.453 mop of a gas confined to a 15.0 L container exerts a pressure at 125.6 kPa on the walls of the container. What is the temperature of the gas?
The temperature of the gas can be gotten from the ideal gas equation as 229.7°C.
What is the ideal gas law?PV = nRT
where:
P is the pressure of the gas in units of pascals (Pa)
V is the volume of the gas in units of cubic meters (m³)
n is the amount of substance of the gas in units of moles (mol)
R is the ideal gas constant with a value of 8.314 J/(mol·K)
T is the absolute temperature of the gas in units of kelvins (K)
We know that;
PV = nRT
Then;
T = PV/nR
T = ?
P = 125.6 kPa or 1.24 atm
n = 0.453 moles
R = 0.082 atmL/K-1mol-1
Then;
T = 1.24 * 15/0.082 * 0.453
= 18.6/0.037
= 502.7 K or 229.7°C
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how many grams of MgCl2 are contained in 0.50 L kc a 1.5 m solution?
0.50 L of a 1.5 m solution contains 71.4 grams of MgCl2.
To solve this problemThe equation moles of solute = molarity x volume (in liters) can be used.
When converting to grams, we may use the molar mass of the MgCl2 to determine how many moles there are in the solution.
MgCl2 has a molar mass of roughly 95.21 g/mol.
The volume must first be changed from liters to milliliters:
0.50 L = 500 mL
Next, we may determine how many moles of MgCl2 are present in the solution:
moles of MgCl2 = molarity x volume (in liters)
moles of MgCl2 = 1.5 mol/L x 0.50 L
moles of MgCl2 = 0.75 moles
Finally, we can figure out how much MgCl2 is present in the solution:
mass of MgCl2 = moles of MgCl2 x molar mass
mass of MgCl2 = 0.75 moles x 95.21 g/mol
mass of MgCl2 = 71.4 grams
Therefore, 0.50 L of a 1.5 m solution contains 71.4 grams of MgCl2.
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Can anyone please name this compound?
Answer:
Fluorobenzene
Explanation: Fluorobenzene is an organic compound, which is a derivative of benzene. It has a fluorine atom attached to one of the carbon atoms in the benzene ring. It appears as a colorless liquid and has a slightly sweet odor. Fluorobenzene is used as a solvent and catalyst in various chemical reactions. It is also used in the production of agrochemicals and pharmaceuticals. Due to its high solubility in water, it can contaminate groundwater and pose a risk to human health and the environment.
105.0 mL of alcohol is dissolved in water and the solution is diluted to a
total final volume of 250 mL. What is the final concentration of ethanol?
105.0 mL of alcohol is dissolved in water and the solution is diluted to a total final volume of 250 mL the final concentration of ethanol in the solution is 0.00722 M.
To calculate the final concentration of ethanol in the solution, we need to know the amount of ethanol present in the solution before and after dilution. We can use the following formula to calculate the final concentration:
Cfinal = Cinitial x (Vinitial / Vfinal)
where Cinitial is the initial concentration of ethanol, Vinitial is the initial volume of the solution, Vfinal is the final volume of the solution, and Cfinal is the final concentration of ethanol.
First, let's find the initial concentration of ethanol in the solution. We know that 105.0 mL of alcohol is dissolved in water, but we don't know the concentration of the alcohol.
Let's assume that the alcohol is pure ethanol (which is not always the case in reality), which has a density of 0.789 g/mL at room temperature. Therefore, the mass of ethanol in 105.0 mL of alcohol is:
mass of ethanol = volume of alcohol x density of ethanol
= 105.0 mL x 0.789 g/mL
= 82.845 g
The molar mass of ethanol is 46.07 g/mol, so the number of moles of ethanol in 82.845 g of ethanol is:
moles of ethanol = mass of ethanol / molar mass of ethanol
= 82.845 g / 46.07 g/mol
= 1.797 mol
The initial volume of the solution is 105.0 mL, so the initial concentration of ethanol is:
Cinitial = moles of ethanol / initial volume of solution
= 1.797 mol / 105.0 mL
= 0.0171 M
Now, let's calculate the final concentration of ethanol. We know that the final volume of the solution is 250 mL. Using the formula above, we get:
Cfinal = Cinitial x (Vinitial / Vfinal)
= 0.0171 M x (105.0 mL / 250 mL)
= 0.00722 M
Therefore, the final concentration of ethanol in the solution is 0.00722 M.
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The numerical value of the equilibrium constant, Kc, for the following reaction is 1.7×102. If the equilibrium mixture contains 0.21 M H2 and 0.015 M N2, what is the molar concentration of NH3?
3H2(g)+N2(g)⇌2NH3(g)
The molar concentration of NH₃ is 0.17 M.
The equilibrium constant expression for the given reaction is:
Kc = [NH₃]² / ([H₂]³[N₂])We are given Kc = 1.7 × 10² and the molar concentrations of H₂ and N₂ in the equilibrium mixture as 0.21 M and 0.015 M, respectively. Let x be the molar concentration of NH₃ at equilibrium. Then, we can write:
Kc = (x)² / (0.21)³(0.015)1.7 × 10² = x² / 1.84 × 10⁻⁶x² = 3.128 × 10⁻⁴x = 0.017 M or 0.17 M (since there are two NH₃ molecules produced for every three H₂ molecules consumed, we take the square root of the calculated value to get the molar concentration of NH₃)
Therefore, the molar concentration of NH₃ in the equilibrium mixture is 0.17 M.
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Does anyone have the Labs: Acid and Bases, lab report. I will give brainliest.
A variety of lichen was used to create the natural acid-base indicator known as litmus.
Since we can't taste everything to determine if it constitutes an acid or a base, litmus paper is essentially an indicator that is used to differentiate acids from bases. It is also known as an acid-base indicator since it can detect the presence of either a base or an acid in a solution.
Litmus is combined with wood cellulose paper to create a litmus paper. The lichen plants that are used to make the purple dye known as litmus are classified as members of the Thallophyta division. A variety of lichen was used to create the natural acid-base indicator known as litmus. You can determine whether a variety of solutions are bases or acids by testing them with red and blue coloured litmus paper.
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Given the thermochemical equation:
4 AlCl3 (s) + 3 O2 (g) ⇒ 2 Al2 03 (s) + 6 Cl2 (g) ; ΔH = -529 kJ
Find ΔH for the following reaction:
1) 3 Al2O3 (s) + Cl2 (g) ⇒ 2/3 AlCl3 (s) + 1/2 O2 (g) ΔH= ?kJ
2) 88.2 kJ
b) 264.5 kJ
c) 529 kJ
d) 176.3 kJ
e) - 176.3 kJ
A thermochemical equation can be written by expressing the heat evolved or absorbed in terms of the enthalpy change ΔH. Here ΔH for the following reaction +88.2 kJ. The correct option is A.
A chemical equation which indicates the heat change occuring during the reaction is defined as the thermochemical equation. In thermochemical equations, physical states of the reactants and products should be specified.
Here the given reaction 4 AlCl₃ (s) + 3 O₂ (g) ⇒ 2 Al₂O₃ (s) + 6 Cl₂ (g) is reversed as 1 /3 Al₂O₃ (s) + Cl₂ (g) ⇒ 2/3 AlCl₃ (s) + 1/2 O₂ (g) and multiplied by 1/6.
So the new enthalpy is +88.16 ≈ 88.2 kJ
Thus the correct option is A.
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I just need # 6,8 and 12 pls
6. 1 [tex]SO_{3}[/tex] + 1 [tex]H_{2}O[/tex] --> 1 [tex]H_{2} SO_{4}[/tex]
8. 1 [tex]K_{2}O[/tex] + 1 [tex]H_{2}O[/tex] --> 2 KOH
12. 1 [tex]CdSO_{4}[/tex] + 1 [tex]H_{2} S[/tex] --> 1 CdS + 1 [tex]H_{2} SO_{4}[/tex]
Exoplanets are usually....
O big, bright and close to the sun.
O gigantic, fireballs that are close to the sun.
O not a planet.
O small, dark, and far from the sun.
Exoplanets are usually gigantic, fireballs that are close to the sun. Therefore, the correct option is option B.
Any planet outside of our solar system is an exoplanet. The majority of exoplanets orbit other stars, while rogue planets—free-floating exoplanets that are unattached to any star—orbit the galactic centre.
The majority of the exoplanets found so far are in the Milky Way, which is a rather tiny area of our galaxy. The Kepler Space Telescope of NASA has revealed that the galaxy has more planets than stars. Exoplanets are usually gigantic, fireballs that are close to the sun.
Therefore, the correct option is option B.
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A 2.26 gg lead weight, initially at 11.1 ∘C∘C, is submerged in 7.45 gg of water at 52.2 ∘C∘C in an insulated container.
Answer:
The temperature of the water and lead weight is 31.0°C.
Explanation:
To solve this problem, we can use the formula:
q = mcΔT
where q is the heat transferred, m is the mass, c is the specific heat, and ΔT is the change in temperature.
First, let's calculate the heat transferred from the water to the lead weight:
q1 = mcΔT = (7.45 g)(4.184 J/g°C)(52.2°C - T) where T is the final temperature of the water and lead weight
q2 = mcΔT = (2.26 g)(0.128 J/g°C)(T - 11.1°C)
Since the container is insulated, the heat transferred from the water to the lead weight is equal to the heat transferred from the lead weight to the water:
q1 = q2
(7.45 g)(4.184 J/g°C)(52.2°C - T) = (2.26 g)(0.128 J/g°C)(T - 11.1°C)
Simplifying and solving for T:
T = 31.0°C
Therefore, the final temperature of the water and lead weight is 31.0°C.
The activation energy, Ea, for a particular reaction is 37.8 kJ/mol. If the rate constant at 280 K is 0.178 M/s, then what is the value of the rate constant at 381 K? (R = 8.314 J/mol • K)
The rate constant that we have at 381 K will be 2.19 M/s.
What is the Arrhenius equation?The Arrhenius equation suggests that the rate of a reaction increases with temperature, because higher temperatures provide more kinetic energy to the reactant molecules, making them more likely to react.
By the use of the Arrhenius equation, we have that;
ln k2/k1 = -Ea/R(1/T2 - 1/T1)
ln k2/0.178 = -37.8 * 10^3/8.314 (1/381 - 1/280)
ln k2/0.178 = - 4647 * (2.62 - 3.57) * 10^-3
lnK2 = 0.786
k2 =e^0.786
k2 = 2.19 M/s
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20. Using the equilibrium expression from Question #19 and based upon the value of K, would you expect there to be more or less product in the equilibrium mixture? Explain why.
The value of K is given as 2344 thus more products are obtained at equilibrium since the equilibrium constant is very large.
The equilibrium constant ( K ) is equal to the rate constant for the forward reaction divided by the rate constant for the reverse reaction.
The balanced equation is -
CaO(s) + CH₄(g) + 2H₂O(g) —> CaCO₃(s) + 4H₂(g)
K = [CaCO₃] [H₂]⁴ / [CaO] [CH₄] [H₂O]²
Since the value of K = 2344, which is large, more of products will be formed in the reaction.
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Assuming that ground water flow does follow the contours of the land, is it possible that there are two sources of contamination? What would you expect to find if all three companies had leaking storage tanks and were actual sources of contamination?o
If ground water flow follows the contours of the land, it is possible that there are two sources of contamination.
This can occur if there are multiple locations of contamination that are not connected through the flow of groundwater. For example, if two companies had leaking storage tanks on opposite sides of a hill, the contaminants from each site could flow in different directions and not mix with each other.
If all three companies had leaking storage tanks and were actual sources of contamination, we would expect to find that the groundwater near each company contained contaminants associated with that company's stored chemicals. The contaminants may be different for each company, depending on the type of chemicals stored.
However, if the contamination has been ongoing for a long time, the chemicals may have mixed together in the groundwater, making it difficult to identify the specific source of contamination for each chemical. In this case, further investigation would be needed to determine the specific sources and extent of contamination from each company.
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In the following reaction, what quantity in moles of CH₃OH are required to give off 6106 kJ of heat?
2 CH₃OH (l) + 3 O₂ (g) → 2 CO₂ (g) + 4 H₂O(g) ∆H° = -1280. kJ
To generate 6106 kJ of heat, 9.5 moles of CH₃OH are required.
The given reaction releases -1280 kJ of heat. We need to find how many moles of CH₃OH are required to release 6106 kJ of heat.
From the given balanced equation, we know that 2 moles of CH₃OH releases 1280 kJ of heat.
Therefore, 1 mole of CH₃OH will release 1280 kJ / 2 = 640 kJ of heat.
To release 6106 kJ of heat, we can use the following proportion:
2 moles CH₃OH / 1280 kJ = x moles CH₃OH / 6106 kJ
Solving for x, we get:
x = (2 moles CH₃OH x 6106 kJ) / 1280 kJ = 9.5 moles CH₃OH
Therefore, 9.5 moles of CH₃OH are required to release 6106 kJ of heat.
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1. Calculate the number of grams of Al in 371 g of Al2O3.
2. Urea [(NH2)2 CO] is used for fertilizer and many other things. Calculate the mass of N, C, O, and H atoms in 1.68 x 10^4 g of urea.
For the following calculations:
371 g of Al₂O₃ contains 196.53 g of Al.
Urea contains 7847.55 g N, 3362.69 g C, 152.18 g H, and 4472.57 g O.
How to calculate contents?1. To calculate the number of grams of Al in 371 g of Al₂O₃, calculate the molar mass of Al₂O₃ and then use stoichiometry to find the mass of Al.
Molar mass of Al₂O₃ = (2 x atomic mass of Al) + (3 x atomic mass of O)
= (2 x 26.98 g/mol) + (3 x 16.00 g/mol)
= 101.96 g/mol
Using stoichiometry to find the mass of Al:
1 mol Al₂O₃ contains 2 mol Al
So, 101.96 g Al₂O₃ contains (2 x 26.98) = 53.96 g Al
Therefore, 371 g of Al₂O₃ contains (53.96/101.96 x 371) = 196.53 g of Al.
2. To calculate the mass of N, C, O, and H atoms in 1.68 x 10⁴ g of urea:
Molar mass of urea = (2 x atomic mass of N) + (1 x atomic mass of C) + (3 x atomic mass of H) + (1 x atomic mass of O)
= (2 x 14.01 g/mol) + (1 x 12.01 g/mol) + (3 x 1.01 g/mol) + (1 x 16.00 g/mol)
= 60.06 g/mol
Using stoichiometry to find the mass of each element:
1 mol urea contains 2 mol N, 1 mol C, 3 mol H, and 1 mol O
So, 60.06 g urea contains 2 x (14.01 g N) = 28.02 g N,
1 x (12.01 g C) = 12.01 g C,
3 x (1.01 g H) = 3.03 g H, and
1 x (16.00 g O) = 16.00 g O.
Therefore, 1.68 x 10⁴ g of urea contains:
(28.02/60.06 x 1.68 x 10⁴) = 7847.55 g N,
(12.01/60.06 x 1.68 x 10⁴) = 3362.69 g C,
(3.03/60.06 x 1.68 x 10^4) = 152.18 g H, and
(16.00/60.06 x 1.68 x 10⁴) = 4472.57 g O.
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8. Base your answer to the following question on the equation below.
2H2(g) + O2(g) → 2H₂O()+571.6 kJ
Identify the information in this equation that indicates the reaction is exothermic.
The positive value of 571.6 kJwhich is the enthalpy change tells us that it is exothermic
What is an exothermic reaction?When energy is released into the surrounding area during a chemical reaction, it's considered an exothermic reaction with ΔH < 0. Conversely, endothermic reactions result from heat absorption with ΔH > 0.
The provided equation's positive value of 571.6 kJ reveals that heat is pouring out of the reaction, causing negative changes in enthalpy and firmly placing this chemical event as exothermic in nature.
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PROBLEM 19.12 Draw the structure of a triacylglycerol that fits each description: a. a saturated triacylglycerol formed from three 12-carbon fatty acids b. an unsaturated triacylglycerol that contains three cis double bonds c. a trans triacylglycerol that contains a trans double bond in each hydrocarbon chain
Ice actually has negative caloric content. How much energy, in each of the following units, does your body lose from eating (and therefore melting) 70 g
of ice? Heat of fusion for water is 6.02 kJ/mol
.
We've figured out what part of the salt causes the flame to change color, so now let's measure the wavelengths created with four metals.
Use the ruler under the "tools" icon in the upper right of the video player to measure the wavelengths of light released by each compound.
Around 450 nm is the wavelength of the spectral line for potassium chloride. The distance among identical spots between two succeeding waves is known as the wavelength.
The distance among identical spots between two succeeding waves is known as the wavelength, which is a feature of waves. The wavelength of a wave is the distance across one wave's peak (or trough) and the next. In mathematics, the Greek symbol lambda () is used to denote wavelength. The colour of light is determined by its wavelength, and the pitch of sound is determined by its wavelength. Around 450 nm is the wavelength of the spectral line for potassium chloride.
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The Galapagos Islands are a series of islands near the coast of South America. The finches (a type of bird) on the Galapagos look similar to the finches in South America, but the birds on each island have some variations, including unique beak structures. Explain what differences on the islands would have led to this beak structure variation.
This archipelago and its enormous marine reserve are referred to be a "living museum and showcase of evolution" for their singularity.
Thus, The 127 islands, islets, and rocks that make up the Galapagos archipelago, 19 of which are major and 4 of which are inhabited, are located roughly 1,000 kilometres from the Ecuadorian mainland. In 1959, national parks were established on 7,665,100 ha, or 97% of the entire emergent surface.
On the remaining three percent of the islands, only human habitation is permitted in designated rural and urban zones (the fifth island only includes an airport, a tourist dock, a fuel containment system, and military installations).
The Galapagos Marine Reserve, one of the largest marine reserves in the world, was established in 1986 and expanded to its current area in 1998. It surrounds the islands. Inland waters of the archipelago are also a part of the marine reserve.
Thus, This archipelago and its enormous marine reserve are referred to be a "living museum and showcase of evolution" for their singularity.
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0.152 mol of sucrose in 602 mL of solution
Answer:
To convert from moles to molarity, we need to divide the number of moles of solute by the volume of solution in liters.
First, we need to convert the volume of solution from milliliters to liters:
602 mL = 0.602 L
Now, we can calculate the molarity:
Molarity = moles of solute / liters of solution
Molarity = 0.152 mol / 0.602 L
Molarity = 0.252 M
Therefore, the molarity of the sucrose solution is 0.252 M.
Explanation:
What volume in milliliters of a 1.00 M solution of sodium hydroxide is required to
make 125 mL of a 0.0600 M solution?
7.50 mL
12.5 mL
16.7 mL
208 mL
What is the density (in g/L) of CO2 in a 5.20 L tank at 760.0 torr and 39.0°C .
The tank's CO₂ density is 1.84 g/L.
How to calculate density?Use the ideal gas law to solve for the density of CO₂:
PV = nRT
where:
P = pressure = 760.0 torr
V = volume = 5.20 L
n = moles of CO2 (we don't know this yet)
R = gas constant = 0.08206 L·atm/K·mol
T = temperature = 39.0°C + 273.15 = 312.15 K
First, convert torr to atm:
760.0 torr ÷ 760 torr/atm = 1 atm
Rearrange the ideal gas law to solve for n:
n = PV/RT
n = (1 atm)(5.20 L)/(0.08206 L·atm/K·mol)(312.15 K)
n = 0.217 mol
Use the mass of CO₂ and the volume of the tank to find the density:
mass = n × molar mass
mass = 0.217 mol × 44.01 g/mol
mass = 9.57 g
density = mass/volume
density = 9.57 g/5.20 L
density = 1.84 g/L
Therefore, the density of CO₂ in the tank is 1.84 g/L.
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What mass in grams of oxygen gas are produced when 2.43 x 10-4 g of KCIO,
are completely reacted according to the following chemical equation:
2 KCIO₂ (s) → 2 KCI (s) + 3 0₂(g)
The unit you work with is leaving a forward area rearming/refueling point and has unused ammunition. The ammunition should be
It is necessary to account for all unused ammunition and return it to a safe storage place or to the concerned authorities for disposal.
What is Ammunition?Bullets, shells and explosives are examples of physical objects that serve as ammunition to project force at a target. These objects are intended to be fired from a weapon such as a gun, rifle, or artillery piece and may be made of a variety of materials such as metal, plastic, or composite materials.
Depending on the weapon and the purpose of the attack, such as whether it is intended for training, target shooting, hunting or fighting, the ammo used will vary. Governments around the world have strict regulations governing ammunition, to ensure its safe handling, movement and application.
Therefore, it is necessary to account for all unused ammunition and return it to a safe storage place or to the concerned authorities for disposal.
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Read the passage and answer the question.
In 2000, when paleontologist Paul Sereno led an expedition into Ténéré desert in Niger looking for the fossils of dinosaurs and ancient crocodiles, the photographer Mike Hettwer wandered away from the group to take pictures of some dunes near the main dig site. The photographer quickly found bones sticking out of the dunes, but they were human bones, not the prehistoric reptile bones the group had been looking for. However, the group was not going to pass up such an amazing discovery.
During the site excavation, the paleontologists found dozens of gravesites. Some of them held skeletal remains and potsherds with wavy lines etched in them. The scientists named this group the Kiffians. The others held skeletal remains that indicated a taller group of people, and their potsherds were decorated with patterns of dots. The scientists named this group the Tenerians. The graves also contained tools and beads made from stones or bones, as well as refuse heaps containing the bones of the animals the people living in the area had consumed. Some of the graves didn't contain any pottery so it was a mystery which group they belonged to. It was also not known when these people lived in the desert or how they survived.
The carbon-14 dating revealed that the Kiffians lived in the area around 9,700 years ago, and then the area was abandoned until the Tenerians lived in the area 7,000 years ago. The scientists want to reconstruct what the land looked like when it was inhabited, as well as understand why there was about a 2,000-year-gap during which nobody lived in the area.
The scientists hypothesize that the two civilizations lived around a lake that dried up during periods of drought and then eventually reformed. The changing lake led people to move, depending on if they had a water source or not.
How can scientists use radiometric dating to reconstruct the geologic history of the area to support or reject their hypothesis of a disappearing and reappearing lake? What is a tool that scientists can use to ensure that their radiometric dating is accurate?
Answer & Explanation:
Scientists can use radiometric dating to reconstruct the geologic history of the area and support or reject their hypothesis of a disappearing and reappearing lake by analyzing the sediments and other geological features present in the area. Radiometric dating, such as carbon-14 dating, can determine the age of the sediments, fossilized remains, and other materials. By analyzing the age and composition of these materials, scientists can track changes in the environment over time, including periods of drought or increased precipitation that could cause a lake to dry up or reform.
To ensure the accuracy of their radiometric dating, scientists can use calibration methods, such as cross-dating with other dating techniques like dendrochronology (tree-ring dating), or comparing their results with well-dated samples from similar or nearby environments. This can help validate the radiometric dating results and provide more confidence in the reconstructed geologic history and any conclusions drawn from it regarding the presence or absence of a lake in the area at different times.
A scientist has a 2-gram sample of a radioactive element. It has a half-life of 1 hour. How much of the sample will decay in one hour?
The amount of the sample that has decayed in one hour is 1 grams
How do I determine the amount that will decay in one hour?First, we must obtain the number of half lives that has elapsed after one hour. This is shown below:
Half-life (t½) = 1 hourTime (t) = 1 hourNumber of half-lives (n) =?n = t / t½
n = 1 / 1
n = 1
Finally, we shall determine the amount remaining after 1 hour. Details below:
Original percentage (N₀) = 2 gramsNumber of half-lives (n) = 1Amount remaining (N) = ?N = N₀ / 2ⁿ
N = 2 / 2¹
N = 2 / 2
N = 1
Finally, we shall obtain the amount that has decayed in one hour. Details below:
Original percentage (N₀) = 2 gramsAmount remaining (N) = 1 gramAmount that decay =?Amount that decay = N₀ - N
Amount that decay = 2 - 1
Amount that decay = 1 gram
Thus, the amount that has decayed in one hour is 1 grams
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The gas pressure in a can is 2.5 atm at 25 °C. Assuming that the gas obeys the ideal-gas equation, what is the pressure (in atm) when the can is heated to 525 °C?
The concept combined gas law is used here to determine the new pressure of the gas. This law states that the ratio between the product of pressure-volume and temperature of a system remains constant.
The combined gas law is the combination of Boyle's law, Charles's law and the Avogadro's law. These laws relate one thermodynamic variable to another holding everything else constant.
Here volume is constant, so the equation is:
P₁ / T₁ = P₂ / T₂
T₁ = 298 K
T₂ = 798 K
Pressure is:
P₂ = P₁ T₂/T₁
P₂= 2.5 × 798 / 298
P₂ = 6.69 atm
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1b. Suppose that you were titrating a 100 mL acid solution with the 0.1 M NaOH solution that you made. You performed the titration multiple times and obtained the data below. Complete the data table below. Show work on a separate piece of paper/ the back of this paper.
Step 1: Write and balance the chemical equation (only need to do this once for each titration)
Step 2: Use the molarity and mL of base used to find the moles of base it took to neutralize the acid
Step 3: Calculate moles of acid neutralized
Step 4: Calculate molarity of acid
Step 5: Calculate pH
1c. Calculate the most likely pH of the acid solution by finding the average of all the pH's you found in each of your multiple titrations. We find the average to minimize human errors made while titrating.
The moles of NaOH used is 0.0008 moles
The molarity of the acid is 0.008 M
What is the molarity of the acid?The molarity of the acid is found as follows:
Moles of NaOH used = concentration of NaOH × volume of NaOH used
the average volume of NaOH used = 8.0 mL
moles of NaOH = 0.1 M × 8.0 mL
moles of NaOH = 0.0008 moles
Molarity of acid:
Assuming the acid is monobasic, the mole ratio of acid to base is 1 : 1
The volume of acid used is 100 mL
The molarity of acid = moles of acid / volume of acid in liters
The molarity of acid = 0.0008 moles / 0.1 L
The molarity of acid = 0.008 M
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Find the critical value(s) and rejection region(s) for the indicated t-test, level of significance , aand sample size n.
right-tailed test, a= 0.10, n=9 The critical value is
The critical value of the indicated t-test is 1.397
t > 1.397 is the rejection region.
How to find critical value and rejection region?A t-distribution table or calculator is required to calculate the critical value for a right-tailed t-test with a significance level of 0.10 and a sample size of 9. The critical value is 1.397 when using a t-distribution table with 8 degrees of freedom (n - 1 = 9 - 1 = 8) and a significance level of 0.10.
t > 1.397 is the rejection zone for a right-tailed t-test with a significance level of 0.10 and a sample size of 9. In other words, if the estimated t-value is bigger than 1.397, the null hypothesis may be rejected.
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