.00mL to 10.00mL, that is a dilution of 10
In the first rinse the concentration must be of 0.9M 10 = 0.09M
2nd = 0.009M
3rd = 0.0009M
4th = 0.00009M
5th = 0.000009M →
In the 5th cycle rinse, the residual concentration of the solution is < 0.00001MJoseph Priestly is frequently credited with the discovery of oxygen, and was reported to have produced molecular oxygen from the decomposition reaction of mercury(II) oxide, which is the reverse of the synthesis of HgO depicted in the following equation. 4 Hg(l) + 2 O2(g) LaTeX: \rightarrow → 4 HgO(s) Determine the value of LaTeX: \Delta ΔH°rxn for the synthesis, given that
Answers
Joseph Priestly is frequently credited with the discovery of oxygen, and was reported to have produced molecular oxygen from the decomposition reaction of mercury(II) oxide, which is the reverse of the synthesis of HgO depicted in the following equation. [tex]4Hg(l)+2O_2(g)\rightarrow 4 HgO(s) [/tex]Determine the value of [tex]\Delta ΔH°rxn[/tex] for the synthesis, given that [tex]\Delta H_f^0[/tex] for HgO is -90.7 kJ/mol.
Answer: The enthalpy change for this reaction is, -362.8 kJ
Explanation:
The balanced chemical reaction is,
[tex]4Hg(l)+2O_2(g)\rightarrow 4HgO(s)[/tex]
The expression for enthalpy change is,
[tex]\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]
[tex]\Delta H=[(n_{HgO}\times \Delta H_{HgO})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{Hg}\times \Delta H_{Hg})][/tex]
where,
n = number of moles
[tex]\Delta H_{O_2}=0[/tex] (as heat of formation of substances in their standard state is zero
[tex]\Delta H_{Hg}=0[/tex] (as heat of formation of substances in their standard state is zero
Now put all the given values in this expression, we get
[tex]\Delta H=[(4\times -90.7)]-[(2\times 0)+(4\times 0)][/tex]
[tex]\Delta H=-362.8kJ[/tex]
Therefore, the enthalpy change for this reaction is, -362.8 kJ
You want to compare the malleability of
two metals. Plan an investigation that would allow you to determine
which metal is more malleable .
Answers
The temperature and the strength of the metallic link are the two most crucial variables that can impact how malleable a metal or alloy is.
What is metal ?A metal is a substance that has a shiny look when freshly processed, polished, or shattered, and conducts electricity and heat rather effectively. Generally speaking, metals are malleable and ductile.
The amount of pressure that a metal can sustain without breaking can be used to gauge its malleability. Varied metals have different degrees of malleability because of variations in their crystal structures.
The temperature of the metal and the strength of the metallic connection are the two parameters that define how malleable a metal or metal alloy will be.
Thus, The kind of the metallic connection can have a significant impact on how easily metal atoms can rearrange themselves.
To learn more about metal, follow the link;
https://brainly.com/question/18153051
#SPJ2
How do the valence electrons of an element determine how they will combine with other elements to produce a compound? Please help this is urgent :)
Answers
Answer:
See explanation
Explanation:
The valence electrons are electrons found on the valence (outermost) shell of an atom.
When an atoms form compounds, there is an exchange of valence electrons between the atoms of one element and the atoms of another element.
Let us consider a typical example, sodium has one valence electron and chlorine has seven valence electrons. This means that chlorine needs one electron to complete its octet while sodium needs to release one electron in order to attain the octet structure.
So, sodium gives out its one electron and becomes a stable sodium ion and chlorine accepts that electron and becomes a stable chloride ion. This is how the compound sodium chloride is formed.
A change of state is a(n)
process.
A. irreversible
B. reversible
Answers
Answer:
Changes of states are reversible, you can go from a solid to liquid and liquid to solid.Answer:
Reversible
Explanation:
Changes of state are physical changes in matter. Common changes of the state include melting, freezing, sublimation, deposition, condensation, and vaporization.
How many orbitals in an atom can have each of the following designations:
(a) 1s;
(b) 4d;
(c) 3p;
(d) n=3?
Answers
Answer:
(a) 1s; has one orbital
(b) 4d; has five orbitals
(c) 3p; has three orbitals
(d) n=3 has nine orbitals
Explanation:
Electrons in an atom are always in constant motion, making it hard to predict there exact position. However, the most probable locations electrons can be be found are described with the terms shells, subshells and orbitals. A shell contains subshells and orbitals are found within subshells. The shells are given names such as K, L, M, N, which correspond to the principal quantum numbers, n = 1, 2, 3, and 4 respectively. There are 4 major types of subshells that can be found in a shell. They are named as s, p, d, f. Each subshell is composed of several orbitals.
a. 1s; the s subshell has only one orbital. Therefore, the 1s subshell has one orbital
b. 4d; the d subshell has five orbitals. Therefore, the 4d subshell has five orbitals
c. 3p; the p subshell has three orbitals. Therefore, the 3d subshell has three orbitals
d. n = 3; the shell with n = 3 has the following subshells, 3s, 3p, 3d.the number of orbitals will be 1 + 3 + 5 = 9 orbitals. Therefore, the number of orbitals in n = 3 is nine orbitals
An atom has 81 electrons, 84 neutrons, and 82 protons. What element is this atom?
Answers
Answer:
Lead
Explanation:
The subatomic particles within an atom can be used to know the atom or element given.
Of particular interest is the number of protons within the atom.
The periodic table is based on the atomic number of atoms. This atomic number is the number of protons within an atomic space.
So; If we know the number of protons within an atom, we can know the element.
The number of protons given is 82, the element is therefore lead.
Answer:
The atomic number of polonium is 84. The atomic number lead is 82.
Explanation:
Calculate the percent composition (percent by mass of each element) of NH4Cl.
Round to the nearest ONES place ((example: 12.34% = 12%))
Answers
Answer:
[tex]\%N=26.2\%\\\\\%H=7.5\%\\\\\%Cl=66.3\%[/tex]
Explanation:
Hello!
In this case, since the calculation of the percent composition of an element in a chemical compound is computing considering its atomic mass, subscript in the formula and molecular mass of the compound it is; for nitrogen, hydrogen and chlorine we have that ammonium chloride has a molar mass of 53.49 g/mol so the percent compositions are:
[tex]\%N=\frac{14.01*1}{53.49}*100\% =26.2\%\\\\\%H=\frac{1.01*4}{53.49}*100\% =7.5\%\\\\\%Cl=\frac{35.45*1}{53.49}*100\% =66.3\%[/tex]
Best regards!
To determine the concentration of citric acid, you will need to titrate this solution with 0.100 M NaOH. You are given a 1.00 M NaOH stock solution and will need to make enough 0.100 M NaOH to perform 3 titrations. For each titration, you will use 20.0 mL of 0.100 M NaOH solution.
Calculate the total volume (in mL) of the diluted solution you will need to prepare for the 3 titrations.
Determine the minimum volume (in mL) of 1.00 M NaOH stock solution needed to prepare the 0.100 M NaOH solution.
Answers
Answer:
60.0mL of the diluted solution are needed
6.00mL of the 1.00M NaOH stock solution is the minimum volume needed to prepare the diluted solution.
Explanation:
As in each titration we need to use 20.0mL of the diluted 0.100M solution. As there are 3 titration, the volume must be:
3 * 20.0mL = 60.0mL of the diluted solution are needed
Now, to prepare a 0.100M NaOH solution from a 1.00M NaOH stock solution the dilution must be of:
1.00M / 0.100M = 10 times must be diluted the solution.
As we need at least 60.0mL, the minimum volume of the stock solution must be:
60.0mL / 10 times =
6.00mL of the 1.00M NaOH stock solution is the minimum volume needed to prepare the diluted solution.A chemist prepares a solution of aluminum sulfate by weighing out of aluminum sulfate into a volumetric flask and filling the flask to the mark with water. Calculate the concentration in of the chemist's aluminum sulfate solution. Be sure your answer has the correct number of significant digits.
Answers
Answer:
25.8 g/dL
Explanation:
A chemist prepares a solution of aluminum sulfate by weighing out 116.0 g of aluminum sulfate into a 450. mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in g/dL of the chemist's aluminum sulfate solution. Be sure your answer has the correct number of significant digits.
Step 1: Given data
Mass of aluminum sulfate (m): 116.0 gVolume of the solution (V): 450. mLStep 2: Convert "V" to dL
We will use the following conversion factors.
1 L = 1000 mL1 L = 10 dL450. mL × 1 L/1000 mL × 10 dL/1 L = 4.50 dL
Step 3: Calculate the concentration (C) of aluminum sulfate if g/dL
We will use the following expression.
C = m/V = 116.0 g/4.50 dL = 25.8 g/dL
water is a unique material in that the density of the solid is lower than the density of the liquid (which is why ice forms at the top of a pond and why ice floats in our drinks). if the density for ice at 0C is .917g/mL and the density for water at 0C is .999 g/mL, what is the calculated free space (as %) for each of these materials. you will need to estimate the volume of water as the sum of 2 H atoms and 1 O atom with radii 37 and 66 pm respectively. note that you will also have to assume a quantity of water to perform this exercise
Answers
Answer:
% Free space in water = [tex]\frac{9.945* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.45%
% Free space in ice = [tex]\frac{9.98* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.8%
Explanation:
As given ,
Density for ice at 0⁰C = 0.917 g/ml
Density for water at 0⁰C = 0.999 g/ml
Radii of H atoms = 37 pm
Radii of O atoms = 66 pm
Now,
Consider 1 ml of water = 1 cm²
As , we know that mass of water in 1 cm² = 0.999 g
Moles of water = [tex]\frac{0.999}{18} = 0.056[/tex]
Volume of H₂O = 1.624×[tex]10^{-31}[/tex] m²
Now,
Volume occupied by water = 0.056×6.022×[tex]10^{23}[/tex]× 1.624×[tex]10^{-31}[/tex] m²
= 5.48×[tex]10^{-9}[/tex] m²
⇒Volume occupied by water = 5.48×[tex]10^{-9}[/tex] m²
Now,
Free space = 1×[tex]10^{-6}[/tex] - 5.48×[tex]10^{-9}[/tex] = 9.95×[tex]10^{-7}[/tex] m²
% Free space = [tex]\frac{9.945* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.45%
Now,
Consider 1 ml of ice = 1 cm²
S.I unit of ice = 1×[tex]10^{-6}[/tex] m²
As , we know that mass of water in 1×[tex]10^{-6}[/tex] m² = 0.917 g
Moles of ice = [tex]\frac{0.917}{18} = 0.012[/tex]
Volume of H₂O = 6.022×[tex]10^{23}[/tex] ×0.012
Volume of ice unit = [tex]\frac{4}{3} \pi (37*10^{-12})^{3} *2 + \frac{4}{3} \pi (66*10^{-12})^{3} = 1.624*10^{-31}m^{3}[/tex]
Now,
Volume occupied by water = 0.012×6.022×[tex]10^{23}[/tex]× 1.624×[tex]10^{-31}[/tex] m²
= 1.17×[tex]10^{-9}[/tex] m²
⇒Volume occupied by water = 1.17×[tex]10^{-9}[/tex] m²
Now,
Free space = 1×[tex]10^{-6}[/tex] - 1.17×[tex]10^{-9}[/tex] = 9.98×[tex]10^{-7}[/tex] m²
% Free space = [tex]\frac{9.98* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.8%
A 1.0 mol sample of he(g) at 25 is mixed with a 1.0 mol sample of Xe(g) at 50 C. What would be the changes in average kineeteic energy and the average speed of the Xe atoms that will occur as the mixture approaches thermal equilibrium?
Answers
Answer:
Explanation:
The average kinetic energy for an ideal gas is directly proportional to the temperature. The average kinetic energy of the gas is a measure of the temperature of the gas molecule
Also, the average speed is usually proportional to the square root of temperature.
Similarly, there is a noticeable increase in K.E and speed in regard to temperature but sometimes it is not usually proportional.
However, provided that there is more temperature in Xe as compared to He, then after the mixture of both takes place at equilibrium; the temperature tends to fluctuate between (25 - 50)°C
Thus, since there is a decrease in temperature in Xe, both the average kinetic energy as well as the speed too will also decrease.
In a space shuttle, the Carbon dioxide, CO2 that the crew exhales is removed from the air by a reaction within canisters of Lithium Hydroxide, LiOH. The LiOH is only 85% efficient. On average, each astronaut exhales around 20.0 mol of CO2 every day. What volume of water is produced when the CO2 reacts with the excess LiOH
Answers
Answer:
What volume of water is produced when the CO2 reacts with the excess LiOH
X = 360 mL H2O
Explanation:
CO2 (g) + 2 LiOH(s) ⇒ Li2CO3 (aq) + H2O(l)
20.0 mol excess x g
X = 360 mL H2O
x mL H20 = 20.0 mol CO2 (1 mol H2O /1 mol CO2)(18 g H2O/1 mol H2O)
(1 mL H2O /1 g H2O)
X = 360 mL H2O
A reaction between liquid reactants takes place at in a sealed, evacuated vessel with a measured volume of . Measurements show that the reaction produced of sulfur hexafluoride gas. Calculate the pressure of sulfur hexafluoride gas in the reaction vessel after the reaction. You may ignore the volume of the liquid reactants. Round your answer to significant digits.
Answers
Answer:
0.41 atm
Explanation:
A reaction between liquid reactants takes place at 10.0 °C in a sealed, evacuated vessel with a measured volume of 5.0 L. Measurements show that the reaction produced 13. g of sulfur hexafluoride gas. Calculate the pressure of sulfur hexafluoride gas in the reaction vessel after the reaction. You may ignore the volume of the liquid reactants. Round your answer to 2 significant digits.
Step 1: Given data
Temperature (T): 10.0 °CVolume of the vessel (V): 5.0 LMass of sulfur hexafluoride gas (m): 13. gStep 2: Convert "T" to Kelvin
We will use the following expression.
K = °C + 273.15
K = 10.0 °C + 273.15 = 283.2 K
Step 3: Calculate the moles (n) of SF₆
The molar mass of SF₆ is 146.06 g/mol.
13. g × 1 mol/146.06 g = 0.089 mol
Step 4: Calculate the pressure (P) of SF₆
We will use the ideal gas equation.
P × V = n × R × T
P = n × R × T/V
P = 0.089 mol × (0.0821 atm.L/mol.K) × 283.2 K/5.0 L = 0.41 atm
A sample of PCl5 weighting 2.69 gram was placed in 1.00 Litter container and completely vaporized at 250C. The pressure observed at that temperature was 1.00 atm. The possibility exists that some of the PCl5 dissociated according to PCl5 (g) ! PCl3 (g) Cl2 (g) . What must be the partial pressures of PCl5 PCl3 and Cl2 under these experimental conditions
Answers
Answer:
Partial pressures:
PCl₅ = 0.558 atm
PCl₃ = 0.22 atm
Cl₂ = 0.22 atm
Explanation:
From the given information:
The number of moles of PCl₅ associated with the evaporation is:
[tex]n_{PCl_5}= \dfrac {weight \ of \ PCl_5} {M.Wt. \ of \ PCl_5}[/tex]
[tex]n_{PCl_5}= \dfrac {2.69 \ g} {208.5 \ g/mol}[/tex]
[tex]n_{PCl_5}= 0.013 \ mol[/tex]
Temperature of the gas = 250° C = (250 + 273.15) K
= 523.15 K
Using the Ideal gas equation to determine the pressure exerted by the completely vaporized PCl₅
PV = nRT
[tex]P = \dfrac{nRT}{V}[/tex]
[tex]P = \dfrac{0.0013 \ mol \times 0.082 \ Latm^0 K^{-1} . mol ^{-1} \times 523.15 \ K}{1.0 \ L}[/tex]
P = 0.558 atm
Thus, at 250° C, decomposition of PCl₅ occurs.
In the container, PCl₅ decomposes to PCl₃ and Cl₂.
i.e.
[tex]PCl_{5(g)} \to PCl_{3(g)}+ Cl_{2(g)}[/tex]
Using Dalton's Law:
[tex]P_{total } =P_1 + P_2+P_3 +...[/tex]
[tex]P_1 = P_{Total} \times X_1[/tex]
where;
X = mole fraction
Then, the total no. of moles in the container is:
[tex]n = \dfrac{PV} {RT}[/tex]
[tex]n = \dfrac{1\ atm \times 1.0\ L}{0.0821 \ L \ atm \ K^{-1}.mol \times 523.15\ K}[/tex]
n = 0.023 mol
Now, the container contains a total amount of 0.023 mol where initially 0.013 mol are that of PCl₅ and remaining 0.005 mol of PCl₃ and 0.005 mol of Cl₂.
Thus, the partial pressure of PCl₃ is:
[tex]P__{PCL_3} }= P_{total} \times \dfrac{no. \ of \ moles \ of PCl_5}{total \ no. \ of \ moles}[/tex]
[tex]P__{PCL_3}} = 1 \ atm \times \dfrac{0.005}{0.023}[/tex]
[tex]P__{PCL_3}} = 0.22 \ atm[/tex]
Thus, since the no of moles of PCl₃ and Cl₂ are the same, then the partial pressure for Cl₂ is = 0.22 atm
The reactants of two chemical equations are listed.
Equation 1: AgNO3 + Zn
Equation 2: AgNO3 + MgCl2
Based on the type of reaction, which reaction can be used to extract silver metal from silver nitrate solution?
Answers
Answer: Equation 1, because Zn being more reactive, replaces Ag from AgNO3
Explanation: I got it right on the quiz and it replaces it
A particular term in an atom in which LS coupling is a good approximation splits into three levels, each having the same L and same S but different J. If the relative spacings between the levels are in the proportion 5:3, find L and S.
Answers
Answer:
Explanation:
From the information given;
Consider using Lande's Interval rule which can be expressed as:
[tex]\Delta E = E_{j+1} - E_jj \ = \alpha (j+1)[/tex]
here;
[tex]j+1[/tex] = highest level of j
and
[tex]\dfrac{\Delta E_1}{\Delta E_2} = \dfrac{(j+2)}{(j+1)}[/tex]
[tex]\dfrac{5}{3} = \dfrac{(j+2)}{(j+1)}[/tex]
[tex]5(j+1) = 3(j+2)[/tex]
[tex]5j+5 = 3j+6[/tex]
[tex]2j = 1\\ \\ j = \dfrac{1}{2}[/tex]
recall that:
[tex]j = |S-L| \ \to \ |S+L |[/tex]
So;
[tex]S-L = \dfrac{1}{2} --- (1)[/tex]; &
[tex]S+L = \dfrac{5}{2} --- (1)[/tex]
Using the elimination method, we have:
[tex]2S = \dfrac{6}{2}[/tex]
[tex]S = \dfrac{3}{2}[/tex]
Since [tex]S = \dfrac{3}{2}[/tex]; then from (1)
[tex]\dfrac{3}{2} -L = \dfrac{1}{2}[/tex]
[tex]L = \dfrac{2}{2}[/tex]
[tex]L = 1[/tex]
Vinegar is insoluble in vegatable oil. Does this mean that vinegar is a totally insoluble substance?
Answers
Answer:
No
Explanation:
This does not mean that vinegar is insoluble totally. In fact, vinegar is soluble in water because water is a polar solvent.
For a substance to be soluble in another, it must obey the rule of solubility.
The rule states that "like dissolves like"
It implies that polar solvent will only dissolve polar solute.
Also, non-polar solvent will only dissolve non-polar solute.
Vegetable oil is a non-polar solventIt cannot dissolve a polar solute such as vinegarTherefore, the answer is no, vinegar will dissolve in water.
Which of the following choices is not evidence supporting the theory of plate tectonics?
Answers
Answer:
B
Explanation:
PLZ HELP ASAP WILL GIVE BRAINLISTS TO RIGHT ANSWER
How many molecules of carbon dioxide are in 12.2 L of the gas at STP?
A) 3.28 x 10^23 molecules
B) 5.01 X 10^23 molecules
C)2.24 x 10^23 molecules
D)8.12 x 10^22 molecules
Answers
Answer:
c
Explanation:
ok than not c than b maybe
0
Which is not one of Earth's layers?
A А
crust
B)
inner core
mantle
D
ocean
Answers
The ocean is not a part of Earth's layers.
Answer:
Ocean
Explanation:
Explain the differences between an ideal gas and a real gas.
Answers
Answer:
Ideal Gas
The ideal gas is extremely small and the mass is almost zero and no volume Ideal gas is also considered as a point mass.
Real Gas
The molecules of real gas occupy space though they are small particles and also have volume.
anation:
The differences between an ideal gas and a real gas are that the ideal gas follows the gas laws perfectly under all conditions. Whereas a real gas deviates from ideal gas behaviors.
The ideal gas law, also known as the general gas equation, is a fundamental principle in thermodynamics and relates the pressure, volume, temperature, and number of moles of an ideal gas.
An ideal gas is a theoretical gas that follows the gas laws perfectly under all conditions of temperature and pressure. It is assumed to have no volume, no intermolecular forces, and elastic collisions between its particles. An ideal gas also obeys the ideal gas law.
On the other hand, a real gas is a gas that does not follow the gas laws perfectly under all conditions of temperature and pressure. Real gases have volume and intermolecular forces that affect their behavior. These forces cause deviations from ideal gas behavior, especially at high pressures and low temperatures.
In summary, while an ideal gas is a theoretical gas that follows the gas laws perfectly under all conditions, a real gas is a gas that deviates from ideal gas behavior due to its volume, intermolecular forces, and non-elastic collisions between its particles.
Learn more about ideal gas law here:
https://brainly.com/question/33342075
#SPJ2
balance the following equation by oxidation reduction method FeSO4
+
KMnO4+ H2SO4 → Fe2 (SO4)3+ k2SO4+MnSO4+H2O
Answers
Answer:
[tex]10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + 8\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + K_2SO_4 + 2\, {Mn}SO_4 + 8\, H_2O[/tex].
Explanation:
Identify the elements with oxidation state changes:
Oxidation states of iron, [tex]\rm Fe[/tex]:
[tex]+2[/tex] in [tex]\rm FeSO_4[/tex] among the reactants.[tex]+3[/tex] in [tex]\rm Fe_2(SO_4)_3[/tex] among the products.Change to the oxidation state: [tex]+1[/tex] (oxidation) for each [tex]\rm Fe[/tex] atom.Oxidation state of manganese, [tex]\rm Mn[/tex]:
[tex]+7[/tex] in [tex]\rm KMnO_4[/tex] among the reactants.[tex]+2[/tex] in [tex]\rm MnSO_4[/tex] among the products.Change to the oxidation state: [tex](-5)[/tex] (reduction) for each [tex]\rm Mn[/tex] atom.The change in the oxidation state of [tex]\rm Mn[/tex] is five times the opposite of the change to the oxidation state of [tex]\rm Fe[/tex]. If there are one mole of [tex]\rm Mn\![/tex] atoms in each mole of this reaction, there would be five times as many [tex]\rm Fe\![/tex] atoms per mole reaction. In other words:
[tex]\displaystyle 5\, \overset{+2}{\rm Fe}\rm SO_4 + 1\, \rm K \overset{+7}{Mn} O_4 + ?\, H_2SO_4\\ \to \frac{5}{2}\, \overset{+3}{Fe} (SO_4)_3 + ?\, K_2SO_4 + 1\, \overset{+2}{Mn}SO_4 + ?\, H_2O[/tex].
(Notice that each mole of this reaction would include five times as many [tex]\rm Fe[/tex] atoms as [tex]\rm Mn[/tex] atoms.)
Multiply the coefficients by [tex]2[/tex] to eliminate the fraction:
[tex]\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + ?\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + ?\, K_2SO_4 + 2\, {Mn}SO_4 + ?\, H_2O[/tex].
Find the unknown coefficients using the conservation of atoms.
Reactants:
[tex]2[/tex] potassium [tex]\rm K[/tex] atoms in two [tex]\rm K_2SO_4[/tex] formula units.Therefore, among the products:
[tex]2[/tex] potassium [tex]\rm K[/tex] atoms in one [tex]\rm K_2SO_4[/tex] formula unit.[tex]\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + ?\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + {1}\, K_2SO_4 + 2\, {Mn}SO_4 + ?\, H_2O[/tex].
Products:
[tex]5 \times 3 + 2 + 1 = 18[/tex] sulfur [tex]\rm S[/tex] atoms in five [tex]\rm Fe_2(SO_4)_3[/tex] formula units, two [tex]\rm K_2 SO_4[/tex] formula units, and one [tex]\rm MnSO_4[/tex] formula unit.Reactants:
There are already ten [tex]\rm S[/tex] atoms in that ten [tex]\rm Fe(SO_4)_2[/tex] formula units. The other [tex]18 - 10 = 8[/tex] formula units would correspond to eight [tex]\rm H_2SO_4[/tex] molecules among the reactants of this reaction.[tex]\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + 8\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + {1}\, K_2SO_4 + 2\, {Mn}SO_4 + ?\, H_2O[/tex].
Products:
There are [tex]8 \times 2 = 16[/tex] hydrogen [tex]\rm H[/tex] atoms in that eight [tex]\rm H_2SO_4[/tex] molecules.Therefore, among the products:
There would be [tex]16 / 2 = 8[/tex] molecules of [tex]\rm H_2O[/tex], with two [tex]\rm H[/tex] atoms in each [tex]\rm H_2O\![/tex] molecule.[tex]\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + 8\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + {1}\, K_2SO_4 + 2\, {Mn}SO_4 + 8\, H_2O[/tex].
When measuring the volume of a liquid, how would sample size (e.g., using a 10 mL graduated cylinder vs. a 100 mL graduated cylinder to measure out 70 mL of a liquid) affect the absolute error and percentage error in the measured values of mass and volume and therefore the density
Answers
Answer:
Explanation:
From the given information:
The accuracy depends on the internal diameter of the cylinder. The cylinder with the least internal diameter is obviously more precise.
Let's assume 1% is the error of measurement.
Then, to measure 70 mL from 10 mL cylinder
The error = [tex]10 \times \dfrac{1}{100} \times 7[/tex]
= 0.7 mL
However; for a 100 mL cylinder, the error = 1 mL
Now,
The total volume for 10 mL = (70 + 0.7) = 70.7 mL
The total volume for 100 mL = (70 + 1 ) = 71 mL
Suppose the density (d) is same for both
Then;
the mass of 10 mL = ( d × 70.7) g
the mass pf 100 mL = (d × 71) g
Thus, the mass of 100 mL is greater than that of 10 mL.
Someone please help i don’t have much time left
Answers
Answer: Energy of reactants = 30, Energy of products = 10
Exothermic
Activation energy for forward reaction is 10.
Explanation:
Exothermic reactions are defined as the reactions in which energy of the product is lesser than the energy of the reactants. The total energy is released in the form of heat and [tex]\Delta H[/tex] for the reaction comes out to be negative.
Energy of reactants = 30
Energy of products = 10
Thus as energy of the product < energy of the reactant, the reaction is exothermic.
Activation energy [tex](E_a)[/tex] is the extra energy that must be supplied to reactants in order to cross the energy barrier and thus convert to products.
[tex]E_a[/tex] for forward reaction is (40-30) = 10.
