15. A 15.30 g sample of calcium (Ca) combines with 7.96 g of sulfur(S) to form calcium sulfide
(Cas). How many grams of CaS are formed?

D

This is when very large or very small numbers must be written. They come in the format or 5.43 x 10^3 or can be written as 8.26E6. The numbers will vary depending on your number.

Explanation:

Identify the substance: Determine the identity of the substance that is being measured in moles.

Determine the molar mass: Look up the molar mass of the substance in a periodic table or a reference book. The molar mass is expressed in grams per mole.

Set up the conversion factor: Use the molar mass to set up a conversion factor. The conversion factor is a ratio that relates the number of moles to the number of grams.

Example: If the molar mass of the substance is 50 g/mol, the conversion factor would be:

1 mol / 50 g

This means that one mole of the substance is equal to 50 grams.

Apply the conversion factor: Multiply the given quantity, expressed in moles, by the conversion factor. The moles unit will cancel out, leaving the unknown quantity in grams.

Example: If the given quantity is 2 moles of the substance, the calculation would be:

2 mol x (1 mol / 50 g) = 0.04 g

Therefore, the unknown quantity is 0.04 grams.

Check the units: Always double-check that the units of the final answer are correct. In this case, the units should be in grams

Answer:

Divide the distance the object traveled by the time it took to get there.

Explanation:

To calculate the speed on an object, start by determining how far the object has traveled. Next, figure out the amount of time that the object took to cover that distance. Finally, divide the distance the object traveled by the time it took to get there. Don't forget to label the speed with the correct units of measurement.

Answer:

1) The mole ratio of KClO₃ to 3O₂ is 2:3

The reciprocal of the mole ratio of KClO₃ to 3O₂ is 3:2

2) The mole ratio of K to NaNO₃ is 1:1

The reciprocal of the mole ratio of K to NaNO₃ is 1:1

Explanation:

The mole ratio (also known as mole-mole ratio) of two compounds involved in a chemical reaction, is the ratio of the number of moles of the compounds in the reaction which can be obtained from the coefficients of the compounds in the balanced chemical reaction

The mole ratio of compound A to compound B = A:B = A/B

1)The given chemical reaction can be expressed as follows;

2KClO₃ → 2KCl + 3O₂

Therefore;

\(The \ mole \ ratio \ of \ KClO_3 \ to \ O_2 = \dfrac{2 \ moles \ of \ KClO_3}{3 \ moles \ of \ O_2} = \dfrac{2}{3} = 2:3\)

\(The \ reciprocal \ of \ the \ mole \ ratio \ of \ KClO_3 \ to \ O_2 = \dfrac{3 \ moles \ of \ O_2}{2 \ moles \ of \ KClO_3} = \dfrac{3}{2} = 3:2\)

2) In the reaction NaNO₃ + K → Na + KNO₃

The mole ratio of K to NaNO₃ = 1:1

Therefore;

The reciprocal of the mole ratio of K to NaNO₃ = 1:1

The endpoint coordinates of the mid-segment of triangle JKL is (5.06, 4.04).

Given the triangle JKL in which mid-segment of triangle JKL is parallel to JL.

Now we need to find the endpoint coordinates of the mid-segment of triangle JKL.

Hence the midpoint of JK = M (3, 2.5) and the length of \(JK = JL/2= \sqrt{((5 - 1)^{2} + (4 - 1)^{2} )} / 2 = 4.12/2 = 2.06\)

Now we know the length of mid-segment JK = 2.06 and its midpoint, we can find the endpoint coordinates by using the slope formula

Endpoint of JK, K is(1, 3)

hence the coordinates are \((3 + 2.06, 2.5 + 1.54) = (5.06, 4.04)\)

Therefore, the endpoint coordinates of the mid-segment of triangle JKL is (5.06, 4.04).

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Iron ore are rock rocks

hope this helps

please mark as brainlist

In Niels Bohr’s model of the atom, electrons are configured in a series of concentric shells around the nucleus. The shells are numbered, with the shell closest to the nucleus being numbered one, and each succeeding shell numbered two, three, and so on.

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No its a physical change .

\(\\ \sf\longmapsto NaCl+Lattice\:Energy\longrightarrow Na^++Cl^-+Hydration\:Energy\)

Answer:

pOH = 3.18

Explanation:

The equilibrium of a weak base as NX3 is:

NX3(aq) + H2O(l) ⇄ HNX3⁺(aq) + OH-(aq)

Where the equilibrium constant, Kb, is:

Kb = 2.5x10⁻⁶ = [HNX3⁺] [OH-] / [NX3]

As both HNX3⁺ and OH- ions comes from the same equilibrium, their concentrations are the same, that is:

[HNX3⁺] = X

[OH-] = X

And: [NX3] = 0.175M:

2.5x10⁻⁶ = [X] [X] / [0.175M]

4.375x10⁻⁷ = X²

X = 6.61x10⁻⁴M = [OH-]

As pOH = -log [OH-]

The minimum value of the partition coefficient (k) to get an efficient extraction with only one extraction step is 1.8.

To answer your question, we need to understand the concept of partition coefficient (k). Partition coefficient (k) is the ratio of the concentration of a solute in the organic phase to its concentration in the aqueous phase at equilibrium. It is a measure of the solubility of a solute in a particular solvent system.
Now, to get an efficient extraction with only one extraction step, we need to ensure that more than 90% of the desired compound is recovered. Given that the desired compound is soluble in the organic solvent and we use equal volumes of the organic solvent and water, the minimum value of the partition coefficient (k) can be calculated using the following formula:
k = [concentration of the desired compound in the organic phase] / [concentration of the desired compound in the aqueous phase]
To achieve an efficient extraction with only one extraction step, we need to ensure that more than 90% of the desired compound is extracted into the organic phase. This means that the concentration of the desired compound in the organic phase should be at least 90% of the initial concentration of the compound. Assuming equal volumes of the organic solvent and water are used, this can be represented as:
[concentration of the desired compound in the organic phase] >= 0.9 x [initial concentration of the desired compound]
Similarly, the concentration of the desired compound in the aqueous phase can be represented as:
[concentration of the desired compound in the aqueous phase] = [initial concentration of the desired compound] / 2
Substituting these values in the formula for k, we get:
k >= (0.9 x [initial concentration of the desired compound]) / ([initial concentration of the desired compound] / 2
Simplifying the expression, we get:
k >= 1.8
In summary, for an efficient extraction with only one extraction step, we need to ensure that the desired compound is soluble in the organic solvent and the partition coefficient (k) is equal to or greater than 1.8.

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(i) The initial-boundary value problem for the given scenarios are as follows:

a) Scenario a:

PDE: ∂u/∂t = k * ∂²u/∂x²

Initial condition: u(x, 0) = 100 (boiling water temperature)

Boundary conditions: u(0, t) = 10, u(L, t) = 10 (constant temperature at the ends)

b) Scenario b:

PDE: ∂u/∂t = k * ∂²u/∂x²

Initial condition: u(x, 0) = 100 (boiling water temperature)

Boundary conditions: u(0, t) = 0 (temperature at x=0), ∂u/∂x(L, t) = 0 (thermal insulation at x=L)

(iii) The solution for the temperature distribution as time approaches infinity can be found by solving the appropriate steady state equation.

(i) The initial-boundary value problem formulation states the partial differential equation (PDE) governing the temperature distribution in the aluminum bar, along with the initial condition and boundary conditions.

In scenario (a), both ends of the bar are immersed in a medium with a constant temperature of 10 degrees Celsius, while in scenario (b), the end at x=0 is immersed in a medium with temperature 0 degrees Celsius and the end at x=L is insulated.

(ii) As time approaches infinity, the temperature distribution in the bar tends to reach a steady state.

In scenario (a), the temperature throughout the bar will eventually approach a constant value of 10 degrees Celsius, since both ends are immersed in a medium with that temperature.

In scenario (b), the temperature at x=0 will approach 0 degrees Celsius, while the temperature at x=L will remain constant due to thermal insulation.

(iii) To find the solution as time approaches infinity, we need to solve the appropriate steady state equation.

In scenario (a), the steady state equation is ∂²u/∂x² = 0, which implies that the temperature gradient is zero throughout the bar, resulting in a constant temperature of 10 degrees Celsius.

In scenario (b), the steady state equation is ∂²u/∂x² = 0 with the boundary condition u(0) = 0, which implies a linear temperature distribution from 0 degrees Celsius at x=0 to a constant temperature at x=L due to insulation.

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Part A: Ru3+ The electron configuration for Ru is [Kr]5s^24d^6. To form Ru3+, three electrons are removed. Therefore, the electron configuration for Ru3+ is [Kr]4d^5.

Part B: As3-

The electron configuration for As is [Ar]4s^23d^104p^3. To form As3-, three electrons are gained. Therefore, the electron configuration for As3- is [Ar]4s^23d^104p^6.

Part C: Y3+

The electron configuration for Y is [Kr]5s^24d^15p^1. To form Y3+, three electrons are removed. Therefore, the electron configuration for Y3+ is [Kr]4d^0.

Part D: Pd2+

The electron configuration for Pd is [Kr]5s^04d^10. To form Pd2+, two electrons are removed. Therefore, the electron configuration for Pd2+ is [Kr]4d^8.

Part A: Ru3+

The electron configuration for Ru is [Kr]5s^24d^6. To form Ru3+, three electrons are removed. This means that the 5s orbital is emptied and one electron is removed from the 4d orbital. Therefore, the electron configuration for Ru3+ is [Kr]4d^5. This configuration indicates that there are 5 electrons in the 4d orbital.

Part B: As3-

The electron configuration for As is [Ar]4s^23d^104p^3. To form As3-, three electrons are gained, filling up the 4p orbital. Therefore, the electron configuration for As3- is [Ar]4s^23d^104p^6. This configuration indicates that there are 6 electrons in the 4p orbital.

Part C: Y3+

The electron configuration for Y is [Kr]5s^24d^15p^1. To form Y3+, three electrons are removed. This means that the 5s and 4d orbitals are emptied. Therefore, the electron configuration for Y3+ is [Kr]4d^0. This configuration indicates that there are no electrons in the 4d orbital.

Part D: Pd2+

The electron configuration for Pd is [Kr]5s^04d^10. To form Pd2+, two electrons are removed, leaving the 5s and 4d orbitals empty. Therefore, the electron configuration for Pd2+ is [Kr]4d^8. This configuration indicates that there are 8 electrons in the 4d orbital.

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Answer:

Pangaea or Pangea ( /pænˈdʒiːə/) was a supercontinent that existed during the late Paleozoic and early Mesozoic eras. It assembled from earlier continental units approximately 335 million years ago, and it began to break apart about 175 million years ago.

Explanation:

The value of the Ksp for silver (I) carbonate is 8.39 × 10⁻¹¹M⁴ if the saturated solution is prepared by dissolving silver (I) carbonate in water.

The solubility product constant (Ksp) of silver (I) carbonate can be calculated by using the concentration of the silver ion present in the solution. The given concentration of silver ion, [Ag+1] is 2.56 × 10⁻⁴ M. The balanced chemical equation for the dissolution of Ag₂CO₃ is given below;

Ag₂CO₃ → 2Ag⁺ + CO₃²⁻

According to the above equation, each mole of silver carbonate that dissolves produces two moles of Ag⁺ ions. Therefore, if ‘s’ moles of silver carbonate dissolve, then the concentration of Ag⁺ ions would be equal to 2s. Thus;

2s = [Ag+1]s = 2.56 × 10⁻⁴ M

The molar solubility of Ag₂CO₃ is therefore equal to;S = s / Vwhere V is the volume of the solution in liters, and s is the molar solubility of Ag₂CO₃.Substituting the value of s, we have;

S = 1.28 × 10⁻⁴ / VS = 1.28 × 10⁻⁴ M

The Ksp of Ag₂CO₃ is given by the product of the concentrations of the silver ion and carbonate ion in the solution, raised to their respective stoichiometric coefficients. That is;Ksp = [Ag+1]²[CO₃²⁻]

Now, we know the concentration of silver ion, and since Ag₂CO₃ is a sparingly soluble salt, it dissociates to a negligible extent. Therefore, we can assume that the concentration of carbonate ion is equal to the molar solubility of the salt. Therefore; Ksp = (2.56 × 10⁻⁴)² (1.28 × 10⁻⁴)Ksp = 8.39 × 10⁻¹¹M⁴

Therefore, the value of the Ksp for silver (I) carbonate is 8.39 × 10⁻¹¹M⁴.

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The amount of concentrated solution needed is (0.286 M)(65 mL) / C M, and the amount of water needed is 65 mL minus the volume of the concentrated solution.

To calculate the molarity of Kool Aid desired, we need to determine the number of moles of Kool Aid powder and sugar in the package. Since the molecular formula for sugar is C12H22O11, we can calculate its molar mass as follows:

Molar mass of C12H22O11 = (12 * 12.01) + (22 * 1.01) + (11 * 16.00)

= 144.12 + 22.22 + 176.00

= 342.34 g/mol

Given that the package contains 4.25 grams of Kool Aid powder, we can calculate the number of moles of Kool Aid powder using its molar mass:

Number of moles of Kool Aid powder = Mass / Molar mass

= 4.25 g / 342.34 g/mol

≈ 0.0124 mol

Similarly, for the sugar, which has a molar mass of 342.34 g/mol, we can calculate the number of moles of sugar using its mass:

Number of moles of sugar = Mass / Molar mass

= 192.00 g / 342.34 g/mol

≈ 0.5612 mol

Now, to calculate the molarity of the desired Kool Aid solution, we need to determine the volume of water. Given that 1.06 quarts is equal to 1 liter, and we have 2.00 quarts of water, we can convert it to liters as follows:

Volume of water = 2.00 quarts * (1.06 liters / 1 quart)

= 2.12 liters

To find the molarity, we use the formula:

Molarity (M) = Number of moles / Volume (in liters)

Molarity of Kool Aid desired = (0.0124 mol + 0.5612 mol) / 2.12 L

≈ 0.286 M

To prepare 65 mL of the desired molarity, we can use dilution calculations. We need to determine the volume of concentrated solution and the volume of water needed.

Let's assume the concentration of the concentrated Kool Aid solution is C M. Using the dilution formula:

(C1)(V1) = (C2)(V2)where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.

Given that C1 = C M and V1 = V mL, and we want to prepare a final volume of 65 mL (V2 = 65 mL) with a final concentration of 0.286 M (C2 = 0.286 M), we can rearrange the formula to solve for the volume of the concentrated solution:

(C M)(V mL) = (0.286 M)(65 mL)

V mL = (0.286 M)(65 mL) / C M

So, the amount of concentrated solution needed is (0.286 M)(65 mL) / C M, and the amount of water needed is 65 mL minus the volume of the concentrated solution.

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Answer:

gamma rays

Explanation:

gamma Ray's have the highest frequency

One:

plants make their own food through a process known as photosynthesis

Two:

Photosynthesis is a process used by plants and other organisms to convert light energy into chemical energy and stored in the form of starch which can be used later

Three:

Photosynthesis converts carbon dioxide and water into oxygen and glucose. Glucose is used as food by the plant and oxygen is a by-product.

Answer:

1. b

2. a

3. b

Explanation:

Answer : When mixing 538 grams of a substance into 647 ml of water, the concentration of the resulting solution in g/L is 0.83.

The concentration of the resulting solution in g/L can be calculated by dividing the mass of the substance (538 g) by the total volume of the solution (647 ml). This gives us a result of 0.83 g/L.

To further explain this calculation, we must first understand the concepts of mass and volume. Mass is a measure of the amount of matter an object contains. Volume, on the other hand, is the amount of space occupied by a given object. When mixing 538 grams of a substance into 647 ml of water, we are creating a solution with a certain concentration of the substance.

To calculate the concentration of the resulting solution, we must divide the mass of the substance (538 g) by the total volume of the solution (647 ml). This gives us a result of 0.83 g/L.

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Answer:

Your answer is ATOM

Explanation:

Atoms consist of a nucleus made of protons and neutrons orbited by electrons. Atoms are the basic units of matter and the defining structure of elements.

Basically, they are the building block of life because everything made of matter has atoms

Skeletal muscle was the Effector in the reaction time test and motor response reflects the muscular component of reaction time.

The Effector in the reaction time test was the skeletal muscle in the finger which is used to press the button. muscle and glands produces a specific response to a stimuli's and the motor response was reaction time test is the time between electromyographic activity and movement and the motor response is the response which reflects the skeletal muscle component of reaction time.

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Net ionic equations for the following chemical reactions: Pb(NO3)2 (aq) + H2SO4 (aq) → PbSO4 (s) + 2 HNO3 (aq)

Ionic Equation: Pb2+ (aq) + 2 NO3- (aq) + 2 H+ (aq) + SO42- (aq) → PbSO4 (s) + 2 H+ (aq) + 2 NO3- (aq)

Net Ionic Equation: Pb2+ (aq) + SO42- (aq) → PbSO4 (s)Pb(C2H3O2)2 (aq) + 2 HCl (aq)

→ PbCl2 (s) + 2 HC2H3O2 (aq)

Ionic Equation: Pb2+ (aq) + 2 C2H3O2- (aq) + 2 H+ (aq) + 2 Cl- (aq) → PbCl2 (s) + 2 HC2H3O2 (aq)

Net Ionic Equation: Pb2+ (aq) + 2 Cl- (aq) → PbCl2 (s) Na3PO4 (aq) + FeCl3 (aq) → 3 NaCl (aq) + FePO4 (s)

Ionic Equation: 3 Na+ (aq) + PO43- (aq) + Fe3+ (aq) + 3 Cl- (aq) → 3 Na+ (aq) + 3 Cl- (aq) + FePO4 (s)

Net Ionic Equation: PO43- (aq) + Fe3+ (aq) → FePO4 (s)(NH4)2S (aq) + Co(NO3)2 (aq)

→ CoS (s) + 2 NH4NO3 (aq)

Ionic Equation: 2 NH4+ (aq) + S2- (aq) + Co2+ (aq) + 2 NO3- (aq) → CoS (s) + 2 NH4+ (aq) + 2 NO3- (aq)

Net Ionic Equation: S2- (aq) + Co2+ (aq) → CoS (s)

In the first equation, H2SO4 (aq) was written as H+ (aq) + SO42- (aq) because H2SO4 (aq) is a strong acid. Hence, it completely dissociates in H+ (aq) and SO42- (aq). Similarly, in the second equation, HCl (aq) was written as H+ (aq) + Cl- (aq) because it is also a strong acid.

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Answer: Positive and 13.

Explanation:

Answer:

0.9M

Explanation:

molarity = mols/L

M=4.7mol/5.2L

= 0.9M

Conversion of 7g/dm³ of H₂O to 0.21 mol/dm³ is

Conversion is the act or process of changing something into a different state or form

Here given data is

7g/dm³ of H₂O we have to convert it into mol/dm³ = ?

Then divide by the relative formula mass

We get  H₂O then H = 1 here 2 hydrogen = 1×2 = 2 and 1 oxygen i.e 1×16 = 16 =

2×16 = 32

So,  7g/dm³/32

= 0.21 mol/dm³

7g/dm³ of H₂O to 0.21 mol/dm³

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If a reaction occurs between K₂2CO₃ and Cu(NO₃)₂ and the equation is properly balanced with the smallest whole-number coefficients, the sum of the coefficients of the reactants and products will be 5.

The reaction between K₂2CO₃ and Cu(NO₃)₂ is represented by the following chemical equation:

K₂2CO₃ and Cu(NO₃)₂ = CuCO₃ + KNO₃

However, the above equation is yet to be balanced. A balanced equation will have the same number of atoms of different elements in the reactants and in the products. The balanced version of the equation is written as:

Cu(NO₃)₂ + K₂CO₃ = --- CuCO₃ + 2KNO₃

The reactants/products with their respective coefficients are as follows:[Cu(NO₃)₂ - 1K₂CO₃ - 1[CuCO₃] - 1[KNO₃ - 2

Thus, the sum of all the coefficients for reactants and products will be:

1 + 1 + 1 + 2 = 5

In other words, if a reaction occurs between K₂CO₃ and Cu(NO₃)₂ and the equation is properly balanced with the smallest whole-number coefficients, the sum of the coefficients of the reactants and products is 5.

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Only III is the correct answer as alkyl halide III allows for an E2 elimination to form the desired alkene.

In order to determine which alkyl halide(s) would give a specific alkene as the only product in an elimination reaction, we need to consider the mechanism of the reaction and the conditions under which it takes place.

Elimination reactions typically involve the removal of a leaving group (usually a halogen) and a proton from adjacent carbons to form a new pi bond. The most common types of elimination reactions are E1 and E2.

In an E1 reaction, the leaving group is first dissociated to form a carbocation, followed by the removal of a proton to form the alkene. In an E2 reaction, the leaving group is removed simultaneously with the deprotonation.

Based on the given information that the elimination product is an alkene, we can deduce that the reaction follows an E2 mechanism since E1 reactions generally lead to carbocation rearrangements and the formation of mixtures of products.

Now, let's analyze the options provided:

A) II and III

B) Only II

C) Only III

D) Only I

Since there is no alkyl halide labeled as "I" in the given options, we can eliminate option D.

For the reaction NH2 (2 equivalents) Br Br, it suggests that two equivalents of ammonia (NH2) are used. This indicates that the reaction is likely to be an E2 reaction, where two molecules of ammonia would act as the base to remove the two bromine atoms.

Based on this analysis, the correct answer is option C) Only III, as the alkyl halide labeled as "III" is the only option that allows for an E2 elimination to occur, leading to the formation of the desired alkene as the only product.

It is important to note that a more comprehensive analysis may be required, considering other factors such as steric hindrance, the presence of different leaving groups, and the strength of the base to make a definitive determination.

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Answer:

a. the 3 represents the principal energy level

Explanation:

3 is the principal energy level. The p is the sublevel. 4 is the possible occupying electron.

Answer:

a is the answer I know and if I'm right mark me brainlest

Answer: C

(no guarantees though)

Motion of particles is not one of the main factors that determine the state of matter.

Matter in chemistry, is defined as any kind of substance that has mass and occupies space that means it has volume .Matter is composed up of atoms which may or not be of same type.

Atoms are further made up of sub atomic particles which are the protons ,neutrons and the electrons .The matter can exist in various states such as solids, liquids and gases depending on the conditions of temperature and pressure.

The states of matter are inter convertible into each other by changing the parameters of temperature and pressure.Matter is  always conserved as per law of conservation of matter.It can be interchanged by varying particle size and  particle forces.

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