The volume of O2 that can be released at STP from the given sample of silver oxide is 23.45 mL.
To solve this problem
Creating the balanced chemical equation for the breakdown of silver oxide is the first step in tackling this issue:
2Ag2O(s) → 4Ag(s) + O2(g)
We can deduce from the equation that 2 moles of AgO will result in 1 mole of O2. Since Ag2O has a molar mass of 231.735 g/mol, 0.4856 g of Ag2O is equivalent to:
0.4856 g Ag2O x (1 mol Ag2O/231.735 g Ag2O) = 0.002095 mol Ag2O
Therefore, the number of moles of O2 that can be produced from 0.4856 g of Ag2O is:
0.002095 mol Ag2O x (1 mol O2/2 mol Ag2O) = 0.0010475 mol O2
1 mole of any gas takes up 22.4 L of space at STP As a result, 0.4856 g of Ag2O can generate the following amount of O2 at STP:
0.0010475 mol O2 x 22.4 L/mol = 0.02345 L or 23.45 mL
Therefore, the volume of O2 that can be released at STP from the given sample of silver oxide is 23.45 mL.
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Isoprenol (C₅H₁₀O) can be produced from isobutene (C₄H₈) and formaldehyde (CH₂O) via the following reaction scheme. What is the change in enthalpy in kJ associated with the production of 155.3 g C₅H₁₀O (isoprenol)?
C₄H₈(g) + CH₂O(g) → C₅H₁₀O(l) ∆H = -191.3 kJ
To determine the change in enthalpy associated with the generation of 155.3 g of [tex]C_5H_1_0O[/tex] we must first calculate the moles of [tex]C_5H_1_0O[/tex] produced using its molar mass.
The molar mass of C₅H₁₀O is:
5(12.01 g/mol) + 10(1.01 g/mol) + 16.00 g/mol = 88.15 g/mol
Moles of C₅H₁₀O produced:
155.3 g / 88.15 g/mol = 1.763 mol C₅H₁₀O
The balanced chemical equation states that the formation of 1 mol of C₅H₁₀O results in an enthalpy change of -191.3 kJ.
As a result, the enthalpy change during the formation of 1.763 mol of C₅H₁₀O is: -191.3 kJ/mol x 1.763 mol = -337.8 kJ
The enthalpy change for the production of 155.3 g of C5H10O is -337.8 kJ.
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Anyone know how to solve this?
The ratio of the concentrations at equilibrium is as follows:
3.7 0.85 0.04 21.3 42.6 12212.92 0.81 0.11 7.4 14.8 6012.2 0.63 0.43 1.5 3 274What are reactions in equilibrium?Chemical equilibrium is the point in a chemical reaction where both the forward and backward processes are occurring at the same rate.
The concentrations of the reactants and products are constant at equilibrium because the forward and reverse speeds are equal.
Considering the given statements based on the reaction equilibrium concentrations, the correct options are:
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The mass of calcium release same number of valence electron as same number of 23g Na
20 g of calcium would release the same number of valence electrons as 23 g of sodium.
The atomic number of calcium (Ca) is 20, which means it has 20 electrons in its neutral state. When calcium loses two electrons, it becomes a Ca2+ ion with 18 electrons.
On the other hand, the atomic number of sodium (Na) is 11, which means it has 11 electrons in its neutral state. When sodium loses one electron, it becomes a Na+ ion with 10 electrons.
To release the same number of valence electrons as 23 g of Na, we need to calculate how many moles of Na there are in 23 g:
Molar mass of Na = 23 g/mol
Number of moles of Na = 23 g / 23 g/mol = 1 mol
Since each Na+ ion has lost one electron, 1 mol of Na+ ions has lost 1 mol of valence electrons.
To release the same number of valence electrons as 1 mol of Na+ ions, we need to calculate how many moles of Ca2+ ions are required:
1 mol of Na+ ions = 1 mol of valence electrons
1 mol of Ca2+ ions = 2 mol of valence electrons
Therefore, we need 0.5 mol of Ca2+ ions to release the same number of valence electrons as 1 mol of Na+ ions.
Finally, we can calculate the mass of calcium that would release the same number of valence electrons as 23 g of Na:
Molar mass of Ca = 40 g/mol
Mass of Ca required = 0.5 mol x 40 g/mol = 20 g
Therefore, 20 g of calcium would release the same number of valence electrons as 23 g of sodium.
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Balloons for a New Years Eve party in Fargo, ND, are filled to a volume of 1.90 L at a temperature of 22.0 c and then hung outside. what is the volume of the balloon once they have cooled to the outside temperature of -34.0 c?
The volume of the balloon once they have cooled to the outside temperature of -34.0 c is 1.53 L.
Charles' law predicts the relationship between the volumes and the temperatures of a sample of an ideal gas at different conditions. For this equation to hold true, the number of molecules and the pressure must remain constant despite changes in the environment.
Determine the volume of the balloon outside, V2. We do this by applying Charles' law, such that we relate the volume, V, and the temperature, T, of a sample of gas as
V₁ /T₁ = V₂/ T₂
at two conditions. We are given the following values for the variables:
• V₁ = 1.90 L
T₁ = 22.0+ 273.15 = 295.15 K
T₂= 34.0+273.15= 239.15 K
We proceed with the solution.
V₁/T₁ = V₂/T₂
V₁ /T₁ × T₂ = V₂
1.90 L/295.15 K x 239.15 K = V₂
1.53 L =V₂
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The complete question is
Balloons for a New Year's Eve party in Fargo, ND, are filled to a volume of 1.90 L at a temperature of 22.0 degrees Celsius and then hung outside where the temperature is -34.0 degrees Celsius. What is the volume of the balloons after they have cooled to the outside temperature? Assume that atmospheric pressure inside and outside the house is the same.
The critical point for water lies at 275 °C and 3.2 atm, calculate the DH°vap of water.
The ΔH°vap of water at the critical point is approximately 0.04614 kJ/mol.
To calculate the ΔH°vap (enthalpy of vaporization) of water at the critical point, we can use the Clausius-Clapeyron equation;
ln(P₂/P₁) = ΔH°vap/R [1/T₁ - 1/T₂]
where P₁ and T₁ are the pressure and temperature at which the enthalpy of vaporization is known (usually at standard conditions of 1 atm and 100 °C), P₂ and T₂ are the pressure and temperature at the critical point, ΔH°vap is the enthalpy of vaporization, and R is the gas constant (8.314 J/mol∙K).
Using the given values, we can plug them into the equation and solve for ΔH°vap;
ln(3.2 atm / 1 atm) = ΔH°vap / R [1/373 K - 1/275 K]
Simplifying;
ln(3.2) = ΔH°vap / R [0.0026819]
ΔH°vap / R = ln(3.2) / 0.0026819
ΔH°vap / R = 5.552
Multiplying both sides by R:
ΔH°vap = 5.552 x R
ΔH°vap = 5.552 x 8.314 J/mol∙K
ΔH°vap = 46.14 J/mol
Converting to kJ/mol;
ΔH°vap = 0.04614 kJ/mol
Therefore, the ΔH°vap of water is 0.04614 kJ/mol.
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please provide explanation!! thank you in advance!!
The molarity of the product is 0.00368 M. Option B
What is the reaction equilibrium?When the rates of the forward and reverse reactions in a chemical reaction are equal and the concentrations of reactants and products are stable over time, this condition is referred to as reaction equilibrium.
The equilibrium constant is a measure of the relative concentrations of reactants and products at equilibrium.
[tex]Keq = [H_{2} O] [CO]/[H_{2}] [CO_{2} ]\\0.106 = x^2/(0.0113)^2\\x = \sqrt{} 0.106 (0.0113)^2\\x = 0.00368 M[/tex]
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8. The compound C2H4 has van der Waals constants a = 4.612 atm•L2/mol2 and b = 0.0582 L/mol. Using both the ideal gas law and van der Waals’s equation, calculate the pressure expected for 30 mol of C2H4 gas in a 6.00-L container at 20 °C.
Using the Ideal Gas Law, the pressure expected for 30 mol of [tex]C_2H_4[/tex] gas in a 6.00-L container at 20 °C is 1210.07 atm, and using the van der Waals equation, the pressure is 1179.71 atm.
To calculate the pressure expected for 30 mol of [tex]C_2H_4[/tex] gas in a 6.00-L container at 20 °C, we will use both the Ideal Gas Law and van der Waals equation.
Ideal Gas Law: PV = nRT
P = pressure
V = volume (6.00 L)
n = moles (30 mol)
R = ideal gas constant (0.0821 L•atm/mol•K)
T = temperature (20 °C + 273.15 = 293.15 K)
Solve for P (pressure):
P = nRT / V
P = (30 mol)(0.0821 L•atm/mol•K)(293.15 K) / 6.00 L
P = 1210.07 atm
Van der Waals equation:
(P + a(n/V)²)(V - nb) = nRT
a = 4.612 atm•L²/mol²
b = 0.0582 L/mol
Solve for P (pressure):
(P + (4.612)(30/6)²) (6 - 0.0582 * 30) = (30)(0.0821)(293.15)
P = 1179.71 atm
Using the Ideal Gas Law, the pressure is 1210.07 atm, and using the van der Waals equation, the pressure is 1179.71 atm.
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please help asap!
3. A double replacement reaction occurs between two solutions of lead (II) nitrate and potassium bromide. Write a
balanced equation for this reaction-identifying the product that will precipitate, and the product that will remain in
solution.
a) Write the balanced equation for this double replacement reaction.
b) If this reaction starts with 32.5 g lead (II) nitrate and 38.75 g potassium bromide, how many grams of the
precipitate will be produced? Remember to use the limiting reactant to calculate the amount of precipitate
formed.
c) How many grams of the excess reactant will remain?
Answer:
Explanation:
a) The balanced equation for the double replacement reaction between lead (II) nitrate and potassium bromide is:
Pb(NO₃)₂(aq) + 2KBr(aq) → PbBr₂(s) + 2KNO₃(aq)
In this reaction, lead (II) bromide (PbBr₂) will precipitate, while potassium nitrate (KNO₃) will remain in solution.
b) To determine the amount of precipitate produced, we need to first determine the limiting reactant. We can do this by calculating the number of moles of each reactant and comparing it to the stoichiometry of the balanced equation.
The molar mass of lead (II) nitrate is 331.21 g/mol and the molar mass of potassium bromide is 119.00 g/mol.
The number of moles of lead (II) nitrate is 32.5 g / 331.21 g/mol = 0.0981 mol The number of moles of potassium bromide is 38.75 g / 119.00 g/mol = 0.3256 mol
According to the balanced equation, one mole of lead (II) nitrate reacts with two moles of potassium bromide to produce one mole of lead (II) bromide. This means that if all the lead (II) nitrate were to react, it would require 0.0981 mol * 2 = 0.1962 mol of potassium bromide.
Since we have more than enough potassium bromide (0.3256 mol > 0.1962 mol), lead (II) nitrate is the limiting reactant.
The number of moles of lead (II) bromide produced will be equal to the number of moles of lead (II) nitrate consumed, which is 0.0981 mol.
The molar mass of lead (II) bromide is 367.01 g/mol, so the mass of lead (II) bromide produced will be 0.0981 mol * 367.01 g/mol = 36.0 g.
c) To determine the amount of excess reactant remaining, we need to subtract the amount consumed from the initial amount.
The number of moles of potassium bromide consumed is half the number of moles of lead (II) nitrate consumed, which is 0.0981 mol / 2 = 0.04905 mol.
The mass of potassium bromide consumed is 0.04905 mol * 119.00 g/mol = 5.84 g.
The mass of potassium bromide remaining is 38.75 g - 5.84 g = 32.91 g.
Strychnine, a deadly poison, has a molecular mass of 334 g/mol and a percentage composition of 75.45% carbon, 6.59% hydrogen, 8.38% nitrogen, and the balance oxygen. What is the molecular formula of strychnine?
The molecular formula of the strychnine, given that it is composed of 75.45% carbon, 6.59% hydrogen, 8.38% nitrogen, and the balance oxygen is C₂₁H₂₂N₂O₂
How do i determine the molecular formula?First, we shall obtain the empirical formula of compound. Details below:
Carbon (C) = 75.45%Hydrogen (H) = 6.59%Nitrogen (N) = 8.38%Oxygen (O) = 100 - (75.45 + 6.59 + 8.38) = 9.58%Empirical formula =?Divide by their molar mass
C = 75.45 / 12 = 6.2875
H = 6.59 / 1 = 6.59
N = 8.38 / 14 = 0.5986
O = 9.58 / 16 = 0.59875
Divide by the smallest
C = 6.2875 / 0.5986 = 10.5
H = 6.59 / 0.5986 = 11
N = 0.5986 / 0.5986 = 1
O = 0.59875 / 0.5986 = 1
Multiply through by 2 to express in whole number
C = 10.5 × 2 = 21
H = 11 × 2 = 22
N = 1 × 2 = 2
O = 1 × 2 = 2
Thus, we can conclude that the empirical formula is C₂₁H₂₂N₂O₂
Now, we shall determine the molecular formula of strychnine. Details below
Empirical formula = C₂₁H₂₂N₂O₂Molar mass of compound = 334 g/molMolecular formula =?Molecular formula = empirical × n = mass number
[C₂₁H₂₂N₂O₂]n = 140.22
[(12×21) + (1×22) + (14×2) + (16×2)]n = 334
334n = 334
Divide both sides by 334
n = 334 / 334
n = 1
Molecular formula = [C₂₁H₂₂N₂O₂]n
Molecular formula = [C₂₁H₂₂N₂O₂]1
Molecular formula = C₂₁H₂₂N₂O₂
Thus, we can conclude that the molecular formula of strychnine is C₂₁H₂₂N₂O₂
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Chemistry balance equation
At equilibrium, the value of K = 0.00659 for the reaction: N₂ (g) + 3 H₂ (g) ← → 2 NH₃ (g). Calculate [N₂] when [NH₃] = 0.000123 M and [H₂] = 0.0275 M
The concentration of the ammonia from the calculation is [tex]1.1 * 10^-15[/tex]M.
What is the equilibrium constant?The ratio of the concentrations of products to reactants in chemical equilibrium, for a given chemical reaction at a given temperature, is described by the equilibrium constant, abbreviated as Kc or Keq.
We can see that;
[tex]Keq = [NH_{3} ]^2/[N_{2} ] [ H_{2}]^3\\Keq[N_{2} ] [ H_{2}]^3 = [NH_{3}]^2\\\\N_{2} ] = [NH_{3}]^2/Keq[ H_{2}]^3[/tex]
=[tex](0.000123 )^2/ 0.00659 * (0.0275)^3= 1.1 * 10^-15 M[/tex]
Thus we would have the nitrogen concentration as [tex]1.1 * 10^-15[/tex] M
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Why does the air feel "sticky" on warm summer days? What is in the air that caausses this
Need help matching pairs of structures to diastereomers, enantiomers, constitutional isomers, not isomers, diff representations of the same?
A pair of molecules which exist in two forms that are mirror images of each other but cannot be superimposed one upon the other are called the enantiomers. They are present in pairs and have similar molecular shape.
The compounds with the same molecular formula but are non-superimposable non-mirror images are called diastereomers. They have distinct physical properties and molecular shape.
The constitutional isomers have the same molecular formula but have different bonding atomic organization and bonding patterns.
So here:
1st structure is constitutional isomers (c), 2nd structures are enantiomers (b) and the 3rd are completely different not isomers (d).
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How many grams of magnesium oxide would be formed if 28.2 grams of magnesium was burned?
Mg + O2 --> MgO
When 28.2 grams of Mg is burned, 46.7 grams of MgO will be formed.
How to determine the amount of MgO formed when 28.2 grams of Mg is burnedThe balanced chemical equation for the combustion of magnesium is:
2 Mg + O2 --> 2 MgO
This equation shows that 2 moles of Mg react with 1 mole of O2 to produce 2 moles of MgO.
To determine the amount of MgO formed when 28.2 grams of Mg is burned, we first need to convert the given mass of Mg to moles:
molar mass of Mg = 24.31 g/mol
moles of Mg = mass of Mg / molar mass of Mg
moles of Mg = 28.2 g / 24.31 g/mol
moles of Mg = 1.16 mol
According to the balanced chemical equation, 2 moles of Mg produce 2 moles of MgO. Therefore, we can use the mole ratio to calculate the moles of MgO formed:
moles of MgO = moles of Mg x (2 moles of MgO / 2 moles of Mg)
moles of MgO = 1.16 mol x 1
moles of MgO = 1.16 mol
Finally, we can convert the moles of MgO to grams using its molar mass:
molar mass of MgO = 40.31 g/mol
mass of MgO = moles of MgO x molar mass of MgO
mass of MgO = 1.16 mol x 40.31 g/mol
mass of MgO = 46.7 g
Therefore, when 28.2 grams of Mg is burned, 46.7 grams of MgO will be formed.
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What mass (grams) of oxygen will be released when 268.9 grams of Potassium Chlorate is thermally decomposed?
KClO3 --> KCl + O2
The centripetal acceleration experienced by the object can be calculated using the formula a = v^2/r, where v is the speed of the object and r is the radius of the circle. Substituting the given values, we get:
a = (50 cm/s)^2 / (250 cm)
a = 10 cm/s^2
Therefore, the centripetal acceleration experienced by the object is 10 cm/s^2.
To calculate the centripetal acceleration experienced by the object, you can use the formula:
Centripetal acceleration (a_c) = (velocity^2) / radius
Here, the velocity (v) is 50 cm/s and the radius (r) is 250 cm. Plugging in these values, we get:
a_c = (50^2) / 250 = 2500 / 250 = 10 cm/s²
So, the centripetal acceleration experienced by the object is 10 cm/s².
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What is the net ionic charge of an oxygen ion ?
Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction
Pb(NO3)2(aq)+2NH4I(aq)⟶PbI2(s)+2NH4NO3(aq)
What volume of a 0.350 M NH4I solution is required to react with 415 mL of a 0.120 M Pb(NO3)2 solution?
How many moles of PbI2 are formed from this reaction?
1. The volume of 0.350 M NH₄I solution is required is 286 mL
2. The mole of PbI₂ formed is 0.0498
1. How do i determine the volume of NH4I required?First, we shall determine the mole in 415 mL of 0.120 M Pb(NO₃)₂. Details below:
Molarity of Pb(NO₃)₂ = 0.120 M MVolume of Pb(NO₃)₂ = 415 mL = 415 / 1000 = 0.415 LMole of Pb(NO₃)₂ =?Mole = molarity × volume
Mole of Pb(NO₃)₂ = 0.120 × 0.415
Mole of Pb(NO₃)₂ = 0.0498 mole
Next, we shall determine the mole of NH₄I that reacted. Details below:
Pb(NO₃)₂(aq) + 2NH₄I(aq) ⟶ PbI₂(s) + 2NH₄NO₃(aq)
From the balanced equation above,
1 moles of Pb(NO₃)₂ reacted with 2 moles of NH₄I
Therefore,
0.0498 mole of Pb(NO₃)₂ will react with = 0.0498 × 2 = 0.1 mole of NH₄I
Finally, we shall determine the volume of NH₄I required for the reaction. Details below:
Molarity of NH₄I = 0.350 MMole of NH₄I = 0.1 moleVolume of NH₄I =?Volume = mole / molarity
Volume of NH₄I = 0.1 / 0.350
Volume of NH₄I = 0.286 L
Multiply by 1000 to express in mL
Volume of NH₄I = 0.286 × 1000
Volume of NH₄I = 286 mL
2. How do i determine the mole of PbI₂ formed?The mole of PbI₂ formed can be obtain as follow:
Pb(NO₃)₂(aq) + 2NH₄I(aq) ⟶ PbI₂(s) + 2NH₄NO₃(aq)
From the balanced equation above,
1 mole of Pb(NO₃)₂ reacted to produced 1 mole of PbI₂
Therefore,
0.0498 mole of Pb(NO₃)₂ will also react to produce 0.0498 mole of PbI₂
Thus, the number of mole of PbI₂ formed is 0.0498 mole
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A sample of an ideal gas has a volume of 2.31 L
at 279 K
and 1.01 atm.
Calculate the pressure when the volume is 1.09 L
and the temperature is 308 K.
The pressure of the gas when the volume is 1.09 L and the temperature is 308 K is 2.36 atm.
What is the final pressure of the gas?The final pressure of the gas is calculated by applying ideal gas law as follows;
(P₁V₁)/T₁ = (P₂V₂)/T₂
where
P₁, V₁, and T₁ are the initial pressure, volume, and temperature, P₂, V₂, and T₂ are the final pressure, volume, and temperature,P₂ = (P₁V₁ x T₂)/(V₂ x T₁)
P₂ = (1.01 atm x 2.31 L x 308 K) / (1.09 L x 279 K)
P₂ = 2.36 atm
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what is the volume and letters of a solution that contains 0.50 moles of NaOH dissolved in enough distilled water to make 3.0 mm of NaOH solution
what is the molarity of a solution that contains 60.0 G of caoh dissolved in 150 mm solution
1. To find the volume and units of a solution that contains 0.50 moles of NaOH dissolved in enough distilled water to make 3.0 M NaOH solution:
We first need to use the formula:
moles = concentration (in moles/L) x volume (in L)
Rearranging the formula to solve for volume, we get:
volume = moles / concentration
Substituting the given values, we get:
volume = 0.50 moles / 3.0 M = 0.17 L
Since the volume is given in liters, the units of the solution are L. Therefore, the solution contains 0.50 moles of NaOH dissolved in 0.17 L of distilled water, which makes a 3.0 M NaOH solution.
2. To find the molarity of a solution that contains 60.0 g of Ca(OH)2 dissolved in 150 mL of solution:
We first need to convert the mass of Ca(OH)2 to moles using the molar mass:
molar mass of Ca(OH)2 = 40.08 g/mol + 2 x 16.00 g/mol + 2 x 1.01 g/mol = 74.10 g/mol
moles of Ca(OH)2 = 60.0 g / 74.10 g/mol = 0.810 moles
Next, we need to convert the volume of the solution from milliliters to liters:
volume of solution = 150 mL / 1000 mL/L = 0.150 L
Finally, we can use the formula:
molarity = moles / volume
Substituting the given values, we get:
molarity = 0.810 moles / 0.150 L = 5.4 M
Therefore, the molarity of the solution is 5.4 M.
If the amount of solute present in a solution at a given temperature is less than the maximum amount that can be dissolved at that tempature the solution is said to be
Answer:
Unsaturated
Explanation:
A solution is unsaturated when it contains less than the maximum amount of solute that is capable of being dissolved.
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:
Three of the primary components of air are
carbon dioxide, nitrogen, and oxygen. In a
sample containing a mixture of only these
gases at exactly one-atmosphere pressure, the partial pressures of carbon dioxide and nitrogen are given as PCO2 = 0.285 torr and
PN2 = 580.502 torr. What is the partial pressure of oxygen?
Answer in units of torr.
The partial pressure of the oxygen is 0.236 atm.
What is partial pressure?The pressure that one gas component in a mixture of gases exerts is known as partial pressure. It is the pressure that the gas would experience if it took up the same amount of space in the mixture at the same temperature on its own.
We know that;
P[tex]CO_{2}[/tex] = 0.285 torr or 0.000375 atm
P[tex]N_{2}[/tex] = 580.502 torr or 0.764 atm
P[tex]O_{2}[/tex] = ?
Total pressure = 1 atm
Then we have that;
PT =P[tex]CO_{2}[/tex] +P[tex]N_{2}[/tex]+ P[tex]O_{2}[/tex]
P[tex]O_{2}[/tex] = PT - (P[tex]CO_{2}[/tex] + P[tex]N_{2}[/tex])
P[tex]O_{2}[/tex] = 1 - (0.000375 + 0.764)
P[tex]O_{2}[/tex]= 0.236 atm
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100 POINTS PLEASE HELP!!! 1. Explain how you would determine the enthalpy of reaction for the hypothetical reaction A2X4(l) + X2(g) → 2AX3(g) using the following information. You do not need to calculate an answer. Respond to the prompt with a minimum response length of 50 words.
To determine the enthalpy of reaction for the given reaction, use Hess's Law, which states that the total enthalpy change of a reaction is independent of the pathway between the initial and final states.
How to determine the enthalpy of reaction?Break the given reaction into a series of steps for which the enthalpy changes are known or can be measured experimentally.
Add the enthalpy changes of each step to determine the overall enthalpy change for the reaction.
For example, determine the enthalpy of formation for A₂X₄(l), X₂(g), and AX₃(g) and use them to calculate the enthalpy of reaction for the given equation.
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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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Which statements are TRUE about fossil fuels? (Select all that apply.)
They are in a limited supply.
They do not replenish themselves.
They are expensive to extract compared to other forms of energy.
They release large amounts of carbon dioxide when burned
✎help its an exam✎ ☕︎if any links I WILL REPORT☕︎
Answer:
All is Correct
Explanation:
Fossil fuels have the following properties:
They are in a limited supply. Fossil fuels are non-renewable resources, meaning that they cannot be replenished at the same rate as they are consumed. Once they are used up, they are gone forever.They do not replenish themselves. Fossil fuels take millions of years to form under specific geological conditions. They cannot be regenerated by natural processes in a human timescale.They are expensive to extract compared to other forms of energy. Fossil fuels require complex and costly methods to locate, drill, mine, transport, and refine. They also have negative externalities, such as environmental damage, health risks, and social conflicts, that are not reflected in their market prices.They release large amounts of carbon dioxide when burned. Fossil fuels contain carbon that was stored underground for millions of years. When they are burned, they release carbon dioxide (CO2) into the atmosphere, which is a greenhouse gas that contributes to global warming and climate change.Therefore, the answer is to select all
Answer:
It's A, B, and D
Explanation:
Maybe not D, because that is burning wood like charcoal. Not sure about that. Hope this helps!
Determine the rate constant for the reaction at 350.0 K
_____M-¹S-¹
The rate constant (k) for a chemical reaction can be determined experimentally by measuring the reaction rate (v) at different concentrations of reactants and plotting the data using a suitable rate law equation.
How to explain the reactionThere are different methods to determine the rate constant depending on the type of reaction, but a general approach is as follows:
Conduct the reaction under different initial concentrations of reactants while keeping other variables constant such as temperature, pressure, and pH.
Measure the reaction rate at each concentration by monitoring the change in concentration of reactants or products over time.
An overview was given based on incomplete information.
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PLEASE HELP!!
351.6g of Chromium-63 is allowed to decay for 128.8 years, how much Chromium-63
is left? (The half life of 63 Cr is 32.2 days.) Please, enter your answer as a one decimal
place number with no units.
A crucial trace mineral is chromium. Trivalent chromium, which is safe for people, and hexavalent chromium, which is toxic, are the two types.
Thus, Foods and dietary supplements both contain trivalent chromium. It might assist maintain normal blood sugar levels by enhancing the body's utilization of mineral.
Chromium is used by people to treat deficiencies. Additionally, it is used to treat bipolar disorder, diabetes, high cholesterol, and a variety of other conditions, but the majority of these uses are not well-supported by science.
Chromium by mouth doesn't help control blood sugar levels in people with prediabetes. Schizophrenia. Taking chromium by mouth doesn't affect weight or mental health in people with schizophrenia.
Thus, A crucial trace mineral is chromium. Trivalent chromium, which is safe for people, and hexavalent chromium, which is toxic, are the two types.
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A flexible container at an initial volume of 5.12 L
contains 8.51 mol
of gas. More gas is then added to the container until it reaches a final volume of 13.3 L.
Assuming the pressure and temperature of the gas remain constant, calculate the number of moles of gas added to the container.
Step-by-step Explanation:
8.51 moles is to 5.12 L as 'x' moles is to (13.3-5.12) L
8.51 moles / 5.12 L = x / ( 13.3-5.12)
x = 13.6 moles
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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d. Given this law, 4 of 4.
Select Choice
of hydrogen (H2) is produced in the following reaction.
Zn + 2HCl → ZnCl2 + H2
65 g 72 g 135 g ?
The mass of hydrogen produced in the reaction is 2g.
What is Mole?The mole is an amount unit similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of the number of atoms or molecules in a bulk sample of matter.
A mole is defined as the amount of substance containing the same number of atoms, molecules, ions, etc. as the number of atoms in a sample of pure 12C weighing exactly 12 g.
Given,
Mass of Zn = 65g
Mass of HCl = 72g
Moles of Zn = mass / molar mass
= 65 / 65 = 1 mole
Moles of HCl = 72 / 36.5
= 1.97 moles
Since moles of Zn is lesser, therefore it is the limiting reagent.
From the reaction, 1 mole of Zn gives 1 mole of hydrogen
Moles of hydrogen = 1 mole
mass of hydrogen = moles × molar mass
= 1 × 2 = 2g
Therefore, the mass of hydrogen produced in the reaction is 2g.
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A 192 gram piece of copper was heated to 100.°C in a boiling water bath, then it was dropped into a beaker containing 850. mL of water at 4.00°C. What is the final temperature of the copper and water after they come to thermal equilibrium?
Note: The specific heat of copper is 0.385 J/g °C.
Do not round your answer in the middle of the problem. Round at the very end.
Round your answer to the correct number of sig figs. Your units should be degrees Celsius.
The final temperature of the copper and water after they come to thermal equilibrium is 109.8°C.
What is temperature?Temperature is a measure of the amount of thermal energy present in a substance or object. It is measured in degrees on a scale such as Fahrenheit, Celsius, or Kelvin. Temperature is important in determining the physical and chemical properties of a substance, such as its melting point, boiling point, and specific gravity. Temperature also affects the rate of a chemical reaction and the speed of diffusion.
The change in temperature of the copper can be calculated using the equation
ΔT = (Q/mc), where Q is the heat transferred, m is the mass of the copper, and c is the specific heat of copper.
Q = mcΔT = (192 g)(0.385 J/g °C)(100°C) = 74080 J
The heat transferred from the copper must equal the heat transferred to the water. Therefore,
(74080 J) = (0.85 L)(4.184 J/g°C)(ΔT)
ΔT = (74080 J)/[(0.85 L)(4.184 J/g°C)] = 109.8°C
The final temperature of the copper and water after they come to thermal equilibrium is 109.8°C.
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