The main difference between practical work inside and outside a laboratory is the environment and tools used for experimentation.
Practical work inside and outside the laboratoryInside a laboratory, experiments are conducted in a controlled environment with specialized equipment and instruments designed to facilitate experimentation, record data, and ensure safety.
On the other hand, outside the laboratory, experiments are often conducted in a less controlled environment, which can make it more challenging to control variables and obtain accurate results.
Also, experiments outside the laboratory often require different tools and techniques to account for environmental factors such as weather conditions. However, outside the laboratory, there is often more opportunity for real-world applications of experimental findings.
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A solution of thickness 3cm transmits 30%. calculate the concentration of the solution. E= 400dm/mol/cm
The concentration of the solution is 0.000435 mol/dm³.
What is the concentration of the solution?The concentration of a solution is calculated as follows;
Concentration = (Absorbance) / (Molar absorptivity x path length)
the path length = 3cm
the molar absorptivity (E) = 400 dm/mol/cm.
if the solution transmits 30% of the light, it absorbs 70% of the incident light.
Absorbance = log (1/Transmittance)
Absorbance = log (1/0.3)
Absorbance = 0.523
Concentration = (0.523) / (400 dm/mol/cm x 3 cm)
= 0.000435 mol/dm³
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The following first-order reaction occurs in CCL4(l) at 45°C: N2O5》N2O4+1÷2O2. The rate consast is k=6.2×10^-4 s^-1 an 80.0 g sample of N2O5 in CCL4 is allowed to decompose at 45°C
a) how long does it take for the quantity of N2O5 to be reduced yo 2.5 g ?
b) how many liters of O2 measured at 745 mmHg and 45°C, are produced up to this point ?
a) The amount of N₂O₅ is lowered to 2.5 g during the course of around 4.41 × 10⁴ seconds or 12.25 hours.
b) 9.71 L of O₂ are generated at 745 mmHg and 45 °C.
How to find quantity?a) To solve for the time required for the quantity of N₂O₅ to be reduced to 2.5 g, use the first-order integrated rate law:
ln[N₂O₅]t/[N₂O₅]0 = -kt
where [N₂O₅]t = concentration of N₂O₅ at time t, [N₂O₅]0 = initial concentration of N₂O₅, k = rate constant, and t = time.
Find the initial concentration of N₂O₅:
n(N₂O₅) = m/M = 80.0 g / 108.01 g/mol = 0.7413 mol
[N₂O₅]0 = n/V = 0.7413 mol / 0.153 L = 4.846 M
where M = molar mass of N₂O₅ and V = volume of the solution.
Substituting the given values into the equation:
ln([N₂O₅]t / 4.846 M) = -6.2×10⁻⁴ s⁻¹ × t
When the quantity of N₂O₅ is reduced to 2.5 g, the concentration is:
n(N₂O₅) = m/M = 2.5 g / 108.01 g/mol = 0.02314 mol
[N₂O₅]t = n/V = 0.02314 mol / 0.153 L = 0.151 M
Substituting this concentration into the equation and solving for t:
ln(0.151 M / 4.846 M) = -6.2×10⁻⁴ s⁻¹ × t
t = 4.41 × 10⁴ s
Therefore, it takes approximately 4.41 × 10⁴ seconds or 12.25 hours for the quantity of N₂O₅ to be reduced to 2.5 g.
b) The balanced equation for the reaction shows that 1 mole of N₂O₅ produces 1/2 mole of O₂:
N₂O₅ → N₂O₄ + 1/2 O2
Therefore, the number of moles of O₂ produced can be calculated using the stoichiometry:
n(O₂) = 1/2 × n(N₂O₅) = 1/2 × 0.7413 mol = 0.3707 mol
The ideal gas law can be used to calculate the volume of O₂ produced at 745 mmHg and 45°C:
PV = nRT
where P = pressure, V = volume, n = number of moles, R = gas constant, and T = temperature in Kelvin.
Convert the pressure to atm and the temperature to Kelvin:
P = 745 mmHg / 760 mmHg/atm = 0.980 atm
T = 45°C + 273.15 = 318.15 K
Substituting the values and solving for V:
V = nRT/P = (0.3707 mol) × (0.08206 L·atm/mol·K) × (318.15 K) / (0.980 atm) = 9.71 L
Therefore, the volume of O₂ produced at 745 mmHg and 45°C is 9.71 L.
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For the reaction:
S8(s) + 8 O2(g)⟶8 SO2(g) ΔH = –2368 kJ
How much heat is evolved when 25.0 moles of sulfur is burned in excess oxygen?
The amount of heat evolved when 25 moles of Sulfur is burned in excess oxygen is -74000 kJ.
The balanced reaction is given that is:
[tex]S_8(s) + 8 O_2(g) \rightarrow 8 SO_2(g)[/tex]
We can see that 1 mole of [tex]S_8[/tex] reacts with 8 moles of [tex]O_2[/tex] to produce 8 moles of [tex]So_2[/tex].
If 25.0 moles of [tex]S_8[/tex] reacts with excess Oxygen, then the amount of [tex]O_2[/tex] which is required in the reaction will be:
8 moles [tex]O_2[/tex] / 1 mole S8 × 25.0 moles S8 = 200 moles [tex]O_2[/tex]
We can use the enthalpy change and calculate the amount of heat evolved:
[tex]\Delta H[/tex] = -2368 kJ/ 8 moles [tex]SO_2[/tex]
The heat evolved = [tex]\Delta H[/tex] × moles of [tex]SO_2[/tex] produced
Moles of [tex]SO_2[/tex] produced = 8 moles [tex]SO_2[/tex] / 1 mole [tex]S_8[/tex] × 25.0 moles [tex]S_8[/tex]
= 200 moles [tex]SO_2[/tex].
Therefore, Heat evolved= -2368 kJ/ 8 moles [tex]SO_2[/tex] × 200 moles [tex]SO_2[/tex]
= -74000 kJ
The amount of heat evolved when 25 moles of Sulfur is burned in excess oxygen is -74000 kJ, the negative sign here indicates that the reaction is exothermic.
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A student has a 2.97 L
bottle that contains a mixture of O2
, N2
, and CO2
with a total pressure of 5.68 bar
at 298 K
. She knows that the mixture contains 0.225 mol N2
and that the partial pressure of CO2
is 0.309 bar
. Calculate the partial pressure of O2
.
What is the molar mass of potassium hydroxide, KOH?
Answer:
56.11 g/mol
Explanation:
To determine the molar mass of potassium hydroxide, we need to find the atomic mass of each element in the compound and add them up.
The atomic mass of potassium (K) is 39.10 g/mol, the atomic mass of oxygen (O) is 16.00 g/mol, and the atomic mass of hydrogen (H) is 1.01 g/mol.
So, the molar mass of potassium hydroxide (KOH) is:
Molar mass of K = 39.10 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of KOH = Molar mass of K + Molar mass of O + Molar mass of H
= 39.10 g/mol + 16.00 g/mol + 1.01 g/mol
= 56.11 g/mol
Therefore, the molar mass of potassium hydroxide (KOH) is 56.11 g/mol.
We wish to determine the mass of Mg required to react completely with 250mL of 1.0 M HCI. HCI reacts with Mg according to the equation below.
2HCl(aq) + Mg(s) → MgCl₂(aq) + H₂(g)
How many moles of HCI are present in 250. mL of 1.0 M HCl?
There are 0.25 moles of HCl present in 250 mL of 1.0 M HCl.
We have to calculate the number of moles of HCl present in some mL of 1.0M HCl. A mole is defined as the amount of substance in a system that contains as many elementary entities as there are atoms in 0.012 kg of carbon 12. We represent mole by the symbol 'mol'. Now, we will see how to calculate the number of moles.
We can calculate the number of moles of a substance using the following expression;
Molarity = no of moles of an element/volume
According to this question, we were given 250. mL of 1.0 M HCl. The number of moles will be calculated by the formula as follows;
no of moles of HCl = 0.250L × 1.0M
no of moles of HCl = 0.250 moles.
Therefore, 0.25 moles are present in 250 mL of 1.0 M HCl.
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4. A silver bar with a mass of 300 grams is heated from 30 °C to 55 °C. How much heat does the silver ber absorb in joules? In kilojoules? The specific heat of silver is 0.235 g C
A silver bar with the mass of the 300 grams is heated from the 30 °C to 55 °C. The amount heat does the silver bar absorb in the joules is 1762.5 J.
The mass of the silver bar = 300 g
The initial temperature = 30 °C
The final temperature = 55 °C
The heat energy is expressed as :
Q = mc ΔT
Where,
The m is mass of the silver bar = 300 g
The c is the specific heat capacity = 0.235 J/g °C
The ΔT is the change in the temperature = final temperature - initial temperature
The ΔT is the change in the temperature = 55 °C - 30 °C
The ΔT is the change in the temperature = 25 °C
The heat energy, Q = 300 × 0.235 × 25
The heat energy, Q = 1762.5 J
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What volume of oxygen gas can be collected
at 1.05 atm pressure and 44.0◦C when 42.5 g
of KClO3 decompose by heating, according to
the following equation?
2 KClO3(s) ∆
−−−−→
MnO2
2 KCl(s) + 3 O2(g)
Answer in units of L.
005 1.0 points
The volume of oxygen gas, O₂ collected at 1.05 atm pressure and 44.0 °C when 42.5 g of KClO₃ decomposed is 13.01 L
How do i determine the volume of oxygen gas collected?We shall begin by obtaining the mole in 42.5 g of KClO₃. Details below:
Mass of KClO₃ = 42.5 g Molar mass of KClO₃ = 122.5 g/mol Mole of KClO₃ =?Mole = mass / molar mass
Mole of CaC₂ = 42.5 / 122.5
Mole of CaC₂ = 0.35 mole
Next, we shall determine the mole of oxygen gas, O₂. produced. Details below:
2KClO₃ -> 2KCl + 3O₂
From the balanced equation above,
2 moles of KClO₃ decomposed to produced 3 mole of O₂
Therefore,
0.35 mole of KClO₃ will decompose to produce = (0.35 × 3) / 2 = 0.525 mole O₂
Finally, we shall determine the volume of oxygen gas, O₂ collected. Details below:
Pressure (P) = 1.05 atmTemperature (T) = 44 °C = 44 + 273 = 317 KGas constant (R) = 0.0821 atm.L/mol KNumber of mole (n) = 0.525 moleVolume of gas (V) =?PV = nRT
1.05 × V = 0.525 × 0.0821 × 317
Divide both sides by 1.05
V = (0.525 × 0.0821 × 317) / 1.05
Volume of oxygen gas = 13.01 L
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What are four methods of separating mechanical mixture?
Answer: Mixtures can be physically separated by using methods that use differences in physical properties to separate the components of the mixture, such as
evaporation, distillation, filtration and chromatography.Explanation:
How many grams of Al are needed to react with 352 mL of a 1.65 M HCl solution? Given the equation 2Al + 6HCl yields to form 2AlCl3 + 3H2
5.221 grams of Al are required to react with 352 mL of 1.65 M HCl solution.
What is meant by molarity?Molarity (M) is defined as the moles of solute per liter of the solution.
Balanced chemical equation is : 2Al + 6HCl → 2AlCl₃ + 3H₂
From the equation, we can see that 2 moles of Al react with 6 moles of HCl to produce 2 moles of AlCl₃ and 3 moles of H₂.
As moles of HCl = Molarity × Volume
moles of HCl = 1.65 mol/L × 0.352 L
moles of HCl = 0.58128 mol
and moles of Al = (2/6) × moles of HCl
moles of Al = (1/3) × 0.58128 mol
moles of Al = 0.19376 mol
mass of Al = moles of Al × molar mass of Al
mass of Al = 0.19376 mol × 26.98 g/mol
mass of Al = 5.221 g
So, 5.221 grams of Al are required to react with 352 mL of 1.65 M HCl solution.
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Iron pyrite (FeS2) is the form in which much of
the sulfur exists in coal. In the combustion of
coal, oxygen reacts with iron pyrite to produce
iron(III) oxide and sulfur dioxide, which is a
major source of air pollution and a substantial
contributor to acid rain. What mass of Fe2O3
is produced from 74 L of oxygen at 2.97 atm
and 161◦C with an excess of iron pyrite?
Answer in units of g
The mass of Fe₂O₃ produced is 101.9 g.
How to calculate mass ?The balanced chemical equation for the combustion of iron pyrite is:
4FeS₂(s) + 11O₂(g) → 2Fe₂O3(s) + 8SO₂(g)
From the equation, 11 moles of oxygen are required to produce 2 moles of Fe₂O₃. Convert the given volume of oxygen to moles:
n(O2) = PV/RT = (2.97 atm)(74 L)/(0.0821 L·atm/mol·K)(161 + 273 K) = 3.51 mol
Since the reaction requires 11 moles of O₂ for every 2 moles of Fe₂O₃, calculate the moles of Fe₂O₃ produced:
n(Fe₂O₃) = (2/11) × n(O₂) = (2/11) × 3.51 mol = 0.638 mol
Finally, use the molar mass of Fe₂O₃ to convert moles to grams:
m(Fe₂O₃) = n(Fe₂O₃) × M(Fe₂O₃) = 0.638 mol × 159.69 g/mol = 101.9 g
Therefore, the mass of Fe₂O₃ produced is 101.9 g.
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Using the equations
N₂ (g) + O₂ (g) → 2 NO (g) ∆H° = 180.6 kJ/mol
N₂ (g) + 3 H₂ (g) → 2 NH₃ (g) ∆H° = -91.8 kJ/mol
2 H₂ (g) + O₂ (g) → 2 H₂O (g) ∆H° = -483.7 kJ/mol
Determine the molar enthalpy (in kJ/mol) for the reaction
4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g).
The molar enthalpy for the reaction 4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g) is 266.4 kJ/mol.
What is the molar enthalpy for the reaction?The molar enthalpy is determined from Hess's law as follows:
Equation 1 x2:
2 N₂ (g) + 2 O₂ (g) → 4 NO (g) ∆H° = 361.2 kJ/mol
Equation 3 x3, :
6 H₂ (g) + 3 O₂ (g) → 6 H₂O (g) ∆H° = -1451.1 kJ/mol
Equation 2 x -4:
-8 N₂ (g) - 12 H₂ (g) → -8 NH₃ (g) ∆H° = 367.2 kJ/mol
Adding the equations together:
-6 N₂ (g) - 6 H₂ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g) - 8 NH₃ (g) ∆H° = 266.3 kJ/mol
Multiplying the equation above by -1/2:
3 N₂ (g) + 3 H₂ (g) - 5/2 O₂ (g) → -2 NO (g) - 3 H₂O (g) + 4 NH₃ (g) ∆H° = -133.2 kJ/mol
Multiplying the above equation by -2:
4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g) ∆H° = 266.4 kJ/mol
This is the molar enthalpy of the given reaction
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PLEASE HELP!!
416 g of Xenon-146 was placed in a container to decay, if there is now 13g of Xenon- 146 left, how long has the Xenon been decaying? (the half-life of ^ 146 Xe is 37 hours) Please enter your answer as with one decimal place and with no units.
The decay of Xenon-146 follows an exponential decay model, where the amount of Xenon-146 remaining after time t is given by:
N(t) = N0 * (1/2)^(t/T)
where N0 is the initial amount of Xenon-146, T is the half-life of Xenon-146, and t is the time that has passed.
We are given that N0 = 416 g, N(t) = 13 g, and T = 37 hours. We can solve for t as follows:
N(t) = N0 * (1/2)^(t/T)
13 = 416 * (1/2)^(t/37)
Taking the natural logarithm of both sides:
ln(13) = ln(416) + (t/37) * ln(1/2)
Solving for t:
t = 37 * [ln(13/416) / ln(1/2)]
t ≈ 111.2 hours
Therefore, the Xenon-146 has been decaying for approximately 111.2 hours.
As the hour of her new job approached, Emma could feel her excitement blank
As the hour of her new job approached, Emma could feel her excitement peak.
What word can replace intensify ?"Peak" is a synonym to "intensify" in this context because it means to reach the highest point or level of something. In the given passage, Emma's excitement is growing stronger and stronger as the time for her volunteer job approaches.
When her excitement "peaks," it means that it has reached the highest point of intensity, just like when something is intensified, it becomes stronger or more intense.
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A titration setup was used to determine the unknown molar concentration of a solution of NaOH. A1.2 M HCl solution was used as the
titration standard. The following data were collected.
Trial 1
Amount of HCI
Standard Used 10.0 mL
0.0 mL
Initial NaOH
Buret Reading
Final NaOH
Buret Reading 12.2 mL
Trial 2
10.0 mL
12.2 mL
23.2 mL
Trial 3 Trial 4
10.0 mL 10.0 mL
23.2 mL 35.2 mL
35.2 mL 47.7 mL
79) Calculate the volume of NaOH solution used to neutralize 10.0 ml. of the standard HCl solution in trial 3 in the given diagram.
[Show your work.]
What is the volume of a 1.0 M solution that has 4.0 moles of solute?
The volume of a 1.0 M solution that has 4.0 moles of solute is 4.0 liters.
What is mole ?A mole is a unit of measurement used in chemistry to represent the quantity of a chemical.
We can use the following formula to get the volume of a 1.0 M solution containing 4.0 moles of solute:
moles of solute = molarity x volume of solution
To determine the volume of the solution, we can rearrange this formula as follows:
Volume of solution = molarity / moles of solute.
By entering the specified values, we obtain:
4.0 moles / 1.0 M is the solution's volume.
Solution volume = 4.0 L
Therefore, the volume of a 1.0 M solution that has 4.0 moles of solute is 4.0 liters.
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Hydrazine, N2H4
, reacts with oxygen to form nitrogen gas and water.
N2H4(aq)+O2(g)⟶N2(g)+2H2O(l)
If 3.55 g
of N2H4
reacts with excess oxygen and produces 0.850 L
of N2
, at 295 K
and 1.00 atm,
what is the percent yield of the reaction?
Hydrazine, reacts with the oxygen to form the nitrogen gas and the water. The percent yield of the reaction is 3.18 %.
The balanced reaction is :
N₂H₄ + O₂ ---> N₂ + 2H₂O
The mass of the N₂H₄ = 3.55 g
The moles of N₂H₄ = mass / molar mass
The moles of N₂H₄ = 3.55 / 32
The moles of N₂H₄ = 0.110 mol
The theoretical yield = 0.110 mol × 28 g/mol
The theoretical yield = 3.08 g
The gas equation is :
P V = n R T
n = P V / R T
n = (1 × 0.850 ) / ( 0.0823 ×295 )
n = 0.0035 mol
The actual yield = 0.0035 × 28
The actual yield = 0.098 g
The percent yield = ( 0.098 / 3.08 ) × 100 %
The percent yield = 3.18 %.
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A compound has the formula X2Fe(CN)6 ∙ 12H2O, where X is an unknown element.
If the compound is 45.34% water by mass, what is the identity of element X?
The identity of element X in the compound X2Fe(CN)6 · 12H2O is sodium (Na).
To find the identity of element X in the compound X2Fe(CN)6 · 12H2O, we can start by determining the molar mass of the compound.
The molar mass of X2Fe(CN)6 is:
2 × molar mass of X + molar mass of Fe + 6 × molar mass of C + 6 × molar mass of N
= 2 × atomic mass of X + atomic mass of Fe + 6 × 12.01 g/mol + 6 × 14.01 g/mol
= 2 × atomic mass of X + 55.85 g/mol + 432.72 g/mol + 84.06 g/mol
= 2 × atomic mass of X + 572.63 g/mol
The molar mass of 12H2O is:
12 × (atomic mass of H + atomic mass of O) = 12 × (1.01 g/mol + 16.00 g/mol) = 216.24 g/mol
The total molar mass of the compound is:
2 × atomic mass of X + 572.63 g/mol + 216.24 g/mol = 2 × atomic mass of X + 788.87 g/mol
Now we can use the given information that the compound is 45.34% water by mass. This means that the mass of water in the compound is 45.34% of the total mass of the compound, and the mass of the rest of the compound (X2Fe(CN)6) is 100% - 45.34% = 54.66% of the total mass of the compound.
Let's assume we have 100 g of the compound. Then the mass of water in the compound is:
45.34 g water = 0.4534 × 100 g compound
The mass of the rest of the compound (X2Fe(CN)6) is:
54.66 g rest of the compound = 0.5466 × 100 g compound
We can now use the mass of the rest of the compound (X2Fe(CN)6) to find the number of moles of the compound:
moles of X2Fe(CN)6 = (54.66 g) / (2 × atomic mass of X + 572.63 g/mol)
We can also use the mass of water to find the number of moles of water:
moles of H2O = (45.34 g) / 18.02 g/mol
Since the compound has 12 moles of water per mole of X2Fe(CN)6, we have:
moles of X2Fe(CN)6 = 1/12 × moles of H2O
We can now set these two expressions for moles of the compound equal to each other and solve for the atomic mass of X:
(54.66 g) / (2 × atomic mass of X + 572.63 g/mol) = 1/12 × (45.34 g) / 18.02 g/mol
Simplifying this equation and solving for the atomic mass of X gives:
atomic mass of X = 22.99 g/mol
The atomic mass of X is very close to the atomic mass of sodium (22.99 g/mol), so it is likely that X is sodium. Therefore, the identity of element X in the compound X2Fe(CN)6 · 12H2O is sodium (Na).
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Calculate the cell potential for the galvanic cell in which the given reaction occurs at 25 °C, given that [Ti2+]=0.00140 M and [Au3+]=0.887 M .
3Ti(s)+2Au3+(aq)↽−−⇀3Ti2+(aq)+2Au(s)
Under the specified conditions, the cell potential of the given galvanic cell is 3.11 V.
How to determine cell potential?The standard reduction potentials for the half-reactions involved in the cell reaction are:
Ti²⁺(aq) + 2e- ⇆ Ti(s) E° = -1.63 V
Au³⁺(aq) + 3e- ⇆ Au(s) E° = +1.50 V
The cell potential (Ecell) is given by:
Ecell = E°cathode - E°anode
where E°cathode = standard reduction potential of the cathode (reduction half-reaction) and E°anode = standard reduction potential of the anode (oxidation half-reaction).
In this case, Ti²⁺ is oxidized (anode) and Au³⁺ is reduced (cathode). Therefore:
E°anode = -1.63 V
E°cathode = +1.50 V
So, Ecell = +1.50 V - (-1.63 V) = +3.13 V
The Nernst equation can be used to calculate the cell potential (Ecell) under non-standard conditions:
Ecell = E°cell - (RT/nF) ln(Q)
where R = gas constant (8.314 J/(molK)), T = temperature in Kelvin (25 °C = 298 K), n = number of electrons transferred in the balanced equation (3 in this case), F = Faraday constant (96,485 C/mol), and Q = reaction quotient.
For the given concentrations:
[Ti²⁺] = 0.00140 M
[Au³⁺] = 0.887 M
The reaction quotient Q can be written as:
Q = ([Ti²⁺]³/[Au³⁺]²)
Substituting the values into the Nernst equation:
Ecell = E°cell - (RT/nF) ln([Ti²⁺]³/[Au³⁺]²)
Ecell = 3.13 V - (8.314 J/(molK) × 298 K / (3 × 96,485 C/mol)) ln(0.00140³/0.887²)
Ecell = 3.13 V - 0.0217 V
Ecell = 3.11 V
Therefore, the cell potential for the given galvanic cell under the given conditions is 3.11 V.
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What will happen when pressure on a reactant mixture at equilibrium and with fewer moles on the reactant side is increased
when pressure of the reactant mixture at the equilibrium and with the fewer moles in reactant side will be increased and the equilibrium will be shift to the side in the reaction where the fewer moles of the gas.
According to the Le Chartelier, when the reaction is in the equilibrium phase and the one of the constraints which will affect the rate of the reactions, and the equilibrium will be shift to the cancel out this effect that the constraint had.
Therefore, If the pressure of the system or the reaction is in the equilibrium is change, the equilibrium of the reaction will be change that is depending on the side of the reaction with the highest number of the molecules.
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If an aqueous solution is 5.321m, which of the following statements is incorrect?
A) Freezing point of solution will lower by 10 C
B) Boiling point of solution will increase by 2.72 C
C) Boiling point of solution will be 100 C
D) Osmotic pressure of solution will be higher than water
Boiling point of solution will be 100 C. The incorrect statement is C)
What is aqueous solution ?An aqueous solution is one in which water serves as the solvent. One or more substances are dissolved in water to create such a solution, and the water molecules surround and separate the individual solute particles to create a homogeneous mixture.
Therefore, A solvent's boiling point and freezing point change when a solute is dissolved in it, respectively. Boiling point elevation and freezing point depression are two terms used to describe this occurrence. The concentration of the solute determines how much of an impact it has.
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Chemistry balance QUESTION
reaction: AB₂C (g) → B₂ (g) + AC (g), find the value of K. At equilibrium, [AB₂C]=0.0168 M, [B₂]= 0.007 M, and [AC] = 0.0118 M
The value of the equilibrium constant, K, for the reaction is 0.0049
How do i determine the equilibrium constant?The following data were obtained from the quesion:
AB₂C (g) ⇌ B₂(g) + AC(g) Concentration of AB₂C, [AB₂C] = 0.0168 MConcentration of B₂, [B₂]= 0.007 MConcentration of AC, [AC] = 0.0118 MEquilibrium constant (K) =?The equilibrium constant for the reaction can be obtain as illustrated below:
Equilibrium constant = [B₂][AC] / [AB₂C]
Equilibrium constant = (0.007 × 0.0118) / 0.0168
Equilibrium constant = 0.0049
Thus, from the above calculation, the equilibrium constant, K for the reaction is 0.0049
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What is the mass (in g) of a solid piece of iron which has a specific heat of 0.449 J/g°C if when it absorbed 948.0 J of heat the temperature rose from 24.0°C to
82.1°C? Give your answer in 3 sig figs.
Answer:
Explanation:
We can use the formula:
q = mcΔT
where q is the heat absorbed, m is the mass, c is the specific heat, and ΔT is the change in temperature.
Given:
specific heat of iron (c) = 0.449 J/g°C
initial temperature (T1) = 24.0°C
final temperature (T2) = 82.1°C
heat absorbed (q) = 948.0 J
Substituting the given values into the formula, we get:
q = mcΔT
948.0 J = m(0.449 J/g°C)(82.1°C - 24.0°C)
948.0 J = m(0.449 J/g°C)(58.1°C)
m = 948.0 J ÷ (0.449 J/g°C × 58.1°C)
m = 33.1 g
Therefore, the mass of the iron piece is 33.1 g (to three significant figures)
Why are leaves green
Answer:
Leaves are green due to the presence of an organelle chloroplast (in abundance) which contains the pigment chlorophyll
Explanation:
Now saying chlorophyll pigment is a green pigment might be slightly incorrect. The two famous types (Chlorophyll a, Chlorophyll b) only absorb red and blue light from the atmosphere and reflect green light hence giving the pigment a green appearance and lastly giving the leaves a green color too
Answer:
Chlorophyll
Explanation:
Plants are often seen as green to the human eye due to the presence of chlorophyll, which is the primary pigment used in photosynthesis. Chlorophyll absorbs light in the red and blue-violet parts of the spectrum, but reflects or transmits green light, resulting in the characteristic green color of leaves.
A 0.4 kg piece of ice at -10 ∘C is dropped from a height h. Upon impact, 3.0 % of its kinetic energy is converted into heat energy. If the impact transforms all of the ice into water that has a final temperature of 0 ∘C , find h .
The height of fall of the ice is determined as 313.25 m.
What is the heat energy of the ice?The total heat energy of the ice during the fall is calculated as follows;
Q = ml + mcΔT
where;
c is the specific heat of waterl is latent heat of fusionΔT is change in temperatureQ = 334000 x 0.4 + 0.4 x 4200 x (10)
Q = 150,400 J
The energy converted into potential energy is calculated as;
3%K.E = 150,400 J
0.03K.E = 150,400 J
K.E = 5,013,333.33 J
¹/₂mv² = 5,013,333.33 J
v = √(2 x 5,013,333.33)/(0.4)
v = 5,006.67 m/s
The height of fall is calculated as;
h = √2gh
h = √(2 x 5,006.67 x 9.8)
h = 313.25 m
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N2(g)+3H2(g)->2NH3(g), ΔH=-92.40kJ 1. How many grams of H2 are needed to involve 150.9kJ of heat? 2. How many moles of NH3 were produced in the process?
1. To solve for the grams of H2 needed, we need to use the given ΔH value to calculate the amount of moles of N2 that reacted. From the balanced chemical equation, we know that for every 3 moles of H2 that reacts, 1 mole of N2 reacts. Therefore, we can use the mole ratio to convert the moles of N2 to moles of H2 and then use the molar mass of H2 to convert to grams.
First, we need to calculate the moles of N2 that reacted to produce 150.9kJ of heat:
ΔH = -92.40 kJ/mol N2
150.9 kJ = (1 mol N2 / -92.40 kJ) x (-150.9 kJ)
mol N2 = 1.63 mol
Using the mole ratio from the balanced chemical equation:
1 mol N2 : 3 mol H2
We can calculate the moles of H2 needed:3 mol H2 = 1 mol N2
3 mol H2 = 1.63 mol N2
mol H2 = 0.543 mol
Finally, we can convert moles of H2 to grams:
mol H2 = 0.543 mol
molar mass of H2 = 2.02 g/mol
grams of H2 = (0.543 mol) x (2.02 g/mol)
grams of H2 = 1.10 g
Therefore, 1.10 grams of H2 are needed to involve 150.9kJ of heat.
2. To solve for the moles of NH3 produced, we can use the same mole ratio from the balanced chemical equation:
1 mol N2 : 2 mol NH3
From the moles of N2 that reacted calculated in part 1, we can calculate the moles of NH3 produced:
1 mol N2 = 2 mol NH3
1 mol N2 = 1.63 mol N2
mol NH3 = (2 mol NH3 / 1 mol N2) x (1.63 mol N2)
mol NH3 = 3.26 mol
Therefore, 3.26 moles of NH3 were produced in the process.
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If 8.25
mol of C5H12
reacts with excess O2,
how many moles of CO2
will be produced by the following combustion reaction?
C5H12+8O2⟶6H2O+5CO2
The given reaction equation tells us that for every 1 mol of C₅H₁₂, 5 moles of CO₂ will be produced. Since 8.25 mol of C₅H₁₂ is given, 8.25 mol C₅H₁₂ x 5 moles CO₂/1 mol C₅H₁₂ = 41.25 moles CO₂ will be produced.
What is reaction?Reaction is the process of responding to an event or stimulus in a particular way. It can occur at the physical, cognitive, or emotional level. Physically, a reaction could be as simple as a reflex or as complex as a multi-step process. Cognitively, it could involve forming a judgment or understanding. Emotionally, it could involve feelings of fear, shock, anger, or joy. In the context of science, reactions are often chemical or physical processes that involve the conversion of one set of substances into another.
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A rock is placed on a scale and gives a reading of 76.89 grams. The rock is then placed in a graduated cylinder with 63.12 mL of water, the water rises to a volume of 73.54mL What is the density of the rock? (you answer must have a total of 2 decimals)
What is the limiting reagent in the reaction of 0.150 g of salicylic acid with 0.350 mL of acetic anhydride (d=1.082 g/mL)? Show your work.
The limiting reagent for the reaction between 0.150 g of salicylic acid and 0.350 mL of acetic anhydride is salicylic acid, C₇H₆O₃
How do i determine the limiting reagent?First, we shall determine the mass of the acetic anhydride. Details below:
Volume of acetic anhydride = 0.350Density of acetic anhydride = 1.082 g/mLMass of acetic anhydride =?Mass = density × volume
Mass of acetic anhydride, C₄H₆O₃ = 1.082 × 0.350
Mass of acetic anhydride, C₄H₆O₃ = 0.3787 g
Finally, we shall determine the limiting reagent. Details below:
C₇H₆O₃ + C₄H₆O₃ -> C₉H₈O₄ + CH₃COOH
Molar mass of C₇H₆O₃ = 138.121 g/molMass of C₇H₆O₃ from the balanced equation = 1 × 138.121 = 138.121 g Molar mass of C₄H₆O₃ = 102.09 g/molMass of C₄H₆O₃ from the balanced equation = 1 × 102.09 = 102.09 gFrom the balanced equation above,
138.121 g of C₇H₆O₃ reacted with 102.09 g of C₄H₆O₃
Therefore,
0.150 g of C₇H₆O₃ will react with = (0.150 × 102.09) / 138.121 = 0.11089 g of C₄H₆O₃
We can see from the above that only 0.11089 g of acetic anhydride, C₄H₆O₃ out of 0.3787 g is needed to react with 0.150 g of salicylic acid, C₇H₆O₃
Thus, the limiting reagent is salicylic acid, C₇H₆O₃
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What is the S-P difference (sec)?
What is the amplitude (mm)?
What is the distance (km)?
What is the magnitude (M)?
The S-P difference (sec) is used to calculate the distance (km) between an earthquake epicenter and a seismic station, while the magnitude (M) is a measure of the energy released during the earthquake.
These parameters are important for understanding the severity and impact of an earthquake, as well as for predicting future seismic activity.
The S-P difference (sec) refers to the time difference between the arrival of the primary (P) waves and the secondary (S) waves at a seismic station. This time difference is used to calculate the distance (km) between the earthquake epicenter and the seismic station, using the equation: distance (km) = S-P difference (sec) x 8 km/sec. This calculation assumes that the waves travel at a constant speed through the Earth's interior.
The magnitude (M) of an earthquake is a measure of the energy released during the earthquake, and is usually determined using a seismometer. The magnitude scale is logarithmic, meaning that each increase of one unit represents a tenfold increase in seismic energy. For example, an earthquake with a magnitude of 5.0 is ten times more powerful than one with a magnitude of 4.0, and 100 times more powerful than one with a magnitude of 3.0.
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