Answers
Answer:
A. 1, 1, 1, 1
B.1,1,1
C. 2, 1, 2, 1
D. 2, 1, 1, 1
E.1,2,1,1
F.1, 1, 1, 2
G. 1, 6, 3, 4
H. 2, 3, 2, 3
Related Questions
16. Keesha performed a chemical reaction, and the products looked quite
different from the reactants. She knew the amount of matter had not
changed due to the law of conservation of mass. According to the law of
conservation of mass, what happens in a chemical reaction? *
Answers
Answer:
it's the third one
Explanation:
matter can't change, but mass can differ
A sample of oxygen gas has a volume of 130.0 ml when its pressure is 0.952 atm. What will the volume of the gas be
at a pressure of 0.974 atm if the temperature remains constant?
Answers
Answer:
the volume of the gas when pressure 0.974atm is equal to 127.06ml
can someone help me convert these please !
Answers
Why do I always resort to brainly when I know it's kind of wrong-ish........?
The answers are rarely correct and the system is pretty faulty. Is anyone else addicted for no reason? This is for a "Science research project"
Answers
Answer:
Because the thought of self-doubt and lack of confidence is getting to you. Other people most likely have the same issue that you do.
Explanation:
Answer:
i like to use it so i can find out new things, my dad thought me that by teaching you learn more, but when it comes to getting answers i put out a few of the same question and if 2 ppl give the same answer i use that one. the reason why i use brainly is because i have looked everywhere and cant find the answer this way even if its wrong it looks like i tried. i hopes that helps
A large fish tank is initially filled with 30 litres of fresh water. You begin to fill the tank by slowly pouring in water with salt concentration of 35 grams per litre (approximate salinity of sea water) at a rate of 2 litres per minute. At the same time, the (perfectly mixed) fluid in the tank is drained from the bottom at a rate of 1 litre per minute. 1. Determine the volume of water in the tank at time t. [1 mark] 2. Let S(t) denote the amount of salt in the fish tank at time t in grams. Show that S(t) satisfies the ODE S
′
(t)=70−
t+30
S
. Write down the appropriate initial condition for the ODE as well. [2 marks] 3. What order is this ODE? Is it linear? Is it separable? [1 mark] 4. Solve the initial value problem to find S(t) using the method of integrating factors. [3 marks] 5. What is the salt concentration in the tank as t→[infinity] ? [1 mark] Part B: Double tanks Next you hook up two fish tanks in a loop so that there is a pipe from tank A to tank B, and also a pipe from tank B back to tank A. Two pumps are added so that you can control the flow rate in each pipe. Initially tank A contains 80 litre of fresh water and tank B 60 litres of fresh water. You begin to pour salt water with concentration 35 grams per litre into tank A at a rate of 2 litres per minute. To keep the tanks from overflowing, you set your pumps so that water is flowing at a constant rate of 4 litres per minute from tank A to tank B, and 2 litre per minute from tank B to tank A. You also put a drain in tank B so that fluid is draining at a rate of 2 litres per minute. 1. Sketch a diagram of the tank setup with arrows for flows entering and leaving each tank. [
1 mark]
2. Let P(t) and Q(t) denote the amount of salt in tank A and tank B respectively. Show that P and Q satisfy a system of ODE's in the form of
P
′
(t)
Q
′
(t)
=c
1
P(t)+c
2
Q(t)+c
3
=c
4
P(t)+c
5
Q(t)
where c
1
,c
2
,c
3
,c
4
and c
5
are constants. Determine the constant c
1
,c
2
,c
3
,c
4
,c
5
and write down appropriate initial conditions. [2 marks] 3. Show that the system of ODE's can be converted into the following second order ODE for P(t) P
′′
(t)=−
60
7
P
′
(t)−
600
1
P(t)+
3
14
State the initial conditions for this ODE. [2 marks] 4. Solve this second order ODE to find P(t), and hence Q(t) as well.
Answers
1. The volume of water in the tank at time t is given by the equation
Volume(t) = 30 + t.
2.The appropriate initial condition for the ODE is S(0) = 0, as there is no salt initially in the tank.
3. It is not separable because the variables S(t) and t are not separable on opposite sides of the equation.
4. The solution can be expressed in terms of the integral as:
\(S(t) = (70 * \int e^{(t^{2/2} + 30t)} dt) / e^{(t^{2/2} + 30t)})\)
5. the salt concentration in the tank as t→infinity is zero.
1. To determine the volume of water in the tank at time t, we need to consider the rate at which water is being added and drained. The tank is being filled at a rate of 2 liters per minute and drained at a rate of 1 liter per minute.
Since the tank starts with an initial volume of 30 liters, the volume of water in the tank at time t can be calculated using the equation:
Volume(t) = Initial volume + (Rate of filling - Rate of draining) * t
Volume(t) = 30 + (2 - 1) * t
So, the volume of water in the tank at time t is given by the equation
Volume(t) = 30 + t.
2. Let S(t) denote the amount of salt in the fish tank at time t in grams.
To show that S(t) satisfies the ODE S'(t) = 70 - (t+30)S(t),
we need to take the derivative of S(t) with respect to t and substitute it into the given ODE.
Taking the derivative of S(t), we have:
S'(t) = 0 - (1+0)S(t) + 0
S'(t) = -S(t)
Substituting this into the given ODE, we get:
-S(t) = 70 - (t+30)S(t)
Simplifying the equation, we have:
S'(t) = 70 - (t+30)S(t)
Therefore, S(t) satisfies the ODE S'(t) = 70 - (t+30)S(t).
The appropriate initial condition for the ODE is S(0) = 0,
as there is no salt initially in the tank.
3. This ODE is a first-order linear ordinary differential equation. It is not separable because the variables S(t) and t are not separable on opposite sides of the equation.
4. To solve the initial value problem for S(t) using the method of integrating factors, we first rewrite the ODE in standard form:
S'(t) + (t+30)S(t) = 70
The integrating factor is given by:
\(\mu(t) = e^{(\int (t+30) dt)} = e^{(t^2/2 + 30t)\)
Multiplying both sides of the equation by μ(t), we have:
\(e^{(t^2/2 + 30t)} * S'(t) + e^{(t^2/2 + 30t)} * (t+30)S(t) = 70 * e^{(t^2/2 + 30t)\)
Applying the product rule to the left side of the equation, we get:
\((e^{(t^{2/2} + 30t) * S(t))' = 70 * e^{(t^{2/2} + 30t)})\)
Integrating both sides of the equation with respect to t, we have:
\(\int (e^{(t^2/2 + 30t)} * S(t))' dt = \int (70 * e^{(t^2/2 + 30t))} dt\)
Using the fundamental theorem of calculus, the left side becomes:
\(e^{(t^2/2 + 30t)} * S(t) = \int (70 * e^{(t^2/2 + 30t))} dt\)
Simplifying the right side by integrating, we get:
\(e^{(t^2/2 + 30t)} * S(t) = 70 * \int e^{(t^2/2 + 30t)} dt\)
At this point, the integration of \(e^{(t^2/2 + 30t)\) becomes difficult to express in terms of elementary functions.
Hence, the solution can be expressed in terms of the integral as:
\(S(t) = (70 * \int e^{(t^2/2 + 30t)} dt) / e^{(t^2/2 + 30t)\)
5. As t approaches infinity, the exponential term \(e^{(t^2/2 + 30t)\) becomes very large, causing the salt concentration S(t) to approach zero. Therefore, the salt concentration in the tank as t→infinity is zero.
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The salt concentration in the tank as t approaches infinity is 70/3.
1. To determine the volume of water in the tank at time t, we need to consider the rate at which water is being poured into the tank and the rate at which water is being drained from the bottom.
At a rate of 2 litres per minute, water is being poured into the tank. So after t minutes, the amount of water poured into the tank is 2t litres.
At a rate of 1 litre per minute, water is being drained from the tank. So after t minutes, the amount of water drained from the tank is t litres.
Since the tank was initially filled with 30 litres of fresh water, the volume of water in the tank at time t is given by:
Volume(t) = 30 + 2t - t
Volume(t) = 30 + t
2. Let S(t) denote the amount of salt in the fish tank at time t. To determine the ODE for S(t), we need to consider the salt being poured into the tank and the salt being drained from the tank.
The salt concentration in the water being poured into the tank is 35 grams per litre. So the amount of salt being poured into the tank per minute is 35 * 2 = 70 grams.
The amount of salt being drained from the tank per minute is S(t)/Volume(t) * 1.
Therefore, the ODE for S(t) is:
S'(t) = 70 - S(t)/Volume(t)
The initial condition for this ODE is S(0) = 0, since there was no salt in the tank initially.
3. The ODE S'(t) = 70 - S(t)/Volume(t) is a first-order linear ODE. It is not separable since the variables S(t) and Volume(t) are mixed together.
4. To solve the initial value problem for S(t), we can rewrite the ODE as:
Volume(t) * S'(t) + S(t) = 70 * Volume(t)
This is a linear ODE of the form y'(t) + p(t)y(t) = g(t), where p(t) = 1/Volume(t) and g(t) = 70 * Volume(t).
To solve this type of ODE, we can multiply both sides by an integrating factor, which is the exponential of the integral of p(t).
The integrating factor is exp(integral of 1/Volume(t) dt) = exp(ln(Volume(t))) = Volume(t).
Multiplying both sides of the ODE by the integrating factor, we get:
Volume(t) * S'(t) + S(t) = 70 * Volume(t)
Volume(t) * S'(t) + Volume(t) * S(t) = 70 * Volume(t)^2
( Volume(t) * S(t) )' = 70 * Volume(t)^2
Integrating both sides with respect to t, we get:
Volume(t) * S(t) = 70/3 * Volume(t)^3 + C
S(t) = 70/3 * Volume(t)^2 + C/Volume(t)
Using the initial condition S(0) = 0, we can solve for C:
0 = 70/3 * 30^2 + C/30
C = -70000
Therefore, the solution for S(t) is:
S(t) = 70/3 * Volume(t)^2 - 70000/Volume(t)
5. As t approaches infinity, the volume of water in the tank becomes very large. In this case, we can approximate the volume of the tank as t, since the rate at which water is being poured in is 2 litres per minute. So the salt concentration in the tank as t approaches infinity is given by:
S(t)/Volume(t) = (70/3 * t^2 - 70000/t) / t
As t approaches infinity, the second term (-70000/t) approaches 0, so the salt concentration in the tank as t approaches infinity is:
S(t)/Volume(t) = 70/3 * t^2 / t = 70/3 * t
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What were Corrine Samuel and Nettie doing in Africa?
Answers
Corrine Samuel and Nettie were both involved in the cocoa industry in Africa. Corrine Samuel was the managing director of Armajaro, a cocoa trading company, and Nettie was a cocoa farmer.
They were likely working to promote sustainable and ethical cocoa farming practices in the region.
Melissa, whose licence was just suspended, working for licensee Corrine. What brokerage chores are still left for Corrine to finish She is not allowed to do any brokerage business.
A number of commonly used synonyms for the verb "perform" include "accomplish," "achieve," "discharge," "effect," and "execute." All of these verbs mean "to carry out or to put into effect," but perform implies action that follows predefined patterns or processes or criteria and typically implies a certain level of expertise. putting up a play, concert, or other sort of entertainment noun 1[countable] gymnastics The show is going to start at seven.
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Which of the following best represents Boyle's Law?A) The pressure of a gas exerted on the walls of its container is directly proportional to the volume of the gas when the temperature of the gas remains constant.B) The volume of a gas inside a container is inversely proportional to the temperature of the gas when the pressure of the gas remains constant.C) The volume of a gas inside a container is directly proportional to the temperature of the gas when the pressure of the gas remains constant.D) The pressure of a gas exerted on the walls of its container is directly proportional to the temperature of the gas when the volume of the gas remains constant.E) The pressure of a gas exerted on the walls of its container is inversely proportional to the temperature of the gas when the volume of the gas remains constant.F)The pressure of a gas exerted on the walls of its container is inversely proportional to the volume of the gas when the temperature of the gas remains constant.
Answers
Answer
F) The pressure of a gas exerted on the walls of its container is inversely proportional to the volume of the gas when the temperature of the gas remains constant.
Explanation
Boyle's law states that the volume of a given mass of gas varies inversely with the pressure when the temperature is kept constant. An inverse relationship is described in this way. As one variable increases in value, the other variable decreases. He discovered that doubling the pressure of an enclosed sample of gas, while keeping its temperature constant, caused the volume of the gas to be reduced by half.
How does the seed of a gymnosperm differ from the seed of an angiosperm?
Answers
Answer:
The seed of a gymnosperm is typically covered by a protective coat, such as a cone or a scale, while the seed of an angiosperm is typically enclosed in a fruit or other protective covering. Additionally, gymnosperm seeds are typically not as well-developed as angiosperm seeds, and they do not contain an embryo or endosperm.
can someone help me with this
Answers
Answer:
phosphorous- 5
calcium- 2
nitrogen- 3 or 5
iron- 8 (transition metals use subshells as valence electrons)
argon- 8
potassium- 1
helium- 2
magnesium- 2
sulfur- 6
lithium- 1
iodine- 7
oxygen- 6
barium- 2
aluminum- 3
hydrogen- 1
xenon- 8
copper- 1
Source: my own chemistry notes
I need help pleaseeee
Answers
The chemical formula for emerald is Be3Al2(SiO3)6. An emerald can be described as
Answers
Answer:
An emerald can be described as a compound because it is made up of many different elements.
Explanation:
There are two types of chemical compound one is covalent compound and other is ionic compound, covalent compound formed by sharing of electron and ionic compound formed by complete transfer of electron. Therefore, an emerald can be described as Compound.
What is chemical Compound?Chemical Compound is a combination of molecule, Molecule forms by combination of element and element forms by combination of atoms in fixed proportion.
An ionic compound is a metal and nonmetal combined compound. Ionic compound are very hard. They have high melting and boiling point because of strong ion bond. The chemical formula for emerald is Be\(_3\)Al\(_2\)(SiO\(_3\))\(_6\). An emerald can be described as Compound.
Therefore, the chemical formula for emerald is Be\(_3\)Al2(SiO\(_3\))\(_6\). An emerald can be described as Compound.
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The Ka of NH4+ is 5.6 × 10−10. The Kb of CN− is 2 × 10−5. The pH of a salt solution of NH4CN would be:
Hints
The Ka of NH4+ is 5.6 x 10-10. The Kb of CN- is 2 x 10-5. The pH of a salt solution of NH4CN would be:
Greater than 7 because CN− is a stronger base than NH4+ is an acid
Less than 7 because CN− is a stronger base than NH4+ is an acid.
Greater than 7 because NH4+ is a stronger acid than CN− is a base.
Less than 7 because NH4+ is a stronger acid than CN− is a base.
Answers
The pH οf a salt sοlutiοn οfNH₄CN wοuld be Greater than 7 because CN− is a strοnger base than NH₄+ is an acid
Define pHWater's pH level indicates hοw acidic οr basic it is. The range is 0 tο 14, with 7 acting as a neutral value. A pH οf greater than 7 denοtes a base, while οne οf less than 7 suggests acidity. The pH scale really measures the prοpοrtiοn οf free hydrοgen and hydrοxyl iοns in water.
NH₄⁺ ⇄ NH₃ + H⁺ with a ka οf 5,6 x \(10^{-10\)
CN⁻ + H₂O → HCN + OH⁻ with a kb οf 2 x10⁻⁵
OH is prοduced frοm CN at a greater rate than H+ is prοduced frοm NH₄+. The base CN is mοre pοwerful than the acid NH₄+.
Thus, since CN is a strοnger base than NH₄+ is an acid, the pH οf a salt sοlutiοn οf NH₄CN wοuld be greater than 7.
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which of the molecular orbital diagrams is correctly filled for the diatomic molecule r2? (each atom of r has six valence electrons in ns and np orbitals.)
Answers
Atomic orbitals are the areas to the left and right of the dashed lines. The possible molecular orbitals that they can form are indicated by the dashed lines.
Normally, in diatomic molecular orbitals, the atomic orbitals with the closest energy level can overlap with each other and form molecular orbitals. Therefore, the atomic orbitals generally tend to overlap one by one from the lowest potential energy to the highest potential energy. For example, in a homonuclear diatomic molecule, which means that both atoms are the same element, the same orbitals will overlap together and form molecular orbitals.
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what is the oxidation state of zn in [zn(nh3)4]2 ?
Answers
The oxidation state of Zn in [Zn(NH₃)₄]₂ is +2. This is because NH₃ is a neutral ligand and each NH₃ molecule donates one electron pair to Zn.
Since there are four NH₃ ligands, the total electron pairs donated to Zn is 4. Since Zn needs 2 more electrons to fill its valence shell, it has an oxidation state of +2 in this compound.
The oxidation state of Zn in [Zn(NH₃)₄]²⁺ is +2. In this complex, Zn is the central atom and NH₃ is a neutral ligand, which does not affect the oxidation state of the metal ion. Therefore, the overall charge of the complex (+2) is solely due to the oxidation state of Zn.
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is Jupiter's volume more than 1,000 times that of Earth
Answers
Answer:
cómo lo ahí no sé amigo blue the rous
Answer:
yes it is i did it on ixl
which mineral would most likely be found in a necklace? graphite, halite, sulfur, or emerald?
Answers
Answer:
D is the answer because I think it is right plus I know they don't use two off them
Answer:Emerald is a gemstone that might be found in a necklace.
Explanation:Trust me i just got it right and i get all my other questions right.
Using the given data, calculate the rate constant of this reaction.
A
+
B
→
C
+
D
Trial [A] (M) [B] (M) [C] (M)
1 0.38 0.39 0.0227
2 0.38 0.897 0.12
3 0.456 0.39 0.0272
Answers
The rate cοnstant οf the reactiοn is apprοximately 0.0582 s⁻¹.
How tο calculate the rate cοnstant οf the reactiοn?
Tο calculate the rate cοnstant οf the reactiοn, we can use the rate equatiοn:
rate = k *\([A]^x * [B]^y\)
where k is the rate cοnstant, [A] and [B] are the cοncentratiοns οf reactants A and B, respectively, and x and y are the respective οrders οf the reactiοn with respect tο A and B.
Tο determine the οrders οf the reactiοn with respect tο A and B, we can cοmpare the initial rates οf the reactiοn fοr different trials.
Let's analyse the given data:
Trial 1: [A] = 0.38 M, [B] = 0.39 M, rate = 0.0227 M/s
Trial 2: [A] = 0.38 M, [B] = 0.897 M, rate = 0.12 M/s
Trial 3: [A] = 0.456 M, [B] = 0.39 M, rate = 0.0272 M/s
Cοmparing trials 1 and 2, we can see that the cοncentratiοn οf B is increased while the cοncentratiοn οf A remains cοnstant. As the rate increases frοm 0.0227 M/s tο 0.12 M/s, it suggests that the reactiοn is mοre sensitive tο changes in the cοncentratiοn οf B. Therefοre, we can determine that the reactiοn is first οrder with respect tο B.
Cοmparing trials 1 and 3, we can see that the cοncentratiοn οf A is increased while the cοncentratiοn οf B remains cοnstant. As the rate increases frοm 0.0227 M/s tο 0.0272 M/s, it suggests that the reactiοn is less sensitive tο changes in the cοncentratiοn οf A. Therefοre, we can determine that the reactiοn is zerο οrder with respect tο A.
Nοw that we have determined the οrders οf the reactiοn, we can calculate the rate cοnstant using any οf the trials. Let's use trial 1:
rate = \(k * [A]^0 * [B]^1\)
0.0227 M/s = \(k * (0.38 M)^0 * (0.39 M)^1\)
0.0227 M/s = k * 0.39 M
Sοlving fοr k, we have:
k = 0.0227 M/s / 0.39 M
k ≈ 0.0582 s⁻¹=
Therefοre, the rate cοnstant οf the reactiοn is apprοximately 0.0582 s⁻¹.
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The amount of energy transferred by the wave in the pan is related to its
1. frequency
2. amplitude
3. frequency and amplitude
Answers
How many moles are in 8.11 X 1020 molecules of CH4?
< >
Answers
Answer:
6.022⋅1023
Explanation:
There are 0.00135 moles in \(8.11*10^{20}\) molecules of CH4.
The number of molecules in 1 mole of any compound = \(6.02*10^{23}\) molecules.
This number is called Avogadro's number.
The same goes for CH4. It also contains \(6.02*10^{23}\) molecules in 1 mole.
So, for \(8.11*10^{20}\) molecules, the number of moles would be:
\(n=\frac{8.11*10^{20}}{6.02*10^{23}} \\\\n=0.00135\) moles of CH4.
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The equation for the saturated solution equilibrium of potassium nitrate (KNO3 ) is shown below.
KNO (s) + energy K+ (aq) + NO – (aq) 33
Compare the rate of dissolving KNO3 with the rate of recrystallization of KNO3 for the saturated solution.
Answers
Answer:
In a saturated solution of potassium nitrate (KNO3), the rate of dissolution and the rate of recrystallization are equal. This equilibrium state is reached when the amount of KNO3 that dissolves equals the amount that recrystallizes from the solution.
The equilibrium can be influenced by various factors such as temperature and pressure. For example, if you increase the temperature of the solution, the solubility of potassium nitrate increases, meaning that more KNO3 can dissolve before reaching saturation. This would momentarily increase the rate of dissolution until a new equilibrium is reached where the rates of dissolution and recrystallization are equal again, but at a higher concentration of KNO3.
To summarize, in a saturated solution of KNO3, the rate of dissolving KNO3 is equal to the rate of recrystallization of KNO3. This is a characteristic of dynamic equilibrium in solutions.
When electrons are in the lowest-energy orbitals available, the atom is
a. unstable
b. in an excited state
c. in the ground state
d. chemically unreactive
Answers
When electrons are in the lowest-energy orbitals available, the atom is in the ground state and is chemically reactive.
When electrons are in the lowest-energy orbitals available, the atom is in the ground state. This means that the electrons occupy the lowest possible energy levels and the atom is in its most stable and chemically reactive state. Excitation occurs when electrons are promoted to higher-energy orbitals, resulting in an unstable or excited state.
The electronic structure of an atom consists of different energy levels or orbitals where electrons can reside. These energy levels are quantized, meaning that electrons can only occupy specific energy states. The lowest-energy orbitals, often referred to as the ground state, are the most stable configuration for an atom.
When electrons are in the ground state, they fill the lowest-energy orbitals available according to the Aufbau principle and the Pauli exclusion principle. The atom in this state is chemically reactive because the electrons can participate in bonding or undergo chemical reactions with other atoms or molecules to achieve a more stable electron configuration.
In contrast, when electrons are excited to higher-energy orbitals by absorbing energy, the atom enters an excited state. In this state, the atom is less stable and more reactive. Excited atoms can release the excess energy by emitting light or undergoing chemical reactions to return to the ground state.
Therefore, when electrons are in the lowest-energy orbitals available, the atom is in the ground state and is chemically reactive.
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you consider starting material a. you know that a can undergo two irreversible reactions as shown in the below reaction coordinate diagram with one reaction pathway labeled in red and one reaction pathway labeled in blue. the red path leads to product b, while the blue path leads to product c. assuming both reaction pathways occur simultaneously in competition with each other, what is the major product, and why?
Answers
Product B because it has a lower energy level than Product C's transition state, which leads to Product C.
What are reaction pathways?The series of reactions required to create a desired product are described by a reaction pathway. The distribution strategy for a product is determined by things like percentage yield. Atomic economics. reaction time. is a connected graph with chemical species as its nodes. If a reaction transfers material from one species to the other, an edge unites the two. An vector from reactant toward the product is depicted as the edge.
What role do reactions pathway ?Energy, or ATP, is created by chemical reactions within our cells. All living things require energy to survive, and Adp would be a reactant that fuels a number of other chemical reactions inside cells. Cells generate energy through a process called cellular respiration.
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Be sure to answer all parts. Radon (Rn) is the heaviest and the only radioactive member of Group 8A(18), the noble gases. It is a product of the disintegration of heavier radioactive nuclei found in minute concentrations in many common rocks used for building and construction. In recent years, health concerns about the cancers caused from inhaled residential radon have grown. If 1. 00 × 1015 atoms of radium (Ra) produce an average of 1. 373 × 104 atoms of Rn per second, how many liters of Rn, measured at STP, are produced per day by 1. 08 g of Ra?
Answers
The final answer is 1.8 × 102 L of Rn, measured at STP, are produced per day by 1.08 g of Ra.
In order to determine the amount of radon that is produced per day by 1.08 g of Ra, it is essential to calculate the number of Rn atoms that are produced per second.
In the given question, it is known that 1.00 × 1015 atoms of radium produce an average of 1.373 × 104 atoms of Rn per second.
Therefore, we can use this ratio to calculate the number of Rn atoms that are produced per day by 1.08 g of Ra. Once we know the number of Rn atoms produced per day, we can calculate the volume of Rn produced by using the ideal gas law.
The final answer is 1.8 × 102 L of Rn, measured at STP, are produced per day by 1.08 g of Ra.
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For an atom of sulfur, there are?
1. two electron shells with 6 valence electrons
2. three electron shells with 6 valence electrons
3. four electron shells with 6 valence electrons
4. five electron shells with 6 valence electrons
Answers
Answer:
I think its b, but ik not completely sure.
Answer:
I think the second one...
What is the shape of salicylic acid structure?
Answers
The carboxyl group is joined to the second carbon atom in the benzene ring, which has six carbon atoms and one hydrogen atom. In the orthogonal position to the carboxyl group, the hydroxyl shape
The spatial arrangement of atoms in a molecule is often referred to as the "shape" of the molecule. A molecule's form is defined by how its atoms are arranged and by the bonds that hold them together. The physical and chemical characteristics of a molecule, such as its reactivity, solubility, and interactions with other molecules, can be significantly influenced by its shape. In rare circumstances, such as with enzymes and receptors, a molecule's structure can even dictate its biological function. The form of molecules may be ascertained, and their characteristics and behaviour can be understood, using a variety of approaches, such as X-ray crystallography and molecular modelling. "shape"
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314.5g HBr (MMHBr = 80.91g/mol), is mixed into 4.50L solution. Find Molarity (M) of the
solution.
Answers
Answer:
\(\boxed {\boxed {\sf M \approx0.864 \ M \ HBr}}\)
Explanation:
The molarity of a solution is the moles per liter. It is found using this formula:
\(M=\frac { moles \ of \ solute}{liters \ of \ solution}\)
First we must calculate moles, since we are given grams. To convert from grams to moles, the molar mass is used. This is also given to us, it is 80.91 grams per mole. We can use it as a ratio.
\(\frac {80.91 \ g \ HBr}{1 \ mol \ HBr}\)
Multiply by the given number of grams.
\(314.5 \ g \ HBr *\frac {80.91 \ g \ HBr}{1 \ mol \ HBr}\)
Flip the fraction so the grams of HBr cancel.
\(314.5 \ g \ HBr *\frac {1 \ mol \ HBr}{80.91 \ g \ HBr}\)
\(\frac {314.5 \ mol \ HBr}{80.91 }=3.88703497714 \ mol \ HBr\)
Now we know the moles, and can calculate the molarity.
There are 3.88703497714 moles of solute and 4.50 Liters of solution.
\(M= \frac {3.88703497714 \ mol }{ 4.50 \ L}\)
\(M= 0.863785550476 \ mol/ L\)
The original measurements have 4 and 3 significant figures, so we use the lowest number of 3 for our answer. For the number we calculated, that is the thousandth place.
The 7 in the ten thousandth place tells us to round the 3 to a 4.
\(M \approx 0.864 \ mol/ L\)
\(M \approx 0.864 \ M \ HBr\)
The molarity of the solution is about 0.864 M HBr.
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One mole of copper has a mass of 63.5 grams. Approximately how many atoms of copper are present in one mole of copper?
63 atoms
64 atoms
32 × 1023 atoms
6 × 1023 atoms
Answers
Answer:
6 × 10^23 atoms
Explanation:
This is because Avogadro's number aka 6.02 x 10^23 is applicable to every element and is the number of atoms in any 1 mole of an element. The answer 6 x 10^23 is closest to 6.02 x 10^23 because the 6.02 was rounded to just 6. Hope this helps :)
Answer:
The answer is D)
Explanation:
6 X 10^23
Does Na2 gas posses metallic character? Explain your answer..
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Explanation:
It contains Na2 molecules and the atoms in this molecule are held together by a purely covalent bond because the electronegativity of the two atoms is identical.Metallic bonding would not kick in until you make clusters of quite a few atoms. Such clusters would likely not be very stable because thermodynamically the larger the clump of material the more stable it gets. So they tend to coalesce until you have chunk of metal.Metallic bonding is in a sense a form of covalent bonding, but it is very collective (delocalized over a great many atoms) and electron deficient (there are more states than electrons to fill them up with, leading to conductive properties. This means that “a metallic bond” is a bit of an oxymoron like a forest with only one tree.Reply me in commentsYes, Na2 gas possesses a metallic character.
Does NA contain metallic bonds?In the stable state, metal sodium functions as an array of Na+ ions which can be surrounded by way of a sea of 3s electrons. However, it would be wrong to consider metal sodium as an ion when you consider that the ocean of electrons is shared by using all of the sodium cations, quenching the nice fee.
Sodium most effective has one valence electron. So, in metallic bonding, it is able to only donate one electron to be delocalized at some point of the structure. In steel bonding, the real bonding is the electrostatic force between the effective cations and the delocalized electrons.
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A 10.00 mL diluted chloride sample was titrated with 0.02749 M AgNO3, and 16.51 mL AgNO, was required to reach the endpoint. How would the following errors affect the calculated concentration of CI? a. The student read the molarity of AgNO, as 0.02479 M instead of 0.02749 M. The experimentally calculated moles of Ag would be too! calculated [CI] in the unknown would come out too b. The student was past the endpoint of the titration when the final buret reading was taken. v The experimentally determined moles of Ag would be too | calculated C1 concentration. so the calculated moles of CI would come out too so the calculated moles of CI would come out ✓ The as would the
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The effect of errors on the calculated concentration of CI is significant.
A 10.00 mL diluted chloride sample was titrated with 0.02749 M AgNO3, and 16.51 mL AgNO, was required to reach the endpoint. The effect of errors on the calculated concentration of CI can be explained as follows:a. The student read the molarity of AgNO, as 0.02479 M instead of 0.02749 M. If the student read the molarity of AgNO, as 0.02479 M instead of 0.02749 M, then the experimentally calculated moles of Ag would be too high. Consequently, the calculated [CI] in the unknown would come out too low. b.
The student was past the endpoint of the titration when the final buret reading was taken. If the student was past the endpoint of the titration when the final buret reading was taken, then the experimentally determined moles of Ag would be too low. This would cause the calculated C1 concentration to come out too high. Consequently, the calculated moles of CI would come out too high. Therefore, the effect of errors on the calculated concentration of CI is significant.
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How many elements are in this chemical formula?
2(NH4)2SO4
Answers
Answer: 30
Explanation:
Sulfur S 1
Oxygen O 4
Nitrogen N 2
Hydrogen H 8
15 in total. For 2 moles, 15*2 = 30