Calculate the mass defect of the oxygen nucleus 16 8o 816o . the mass of neutral 16 8o 816o is equal to 15.994914 atomic mass units

Making both reactants in excess and the concentration of Fe(SCN)* can be determined, Making FeSt in excess and SCN as the limiting reactant and the concentration of Fe(SCN)* will be equal to the concenation of SCN.

A reagent, sometimes known as an analytical reagent, in chemistry is a substance or compound supplied to a system to either initiate a chemical reaction or test whether one happens. Although the terms "reactant" and "reagent" are not frequently used interchangeably, "reactant" refers to a material that is consumed during a chemical reaction. The anion [SCN] is thiocyanate. Thiocyanic acid's conjugate base is it. The colorless compounds potassium thiocyanate and sodium thiocyanate are examples of common derivatives. In the past, pyrotechnics employed mercury(II) thiocyanate. Thiocyanic acid is a hydracid, which is cyanic acid in which the oxygen is substituted by a sulfur atom. SCN is a tiny, very acidic [1] pseudohalide thiolate.

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Explanation:

ΛЛらƜƐ尺 :-

A component of John Dalton's atomic theory is The ratio of atoms in a compound is fixed . Atoms combine in a ratio of small whole numbers to form compounds.

According to Daltons atomic theory ,all matter, an element, a compound or a mixture is composed of small particles called atoms.

The postulates of this theory are as follows:

(i) All matter is made of tiny particles called atoms.

(ii) Atoms are indivisible particles which cant be created or destroyed in a chemical reaction.

(iii) Atoms of a given element are identical in mass and chemical properties.

(iv) Atoms of different elements have different masses and chemical properties.

(v) Atoms combine in ratio of small whole numbers to form compounds.

(vi) Atoms can neither be created nor be destroyed

Only statement 1 is correct from the given options.

Statement 2 , 3 and 4 in the given options are wrong.

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Considering the definition of empirical formula, the empirical formula is Ag₁N₁O₃= AgNO₃.

The empirical formula is the simplest expression to represent a chemical compound, which indicates the elements that are present and the minimum proportion in whole numbers that exist between its atoms, that is, the subscripts of chemical formulas are reduced to the most integers. small as possible.

Assuming a 100 grams sample, the percentages match the grams in the sample. So you have 63.5 grams of silver, 8.2 grams of nitrogen and 28.3 grams of oxygen.

Then it is possible to calculate the number of moles of each atom in the molecule, taking into account the corresponding molar mass:

Silver: \(\frac{63.5 g}{107.87 \frac{g}{mole} } =\)0.588 moles

Nitrogen: \(\frac{8.2 g}{14 \frac{g}{mole} } =\) 0.586 moles

Oxygen: \(\frac{28.3 g}{16 \frac{g}{mole} } =\)1.77 moles

The empirical formula must be expressed using whole number relationships, for this the numbers of moles are divided by the smallest result of those obtained. In this case:

Silver: \(\frac{0.588 moles}{0.586 moles}\)≅ 1

Nitrogen: \(\frac{0.586 moles}{0.586 moles}\)= 1

Oxygen: \(\frac{1.77 moles}{0.586 moles}\)≅ 3

Therefore the Ag: N: O mole ratio is 1: 1: 3

Finally, the empirical formula is Ag₁N₁O₃= AgNO₃.

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Answer:

3 feet is in one yard, which is 36 inches.

The ratio of octene to oxygen is 1:12.

To determine the ratio of octene (C8H16) to oxygen (O2) in the given reaction, we need to examine the balanced chemical equation. However, the equation you provided does not seem to be balanced. The coefficients for each compound must be determined to achieve a balanced equation before we can calculate the desired ratio.

Assuming you meant the combustion reaction of octene, a balanced equation would be:

C8H16 + 12O2 → 8CO2 + 8H2O

From the balanced equation, we can see that for every 1 mole of octene (C8H16), we require 12 moles of oxygen (O2) to completely react.

This means that for every 1 mole of octene, we need 12 moles of oxygen to fully combust the octene and produce the corresponding amounts of carbon dioxide (CO2) and water (H2O) as shown in the balanced equation.

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The elements can also be classified into the main-group elements (or representative elements) in the columns labeled 1, 2, and 13–18; the transition metals in the columns labeled 3–12; and inner transition metals in the two rows at the bottom of the table (the top-row elements are called lanthanides and the bottom-row ...

Hello,

Elements are a group of the Periodic Table that shares common valence electron structures. They are arranged in order of increasing Atomic Number.

In modern periodic table they are classified into three parts- metals, non-metals and metalloids.

In detail:-

• Metals- Metals are those elements which donate electrons. Ex- Na, K, Rb, Mg, Ca, etc.

• Non-metals- Non-metals are those elements which accept electrons. Ex- N, O, F, Cl, etc .

• Metalloids- Metalloids are those elements which show property of both metals and non-metals. Ex- B, Si, Ge, etc.

Hope it helps...

(by Benjemin)

Based on the given information, the true statement about the time indicated by the blue arrow is: (c) The rate of the forward reaction (N₂O₄ formation) is equal to the rate of the reverse reaction (NO₂ formation).

The graph shows the relative amounts of NO₂ and N₂O₄ over time, and the point indicated by the blue arrow represents a state of equilibrium. In an equilibrium state, the forward and reverse reactions occur at equal rates.

The concentrations of NO₂ and N₂O₄ reach a constant value, indicating that the conversion of NO₂ to N₂O₄ and the conversion of N₂O₄ to NO₂ are occurring at the same rate.

Option a suggests that NO₂ molecules are no longer reacting, which is incorrect as the reaction is still ongoing at equilibrium. Option b suggests that the reactant has been completely used up, which is not the case in an equilibrium state. Option d refers to the activation energy, which is unrelated to the equilibrium state. Therefore, option c is the correct choice.

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The simplest formula for the compound is \(CO_{2}\) if a 0.478g sample of the compound was burned.

The simplest formula can be termed as the imperial formula which can be determined by using the molar mass to convert the mass of each element to moles. Then each mole value is divided by the smallest calculated number of moles.

First, we find the number of moles of carbon, hydrogen, and oxygen as follows;

Mass of carbon = 1.50 × (12/44) = 0.41 g

Number of moles of carbon = 0.41 / 12 = 0.034 moles

Mass of hydrogen = 0.615 × (2/18) = 0.068 g

Number of moles of hydrogen = 0.068 / 1 = 0.068 moles

Mass of oxygen = 0.478 - (0.41g + 0.068g) = 0 g

Number of moles of oxygen = 0 moles

Now we divide the calculated number of moles of carbon and hydrogen by the smallest calculated number of moles which in this case is of carbon.

Carbon: 0.034 ÷ 0.034 = 1

Oxygen: 0.068 ÷ 0.034 = 2

Therefore, the simplest formula for the compound is determined to be \(CO_{2}\)

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To find the concentration of nh3(aq) if the pH is fixed at 9.25, we need to use the equilibrium constant expression for the reaction of NH3 with water:
NH3 + H2O ⇌ NH4+ + OH-

The equilibrium constant expression for this reaction is:
Kw/Kb = [NH4+][OH-]/[NH3]
Where Kw is the ion product constant of water (1.0 x 10^-14) and Kb is the base dissociation constant of NH3. At a pH of 9.25, the pOH is 4.75 and the Kb for NH3 is 1.8 x 10^-5.

Using the equation above and substituting the values, we get:
1.0 x 10^-14/1.8 x 10^-5 = [NH4+][OH-]/[NH3]
Solving for [NH3], we get:
[NH3] = [NH4+][OH-]/(1.0 x 10^-14/1.8 x 10^-5)

At a pH of 9.25, the [NH4+] concentration can be calculated using the equation:
pH = pKa + log([NH4+]/[NH3])
The pKa of NH4+ is 9.24.

Substituting the values and solving for [NH4+], we get:
9.25 = 9.24 + log([NH4+]/[NH3])
log([NH4+]/[NH3]) = 0.01
[NH4+]/[NH3] = 1.02

Substituting this value in the expression for [NH3], we get:
[NH3] = (1.02)(1.0 x 10^-14/1.8 x 10^-5)
[NH3] = 5.67 x 10^-10 M
Therefore, the concentration of NH3 in the solution is 5.67 x 10^-10 M if the pH is fixed at 9.25.

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The equation shows reactants to a neutralization reaction.

Ca²⁺ + OH⁻ → H⁺ + NO⁻. Here H binds with OH⁻ to make water. So the option c is correct.

As the answer to the question is cleared from these reactions. Firstly calcium hydroxide combines with nitric acid to form the products as follows;

Ca(OH)₂ + 2HNO₃ → Ca(NO₃)₂ + 2H₂O

Now the reaction between ionic species will be as follows

Ca²⁺ + 2OH⁻ + 2H⁺ + 2NO₃⁻ → Ca²⁺ + 2NO₃⁻ + 2H₂O

Hence OH⁻ combines with  H⁺  to make  H₂O(water)  

OH⁻ + H⁺ → H₂O

H⁺ binds with OH⁻ to make water

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The empirical formula for the compound is Ca(OH)₂.

To determine the empirical formula of the compound, we need to find the simplest whole-number ratio of the elements present. We can do this by calculating the number of moles of each element and then finding the ratio between them.

Mass of calcium (Ca) = 13.5 grams

Mass of oxygen (O) = 10.8 grams

Mass of hydrogen (H) = 0.675 grams

First, we need to convert the masses of each element to moles using their respective atomic masses.

Atomic mass of Ca = 40.08 g/mol

Atomic mass of O = 16.00 g/mol

Atomic mass of H = 1.01 g/mol

Number of moles of Ca = 13.5 g / 40.08 g/mol = 0.3367 mol

Number of moles of O = 10.8 g / 16.00 g/mol = 0.675 mol

Number of moles of H = 0.675 g / 1.01 g/mol = 0.6693 mol

Next, we need to find the simplest whole-number ratio by dividing the number of moles of each element by the smallest number of moles obtained.

Dividing by 0.3367 (smallest value):

Number of moles of Ca = 0.3367 mol / 0.3367 mol = 1

Number of moles of O = 0.675 mol / 0.3367 mol = 2.005

Number of moles of H = 0.6693 mol / 0.3367 mol = 1.988

Rounding these values to the nearest whole number, we have:

Number of moles of Ca = 1

Number of moles of O = 2

Number of moles of H = 2

Thus, the empirical formula of the compound is Ca(OH)₂.

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The monobrominated compounds produced when 2-propyl-1-pentene is treated with NBS and UV light irradiation are the compounds that contain a bromine atom bonded to a single carbon atom within the 2-propyl-1-pentene molecule.

When 2-propyl-1-pentene undergoes bromination using NBS (N-bromosuccinimide) and UV light, the reaction occurs via a radical mechanism. This leads to the substitution of a hydrogen atom with a bromine atom, resulting in the formation of monobrominated compounds.

To determine the specific monobrominated compounds and their stereoisomers, it is necessary to analyze the structure of 2-propyl-1-pentene and consider the possible sites of bromination. 2-propyl-1-pentene has multiple carbon atoms that could potentially undergo bromination, leading to different structural isomers.

By introducing a bromine atom at various positions within the 2-propyl-1-pentene molecule, different monobrominated compounds can be formed. These compounds may exhibit stereoisomerism if there are chiral centers present in the original molecule.

To provide a comprehensive answer, the specific structures of the monobrominated compounds and their stereoisomers resulting from the reaction would need to be analyzed and described.

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Answer

b

Explanation:

because I took the test

Each hydrogen shares two electrons with nitrogen.

NH3 is a molecular compound. In molecular compounds, electrons are shared between the bonding atoms in a covalent bond. The number of electrons shared between atoms depends on the respective valencies of the combining atoms.

Nitrogen has a valency of three and hydrogen has a valency of one. Therefore, in NH3, each hydrogen shares two electrons with nitrogen.

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Answer:

Refer to your periodic table. Lewis dot structures are based off the number of valence electrons an atom has.

Looking at the compounds, we can see that Gallium has three valence electrons in its outer shell and oxygen has six. Oxygen and Gallium are going to share electrons with one another, making a V shape in their diagram.

One Oxygen would make a double bond with a Gallium, leaving one valence electron to another oxygen. That oxygen takes that Final electron. It now has 7 in its outer shell. The remaining Gallium and Oxygen do the same double bond as the one before, leaving the 7 valence electron oxygen with one more electron.

The Lewis dot structure of Ga₂O₃ is shown below.

In molecular structure, Lewis structures describe the relationship between molecules' atoms and their lone pairs of electrons. Lewis structures are also used to predict molecular geometry in conjunctions with hybrid orbitals.

Ga₂O₃ is the inorganic compound of gallium (III) oxide. It is a semiconductor with a very wide bandgap. It is widely used in power electronics. It is also used in phosphors and gas sensing.

There are several polymorphs of gallium oxide. The most stable polymorph of gallium oxide is the polymorph of gallium(III)(monoclinic).

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The image for the Lewis dot structure is shown below.

Answer:Is Fruit ripening dependent on the type of light exposed to?

Explanation: Because you can test this theory

Answer:

A stock or standard solution is a solution in which you accurately know its concentration. You can make stock solutions in the chemistry laboratory or buy from chemical manufacturers. Once you have a stock solution, you can prepare solutions of lower concentration by diluting the concentrated stock solution.

Explanation:

To make the correct volume of stock solution, we should add first solid to a volumetric flask.

Stock solution is a standard solution in which exact concentration of solute is present in particular volume. It is mainly used in analytical chemistry during the titration process.

During the preparation of stock solution, we should first add solid in the volumetric flask and then little amount of water to dissolve the solid and then finally add water to the required volume. If we first add total volume and then add solid then dissolution of solid may increase volume.

Hence, first we add solid in the volumetric flask.

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Enantiomers are pairs of compounds with exactly the same connectivity but opposite three-dimensional shapes.

What is a enantiomer?

In chemistry, an enantiomer (also called an optical isomer, enantiomer, or optical antipode) is one of two stereoisomers that are not superimposable on their own mirror image. Enantiomers are similar to the right and left hands and cannot be superimposed when looking at the same face. No amount of reorientation will align the four unique groups (see Chirality (Chemistry)) on the chiral carbons exactly. The number of stereoisomers a molecule has can be determined by the number of chiral carbons it has. Stereoisomers include both enantiomers and diastereomers.

Like enantiomers, diastereomers have the same molecular formula and are not superimposable, but they are not mirror images of each other.

Therefore, Enantiomers are pairs of compounds with exactly the same connectivity but opposite three-dimensional shapes.

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Answer:

i believe lithium

Explanation: i hope this is right.

The relative formula mass of the group 2 carbonate W is 41 g/mol.

Assuming that the gas produced is measured at standard temperature and pressure (STP), which is 0°C and 1 atmosphere pressure, the volume of one mole of gas is 22.4 dm³.

So if 1 mole of Group 2 carbonate W produces 24 dm³ of gas at STP, then the molar volume of the gas produced is:

24 dm³ / 22.4 dm³/mol = 1.07 mol

This means that one mole of Group 2 carbonate W produces 1.07 moles of gas. Since the gas is most likely carbon dioxide (CO₂) which has a formula mass of 44 g/mol, the relative formula mass of Group 2 carbonate W can be calculated as:

Relative formula mass = Mass of 1 mole of CO₂ / 1.07

Relative formula mass = 44 g/mol / 1.07

Relative formula mass ≈ 41 g/mol

Therefore, the relative formula mass of the Group 2 carbonate W is approximately 41 g/mol.

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The complete question is:

24 dm3 of gas is produced when one mole of a Group 2 carbonate is heated.

Determine the relative formula mass of the Group 2 carbonate W.

Use the graph above.

The amount of heat to absorb is 6261 J

The heat capacity is the amount of heat expressed usually in Joules or calories needed to change the system by 1 degree Celsius

Here the calculation for heat is

Heat required to raise the temprature of ice form -20°C to 0°C

Then the formula for specific heat is used to calculate the amount of heat

Q = c×m×ΔT

Where Q = heat exchanged by a body

m = mass of the body

c = specific heat

ΔT = change in temprature

Then the given data is

m = 10g

Specific heat of the ice = 2.1

ΔT =  0°C - (-20°C) = 20°C

Then substituting the value

Q = 10g× 2.1× 20°C

Q = 420 J

Heat required to convert 0°C ice to 0°C water

The heat Q necessary to melt a substances depend on its mass m and this called latent heat of fusion of each substances

Q = m×ΔH fusion

Heat required to raise the temprature of water from 0°C to 60°C

m = 10g

Specific heat of water = 4.18

ΔT = 60°C -(- 0°C) = 60°C

Then substituting

Q = 10g×4.18×60°C

Q = 2508 J

Then Q total = 420J+3333J+2508J

Q total = 6261 J

The amount of heat to absorb is  6261 J

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Answer:

There are 126 neutrons in the bismuth atom.

Explanation:

We are given that a Bi atom has an atomic mass of 209 amu and an atomic number of 83.

We want to find how many neutrons the bismuth atom has.

The atomic mass of an atom is equal to the mass of the protons + mass of neutrons (n.b. the electrons are part of the mass too, however they are very, very small. Therefore, we tend to count the mass of the electrons as 0 when calculating the atomic mass).

Recall that protons and neutrons both have a mass of about 1 amu.

Also recall that the proton number is also the atomic number. Because of this, the bismuth atom has 83 protons. And since that protons have a mass of about 1 amu, the atomic mass of the protons will be:

83 protons × \(\frac{1 amu}{1 proton}\) = 83 amu

That means that if the atomic mass is 209 amu, it is equal to the proton mass (83 amu) + an unknown mass of neutrons (let's say it is x amu).

As an equation, it will be:

83 + x = 209

Subtract 83 from both sides.

x = 126

This means that the neutrons will weigh 126 amu.

Since every neutron is about 1 amu, the amount of neutrons will be:

126 amu × \(\frac{1 neutron}{1 amu}\) = 126 neutrons

Answer:

126

Explanation:

To calculate for neutron #, subtract atomic # from atomic mass.

t hus 09- 83= 126

True. Plasticizers are indeed liquids of low volatility that can be lost through diffusion or evaporation as a plastic material ages.

Plasticizers are substances added to plastics to enhance their flexibility, durability, and workability. They are typically low-molecular-weight compounds with a liquid form and have a relatively low boiling point, which makes them volatile. Over time, as plastic materials age and are exposed to various environmental factors such as heat, light, and mechanical stress, the plasticizers can migrate or evaporate out of the plastic matrix.

Diffusion refers to the process of plasticizers gradually moving from areas of high concentration within the plastic to areas of lower concentration. This diffusion can occur both within the plastic material itself and between the plastic and its surrounding environment. Evaporation, on the other hand, involves the direct conversion of the liquid plasticizers into vapor, which can escape into the air.

The loss of plasticizers can have significant effects on the properties of the plastic material. As plasticizers are responsible for increasing flexibility and reducing brittleness, their loss can lead to increased stiffness, decreased elongation, and potential cracking or degradation of the plastic. Therefore, proper selection and monitoring of plasticizers are crucial in ensuring the long-term performance and integrity of plastic materials.

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Answer:

Virus is living due to reproduction and non-living due to crystal appearance.

Explanation:

Characteristics that viruses display is the ability of reproduction in which they increase in population which is a living character and have DNA or RNA which help them to make exact copies of itself. Virus is parasitic in nature because it causes harm to the living host such as humans, animals and plants. Some characteristics that viruses don't display are that they are not like living cells, have no membrane around them, no organelles such as mitochondria, golgi bodies, endoplasmic reticulum and lysosomes etc. Viruses are present in crystal form outside the cell.

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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Answer:

-3.5g−6

Explanation:

-3.5g−6