Answer:
[tex]k_2=0.504s^{-1}[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the rate constant at 55 °C by using the temperature-variable version of the Arrhenius equation:
[tex]ln(\frac{k_2}{k_1} )=-\frac{Ea}{R}(\frac{1}{T_2} -\frac{1}{T_1} )[/tex]
Thus, we plug in the temperatures, activation energy and universal constant of gases in consistent units to obtain:
[tex]ln(\frac{k_2}{0.0215s^{-1}} )=-\frac{72000\frac{J}{mol}}{8.3145\frac{J}{mol*K}}(\frac{1}{55+273} -\frac{1}{20+273} ) \\\\ln(\frac{k_2}{0.0215s^{-1}} )=3.154\\\\k_2=0.0215s^{-1}exp(3.154)\\\\k_2=0.504s^{-1}[/tex]
Regards!
How is magma formed?
Answer:
“Magma” is exclusively found and formed beneath the earth’s surface. Once magma is on or above the surface of the earth it is referred to as “lava.” Magma is typically formed by extreme temperature melting solid rock within the earth. Pressure and rock composition can also affect magma formation. High pressure can help magma be “squeezed” from partially molten rock. Likewise, as rocks are usually composed of different minerals with different melting points, magma formation from rocks is usually only partial and uneven.
Explanation:
molecular weight of K2SO3
Explanation:
the molecular weight of K2SO3 is 158. 2598 m/s.
A solution is made by dissolving 5.84 grams of NaCl in enough distilled water to give a final volume of 1.00 L. What is the molarity of the solution
Group of answer choices
0.0250 M
0.400 M
0.100 M
1.00 M
Answer:
Explanation:
1. A solution is made by dissolving 5.84g of NaCl is enough distilled water to a give a final volume of 1.00L. What is the molarity of the solution? a. 0.100 M b. 1.00 M c. 0.0250 M d. 0.400 M 2. A 0.9% NaCl (w/w) solution in water is a. is made by mixing 0.9 moles of NaCl in a 100 moles of water b. made and has the same final volume as 0.9% solution in ethyl alcohol c. a solution that boils at or above 100°C d. All the above (don't choose this one) 3. In an exergonic process, the system a. gains energy b. loses energy c. either gains or loses energy d. no energy change at all
Answer:
[tex]\boxed {\boxed {\sf 0.100 \ M }}[/tex]
Explanation:
Molarity is a measure of concentration in moles per liter.
[tex]molarity = \frac{moles \ of \ solute}{liters \ of \ solution}}[/tex]
The solution has 5.84 grams of sodium chloride or NaCl and a volume of 1.00 liters.
1. Moles of SoluteWe are given the mass of solute in grams, so we must convert to moles. This requires the molar mass, or the mass of 1 mole of a substance. These values are found on the Periodic Table as the atomic masses, but the units are grams per mole, not atomic mass units.
We have the compound sodium chloride, so look up the molar masses of the individual elements: sodium and chlorine.
Na: 22.9897693 g/mol Cl: 35.45 g/molThe chemical formula (NaCl) contains no subscripts, so there is 1 mole of each element in 1 mole of the compound. Add the 2 molar masses to find the compound's molar mass.
NaCl: 22.9897693 + 35.45 = 58.4397693 g/molThere are 58.4397693 grams of sodium chloride in 1 mole. We will use dimensional analysis and create a ratio using this information.
[tex]\frac {58.4397693 \ g\ \ NaCl} {1 \ mol \ NaCl}[/tex]
We are converting 5.84 grams to moles, so we multiply by that value.
[tex]5.84 \ g \ NaCl *\frac {58.4397693 \ g\ NaCl} {1 \ mol \ NaCl}[/tex]
Flip the ratio. It remains equivalent and the units of grams of sodium chloride cancel.
[tex]5.84 \ g \ NaCl *\frac {1 \ mol \ NaCl}{58.4397693 \ g\ NaCl}[/tex]
[tex]5.84 *\frac {1 \ mol \ NaCl}{58.4397693 }[/tex]
[tex]0.09993194823 \ mol \ NaCl[/tex]
2. MolarityWe can use the number of moles we just calculated to find the molarity. Remember there is 1 liter of solution.
[tex]molarity= \frac{moles \ of \ solute}{liters \ of \ solution}[/tex]
[tex]molarity= \frac{ 0.09993194823 \ mol \ NaCl}{1 \ L}[/tex]
[tex]molarity= 0.09993194823 \ mol \ NaCl/L[/tex]
3. Units and Significant FiguresThe original measurements of mass and volume have 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandths place. The 9 in the ten-thousandths place tells us to round the 9 to a 0, but then we must also the next 9 to a 0, and the 0 to a 1.
[tex]molarity \approx 0.100 \ mol \ NaCl/L[/tex]
1 mole per liter is 1 molar or M. We can convert the units.
[tex]molarity \approx 0.100 \ M \ NaCl[/tex]
The molarity of the solution is 0.100 M.
Determine the kinds of intermolecular forces that are present in each of the following. Part A Xe Check all that apply. dispersion forces dipole-dipole forces hydrogen bonding Request Answer Part B N2 Check all that apply. dispersion forces dipole-dipole forces hydrogen bonding Request Answer Part C CO Check all that apply. dispersion forces dipole-dipole forces hydrogen bonding Request Answer Part D HF Check all that apply. dispersion forces dipole-dipole forces hydrogen bonding
Answer:
Part A
dispersion forces
Part B
dispersion forces
Part C
dispersion forces
dipole-dipole forces
Part D
dispersion forces
dipole-dipole forces
hydrogen bonding
Explanation:
Dispersion forces occur in all molecules. They result from momentary shifts in the electron cloud of molecules which induces a dipole in another molecule. This induced dipole eventually spreads throughout the molecule.
For Xe which is a noble gas and N2 which is a diatomic molecule, dispersion forces is the only kind of intermolecular force present in the molecule.
CO is a polar molecule hence in addition to dispersion forces, dipole-dipole forces also exist in the molecule.
HF is a polar molecule hence it possesses dipole-dipole forces in addition to dispersion forces. In this molecule, hydrogen is bonded to a highly electronegative atom (fluorine). Hence, hydrogen bonding is a dominant intermolecular interaction in the molecule.
How many moles of (CH3)3NH+ are in 6.0 g of (CH3)3NH+?
Answer:
0.1 mol
Explanation:
6/(15*3+15)
0.1 mol moles of (CH3)3NH+ are in 6.0 g of (CH3)3NH+
What is mole?
The mole, symbol mol, exists as the SI base unit of the amount of substance. The quantity amount of substance exists as a measure of how many elementary entities of a provided substance exist in an object or sample.A mole corresponds to the mass of a substance that includes 6.023 x 1023 particles of the substance. The mole exists the SI unit for the amount of a substance. Its symbol stands mol.
The compound trimethylamine, (CH3 )3N, exists as a weak base when dissolved in water.
A mole exist expressed as 6.02214076 × 1023 of some chemical unit, be it atoms, molecules, ions, or others. The mole exists as a convenient unit to utilize because of the great number of atoms, molecules, or others in any substance.
To find the amount of the substance (CH3)3NH+ to calculate its molar mass:
M((CH3)3NH+) = (12+3)*3 + 14+1 = 60 g/mol
n((CH3)3NH+) = m/M
m((CH3)3NH+) = 6g
Thus,
n((CH3)3NH+) = 6g/60 g/mol = 0.1 mol
Hence,
n((CH3)3NH+) = 0.1 mol
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Different vinegars can be 5-20% acetic acid solutions and have been used for medicinal purposes for thousands of years. If a person takes 2.0 tablespoons of vinegar a day and the Molarity of the vinegar is .84 M, then how many grams of acetic acid (HC2H3O2) will be consumed? 1 Tablespoon is 15 mL.
.013 g
.026 g
.76 g
1.5 g
Answer:
1.5g
Explanation:
Remember that Molarity = (#moles of solute)/(#liters of solution)
This problem informs us that the Molarity of the vinegar is 0.84 and that the solution is 15mL.
First let's get your SI units to the correct ones.
15mL (1L/1000mL) = 0.015L
Molarity = (#moles of solute)/(#liters of solution) ~
(Molarity)(#liters of solution) = #moles of solute
(0.84M)(.015L) = 0.0126moles of acetic acid per tablespoon
2 tablespoons a day = 0.0126moles*2 = 0.0252 moles of acetic acid.
Now that we have the # of moles of acetic acid we need to get our answer into grams. The molecular weight of HC2H3O2 is 60g/mole.
0.0252mole HC2H3O2 (60g HC2H3O2/1mole HC2H3O2) = 1.512g ~ 1.5g HC2H3O2.
The turbines in a hydroelectric plant are fed by water falling from a 50 m height. Assuming 91% efficiency for conversion of potential to electrical endrgy, and 8% loss of the resulting power in transmission, what is the mass flow rate of water required to power a 200 W light bulb?
From the information given;
the height of the water stream = 50 mthe efficiency of conversion from potential energy to electrical energy is 91%loss of power transmission = 8%To determine the mass flow rate, let's start by understanding some concepts and parameters.
The power is known to be the energy per unit of time. Mathematically, it can be written as:
[tex]\mathbf{Power = \dfrac{Energy}{Time}}[/tex]
[tex]\mathbf{P =\dfrac{E_p}{time}}[/tex]
[tex]\mathbf{P =\dfrac{m\times g\times z}{time}}[/tex]
where;
[tex]\mathbf{E_p}[/tex] is the potential energy of the streamm = mass flow rateg = acceleration under gravityz = heightThus;
[tex]\mathbf{E_p}[/tex] = m × 9.81 m/s² × 50 m
[tex]\mathbf{E_p}[/tex] = m × 490.5 (m²/s²)
Recall that:
The power P = 200 W, and;the conversion of the P.E = 91% = 0.91∴
[tex]\mathbf{E_p}[/tex] = 0.91 × 490.5m (m²/s²)
[tex]\mathbf{E_p}[/tex] = 446.355m (m²/s²)
Since the resulting power transmission is said to be 8%
Then;
the loss in the power transmission (P) = 100% - 8% × 446.355m (m²/s²)
the loss in the power transmission (P) = 92% × 446.355m (m²/s²)
the loss in the power transmission (P) = 0.92 × 446.355m (m²/s²)
the loss in the power transmission (P) = 410.65m (m²/s²)
Finally;
P = 410.65m (m²/s²)
[tex]\mathbf{P = 410.65 \times m (\dfrac{m^2}{s^2})}[/tex]
replacing the values, we have:
[tex]\mathbf{200 = 410.65 \times m (\dfrac{m^2}{s^2})}[/tex]
[tex]\mathbf{m = \dfrac{200 watt}{410.65\times (\dfrac{m^2}{s^2})}}[/tex]
[tex]\mathbf{m = \dfrac{200 \dfrac{J}{s}}{410.65\times (\dfrac{m^2}{s^2})}}[/tex]
since 1 J/s = 1 kgm²/s²)
Then:
[tex]\mathbf{m = \dfrac{200 \dfrac{\dfrac{kg\times m^2}{s^2}}{s}}{410.65\times (\dfrac{m^2}{s^2})}}[/tex]
[tex]\mathbf{m = \dfrac{200 \ {kg}}{410.65 \ s}}[/tex]
mass flow rate of the water (m) = 0.487 kg/s
Therefore, we can conclude that the mass flow rate of the water required to power a 200 W bulb light is 0.487 kg/s
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385 x 42.13 x 0.079 is (consider significant figures):
385 x 42.13 x 0.079 = 1281.38395
A solution is prepared by dissolving 6.60 g of an nonelectrolyte in water to make 550 mL of solution. The osmotic pressure of the solution is 1.84 atm at 25 °C. The molecular weight of the nonelectrolyte is ________ g/mol.
Answer:
160 g/mol
Explanation:
Step 1: Calculate the molarity of the solution
We will use the following expression.
π = M × R × T
where,
π: osmotic pressure of a nonelectrolyteM: molarityR: ideal gas constantT: absolute temperature (25 °C = 298 K)M = π / R × T
M = 1.84 atm / (0.0821 atm.L/mol.K) × 298 K = 0.0752 mol/L
Step 2: Calculate the moles of solute in 550 mL (0.550 L)
0.550 L × 0.0752 mol/L = 0.0413 mol
Step 3: Calculate the molecular weight of the nonelectrolyte
0.0413 moles weigh 6.60 g.
6.60 g/0.0413 mol = 160 g/mol
15. In the image given below, magnesium metal is coiled as a thin ribbon. What property of metal is exhibited by it? A Ductility B Lustrous C Sonorous D Malleability
Answer: The property of magnesium that is exhibited by it is DUCTILITY. The correct option is A.
Explanation:
Magnesium is a member of the alkaline earth metals. It occurs in nature, only in the combined state, as Epsom salt, dolomite and in many trioxosilicates( IV) including talc and asbestos. They have the following physical properties:
--> Appearance: they are silvery-white solids
--> Relative density: It has a relative density of 1.74
--> DUCTILITY: it's very ductile in nature
--> melting point: it has a melting point of 660°C.
--> Conductivity: They are good conductor of heat and electricity.
Furthermore, DUCTILITY is the physical property of a metal associated with the ability to be hammered thin or stretched into wire without breaking. A metal such as magnesium can therefore be coiled as a thin ribbon without fracturing due to its ductile physical properties.
A mixture of argon and neon gases at a total pressure of 874 mm Hg contains argon at a partial pressure of 662 mm Hg. If the gas
mixture contains 12.0 grams of argon, how many grams of neon are present?
Answer:
6.684g
Explanation:
Here, we can use the mole ratio of the gases to calculate.
We know that the mole ratio of the gases equate to their number of moles.
Firstly, we calculate the number of moles of the oxygen gas. The number of moles of the oxygen gas is equal to the mass of the oxygen gas divided by the molar mass of the oxygen gas. The molar mass of the oxygen gas is 32g/mol
Thus, the number of moles produced is 5.98/32 = 0.186875
Where do we move from here?
We know that if we place the partial pressure of oxygen over the total pressure, this would be equal to the number of moles of oxygen divided by the total number of moles. Now let’s do this.
449/851 = 0.186875/n
n =(0.186875 * 851)/449
n = 0.3542
Now we do the same for argon to get the number of moles of argon.
Firstly, we use dalton’s partial pressure law to get the partial pressure of argon. In the simplest form, the partial pressure of argon is the total pressure minus the partial pressure of oxygen.
P = 851 - 449 = 402 mmHg
We now use the mole ratio relation.
402/851 = n/0.3542
n = (402 * 0.3542) / 851
n = 0.1673
Since we now know the number of moles of argon, we can use this multiplied by the atomic mass of argon to get the mass.
The atomic mass of argon is 39.948 amu
The mass is thus 39.948 * 0.1673 = 6.684g
Carbon dioxide gas is collected at 27.0 oC in an evacuated flask with a measured volume of 30.0L. When all the gas has been collected, the pressure in the flask is measured to be 0.480atm. Calculate the mass and number of moles of carbon dioxide gas that were collected.
Answer:
[tex]M_{CO_2}= 25.7g[/tex]
Explanation:
From the question we are told that:
Temperature [tex]T=27.0[/tex]
Volume [tex]V=30L[/tex]
Pressure [tex]P=0.480atm[/tex]
Generally the equation for Ideal gas is mathematically given by
PV=nRT
Therefore
[tex]n=\frac{0.480 x 30}{0.08205 x 300}[/tex]
[tex]n=0.59moles[/tex]
Generally Mass of CO2 is given as
[tex]M_{CO_2}= 0.59 * 44 g/mol[/tex]
[tex]M_{CO_2}= 25.7g[/tex]
A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. The metal and water come to the same temperature at 24.6 °C. How much heat did the metal give up to the water?
Answer:
1087.84 J
Explanation:
From the question given above, the following data were obtained:
Mass of metal (Mₘ) = 70 g
Temperature of metal (Tₘ) = 80 °C
Mass of water (Mᵥᵥ) = 100 g
Temperature of water (Tᵥᵥ) = 22 °C
Equilibrium temperature (Tₑ) = 24.6 °C
Heat lost by metal (Qₘ) =?
NOTE: Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gºC
Heat lost by metal (Qₘ) = Heat gained by water (Qᵥᵥ)
Qₘ = Qᵥᵥ
Thus, we shall determine the heat gained by water. This can be obtained as follow:
Qᵥᵥ = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)
Qᵥᵥ = 100 × 4.184 (24.6 – 22)
Qᵥᵥ = 418.4 × 2.6
Qᵥᵥ = 1087.84 J
Thus, the heat gained by water is 1087.84 J.
Heat lost by metal (Qₘ) = Heat gained by water (Qᵥᵥ)
Qₘ = Qᵥᵥ
Qᵥᵥ = 1087.84 J
Qₘ = 1087.84 J
Therefore, the heat lost by the metal is 1087.84 J
A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. After reaching a temperature of 24.6 °C, the heat given up by the metal to the water is -1.08 kJ.
What is a calorimeter?A calorimeter is an object used for calorimetry, or the process of measuring the heat of chemical reactions or physical changes as well as heat capacity.
A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. The final temperature of the system is 24.6 °C.
Let's use the following expression to calculate the heat absorbed by the water.
Qw = c × m × ΔT
Qw = (4.184 J/g.°C) × 100 g × (24.6 °C - 22.0 °C) = 1.08 kJ
where,
Qw is the heat absorbed by the water.c is the specific heat capacity of water.m is the mass of water.ΔT is the change in the temperature for water.According to the law of conservation of energy, the sum of the heat absorbed by the water and the heat released by the metal (Qm) is zero.
Qw + Qm = 0
Qm = -Qw = -10.8 kJ
A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. After reaching a temperature of 24.6 °C, the heat given up by the metal to the water is -1.08 kJ.
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Dung dịch nào sau đây chỉ chứa các ion (bỏ qua sự điện li của nước, các chất điện li mạnh phân li hoàn toàn)?
A. HBr, Na2S, Mg(OH)2, Na2CO3.
B. H2SO4, NaOH, NaCl, HF.
C. HNO3, H2SO4, KOH, K2SiO3.
D. Ca(OH)2, KOH, CH3COOH, NaCl.
Answer:
Dung dịch nào sau đây chỉ chứa các ion (bỏ qua sự điện li của nước, các chất điện li mạnh phân li hoàn toàn)?
A. HBr, Na2S, Mg(OH)2, Na2CO3.
B. H2SO4, NaOH, NaCl, HF.
C. HNO3, H2SO4, KOH, K2SiO3.
D. Ca(OH)2, KOH, CH3COOH, NaCl.
100 mL of 0.2 mol/L sodium carbonate solution and 200 mL of 0.1 mol/L calcium nitrate solution are mixed together. Calculate the mass of calcium carbonate that would precipitate and the concentration of the sodium nitrate solution that will be produced.
Answer:
Explanation:
Na2CO3+Ca(NO3)2=CaCO3+2NaNO3
nNa2CO3=0.02
nCa(NO3)2=0.02
mCaCO3=0.02*100=2 gram
nNaNo3=0.04
Cm=2/15
From the calculation, the mass of the product is 2 g.
What is a reaction?A chemical reaction occurs when two more substances are mixed together. In this case, the reaction is shown by; Ca(NO3)2 + Na2CO3 ----> CaCO3(s) + 2NaNO3.
Number of moles of Na2CO3 = 100/1000 L * 0.2 mol/L = 0.02 moles
Number of moles of Ca(NO3)2 = 200/1000 L * 0.1 mol/L = 0.02 moles
Since the reaction is equimolar, amount of the product = 0.02 moles * 100 g/mol = 2 g
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15.27
The following equilibria were attained at 823 K:
COO(s) + H2() Co(s) + H2O(g) K = 67
COO(s) + CO(8) = Co(s) + CO2(8) K = 490
Based on these equilibria, calculate the equilibrium con-
stant for
H2(g) + CO2(g) = CO(g) + H2O(g) at 823 K.
The equilibrium constant for the reaction is K = 0.137
We obtain the equilibrium constant considering the following equilibria and their constants:
COO(s) + H₂(g) → Co(s) + H₂O(g) K₁ = 67
COO(s) + CO(g) → Co(s) + CO₂(g) K₂ = 490
We write the first reaction in the forward direction because we need H₂(g) in the reactants side:
(1) COO(s) + H₂(g) → Co(s) + H₂O(g) K₁ = 67
Then, we write the second reaction in the reverse direction because we need CO₂(g) in the reactants side. Thus, the equilibrium constant for the reaction in the reverse direction is the reciprocal of the constant for the reaction in the forward direction (K₂):
(2) Co(s) + CO₂(g) → COO(s) + CO(g) K₂ = 1/490
From the addition of (1) and (2), we obtain:
COO(s) + H₂(g) → Co(s) + H₂O(g) K₁ = 67
+
Co(s) + CO₂(g) → COO(s) + CO(g) K₂ = 1/490
-------------------------------------------------
H₂(g) + CO₂(g) → CO(g) + H₂O(g)
Notice that Co(s) and COO(s) are removed that appear in the same amount at both sides of the chemical equation.
Now, the equilibrium constant K for the reaction that is the sum of other two reactions is calculated as the product of the equilibrium constants, as follows:
K = K₁ x K₂ = 67 x 1/490 = 67/490 = 0.137
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A hot pot of water is set on the counter to cool. After a few minutes it has lost 495 J of heat energy. How much heat energy has the surrounding air gained?
_____unit_____
Answer:
495 J
Explanation:
When the hot pot was set on the counter to cool, heat energy was lost from the pot. Note that according to the first law of thermodynamics, heat is neither created nor destroyed.
This implies that, the heat energy lost from the pot must be gained by the surrounding air. Therefore, if 495 J of energy is lost from the pot, then 495 J of energy is gained by the surrounding air.
how can we convert plastic garbage energy into electric energy
Answer:
Unfortunately, we don`t know how to convert plastic material into electricity yet. I suppose an idea is for someone to invent a machine similar to biomass, where dead plants created energy, but here it`s plastic. The only issue is that it could release deadly chemicals.
Sorry if this isn`t much help, but there isn`t really an answer. :/
The half life of radium-226 is 1600 years. If you have 200 grams of radium today how many grams would be present in 8000 years?
Answer:
Half life is the time taken by a radio active isotope to reduce by half of its original amount. Radium-226 has a half life of 1602 years meaning that it would take 1602 years for a mass of radium to reduce by half.
Number of half lives in 9612 years = 9612/1602 = 6 half lives
New mass = Original mass x (1/2)n where n is the number of half lives.
Therefore, New mass= 500 x (1/2)∧6
= 500 x 0.015625
= 7.8125 g
Hence the mass of radium after 9612 years will be 7.8125 grams.
Explanation:
Answer:
[tex]\boxed {\boxed {\sf 6.25 \ grams}}[/tex]
Explanation:
We are asked to find the mass of a sample of radium-226 after half-life decay. We will use the following formula:
[tex]A= A_o *\frac{1}{2}^{\frac{t}{h}}[/tex]
In this formula, [tex]A_o[/tex] is the initial amount, t is the time, and h is the half-life.
For this problem, the initial amount is 200 grams of radium-226, the time is 8,000 years, and the half-life is 1,600 years.
[tex]\bullet \ A_o= 200 \ g \\\\bullet \ t= 8,000 \ \\\bullet \ h= 1,600[/tex]
Substitute the values into the formula.
[tex]A= 200 \ g * \frac{1}{2} ^{\frac{8.000}{1,600}[/tex]
Solve the fraction in the exponent.
[tex]A= 200 \ g * \frac{1}{2}^{5}[/tex]
Solve the exponent.
[tex]A= 200 \ g *0.03125[/tex]
[tex]A= 6.25 \ g[/tex]
In addition, we can solve this another way. First, we find the number of half-lives by dividing the total time by the half-life.
8,000/1,600= 5 half-livesEvery half-life, 1/2 of the mass decays. Divide the initial mass in half, then that result in half, and so on 5 times.
1. 200 g/2= 100 g2. 100 g / 2 = 50 g3. 50 g / 2 = 25 g 4. 25 g / 2 = 12.5 g5. 12.5 g / 6.25 gAfter 8,000 years, 6.25 grams of radium-226 remains.
Oxygen is composed of three isotopes: oxygen-16, oxygen-17 and oxygen-18 and has an average atomic mass of 15.9982 amu. Oxygen-17 has a mass of 16.988 amu and makes up 0.032% of oxygen. Oxygen-16 has a mass of 15.972 amu and oxygen-18 has a mass of 17.970 amu. What is the percent abundance of oxygen-18?
Answer:
The percent abundance of oxygen-18 is 1.9066%.
Explanation:
The average atomic mass of oxygen is given by:
[tex] m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18} [/tex]
Where:
m: is the atomic mass
%: is the percent abundance
Since the sum of the percent abundance of oxygen isotopes must be equal to 1, we have:
[tex] 1 = \%_{16} + \%_{17} + \%_{18} [/tex]
[tex] 1 = x + 3.2 \cdot 10^{-4} + \%_{18} [/tex]
[tex] \%_{18} = 1 - x - 3.2 \cdot 10^{-4} [/tex]
Hence, the percent abundance of O-18 is:
[tex] m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18} [/tex]
[tex]15.9982 = 15.972*x + 16.988*3.2 \cdot 10^{-4} + 17.970*(1 - 3.2 \cdot 10^{-4} - x)[/tex]
[tex] x = 0.980614 \times 100 = 98.0614 \% [/tex]
Hence, the percent abundance of oxygen-18 is:
[tex]\%_{18} = (1 - 3.2 \cdot 10^{-4} - 0.980614) \times 100 = 1.9066 \%[/tex]
Therefore, the percent abundance of oxygen-18 is 1.9066%.
I hope it helps you!
A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 6.90kg of water at 34.7 degrees C . During the reaction 57.1kJ of heat flows out of the bath and into the flask.
Calculate the new temperature of the water bath. You can assume the specific heat capacity of water under these conditions is 4.18J.g^(-1).K^(-1) . Round your answer to significant digits.
Answer:
[tex]T_2= 36.7 \textdegree C[/tex]
Explanation:
Mass of Water [tex]m_w=6.90kg[/tex]
Temperature [tex]T=34.7 degrees[/tex]
Heat Flow [tex]H=57.1kJ[/tex]
Specific heat capacity of water [tex]\mu= 4.18J.g^(-1).K^(-1)[/tex]
Generally the equation for Final Temperature is mathematically given by
[tex]M*\mu *T_1 + Q = M*\mu *T_2[/tex]
[tex]T_2=\frac{M*\mu *T_1 + Q }{M*\mu}[/tex]
Therefore
[tex]T_2=\frac{6.90*4.18*34.7 + 57.1}{6.90*4.18}[/tex]
[tex]T_2= 36.7 \textdegree C[/tex]
convert 100kcals to kilojoules
Answer:
Explanation:
418.4kj is the correct answer
Consider the reaction: NaNO3(s) + H2SO4(l) NaHSO4(s) + HNO3(g) ΔH° = 21.2 kJ
How much heat must absorbed by the reaction system to convert 100g of NaNO3 into NaHSO4(s)?
Answer:
endet nach selam nw
4gh7
Arrange the following compounds in order of increasing reactivity (least reactive first.) to electrophilic aromatic substitution:.
Bromobenzene Nitrobenzene Benzene Phenol
a. Bromobenzene < Nitrobenzene < Benzene < Phenol
b. Nitrobenzene < Bromobenzene < Benzene < Phenol
c. Phenol < Benzene < Bromobenzene < Nitrobenzene
d. Nitrobenzene < Benzene < Bromobenzene < Phenol
Answer:
Nitrobenzene < Bromobenzene < Benzene < Phenol
Explanation:
Aromatic compounds undergo electrophilic aromatic substitution reaction in the presence of relevant electrophiles. Certain substituents tend to increase or decrease the tendency of an aromatic compound towards electrophilic aromatic substitution reaction.
Substituents that increase the electron density around the ring such as in phenol tends to make the ring more reactive towards electrophilic substitution. Halogens such as bromine has a -I inductive effect as well as a +M mesomeric effect.
However the -I(electron withdrawing effect) of the halogens supersedes the +M electron donation due to mesomeric effect.
Putting all these together, the order of increasing reactivity of the compounds towards electrophilic aromatic substitution is;
Nitrobenzene < Bromobenzene < Benzene < Phenol
compared to an atom of C-14, an atom of C-12 has a lesser
atomic number
number of protons
number of electrons
number of neutrons
Answer:
mass number
Explanation:
because the mass
number is the number of protons plus the number of neutron and the number of proton in an elements is always the same , therefore and atom of C-14 has greater mass numberA solution is made by dissolving 0.565 g of potassium nitrate in enough water to make up 250. mL of solution. What is the molarity of this solution?
Please explain and show work.
[tex]\\ \large\sf\longmapsto KNO_3[/tex]
[tex]\\ \large\sf\longmapsto 39u+14u+3(16u)[/tex]
[tex]\\ \large\sf\longmapsto 53u+48u[/tex]
[tex]\\ \large\sf\longmapsto 101u[/tex]
[tex]\\ \large\sf\longmapsto 101g/mol[/tex]
Now
[tex]\boxed{\sf No\:of\:moles=\dfrac{Given\:mass}{Molar\:mass}}[/tex]
[tex]\\ \large\sf\longmapsto No\:of\:moles=\dfrac{0.565}{101}[/tex]
[tex]\\ \large\sf\longmapsto No\:of\:moles=0.005mol[/tex]
We know
[tex]\boxed{\sf Molarity=\dfrac{Moles\:of\:solute}{Vol\:of\:Solution\:in\:L}}[/tex]
[tex]\\ \large\sf\longmapsto Molarity=\dfrac{0.005}{\dfrac{250}{1000}L}[/tex]
[tex]\\ \large\sf\longmapsto Molarity=\dfrac{0.005}{0.250}[/tex]
[tex]\\ \large\sf\longmapsto Molarity=0.02M[/tex]
[tex] \: \: \: \: \: \: \: \: \: [/tex]
What Volume of silver metal will weigh exactly 2500.0g. The density of silver
Answer:
cm3 = 2500.0 g / 10.5 g/cm3 = 238 cm3
There are three isotopes of carbon. They have mass number of 12, 13 and 14. The average atomic mass of carbon is 12.0107 amu. What does this say about the relative abundances of the three isotopes?
Answer:
lots more of the carbon 12 than the others
havent calculated it percentage-wise but you can see its very close to 12 meaning it is of far greater abundance that carbon 13 and 14
Explanation:
Please help fast
All four referenced Greek thinkers: Democritus, Aristotle, Archimedes, and Anaxagoras, observed Nature and argued for his theory of
the composition of matter and natural laws. Only one of them tested his hypothesis and proposed a natural laws based on reproducible
observations, controlled experiments, and mathematical reasoning. All others used logic and thought experiments, as philosophers do,
to support their theories. Who is the experimental scientist in this group?
O Democritus
O Aristotle
O Archimedes
O Anaxagoras
Answer:
Anaxagoras was perhaps the first literate person to attempt to explain physical phenomena rationally, basing his ideas upon careful observations and simple experiments. This is fundamental to modern science and is the sine qua non of environmental study.
19. Which type of chemical process is used to remove salt from ocean water?
O A. Alkylation
O B. Doping
O C. Dehydrogenation
D. Desalination
Answer:
D
Explanation:
Desalination
Removing salt from sea water is known as desalination
