2 M solution of a compound can be made by dissolving 2g of compounds in 1 liter of water. Either by diluting the appropriate volume of a more concentrated solution also known as a stock solution to the desired final volume or by dissolving a known mass of solute in a solvent, solutions of known concentration can be prepared.
For example, when we add one part by weight of powder to 19 parts by weight of the solvent is used to make a 5% solution. For 1% solution, the addition of 1gm of NaCl to 100L of water. which is the final volume of the solution.
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Completely describe the electrolytic cell corresponding to the following equation. (Hint: you may need to combine 2 half reactions from Table 17-1 to make one of the half reactions for this cell)
Cr2O7^2– + I^– → Cr^3+ + IO3^–
With work please
The first half-reaction is the oxidation of Cr2O7^2– to Cr^3+ and the second half-reaction is the reduction of I^– to IO3^–. When combined, the overall reaction is Cr2O7^2– + I^– → Cr^3+ + IO3^–.
The electrolytic cell consists of two electrodes, one anode and one cathode, both of which are immersed in an electrolyte solution. At the anode, the Cr2O7^2– ions are oxidized to Cr^3+ ions, releasing electrons into the external circuit.
At the cathode, the I^– ions are reduced to IO3^– ions, and the electrons from the external circuit are used to drive the reaction. The electrolyte solution must contain both Cr2O7^2– and I^– ions in order to facilitate the transfer of electrons between the electrodes.
The overall reaction is driven by the potential difference between the anode and the cathode, which is created by the flow of electrons through the external circuit.
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Calculating equilibrium concentrations when the net reaction proceeds forward
Consider mixture B, which will cause the net reaction to proceed forward.
Concentration (M)initial:change:equilibrium:[XY]0.500−x0.500−xnet→⇌[X]0.100+x0.100+x+[Y]0.100+x0.100+x
The change in concentration, x, is negative for the reactants because they are consumed and positive for the products because they are produced.
Based on a Kc value of 0.140 and the given data table, what are the equilibrium concentrations of XY, X, and Y, respectively?
We must utilize the reaction's equilibrium equation to determine the equilibrium concentrations:
Kc = [X][Y]/[XY]
replacing the specified values:
0.140 = (0.100+x)(0.100+x)/(0.500-x)
Expanding and condensing:
0.14 = (0.01 + 0.2x + x^2)/(0.5 - x)
0.07 - 0.14x = x^2 + 0.2x + 0.01
Changing the order and applying the quadratic formula to find x
x^2 + 0.34x - 0.06 = 0
x = 0.193 or -0.533
X cannot be negative because it denotes a change in concentration. Consequently, x = 0.193 M.
It is now possible to determine the equilibrium concentrations:
[XY] = 0.500 - x = 0.307 M
[X] = [Y] = 0.100 + x = 0.293 M
Therefore, the equilibrium concentrations of XY, X, and Y are 0.307 M, 0.293 M, and 0.293 M, respectively.
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 If you mixed three plastic cup's contents containing 0,05 L of 1M Kool-Aid, 0.05 L of 2.5M Kool-Aid, and 0.05 L of 0.5M Kool-Aid, what would the molarity be of the resulting solution?
M = 1.183
Molarity is moles/Liter, so setting up some equations gives you
1) 1M = Xmol/0.05L
2) 2.5M = Xmol/0.05L
3) 0.5M = Xmol/0.05L
Solve for each X
1) X = 0.05mol
2) X = 0.125mol
3) X = 0.0025mol
Now add all the moles and Liters (separately)
XM = (0.05+0.125+0.0025)mol/(0.05+0.05+0.05)L
XM = 0.1775mol/0.15L
X = 1.183M
Adding 1.64 g of naphthalene to 80.00 g of cyclohexane, C.H₁2() lowers the freezing point from
5.53°C to 1.37°C. If K, for cyclohexane is 19.9°C/m, calculate the molar mass of the naphthalene.
A cycloalkane with the chemical formula C6H12 is cyclohexane. Non-polar describes cyclohexane.
Thus, Colorless and combustible, cyclohexane has a characteristic detergent-like stench that is reminiscent of cleaning products.
The industrial manufacturing of adipic acid and caprolactam, which are building blocks for nylon, uses cyclohexane primarily.
cyclohexanone and cyclohexanol that serves as the starting point for the production of adipic acid and caprolactam, which are both precursors to nylon. Annual production of cyclohexanone and cyclohexanol totals several million kilos.
Thus, A cycloalkane with the chemical formula C6H12 is cyclohexane. Non-polar describes cyclohexane.
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Readiness Test: Populations, Communities, and Ecosystems
There are unique interactions and relationships which are involved in the transportation of energy. The energy which is produced and captured once is distributed throughout various living organisms. This transfer of energy is called the food web.
The decomposers of the food web are known as the organisms which tend to degrade the dead living organisms of an ecosystem. Some examples of decomposers are fungus, bacterium or invertebrates, etc.
The decomposers help in the energy flow of the ecosystem and help in replenishing the fertility of soil through secreting the digestive enzymes over dead plants and animals and provide nutrients to the soil.
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What volume will 5.00 mol of an ideal gas occupy at 25.0 C. and 153 kPa of pressure?
Answer:
5.00 mol of an ideal gas will occupy 103.6 L at 25.0 C and 153 kPa of pressure.
Explanation:
Using the ideal gas law, PV=nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in Kelvin, we can solve for V.
First, we need to convert the temperature from Celsius to Kelvin by adding 273.15 K. Therefore, the temperature is 25.0 + 273.15 = 298.15 K.
Next, we can plug in the values we know:
PV = nRT
(153 kPa) V = (5.00 mol) (8.31 J/mol*K) (298.15 K)
Simplifying:
V = (5.00 mol) (8.31 J/mol*K) (298.15 K) / (153 kPa)
V = 103.6 L
Therefore, 5.00 mol of an ideal gas will occupy 103.6 L at 25.0 C and 153 kPa of pressure.
Can anyone help me with this??
In the region labelled as H, weather fronts are most likely to be located. A weather front comprises a line dividing air masses with various differences in their attributes,
A weather front comprises a line dividing air masses with various differences in their attributes, including air density, winds, temperatures, or humidity. These variations frequently result in disturbed or unstable weather at the boundary.
For instance, although warm fronts usually arrive accompanied by stratiform precipitation with fog, cold fronts can bring belts of thunderstorms with cumulonimbus precipitation and be preceded by squall lines. Dry lines, which are softer humidity gradients, can cause severe weather in the summer. In the region labelled as H, weather fronts are most likely to be located.
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Luke needs to measure the iron (Fe) content of a 2.470 g steel sample. He starts by completely converting the iron in the steel to Fe2+ and dissolving it in 50.00 mL of water. Then, Luke determines the amount of Fe2+ by reacting it with a standardized aqueous 0.110 M potassium permanganate (KMnO4) solution.
MnO−4+8H++5Fe2+⟶Mn2++5Fe3++4H2O
The titration of the Fe2+ solution requires 39.23 mL KMnO4 solution to reach the end point. What is the percentage by weight of iron (Fe) in the steel sample?
The balanced equation for the reaction between [tex]Fe^+^2[/tex] and KMnO₄ is: MnO₄ + 8H+ + 5 [tex]Fe^+^2[/tex]⟶ [tex]Mn^+^2[/tex] + 5[tex]Fe^3^+[/tex] + 4H₂O. Therefore, the percentage by weight of iron in the steel sample is 1.94%.
Molarity of KMnO₄ = 0.110 M, Volume of KMnO₄ used = 39.23 mL = 0.03923 L
moles of [tex]Fe^+^2[/tex] = 0.110 M x 0.03923 L / 5 = 0.000856 moles
The mass of iron in the steel sample can be calculated from the number of moles of [tex]Fe^+^2[/tex]:
moles of Fe = 0.000856 moles [tex]Fe^+^2[/tex] x 1 mole Fe / 1 mole [tex]Fe^+^2[/tex]
= 0.000856 moles Fe
The mass of Fe can be calculated from the number of moles of Fe:
mass of Fe = 0.000856 moles Fe x 55.85 g/mol = 0.0479 g Fe
Finally, the percentage by weight of iron in the steel sample can be calculated as:
% Fe = (mass of Fe / mass of steel sample) x 100%
mass of steel sample = 2.470 g
% Fe = (0.0479 g / 2.470 g) x 100% = 1.94%
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How does understanding weather patterns help make predicting the weather more accurate?
If the weather seems to have a constant pattern that has barely been or has not been changed, you can assume that there will be a greater chance the weather pattern will remain constant, or at least similar to the pattern.
What mass of oxygen gas, O2, from the air is consumed in the combustion of 702 g of octane, C8H18, one of the principal components of gasoline?
Select one:
A. 13.22 g
O B. 25.0 g/mol
O C. 32.0 g
O D. 33.22 g/mol
O E. 39.0 g
The principal components of gasoline is 702 g of octane.
Thus, Crude oil and other petroleum liquids are used to create gasoline, a fuel. The majority of gasoline is utilized in car engines.
For retail sale at gas stations, finished motor gasoline is produced at petroleum refineries and blending facilities.
In order to create finished motor gasoline, petroleum refineries primarily produce gasoline blendstocks, which must be blended with other liquids. In order to create finished motor gasoline in various grades and formulas, gasoline blendstocks, finished gasoline, and fuel ethanol are combined at blending terminals.
Thus, The principal components of gasoline is 702 g of octane.
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Thank you in advance.
The rate expression is k [BF₃]²[NH₃], overall order is 3, rate constant k is 1.13 × 10⁻³ mol⁻² dm⁶ s⁻¹.
How to calculate rate expression?The rate expression for this reaction can be written as:
Rate = k [BF₃]^m[NH₃]^n
The method of initial rates can be used to calculate the values of m and n. When we compare experiments 1 and 2, notice that halving the concentration of [NH3] reduces the initial rate of reaction. This suggests that the reaction is first order in terms of [NH3], implying that n = 1.
When comparing experiments 1 and 3, notice that increasing the concentration of [BF3] by a factor of 2.5 increases the initial rate of reaction by a factor of 9.39 (i.e., 2.13/0.227). This shows that the reaction is about second order in relation to [BF3], i.e., m 2.
Therefore, the rate expression for the reaction is:
Rate = k [BF₃]²[NH₃]
The overall order of the reaction is m + n = 3.
Using the data from experiment 4, substitute the values and solve for k:
Rate = k [BF₃]²[NH₃]
1.02 × 10-¹ = k (3.00 × 10⁻¹)² (1.00 x 10⁻¹)
k = 1.02 × 10-¹ / (3.00 × 10⁻¹)² (1.00 x 10⁻¹)
k = 1.13 × 10⁻³ mol⁻² dm⁶ s⁻¹
Therefore, the rate constant k is 1.13 × 10⁻³ mol⁻² dm⁶ s⁻¹.
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The products of a gamma decay reaction
are shown below. What was the parent
isotope of this nuclear reaction?
22 0
Ne + Y
10 0
The product of the gamma decay reaction is 22/10 Ne
What is gamma decay ?Gamma radiation, commonly referred to as gamma decay, is a type of radioactive decay in which an unstable atomic nucleus generates gamma rays, a form of extremely high-energy electromagnetic radiation.
Gamma rays are neutral and have no mass, in contrast to alpha and beta particles, which have mass and charge.
The result of the gamma decay would not alter the parent nuclei thus the product would have the same mass and charge as the parent nucleus.
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2.50 g of As2O3 are titrated with 38.5 mL of KMnO4 to reach the end point.
5As2O3(s)+4MnO−4(aq)+9H2O(l)+12H+(aq)⟶10H3AsO4(aq)+4Mn2+(aq)
Calculate the concentration of the KMnO4 solution.
50 g of As[tex]_2[/tex]O[tex]_3[/tex] are titrated with 38.5 mL of KMnO[tex]_4[/tex] to reach the end point. 0.26M is the concentration of the KMnO[tex]_4[/tex] solution.
Concentration in chemistry refers to the quantity of a material in a certain area. The ratio of the solute within a solution to the solvent or whole solution is another way to define concentration. In order to express concentration, mass in unit volume is typically used.
The solute concentration can, however, alternatively be stated in moles or volumetric units. Concentration may be expressed as per unit mass rather than volume.
5As[tex]_2[/tex]O[tex]_3[/tex](s)+4MnO[tex]_4[/tex]⁻(aq)+9H[tex]_2[/tex]O(l)+12H⁺(aq)⟶10H[tex]_3[/tex]AsO[tex]_4[/tex](aq)+4Mn[tex]_2[/tex]⁺(aq)
the stoichiometry ratio between As[tex]_2[/tex]O[tex]_3[/tex] and MnO[tex]_4[/tex]⁻ is 5:4
0.0126 moles of As[tex]_2[/tex]O[tex]_3[/tex] will react with 4/5×0.0126 moles = 0.01008moles
0.01008moles of MnO[tex]_4[/tex]⁻ is present in 38mL
concentration of KMnO[tex]_4[/tex]= moles×volume
= 0.010/38×1000
=0.26M
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6. On a cold day a person takes in a breath of 450.0 mL of air
at 756 mmHg and -10.0°C. Assuming that amount and pressure
remain constant, what is the volume of the air when it warms to
body temperature (37.0°C) in the lungs?
The volume of the air when the breath is warm to body temperature (37.0°C) in the lungs is 530.4 mL
How do I determine the volume at 37 °C?First, we shall list out the given parameters from the question. This is given below:
Initial volume (V₁) = 450 mLPressure = 756 mmHg = ConstantInitial temperature (T₁) = -10 °C = -10 + 273 = 263 KNew temperature (T₂) = 37 °C = 37 + 273 = 310 KNew volume (V₂) =?The volume of the breath when it is warm to body temperature (37.0°C) can be obtained by using Charles' law equation as shown below:
V₁ / T₁ = V₂ / T₂
450 / 263 = V₂ / 310
Cross multiply
263 × V₂ = 450 × 310
263 × V₂ = 139500
Divide both side by 263
V₂ = 139500 / 263
V₂ = 530.4 mL
Thus, the volume is 530.4 mL
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2. Label the different layers of the atmosphere and the separating boundaries between each layer.
The layers of the atmosphere with the boundaries that are between them are shown in the image attached here.
Layers of the atmosphereThe troposphere, which rises up to an altitude of roughly 10-15 km, is the layer that is closest to the Earth's surface. The majority of the Earth's air mass is contained here, which is also where weather happens.
The tropopause marks the transition from the troposphere to the stratosphere, the next layer. It designates the altitude at which the temperature stops dropping.
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Q1-50.0-mL sample of 0.200 M sodium hydroxide is titrated with
0.200 M Hydrochloric acid acid. Calculate:
(a) the pH after adding 30.00 mL of HCl.
(b) the pH at the equivalence point
(c) the pH after you add a total of 60.0 mL of 0.200 M HCl
Round pH values to two decimal places.
Q2- A 40.0-mL sample of 0.100 M HNO2 is titrated with 0.200 M
KOH. Calculate:
(a) the volume required to reach the equivalence point
(b) the pH after adding 5.00 mL of KOH
(c) the pH at one-half the equivalence point
(d) the pH at the equivalence point for the titration between
HNO2 and KOH
The acidity or alkalinity of a solution depends upon its hydronium ion concentration and hydroxide ion concentration. The pH scale is introduced by the scientist Sorensen.
The pH of a solution is defined as the negative logarithm to the base 10 of the hydronium ion concentration in moles per litre.
a. Concentration of NaOH = 0.200 M
Concentration of HCl = 0.200 M
Total volume = 50 + 30 = 80 mL = 0.08 L
The neutralization reaction between NaOH and HCl is:
OH⁻ + H⁺ ----------> H₂O
Then,
n(H⁺) = 0.03 × 0.200 = 0.006 moles
n(OH⁻) = 0.05 × 0.200 = 0.01 moles
Remaining moles of OH⁻ = 0.01 - 0.006 = 0.004
pOH = -log [OH⁻] = 2.39
pH + pOH = 14
pH = 14 - pOH = 14 - 2.39 = 11.61
b. pH at the equivalence point is about pH 7.
c. Total volume = 0.11 L
n(H⁺) = 0.06 × 0.200 = 0.012 moles
n(OH⁻) = 0.05 × 0.200 = 0.01 moles
Remaining moles of H⁺ = 0.012 -0.01 = 0.002
Concentration of H⁺ is 0.002 M / 0.11 L = 0.01818 moles/L
pH = - log [ H⁺ ]
pH = - log [ 0.01818 ]
pH = 1.74
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Help with this
A scientist discovers a new object in the Kuiper Belt and claims it is a dwarf planet. What would have to be true about the object for the scientist to be correct?
A The object would have to be irregularly shaped, have an atmosphere, and be traveling in a path with other objects.
B The object would have to be irregularly shaped, orbit the Sun, and be traveling in a path free from other objects.
C The object would have to be spherical, orbit the Sun, and be traveling in a path with other objects.
D The object would have to be spherical, have an atmosphere, and be traveling in a path free from other objects.
Answer:
C The object would have to be spherical, orbit the Sun, and be traveling in a path with other objects.Explanation:
The sphere is a shape which allows for the most amount of volume with the least amount of surface. It is a very efficient way of enclosing objects and also allows for any point on its surface[tex]\large \boxed{\mathrm{Correct \ answer \ Is \ C.}}[/tex]
[tex]\large \boxed{\mathrm{Sorry \ have \ no \ time \ for \ explanation \ so \ sorry!}}[/tex]
If you have 155 mL solution of a 0.762 M FeCIs solution, how many grams of FeCIs are contained in this sample?
What is the mass in grams of KBr in 0.400 L of a 0.350 M solution?
1. The mass (in grams) of FeCl₂ contained in the sample is 14.96 grams
2. The mass (in grams) of KBr present in the solution is 16.66 grams
How do i determine the mass present?1. The mass of FeCl₂ contained in the sample can be obtained as shown below:
First, we shall obtain the mole
Volume = 155 mL = 155 / 1000 = 0.155 LMolarity = 0.762 MMole of FeCl₂ =?Mole = molarity × volume
Mole of FeCl₂ = 0.762 × 0.155
Mole of FeCl₂ = 0.118 mole
Finally, we shall determine the mass of FeCl₂ present in the sample. Details below:
Mole of FeCl₂ = 0.118 moleMolar mass of FeCl₂ = 126.75 g/molMass of FeCl₂ = ?Mass = Mole × molar mass
Mass of FeCl₂ = 0.118 × 126.75
Mass of FeCl₂ = 14.96 grams
2. The mass of KBr present in the solution can be obtained as shown below:
First, we shall obtain the mole
Volume = 0.4 LMolarity = 0.350 MMole of KBr =?Mole = molarity × volume
Mole of KBr = 0.350 × 0.4
Mole of KBr = 0.14 mole
Finally, we shall determine the mass of KBr in the solution. Details below:
Mole of KBr = 0.14 moleMolar mass of KBr = 119 g/molMass of KBr = ?Mass = Mole × molar mass
Mass of KBr = 0.14 × 119
Mass of KBr = 16.66 grams
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The combustion of octane, C8H18, proceeds according to the reaction shown.
2C8H18(l)+25O2(g)⟶16CO2(g)+18H2O(l)
If 354 mol of octane combusts, what volume of carbon dioxide is produced at 15.0 ∘C
and 0.995 atm?
The concept ideal gas equation is used here to determine the volume of the carbondioxide. Combustion reactions are generally highly exothermic reactions. The volume of CO₂ is
A combustion is a chemical reaction in which a fuel undergoes oxidation as a result of the reaction with an oxidizing agent which causes the release of energy in the form of heat.
15.0 °C = 288 K
The ideal gas equation is:
PV = nRT
V = nRT / P
V = 354 × 0.0821 × 288 / 0.995 = 8412.3 L
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Part D
Now predict the shape of a graph that shows velocity in the x direction (v) versus time
The shape of a graph that shows the velocity in the x direction (v) versus time when the object is accelerating in the x direction would be a curve with a non-zero slope.
What is the velocity of a body?The rate at which an object's displacement changes in relation to a frame of reference is its velocity. The direction of the movement of the body or the object is defined by its velocity.
According to the equation v = s/t, velocity (v) is a vector quantity that quantifies displacement (or change in position, s), over the change in time (t).
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Answer: I expect a horizontal line (slope 0). As time increases, the velocity in the x direction stays about the same.
Explanation: Edmentum sample answer
Determine E° for a galvanic (voltaic) cell if ∆G° = -6.3 kJ/mol and n = 3. (F = 96,500 J/(V･mol))
The E° for a galvanic cell is 0.000217 volts if ∆G° = -6.3 kJ/mol and n = 3. (F = 96,500 J/(V･mol).
A galvanic cell or voltaic cell, named after the scientists Luigi Galvani and Alessandro Volta, respectively, is an electrochemical cell in which an electric current is generated from spontaneous Oxidation-Reduction reactions. A common apparatus generally consists of two different metals, each immersed in separate beakers containing their respective metal ions in solution that are connected by a salt bridge or separated by a porous membrane.
E°=ΔG°/-nF= -6.3/-3×96500=0.000217 V.
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Why does increasing the temperature increase the rate of a reaction?
When temperature increases, the particles have more energy
Increasing temperature prevents products from colliding
Particles moving with more energy will collide more often
Warm temperatures are more comfortable
Particles are more likely to collide with enough energy to form an activated complex
The correct answer is "Particles are more likely to collide with enough energy to form an activated complex," which is the last option, as when the temperature of a reaction increases, the average kinetic energy of the particles increases.
Chemical reactions require a certain amount of energy to occur. This energy is called activation energy. In order for reactant molecules to undergo a chemical reaction, they must overcome the activation energy barrier. This barrier is like a hill that reactants must climb over to reach the product state. One way to provide the energy required to overcome the activation energy barrier is to increase the temperature of the system. This is because temperature is a measure of the average kinetic energy of particles in a system. As the temperature increases, the kinetic energy of the particles also increases. This means that they are moving faster and colliding with greater energy.
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Help with number 17 please
The partial pressure of O₂ is 0.954 atm
The moles O₂ that formed is 0.000859 moles
The mass of NaNO₃ that reacted is 0.146 g.
What is Dalton's law of partial pressures?The balanced chemical equation for the reaction is given below:
2 NaNO₃ (s) → 2 NaNO2 (s) + O₂(g)Using Dalton's Law of partial pressures:
The partial pressure of O₂ = Total pressure - Partial pressure of water vapor
The partial pressure of O₂ = (0.98 atm) - (0.026 atm)
The partial pressure of O₂ is 0.954 atm
b. Using the ideal gas law to determine the moles of oxygen gas formed:
PV = nRT
Making subject of the formula:
n = PV/RT
n = (0.954 * 0.0230)/(0.0821 * 295.15)
n = 0.000859 moles
Moles of O₂ formed = 0.000859 moles
c. the mass of NaNO₃ that reacted is calculated as follows:
mass NaNO₃ = moles NaNO₃ x molar mass NaNO₃
molar mass NaNO = 84.99 g/mol
From the mole ratio of the reaction, moles of NaNO₃ = 0.000859 moles * 2
moles of NaNO₃ = 0.001718 moles
mass NaNO3 = 0.001718 * 84.99
mass NaNO3 = 0.146 g NaNO₃
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Change the following word equations into formula equations AND balance them using the tally method.
*The answers have been provided under each equation, all you have to do is show your work using the tally method!*
a. copper(I) iodate + potassium carbonate → potassium iodate + copper(I)carbonate
2Cu(IO3) + K2(CO3) → 2K(IO3) + Cu2(CO3)
b. iron(III) carbonate + cesium → cesium carbonate + iron
Fe2(CO3)3 + 6Cs → 3Cs2(CO3) + 2Fe
c. sodium chlorate → sodium chloride + oxygen
2Na(ClO3) → 2NaCl + 3O2
The balanced chemical equations with suitable coefficients are 2 Cu(IO₃) + K₂(CO₃) → 2 KIO₃ + Cu₂CO₃, Fe₂(CO₃)₃ + 6 Cs → 3 Cs₂(CO₃) + 2 Fe,
2 NaClO₃ → 2 NaCl + 3 O₂.
Chemical equation is a symbolic representation of a chemical reaction which is written in the form of symbols and chemical formulas.The reactants are present on the left hand side while the products are present on the right hand side.
A plus sign is present between reactants and products if they are more than one in any case and an arrow is present pointing towards the product side which indicates the direction of the reaction .There are coefficients present next to the chemical symbols and formulas .
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What is the difference in energy levels of the sodium atom if emitted light has a wavelength of 589 nm?
To measure the reaction heat more accurately at isobaric condition, what modification(s) would you suggest making on the coffee cup calorimeter? Please justify the suggested change(s).
Cutting or tearing a strip of cardboard sufficiently large to entirely enclose the Styrofoam cup's mouth. This is the modification for making on the coffee cup calorimeter.
A coffee cup calorimeter without a cover serves as the basic component of a coffee cup calorimeter. A thermometer with its bulb below the water's surface can be inserted through the cup's lid after the cup has been partly filled using an established quantity of water.
The quantity of heat that was either absorbed or generated throughout the reaction is determined using the alteration in water temperature. Cutting or tearing a strip of cardboard sufficiently large to entirely enclose the Styrofoam cup's mouth. This is the modification for making on the coffee cup calorimeter.
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What is the pH of 6.00 M H2CO3 if it has 7% dissociation? SHOW YOUR WORK!!!
From the calculation, the pH of the solution can be obtained as 0.39.
What is the pH of the solution?We first have to obtain the dissociation constant of the solution;
α = √Ka/C
Ka = Cα^2
Ka = 4.9 * 10^-3 * 6
Ka = 0.0294
Then we have that;
0.0294 = [x]^2/[6 - x]
0.0294 * [6 - x] = x^2
0.1764 - 0.0294 x = x^2
x^2 + 0.0294 x - 0.1764 =0
x = 0.41 M
Thus the pH of the solution is;
pH = -log[0.41]
= 0.39
pH is an important parameter in chemistry, biology, and many other fields, as it can affect the behavior and properties of substances and reactions.
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Balance equation Pb(C2H3O2)2 + CO2 + H2O= PbCO3 + HC2H3O2
The balanced equation is: Pb(C2H3O2)2 + CO2 + H2O → PbCO3 + 2 HC2H3O2
Balancing chemical equationsIn a balanced chemical equation, the number of atoms of each element is the same on both sides of the equation. Thus, the balanced equation of the reaction is:
Pb(C2H3O2)2 + CO2 + H2O → PbCO3 + 2 HC2H3O2
The coefficients used to balance the equation are:
1 Pb(C2H3O2)2 + 1 CO2 + 1 H2O → 1 PbCO3 + 2 HC2H3O2
After balancing the equation, we can see that one molecule of lead(II) acetate reacts with one molecule of carbon dioxide and one molecule of water to produce one molecule of lead(II) carbonate and two molecules of acetic acid.
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If a sulfuric acid (H₂SO₄, MM = 98.09 g/mol) solution is completely neutralized with 61.37 mL of the NaOH solution above (from question 2) what was the mass in grams of sulfuric acid in the solution. Remember, you need a balanced chemical reaction for stoichiometry. (do not forget about SF)
Taking into account the reaction stoichiometry, 3.00713 grams of H₂SO₄ is completely neutralized with 61.37 mL of the NaOH.
Reaction stoichiometryIn first place, the balanced reaction is:
H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
H₂SO₄: 1 moleNaOH: 2 moles Na₂SO₄: 1 moleH₂O: 2 molesThe molar mass of the compounds is:
H₂SO₄: 98 g/moleNaOH: 40 g/moleNa₂SO₄: 142 g/moleH₂O: 18 g/moleBy reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
H₂SO₄: 1 mole× 98 g/mole= 98 gramsNaOH: 2 moles× 40 g/mole= 80 gramsNa₂SO₄: 1 mole× 142 g/mole= 142 gramsH₂O: 2 moles× 18 g/mole= 36 gramsMass of H₂SO₄ requiredA sulfuric acid solution is completely neutralized with 61.37 mL (0.06137 L) of the NaOH solution. Assuming the concentration of the NaOH solution is 1.0 M, the moles of NaOH that react is calculated as:
Moles of NaOH reacting = concentration× volume in liters
Moles of NaOH reacting = 1 M× 0.06137 L
Moles of NaOH reacting = 0.06137 moles
Now, the following rule of three can be applied: If by reaction stoichiometry 2 moles of NaOH react with 98 grams of H₂SO₄, 0.06137 moles of NaOH react with how much mass of H₂SO₄?
mass of H₂SO₄= (0.06137 moles of NaOH× 98 grams of H₂SO₄)÷2 moles of NaOH
mass of H₂SO₄= 3.00713 grams
Finally, 3.00713 grams of H₂SO₄ is required.
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determine whether each molecule will have hydrogen bonding. defend your answer with evidence from your notes.
A - Has no hydrogen bond since hydrogen is not bonded to an electronegative element
B - Has a hydrogen bond due to the N- H bond
C - Has a hydrogen bond due to the O - H bond
D - Has a hydrogen bond due to the H -I bond
What is the hydrogen bond?In a hydrogen bond, a weak electrostatic connection is created when the positively charged hydrogen atom is drawn to the negatively charged electronegative atom.
This contact is stronger than van der Waals forces but not as strong as covalent or ionic bonds or other intermolecular forces.
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