Discussion:
You can describe the reaction that took place in terms of enthalpy, by writing a fully balanced equation (and net ionic equation) for the reaction, as well as drawing an energy change diagram for the reaction, clearly indicating the measured quantity of heat energy change.Is the reaction exothermic or endothermic? Explain this in terms of bonds breaking and formingDiscuss the method and set up of the experimentWas the experiment accurate (calculate percentage error). [tex]|\frac{theoretical-experimental}{theoretical} |[/tex] × 100%. The theoretical value is -55.84 kJ/mol. error from 0% to ≈30% is accurate.Was the experiment reliable? (are results of each trial close to each other?) (only if applicable)Was the experiment valid? (is it both reliable and accurate)What can be done to improve the experiment? to improve reliability, validity, accuracy?Conclusion: You can use this basic outline, to structure your conclusion, and expand it from there.
By investigating/measuring/using a....... it was determined that........ This is consistent/not consistent with the expected result/theory of...... due to/because of...........
in an endothermic reaction, the total energy at the beginning of the reaction is group of answer choices less than the total energy at the end of the reaction. greater than the total energy at the end of the reaction. equal to the total energy at the end of the reaction. none of the above
The correct option is
In an endothermic response(reaction), the whole vitality(total energy) at the beginning of the response is more noteworthy than the full vitality at the conclusion of the response
because endothermic responses retain warmth from the environment, which implies that the vitality of the framework increments.
An endothermic response may be a chemical response that retains warmth from the environment, which implies that the vitality of the framework increments.
This increment in vitality is utilized to break the bonds between the particles or atoms within the reactants, and the items are shaped from the modification of these iotas or atoms into unused bonds.
As a result, the whole vitality of the framework at the conclusion of the response is more noteworthy than the full vitality at the start of the response. This increment in vitality is ordinarily watched as an increment within the temperature of the framework or its environment.
In an endothermic response, the whole vitality at the beginning of the response is less than the overall vitality at the end of the response
.
Usually, endothermic responses retain warmth from the environment, which implies that the vitality of the framework increments.
As a result, the entire vitality of the framework at the conclusion of the response is greater than the full energy at the start of the response. Subsequently,
The proper reply is "In an endothermic response(reaction), the whole vitality(total energy) at the beginning of the response is more noteworthy than the full vitality at the conclusion of the response ".
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How much energy is radiated by a non-spinning black hole that accretes 10-7 Msun per year? We compute this using L=ηMc2 (where M is the accretion rate). Putting in the numbers, we find L=0.06(10-7 x 6.3 kg/s)(3 x 108 m/s)2 = 3.4 x 1031.
The energy radiated by the black hole would be approximately 3.4 x 1031 Joules per second.This means the black hole radiates 3.4 x 10^31 watts of energy.
The energy radiated by a non-spinning black hole that accretes 10-7 Msun per year can be computed using the formula L=ηMc2, where M is the accretion rate. Putting in the numbers, we find L=0.06(10-7 x 6.3 kg/s)(3 x 108 m/s)2 = 3.4 x 1031. Therefore, the energy radiated by the black hole would be approximately 3.4 x 1031 Joules per second.
To calculate the energy radiated by a non-spinning black hole that accretes 10^-7 Msun per year, you can use the formula L=ηMc^2, where L is the luminosity, η is the efficiency, M is the accretion rate, and c is the speed of light. Plugging in the numbers, L=0.06(10^-7 x 6.3 kg/s)(3 x 10^8 m/s)^2 = 3.4 x 10^31. This means the black hole radiates 3.4 x 10^31 watts of energy.
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which compounds used in this experiment should one be careful with when using a hot plate? 4-methylphenol and diethyl ether 2-methyl-2-propanol and sulfuric acid diethyl ether and tert-butanol 4-methylphenol and glacial acetic acid
One should be careful with diethyl ether and tert-butanol when using a hot plate as they have low flash points and can easily ignite.
It is important to take proper precautions such as using a well-ventilated area and avoiding any sources of ignition. Sulfuric acid and glacial acetic acid are also potentially dangerous as they are corrosive and can cause severe burns if they come into contact with skin. Propanol and butanol have higher flash points and are generally safer to use on a hot plate.
When using a hot plate in an experiment, one should be particularly careful with diethyl ether and tert-butanol. Diethyl ether is highly flammable and volatile, while tert-butanol (2-methyl-2-propanol) can generate flammable vapors when heated. These compounds pose a risk of fire or explosion if not handled properly.
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How would the reading on a barometer change if you were to take one on a trip from los angeles to Lake Tahoe, which is at a much higher altitude?
The reading on a barometer would change if you were to take one on a trip from Los Angeles to Lake Tahoe, which is at a much higher altitude.
1. A barometer measures atmospheric pressure.
2. Atmospheric pressure decreases with an increase in altitude.
3. Los Angeles is at a lower altitude (approximately 305 feet or 93 meters above sea level) compared to Lake Tahoe (about 6,225 feet or 1,897 meters above sea level).
4. As you travel from Los Angeles to Lake Tahoe, the altitude increases.
5. Due to the increase in altitude, the atmospheric pressure decreases.
6. The barometer reading in Lake Tahoe will be lower than the reading in Los Angeles.
In conclusion, the barometer reading would be lower in Lake Tahoe compared to Los Angeles due to the higher altitude and the resulting decrease in atmospheric pressure.
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If you were to take a barometer on a trip from Los Angeles to Lake Tahoe, which is at a much higher altitude, the reading on the barometer would decrease as you ascend to higher altitudes.
A barometer is a device used to measure atmospheric pressure. The pressure at sea level is approximately 1013 hPa (hectopascals) or 29.92 inches of mercury (inHg). As you increase in altitude, the pressure decreases due to the decreased weight of the atmosphere above. For every 1000 feet increase in altitude, there is an approximate decrease of 1 inch of mercury (inHg) or 33 hPa in pressure.
Lake Tahoe has an elevation of approximately 6,225 feet, which is significantly higher than Los Angeles, which is only 233 feet above sea level. As a result, the atmospheric pressure at Lake Tahoe would be lower than the pressure in Los Angeles. Therefore, if you were to take a barometer from Los Angeles to Lake Tahoe, the reading on the barometer would decrease as you ascend to higher altitudes.
In summary, the reading on a barometer would decrease as you ascend to higher altitudes such as Lake Tahoe from Los Angeles.
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which of the following phase changes is exergonic? question 29 options: melting vaporization all phase changes are exergonic condensation
Only the condensation is an exergonic phase change.
Exergonic refers to a process that releases energy. Among the given options, the exergonic phase change is condensation. When a gas turns into a liquid during condensation,
it releases heat energy. This is because the molecules lose kinetic energy and move closer together, forming stronger attractive forces. The energy that was previously used to keep the gas molecules apart is released as heat energy.
On the other hand, melting and vaporization are endergonic phase changes, which require an input of energy to occur. Melting requires heat energy to break the intermolecular bonds between the solid molecules,
while vaporization requires even more energy to overcome the stronger intermolecular forces in the liquid and form a gas.
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the shattered glass case at the scene of a jewelry store robbery was determined to be made of potash borosilicate glass, which has a density of 2.16 g/ml. a 2.573 g glass fragment was recovered from a suspect's clothing. when the fragment was placed into a graduated cylinder filled with water, 1.14 ml of the water was displaced. calculate the density of the glass fragment.
The density of the glass fragment is approximately 2.26 g/ml
What is the density of the fragment?To calculate the density of the glass fragment, we can use the formula:
Density = Mass / Volume
First, let's calculate the volume of the glass fragment using the displacement method. The volume of water displaced when the glass fragment was submerged in the graduated cylinder is given as 1.14 ml.
So, the volume of the glass fragment is 1.14 ml.
Next, we can calculate the density of the glass fragment by dividing the mass of the glass fragment by its volume:
Density = Mass / Volume = 2.573 g / 1.14 ml
Density = 2.573 g / 1.14 ml ≈ 2.26 g/ml
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if a solution originally 0.532 m in acid ha is found to have a hydronium concentration of 0.112 m at equilibrium, what is the percent ionization of the acid?
To find the percent ionization of the acid, we need to first calculate the initial concentration of the acid (HA) before it dissociates.
Since the solution is originally 0.532 M in acid (HA), we can assume that the initial concentration of HA is also 0.532 M.
Next, we need to calculate the concentration of the conjugate base (A-) at equilibrium. We can use the equation for the dissociation of an acid:
HA + H2O ⇌ H3O+ + A-
We know that the hydronium concentration at equilibrium is 0.112 M, so the concentration of the conjugate base is also 0.112 M.
To calculate the percent ionization of the acid, we use the equation:
% ionization = (concentration of dissociated acid / initial concentration of acid) x 100
We can find the concentration of dissociated acid (H3O+) by subtracting the concentration of the conjugate base (A-) from the hydronium concentration:
[H3O+] = 0.112 M - 0 M = 0.112 M
Plugging in the values, we get:
% ionization = (0.112 M / 0.532 M) x 100 = 21.05%
Therefore, the percent ionization of the acid is 21.05%.
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The percent of ionization of an acid in solution of 0.532 M in acid HA i and have a hydronium concentration of 0.112 M is equals to the 21.1%.
The ionization of acids results hydrogen ions, thus, that's why compounds act as proton donors.
Molarity of solution = 0.532 M
At Equilibrium, hydronium concentration = 0.112 M
As we know, concentration is defined as the number of moles of substance in a litre of solution, that most of time concentration is replaced by molarity. So, concentration of acid solution, [ H A] = 0.532 M
Chemical reaction, [tex]HA (aq) + H_2O -> H_3O^{ +}+A^{-}[/tex]
percent of ionization of the acid =
[tex] \frac{ [ H_3O^{+}] }{ [ HA]} × 100 [/tex]
= (0.112/0.532) × 100
= 21.1%
Hence, required value is 21.1%.
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What is the function of a buffer?
if 10 grams of aluminum reacts with 4 grams of oxygen, what is the expected grams of product?
Expected grams of aluminum oxide product from the given masses of reactants are 18.93 g.
What is aluminum?Aluminum is chemical element with symbol Al and atomic number is 13.
4Al + 3O₂ → 2Al₂O₃
10 g Al × 1 mol Al / 26.98 g Al = 0.371 mol Al
4 g O₂ × 1 mol O₂ / 32.00 g O₂ = 0.125 mol O₂
We determine the limiting reactant by comparing the mole ratios of aluminum and oxygen in the balanced equation and reactant that produces smaller amount of product is limiting reactant. In this case, aluminum is the limiting reactant because it produces only 0.1855 moles of aluminum oxide, which is less than the 0.25 moles of aluminum oxide produced by the oxygen:
0.371 mol Al × 2 mol Al₂O₃ / 4 mol Al = 0.1855 mol Al₂O₃
0.125 mol O₂ × 2 mol Al₂O₃ / 3 mol O2 = 0.2083 mol Al₂O₃
0.1855 mol Al₂O₃ × 101.96 g/mol = 18.93 g Al₂O₃
Therefore, expected grams of aluminum oxide product from the given masses of reactants are 18.93 g.
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The breakdown of a certain pollutant X in sunlight is known to follow first-order kinetics. An atmospheric scientist studying the process fills a 20. 0Lreaction flask with a sample of urban air and finds that the partial pressure of X in the flask decreases from 0. 473atm to 0. 376atm over 5. 6hours.
Calculate the initial rate of decomposition of X, that is, the rate at which Xwas disappearing at the start of the experiment.
Round your answer to 2 significant digits
The initial rate of decomposition of X is 0.0013 M/h.
The first-order rate law is given as:
Rate = k [X]
Where, k = rate constant
[X] = concentration of X
Since the partial pressure of X is given in the problem, we need to convert it to concentration using the ideal gas law:
PV = nRT
where:
P = partial pressure of X = 0.473 atm
V = volume of the flask = 20.0 L
n = number of moles of X
R = ideal gas constant = 0.08206 L atm K^-1 mol^-1
T = temperature of the flask (assumed constant) = 298 K
Solving for n,
n = PV/RT = (0.473 atm)(20.0 L)/(0.08206 L atm K^-1 mol^-1)(298 K) = 0.952 mol X
At t = 0, the concentration of X is [X]_0 = n/V = 0.952 mol/20.0 L = 0.0476 M.
Using the given data, we can calculate the rate constant (k) as follows:
ln([X]_0/[X]) = kt
where:
t = time = 5.6 hours
Substituting the given values,
ln(0.0476/0.0376) = k(5.6 hours)
Solving for k, we get:
k = (ln(0.0476/0.0376))/5.6 hours = 0.0263 h^-1
The initial rate of decomposition of X is given by:
Rate = k[X]_0 = (0.0263 h^-1)(0.0476 M) = 0.00125 M/h
Rounding off to 2 significant digits,
Initial rate of decomposition of X = 0.0013 M/h.
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All right! And when that
impetus reduces,
motion also reduces.
When the impetus is
removed, the object
stops moving!
When the impetus driving an object decreases, its motion also decreases. And when the impetus is completely removed, the object stops moving.
When the impetus driving an object decreases, its motion also decreases. The term "impetus" in this context refers to the force that sets an object in motion or maintains its motion. When this force decreases, the object experiences a decrease in its velocity or acceleration. This is due to the fact that the force acting on the object is directly proportional to the rate of change of its motion, as described by Newton's second law of motion.
If the impetus is completely removed, the object stops moving altogether. This is because there is no longer any force acting on the object to maintain its motion, and hence it decelerates and eventually comes to rest. This can be seen in everyday scenarios, such as a ball rolling to a stop when it reaches the bottom of a hill or a car slowing down and stopping when the engine is turned off.
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--The complete question is, What happens to the motion of an object when the impetus driving it decreases, and what happens when the impetus is completely removed?--
what volume would be occupied by 100.0g of oxygen gas at a pressure of 1.5atm and a temperature of 25c?
100.0 g of oxygen gas at a pressure of 1.5 atm and a temperature of 25°C would occupy a volume of 49.2 L.
To solve this problem, we can use the Ideal Gas Law, which states:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
We need to rearrange this equation to solve for the volume V:
V = (nRT) / P
where n is the number of moles of the gas, which we can calculate using the molar mass of oxygen gas:
n = m / M
where m is the mass of the gas and M is the molar mass of oxygen gas (32 g/mol).
n = 100.0 g / 32 g/mol = 3.125 mol
Now we can substitute the given values into the equation to find the volume:
V = (nRT) / P
V = (3.125 mol)(0.0821 L·atm/mol·K)(298 K) / 1.5 atm
V = 49.2 L
Therefore, 100.0 g of oxygen gas at a pressure of 1.5 atm and a temperature of 25°C would occupy a volume of 49.2 L.
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ما
1. Blood plasma contains a total carbonate pool of 0. 0252M.
(a) What is the HCO3"/CO2 ratio
(b) What is the concentration of each buffer component present at pH=7. 4
(c) What would the pH be if 0. 01M H" is added assuming that the excess CO2 is not
released.
(d) What would the pH be if 0. 01M H is added assuming that the excess CO2 is released.
(a) The HCO₃⁻ / CO₂ ratio is 20 : 1.
(b) The concentration of each buffer present at the pH=7. 4 is [CO₂] = 1.20 × 10⁻³ M, [HCO₃⁻] = 0.0240 M.
(c) The pH be if 0. 01M H⁺ is added ,the excess CO2 is not released is 6.20.
(d) The pH be if 0. 01M H⁺ is added , the excess CO2 is released. is 7.17.
(a) pH = pka + log HCO₃ / CO₂
HCO₃⁻ / CO₂ = 10^pH - pka
HCO₃⁻ / CO₂ = 10 ^7.4 - 6.1
HCO₃⁻ / CO₂ = 20 : 1
(b) Total concentration = 0.0252 M
HCO₃⁻ + CO₂ = 0.0252
20 CO₂ + CO₂ = 0.0252
[CO₂] = 1.20 × 10⁻³ M
[HCO₃⁻ ] = 0.0240 M
(c) pH = pka + log HCO₃ / CO₂
pH = 6.1 + log 0.0140 / 0.0112
pH = 6.20
(d) pH = pka + log HCO₃ / CO₂
pH = 6.1 + log 0.0140 / 1.20 × 10⁻³
pH = 7.17
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Observations CuSO4 & NH4Cl Conventional, total ionic, net ionic
Therefore, the net ionic equation for the reaction is Copper(2+) (aq) + 2 chlorine- (aq) → Copper(II) chloride (aq).
What takes place when Copper(II) sulfate and Ammonium hydroxide interact?Ammonium sulphate and Copper hydroxide precipitate are the first products of the reaction between copper sulphate and ammonium hydroxide.
Mixing copper(II) sulphate and ammonium chloride results in the following observations:
Conventional: When copper ions (Copper2+) from Copper(II) sulfate are present, a blue solution develops. The colour of Ammonium Chloride doesn't seem to have changed at all.
Ionic total: While Ammonium Chloride dissociates into Ammonium and Chlorine- ions in solution, Copper(II) sulfate dissociates into Copper2+ and Sulfate 2- ions.
Copper(II) sulfate (aq) + 2 Ammonium Chloride (aq) → Copper(II) Chloride (aq) + 2 Ammonium (aq) + Sulfate 2- (aq)
Net Ionic: The net ionic equation shows only the species involved in the reaction. In this case, the Copper2+ and the Cl- ions combine to form Copper(II) chloride.
Copper2+ (aq) + 2 Chlorine- (aq) → Copper(II) chloride (aq)
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how many grams of n2 are required to completely react with 3.03 grams of h2 for the following balanced chemical equation? A. 1.00 B. 6.00 C. 14.0 D. 28.0
The grams of N2 are required to completely react with 3.03 grams of H2 for the following balanced chemical equation is 14 g.
We may calculate the number of moles of H2 that will be used by dividing the amount of H2 that will be utilised by its molar mass. We may multiply that number by the molar mass of N2 to get how many grammes we should use. We can divide that mole quantity by 3 to determine how many moles of N2 the reaction will consume.
In the reaction 1 mole of N2 react with 3 mole of H2 and give 2 mole of NH3
mass of H2 = 3.03g
No of moles of H2 = 3.03g/2 gmol-1
= 1.51 mole
1.51 mole of H2 require N2 = (1/3)× 1.51 moles
= 0.50 mole N2
molar mass of N2 =28g/mol
Mass of N2 require = 0.50mole ×28g/mol
= 14g
Mass of N2 require = 14g.
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The answer is C. 14.0 grams of N2 are required to completely react with 3.03 grams of H2.
The balanced chemical equation is:
N2 + 3H2 -> 2NH3
From the equation, we can see that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
To find out how many grams of N2 are required to react with 3.03 grams of H2, we first need to convert 3.03 grams of H2 to moles:
moles of H2 = mass of H2 / molar mass of H2
moles of H2 = 3.03 / 2.016
moles of H2 = 1.505
Now, we can use the mole ratio from the balanced equation to find out how many moles of N2 are required to react with 1.505 moles of H2:
moles of N2 = (1.505 mol H2) / (3 mol H2/1 mol N2)
moles of N2 = 0.5017
Finally, we can convert moles of N2 to grams of N2:
mass of N2 = moles of N2 x molar mass of N2
mass of N2 = 0.5017 x 28.02
mass of N2 = 14.04
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how many moles of aluminum nitrate are obtained from the reaction of 0.75 mol of silver nitrate with a sufficient amount of aluminum?
The balanced chemical equation for the reaction between aluminum and silver nitrate is:
2 Al + 3 AgNO3 → 3 Ag + 2 Al(NO3)3
From the equation, we can see that 3 moles of aluminum nitrate (Al(NO3)3) are produced for every 3 moles of silver nitrate (AgNO3) consumed.
Therefore, if 0.75 moles of silver nitrate react, we can calculate the number of moles of aluminum nitrate produced as follows:
0.75 mol AgNO3 x (2 mol Al(NO3)3 / 3 mol AgNO3) = 0.50 mol Al(NO3)3
So, 0.50 moles of aluminum nitrate (Al(NO3)3) are obtained from the reaction of 0.75 mol of silver nitrate with a sufficient amount of aluminum.
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which of the following aqueous solutions has the highest molar concentration of na (aq)?(assume each compound is fully dissolved in water.)group of answer choices3.0m nacl (sodium chloride)3.0m nac2h3o2 (sodium acetate)1.5m na2so4 (sodium sulfate)1.0m na3po4 (sodium phosphate)all of these solutions have the same concentration of na (aq).
All of these solutions have the same concentration of Na⁺ (aq) at 3.0 moles for molar concentration.
The highest molar concentration of Na⁺ (aq) can be determined by calculating the moles of Na⁺ ions in each solution.
1. Identify the number of sodium ions (Na⁺) in each compound:
- NaCl: 1 Na⁺ ion
- NaC₂H₃O₂: 1 Na⁺ ion
- Na₂SO₄: 2 Na⁺ ions
- Na₃PO₄: 3 Na⁺ ions
2. Calculate the moles of Na⁺ ions in each aqueous solution:
- 3.0 M NaCl: 3.0 M * 1 Na⁺ ion = 3.0 moles of Na⁺ ions
- 3.0 M NaC₂H₃O₂: 3.0 M * 1 Na⁺ ion = 3.0 moles of Na⁺ ions
- 1.5 M Na₂SO₄: 1.5 M * 2 Na⁺ ions = 3.0 moles of Na⁺ ions
- 1.0 M Na₃PO₄: 1.0 M * 3 Na⁺ ions = 3.0 moles of Na⁺ ions
3. Compare the moles of Na⁺ ions in each solution to determine the highest concentration.
All of these solutions have the same concentration of Na⁺ (aq) at 3.0 moles.
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Though all the solutions have the same concentration of Na+ (aq), an aqueous solution of NaCl with 3.0 M has the highest molar concentration among the given solutions.
Explanation: To determine the molar concentration of Na+ (aq) in each solution, we need to consider the stoichiometry of the dissociation of each compound in water.
For sodium chloride (NaCl), it dissociates completely into Na+ and Cl- ions, so the molar concentration of Na+ (aq) is equal to the molar concentration of NaCl. Therefore, the molar concentration of Na+ (aq) in 3.0M NaCl is 3.0M.
For sodium acetate (NaC2H3O2), it dissociates into Na+ and C2H3O2- ions, but in a 1:1 ratio. So, the molar concentration of Na+ (aq) is half of the molar concentration of NaC2H3O2. Therefore, the molar concentration of Na+ (aq) in 3.0M NaC2H3O2 is 1.5M.
For sodium sulfate (Na2SO4), it dissociates into 2 Na+ ions and 1 SO4 2- ion. So, the molar concentration of Na+ (aq) is twice the molar concentration of Na2SO4. Therefore, the molar concentration of Na+ (aq) in 1.5M Na2SO4 is 3.0M.
For sodium phosphate (Na3PO4), it dissociates into 3 Na+ ions and 1 PO4 3- ion. So, the molar concentration of Na+ (aq) is three times the molar concentration of Na3PO4. Therefore, the molar concentration of Na+ (aq) in 1.0M Na3PO4 is 3.0M.
Therefore, the solution with the highest molar concentration of Na+ (aq) is 3.0M NaCl (sodium chloride).
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In what way was the reaction of the splint and CO2 different from the reaction of the H2 to the flaming splint
Explain to the kids that since there is essentially no —which is required for fire—if the bag contains only pure carbon dioxide, the splint would burn out right away.
What occurs when a burning splint is placed in hydrogen?H2 - Hydrogen Pure hydrogen gas will burst into flames when a burning splint is added to it, making a popping sound. Oxygen (O2) A smouldering splint will rekindle when exposed to a sample of pure oxygen gas.
The flame goes out as a result of carbon dioxide replacing the oxygen it requires to burn (the effect). A popping sound is produced when a flame is near hydrogen because of how the gas burns.
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What is the difference between a bacteria cell and a
human nervous cell?
most bacteria have flagellum, also nerve cells are larger
At 20°C a gas has a volume of 16.00 L. What will the volume be at 175.0 °C?
The volume of the gas at 175.0 °C will be 24.50 Litres
What will the volume of the gas be at 175.0 °C?Charles's law states that "the volume occupied by a definite quantity of gas is directly proportional to its absolute temperature.
It is expressed as;
V₁/T₁ = V₂/T₂
Where V1 and T1 are the initial volume and temperature, and V2 and T2 are the final volume and temperature.
To use this formula, we need to convert the temperatures to Kelvin by adding 273.15 to them:
T1 = 20°C + 273.15 = 293.15 K
T2 = 175.0°C + 273.15 = 448.15 K
Substituting the values into the formula, we get:
16.00 L / 293.15 K = V2 / 448.15 K
Solving for V2, we get:
V2 = 24.50 L
Therefore, the final volume is 24.50 L.
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a buffer is prepared by adding 1.00 l of 1.0 m hcl to 750 ml of 1.5 m nahcoo. what is the ph of this buffer? [ka (hcooh)
Answer:The pH of a buffer prepared by adding 1.00 L of 1.0 M HCl to 750 ml of 1.5 M NaHCOO is 2.84
What is pH?
pH is a measure of the acidity of a solution.
pH is calculated from the negative logarithm to base ten of the hydrogen ions concentration of the solution.
For weak acids such as those used in the preparation of buffers, the acid dissociation constant, Ka are used to determine the pH of the solution.
Therefore, from the Ka of acetic acid, the pH of a buffer prepared by adding 1.00 L of 1.0 M HCl to 750 ml of 1.5 M NaHCOO is 2.84
the following thermochemical equation is for the reaction of n2 with h2 to form nh3. how many grams of n2 would have to react to produce 31.5 kj of energy?
The grams of N₂ would have to react to produce 31.5 kJ of the energy is 84 g.
The chemical equation is as :
N₂ + 3H₂ ---> 2NH₃ ΔH = -96 kJ
The energy produces = 31.5 kJ
We have multiplied the factor so that the value of the enthalpy has also been multiplied.
The factor = 96 / 31.5 = 3
Thus, the balanced chemical equation is :
3N₂ + 9H₂ ---> 6NH₃
The moles of N₂ = 3 mol
The mass of the N₂ = moles × molar mass
The mass of the N₂ = 3 mol × 28 g/mol
The mass of the N₂ = 84 g
The amount of the N₂ would have to react to produce the 31.5 kJ of the energy is 84 g.
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This question is incomplete, the complete question is :
The following thermochemical equation is for the reaction of n2 with h2 to form nh3. how many grams of n2 would have to react to produce 31.5 kj of energy?
N₂ + 3H₂ ---> 2NH₃ ΔH = -96 kJ
The last 4 miles in the activity series of metals are commonly referred to as the "coinage medals". Why would these metals be chosen over more active metals for the use in coins? Why do you think some more active metals, such as zinc or nickel, or sometimes used in coins?
Coinage metals, which typically include copper, silver, and gold, are chosen over more active metals for use in coins because they are less reactive and more resistant to corrosion.
This ensures durability and preserves the appearance of the coins. Some more active metals like zinc or nickel are sometimes used in coins due to their lower cost and availability, while still maintaining adequate resistance to corrosion and wear for everyday use.
The reason why the last 4 miles in the activity series of metals, which are gold, silver, platinum, and palladium, are commonly referred to as the "coinage medals" is because they are highly resistant to corrosion and have a low reactivity towards other chemicals, making them ideal for use in coins. These metals are also very rare and valuable, which adds to their appeal as a currency.
More active metals such as zinc or nickel are sometimes used in coins because they are more abundant and less expensive than the "coinage metals". However, these metals tend to be more reactive and therefore more prone to corrosion and other chemical reactions, which can affect the appearance and value of the coins over time. Additionally, the use of these metals in coins is often limited to lower denominations or commemorative coins, rather than as a standard currency.
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The "coinage metals" are typically gold, silver, copper, and platinum, which are the last 4 metals in the activity series. These metals are chosen over more active metals for use in coins because they are relatively unreactive and do not corrode easily, making them ideal for coins that need to be durable and long-lasting. Additionally, these metals have been historically valued and used as currency, making them culturally significant as well.
However, some more active metals such as zinc or nickel are sometimes used in coins because they are cheaper and more readily available than the coinage metals. These metals may be used as an alloy with the coinage metals to make coins more affordable, or they may be used as a substitute for the more expensive metals in lower denomination coins. However, these metals are not as durable as the coinage metals and may corrode more easily, leading to shorter lifespans for the coins.
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the structures of d-gulose and d-psicose are shown above. what test could be used to distinguish between solutions of these two carbohydrates? explain your answer by predicting the results of the test for each sugar.
a small amount of Tollens' reagent (ammoniacal silver nitrate) is added to the sugar solution and the mixture is heated. If a reducing sugar is present, it will reduce the silver ions in the Tollens' reagent to metallic silver, which will form a silver mirror on the inside of the test tube.
Based on the structures of D-gulose and D-psicose, it can be predicted that both sugars will give a positive result in the Tollens' test because they both have an aldehyde group that can act as a reducing agent. However, the intensity of the reaction may differ for each sugar.
D-gulose has an aldehyde group at carbon 1, which is in the linear form of the sugar, while D-psicose has an aldehyde group at carbon 2. Since D-gulose can easily convert to its linear form, it is expected to give a stronger positive result in the Tollens' test compared to D-psicose, which may show a weaker positive result due to the steric hindrance of the bulky ketone group at carbon 3.
In summary, the Tollens' test can be used to distinguish between solutions of D-gulose and D-psicose by observing the intensity of the silver mirror formed. D-gulose is expected to give a stronger positive result due to its ability to convert to the linear form, while D-psicose may show a weaker positive result due to steric hindrance.
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what is the most important use of an element's atomic number? what else can we know from a neutral atom's atomic number
The most important use of an element's atomic number is that it determines the identity of an element. From a neutral atom's atomic number, we can also determine the number of electrons in that atom.
The most important use of an element's atomic number is that it determines the element's unique identity and its position on the periodic table. The atomic number is equal to the number of protons in the nucleus of an atom, which also determines the number of electrons in a neutral atom.
From a neutral atom's atomic number, we can also determine the element's symbol, its electron configuration, and its properties such as its atomic mass and the number of isotopes it has. Additionally, the atomic number can provide information about the element's reactivity and its ability to bond with other elements to form compounds. Overall, the atomic number is a fundamental characteristic of an element that is used in many different areas of chemistry and physics.
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The most important use of an element's atomic number is that it determines the element's unique identity and properties.
The atomic number also tells us the number of protons in the nucleus of an atom, which in turn determines the number of electrons in the neutral atom. Additionally, the atomic number can give us information about the element's electron configuration and its position on the periodic table. Overall, the atomic number is a crucial piece of information for understanding an element's properties and behavior.
Hi! The most important use of an element's atomic number is to identify the specific element and its position in the periodic table. The atomic number represents the number of protons in the nucleus of an atom of that element.
From a neutral atom's atomic number, we can also determine the number of electrons, as a neutral atom has an equal number of protons and electrons. This information helps us understand the element's chemical properties and reactivity, as the arrangement of electrons in the atom's electron shells influences its behavior in chemical reactions.
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At 215°C a gas has a volume of 18.00 L. What is the volume of this gas at 23.0°C?
Answer:
using
V1/T1=V2/T2
make V2 subject of formula
V2= V1T2/T1
V2= 1.9L
if you are given three different capacitors C1, C2, and C3, how many different combiations of capacitance can you produce, using all capacitors in your circuits?
Assuming that the capacitors are distinct and not identical, there are eight possible combinations of capacitance that can be produced using all three capacitors in a circuit.
This is because each capacitor can either be included or excluded from the circuit, resulting in two possibilities for each capacitor. With three capacitors, there are 2x2x2 = 8 possible combinations.
For example, if C1 = 1μF, C2 = 2μF, and C3 = 3μF, the eight possible combinations would be 1μF, 2μF, 3μF, 1+2=3μF, 1+3=4μF, 2+3=5μF, 1+2+3=6μF, and no capacitor connected.
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a gaseous product has a mass of 2.34 g and occupies a volume of 0.854 l. the temperature in the laboratory is 302 k, and the air pressure is 1.04 atm. calculate the molar mass of the gas. (3 points) 44.0 g/mol 86.9 g/mol 65.3 g/mol 22.4 g/mol
The molar mass of the gas is approximately 65.3 g/mol. The closest answer choice is 65.3 g/mol, so that is the correct answer.
To calculate the molar mass of the gas, we can use the ideal gas law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First, we need to calculate the number of moles of the gas using the given information:
n = (PV) / (RT)
n = (1.04 atm * 0.854 L) / (0.0821 L·atm/(mol·K) * 302 K)
n = 0.0361 mol
Next, we can calculate the molar mass of the gas by dividing its mass by the number of moles:
molar mass = mass / number of moles
molar mass = 2.34 g / 0.0361 mol
molar mass = 64.9 g/mol
Therefore, the molar mass of the gas is approximately 65.3 g/mol. The closest answer choice is 65.3 g/mol, so that is the correct answer.
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you are preparing a standard aqueous solution for analysis by measuring a property of the solution that is directly related to a solution's concentration. unknown to you, the volumetric flask that you are using to make the solution has some residual water in it from the last time it was used. what effect will this have on the measured property of this solution?
Fill the volumetric flask approximately two thirds full and mix. Carefully fill the flask to the mark etched on the neck of the flask. Use a wash bottle or medication dropper if necessary. Mix the solution wholly by using stoppering the flask securely and inverting it ten to twelve times.
Why volumetric flask is more appropriate to be used in the preparation of the standard solution?A volumetric flask is used when it is imperative to be aware of each precisely and accurately the quantity of the solution that is being prepared. Like volumetric pipets, volumetric flasks come in distinctive sizes, depending on the extent of the answer being prepared.
Firmly stopper the flask and invert multiple times (> 10) to make certain the solution is nicely mixed and homogeneous. When working with a solute that releases warmth or gas all through dissolution, you ought to additionally pause and pull out the stopper once or twice. Use flasks for preparing options only.
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https://brainly.com/question/2088214#SPJ1A solution has a concentration of 3.0 M and a volume of 0.20 L. If the solution is diluted to 4.0 L, what is the new concentration, in molarity?
Your answer should have two significant figures.