A mixture of gases contains 10.25g of F2, 2.83g of H2, and 5.95g of CO2. If the total pressure of the mixture is 2.75 atm, what is the partial pressure of each component?
A piece of metal with a mass of 32.8 g is heated to 100.5 C and dropped into 138.2 g of water at 20.0 C. the final temperature of the system is 30.2 C. What is the specific heat capacity of the metal
Answer:
To solve this problem, we can use the equation:
Q = m * c * ΔT
where Q is the heat transferred, m is the mass of the substance, c is its specific heat capacity, and ΔT is the change in temperature.
In this case, we know that the heat lost by the metal is equal to the heat gained by the water:
Q lost = Q gained
We can calculate the heat lost by the metal using the equation:
Q lost = m * c * ΔT
where m is the mass of the metal, c is the specific heat capacity of the metal (which we are trying to find), and ΔT is the change in temperature of the metal (100.5 C - 30.2 C = 70.3 C).
We can calculate the heat gained by the water using the equation:
Q gained = m * c * ΔT
where m is the mass of the water and ΔT is the change in temperature of the water (30.2 C - 20.0 C = 10.2 C).
Setting the two equations equal to each other, we get:
m * c * ΔT (metal) = m * c * ΔT (water)
Simplifying, we get:
c (metal) = (m * c * ΔT (water)) / (m * ΔT (metal))
Plugging in the values we know:
m (metal) = 32.8 g
ΔT (metal) = 70.3 C
m (water) = 138.2 g
ΔT (water) = 10.2 C
c (metal) = (138.2 g * 4.184 J/g·K * 10.2 C) / (32.8 g * 70.3 C)
c (metal) = 0.192 J/g·K
Therefore, the specific heat capacity of the metal is 0.192 J/g·K.
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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Balance the following reaction by typing in the correct coefficients in front of each reactant and product.
H3PO4(s) -
-->
H₂(g) +
P(s) +
O₂(g)
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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a reaction between 1.7 moles of zinc iodide and excess sodium
Answer: I got you fam
Explanation:
The formula is Na2CO3 + ZnI2 → 2NaI + ZnCO3
Or -5.19
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
A student used a video camera to record another student dropping a marble through water in a graduated cylinder. The students watched the video in slow motion and made the observations shown below. During which part or parts of the marble’s fall did the marble experience unbalanced forces?
Parts B and C of the marble's fall did the marble experience unbalanced forces. Option 4 is correct.
A force is a push or pull (interaction) which changes the momentum of an object, either stationary or in motion when unopposed. All objects experience different forces depending on their environment. When immersed in fluids, unbalanced forces of one upward moving force tends to cancel the gravity force moving downward on a sinking object causing deceleration to a constant sinking speed.
This upward moving force is called as Buoyant force. This is where at part A, the object will experiences a balanced force of gravity which accelerates due to the absence of an opposing force acting upwards on the object. At part B, the speed of the sinking object decreases due to an unbalanced force that cancels the acceleration by the buoyant force. Once the sinking object’s acceleration is cancelled, its sinking speed turns constant at part C.
Hence, 4. is the correct option.
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--The given question is incomplete, the complete question is
"A student used a video camera to record another student dropping a marble through water in a graduated cylinder. The students watched the video in slow motion and made the observations shown below. During which part or parts of the marble's fall did the marble experience unbalanced forces? (1) Part A only (2) Parts A and B only (3) Part C only (4) Parts B and C only."--
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
Suppose that 0.95 g of water condenses on a 75.0 g block of iron that is initially at 22 °C. If the heat released during condensation goes only to warming the iron block, what is the final tempera- ture (in Celsius) of the iron block? (Assume a constant enthalpy ofvaporizationforwaterof44.0kJmol-1.)
Answer:
The temperature of the iron block is 68.5°C.
Explanation:
The heat released during condensation of water is used to warm the iron block:
q = m_H2O * ΔH_vap = m_fe * c_fe * ΔT
where q is the heat released, m_H2O is the mass of water condensed, ΔH_vap is the enthalpy of vaporization for water, m_fe is the mass of iron, c_fe is the specific heat capacity of iron, and ΔT is the change in temperature of the iron block.
Rearranging the equation gives:
ΔT = (m_H2O * ΔH_vap) / (m_fe * c_fe)
Substituting the given values gives:
ΔT = (0.95 g * 44.0 kJ/mol) / (75.0 g * 0.449 J/(g°C))
ΔT = 46.5°C
Therefore, the final temperature of the iron block is:
T_f = T_i + ΔT = 22°C + 46.5°C = 68.5°C.
The final temperature of the iron block is 68.5°C.
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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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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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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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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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.
Calculate the pH of the three solutions described below. (25 pts)
a) 6.0 x 10-3 M HClO4
b) .0009 M Sr(OH)2
c) The solution made by mixing 25.00 mL of 0.121 M HCl + 30.00 mL of 0.100 M KOH
a) HClO₄ is a strong acid, and in water, it will dissociate completely into H+ and ClO₄⁻. The concentration of H⁺ in this solution is therefore equal to the concentration of HClO₄:
[H⁺] = 6.0 x 10⁻³ M
Using the definition of pH, we have:
pH = -log[H⁺]
pH = -log(6.0 x 10⁻³)
pH = 2.22
b) Sr(OH)₂ will dissociate completely into Sr²⁺ and 2OH⁻. The concentration of OH⁻ in this solution is therefore twice the concentration of Sr(OH)₂:
[OH-] = 2 x 0.0009 M
[OH-] = 0.0018 M
Using the definition of pH, we have:
pOH = -log[OH-]
pOH = -log(0.0018)
pOH = 2.74
Since pH + pOH = 14, we have:
pH = 14 - pOH
pH = 14 - 2.74
pH = 11.26
c) Solution made by mixing 25.00 mL of 0.121 M HCl and 30.00 mL of 0.100 M KOH
HCl + KOH → H2O + KCl
The moles of HCl in 25.00 mL of 0.121 M HCl are:
moles HCl = concentration x volume
moles HCl = 0.121 mol/L x 0.025 L
moles HCl = 0.003025 mol
The concentration of HCl in the final solution is:
[HCl] = moles HCl / total volume of solution
[HCl] = 0.003025 mol / (25.00 mL + 30.00 mL)
[HCl] = 0.003025 mol / 0.055 L
[HCl] = 0.055 M
Similarly, the moles of KOH in 30.00 mL of 0.100 M KOH is:
moles KOH = concentration x volume
moles KOH = 0.100 mol/L x 0.030 L
moles KOH = 0.003 mol
The concentration of OH⁻ in the final solution is:
[OH⁻] = moles OH⁻ / total volume of solution
[OH⁻] = 0.006 mol / 0.055 L
[OH⁻] = 0.109 M
Using the definition of pH, we have:
pH = 14 - pOH
pH = 14 - (-log[OH⁻])
pH = 14 - (-log(0.109))
pH = 12.06
Thus, the pH of (a) is 2.22, (b) is 11.26, and (c) is 12.06.
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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:
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.
why is often difficult to identify a highly weathered mineral
Weathering changes the chemical and physical nature of an element that is why it is often difficult to identify a highly weathered mineral.
The breakdown and alteration of rocks and minerals at or near the Earth's surface as a result of exposure to various weathering agents, such as water, wind and temperature changes is known as weathering.
Minerals can undergo physical changes as a result of weathering, such as being broken up into smaller pieces or having their color and texture altered. Additionally, it may result in chemical alterations such as the removal or addition of specific chemical components.
This may lead to the creation of brand-new minerals or the modification of already existing minerals into new ones. Highly weathered minerals might not still possess the same physical and chemical characteristics as their unweathered counterparts.
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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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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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Please help asp and don 't just put random answers please
The value of Tan P as fraction simplest form is 15 / 29
How do determine the value of tan P?The following data were obtained from the question:
Angle θ = POpposite = 15Adjacent = 29Tan P =?Tan θ ratio is express as:
Tan θ = Opposite / Adjacent
Inputting the various parameters obtained from the question, we can obtain Tan P as shown below:
Tan P = Opposite / Adjacent
Tan P = 15 / 29
Thus, from the above calculation, we can conclude that the value of tan P in it's lowest fraction is 15 / 29
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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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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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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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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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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]
PLEASE HELP ASAP
(50 POINTS)
You have 400,000 atoms of a radioactive substance. After 2 half-lives have past, how
many atoms remain?
Remember that you cannot have a fraction of an atom, so round the answer to the
nearest whole number.
Answer:
If 2 half-lives have passed, it means that the radioactive substance has decayed twice, so the number of remaining atoms would be:
1st half-life: 400,000 / 2 = 200,000 atoms remaining
2nd half-life: 200,000 / 2 = 100,000 atoms remaining
Therefore, after 2 half-lives have passed, 100,000 atoms would remain, rounded to the nearest whole number
Explanation:
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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
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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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