The hybridisation of the central atom in ICL-2
a) sp.
b) sp2
c) sp3d d)sp3​

The mass of the asteroid is C. 1.2 x \(10^{12}\) Kg

To find the mass of the other asteroid, we can rearrange the equation for the gravitational force between two objects:

F = (G * m1 * m2) / \(r^{2}\)

where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of the two asteroids, and r is the distance between them.

Given that the distance between the asteroids is 75000 m, the force of gravity between them is 1.14 N, and one asteroid has a mass of 8 x \(10^{7}\) kg, we can substitute these values into the equation and solve for the mass of the other asteroid (m2):

1.14 N = (6.67430 × \(10^{-11}\) N \(m^{2}\)/\(Kg^{2}\) * 8 x \(10^{7}\) kg * \(m2\)) / \((75000 m)^{2}\)

Simplifying and solving the equation, we find that the mass of the other asteroid (m2) is approximately 1.2 x \(10^{12}\) kg. Therefore, Option C is correct.

The question was incomplete. find the full content below:

Two asteroids are 75000 m apart one has a mass of 8 x \(10^{7}\) kg if the force of gravity between them is 1.14 what is the mass of the asteroid

A. 3.4 x \(10^{11}\) kg

B. 8.3 x \(10^{12}\) kg

C. 1.2 x \(10^{12}\) kg

D. 1.2 x \(10^{10}\) kg

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1)The new volume at a pressure of 1 atm is 0.25 L.

2)The new temperature of the container at a volume of 2 L is approximately 398°C.

3)The new pressure at 2.5 L is approximately 4.8 atm.

4)The new volume at a temperature of 60°C is approximately 55.8

1)To solve these gas law problems, we can use the ideal gas law equation, which states:

PV = nRT,

where P is pressure, V is volume, n is the number of moles of gas, R is the ideal gas constant, and T is temperature in Kelvin.

Balloon volume at a pressure of 0.5 atm:\(V_1\) = 0.5 L, \(P_1\)= 0.5 atm.

New volume at a pressure of 1 atm:\(P_2\) = 1 atm.

We can use the relationship\(P_1V_1 = P_2V_2\) to find the new volume (\(V_2\)).

(0.5 atm)(0.5 L) = (1 atm)(\(V_2\))

\(V_2\) = 0.25 L.

Therefore, the new volume at a pressure of 1 atm is 0.25 L.

2)Container volume: \(V_1\) = 2 L, \(T_1\)= 125°C.

New temperature at the same volume: \(V_2\) = 2 L.

We can use the relationship\(V_1\)/\(T_1\) = \(V_2\)/\(T_2\) to find the new temperature (\(T_2\)).

(2 L)/(125 + 273) K = (2 L)/(\(T_2\) + 273) K

Solving for\(T_2\), we get \(T_2\) ≈ 398°C.

Therefore, the new temperature of the container at a volume of 2 L is approximately 398°C.

3)Initial volume: \(V_1\)= 4.0 L, \(P_1\) = 3.0 atm.

Final volume: \(V_2\) = 2.5 L.

Since the temperature (T) is constant, we can use the relationship \(P_1\)\(V_1\) = \(P_2V_2\) to find the new pressure (\(P_2\)).

(3.0 atm)(4.0 L) = (\(P_2\))(2.5 L)

\(P_2\) ≈ 4.8 atm.

Therefore, the new pressure at 2.5 L is approximately 4.8 atm.

4)Initial volume: \(V_1\)= 50 mL, \(T_1\) = 25°C.

New temperature: \(T_2\) = 60°C.

We need to convert the temperatures to Kelvin.

\(T_1\)= 25 + 273 = 298 K, \(T_2\) = 60 + 273 = 333 K.

We can use the relationship \(V_1/T_1 = V_2/T_2\) to find the new volume (\(V_2\)).

(50 mL)/(298 K) = (\(V_2\))/(333 K)

\(V_2\) ≈ 55.8 mL.

Therefore, the new volume at a temperature of 60°C is approximately 55.8

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Answer:

This idea helps students explain why more rain forms over West Ferris than East Ferris. ... Therefore, when students explain that water vapor condenses higher in the atmosphere, they are actually explaining that water vapor condenses high in the troposphere, which is relatively low in the atmosphere.

Explanation:

The paper is soaked and diluted if a paper is left in the container overnight.

To soak something is to submerge it into water

When the hydrogen bonds that hold the cellulose fibers together break apart and weaken, this causes the fibers to separate more easily. Thus, the paper becomes weak and much easier to tear.

Even thick stacks of notebook paper can be torn easily when they're soaking wet.

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Approximately 194.0 grams of glucose (C6H12O6) would be required to prepare a 5000 mL solution with a concentration of 0.215 M.

To determine the mass of glucose (C6H12O6) required to prepare a 0.215 M solution in 5000 mL, we need to use the formula:

Molarity (M) = moles of solute / volume of solution (in liters)

First, let's convert the volume of the solution from milliliters (mL) to liters (L):

5000 mL = 5000/1000 = 5 L

Now, we can rearrange the formula to solve for moles of solute:

moles of solute = Molarity (M) x volume of solution (L)

moles of solute = 0.215 M x 5 Lmoles of solute = 1.075 mol

Since glucose (C6H12O6) has a molar mass of approximately 180.16 g/mol, we can calculate the mass of glucose using the equation:

mass of solute = moles of solute x molar mass of solute

mass of glucose = 1.075 mol x 180.16 g/mol

mass of glucose = 194.0 g (rounded to three significant figures)

Therefore, approximately 194.0 grams of glucose (C6H12O6) would be required to prepare a 5000 mL solution with a concentration of 0.215 M. It's important to note that the molar mass of glucose used in this calculation may vary slightly depending on the level of precision required.

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The substance with the higher freezing point depression is CaCl2.

Water freezes at 0° C in normal conditions but once water is mixed with other substances this temperature can change.

As previously mentioned, the freezing point changes according to the substance water is mixed with, here are the changes for the listed substances:

Based on this, the one that has the greatest freezing point depression is CaCl2.

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The initial concentration of Fe3+ in equilibrium solution 1 is the same as the concentration of the Fe(NO3)3 solution used, which is 0.00200 M. This is because the volume of the solution does not change, so the concentration stays the same. Therefore, the answer is 0.00200.

Here is the step-by-step explanation:
1. Start with the given concentration of the Fe(NO3)3 solution: 0.00200 M
2. Since the volume of the solution does not change (5.00 mL of the original solution is used to make equilibrium solution 1), the concentration of Fe3+ in equilibrium solution 1 is the same as the concentration of the Fe(NO3)3 solution.
3. Therefore, the answer is 0.00200 M, without units or scientific notation, and with the correct number of significant figures.

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Answer:

According to Coulomb’s law, the electric force between two identical negative charges is inversely proportional to the square of the distance between them. This means that as the distance between the two charges decreases, the electric force between them will increase. Since the charges are both negative, they will repel each other, so as they move closer together, the repulsive force between them will become stronger.

Explanation:

Electronegativity is defined as the ability of an element to accept electrons to its shell during a chemical reaction for the formation of a stable bond structure.

A neutral atom is the atom that contains an equal amount of electrons and protons which means that is has a zero net charge. Example is sodium.

Electronegativity is defined as the ability of an element to accept electrons to its shell during a chemical reaction for the formation of a stable bond structure.

During ionic bond formation, the elements involved tries to obtain an octet structure by either donating or accepting electrons.

Alkali metals( such as the neutral element of sodium? have the lowest electronegativities, while halogens have the highest.

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Since HCl is a polar molecule, there will be attraction between the the positive end of the dipole in one molecule and the negative end of the dipole in another molecule.

Polar molecules are those molecules in which there is a high difference in electronegativity between the bonding atoms. This results in the appearance of partial charges in the molecule. One end of the molecule has a partial negative charge and the other end of the molecule has a partial positive charge.

Since HCl is a polar molecule, there will be attraction between the positive end of one molecule and the negative end of another molecule.

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If the reaction yield is 95.7%, then 2.37 grams of lead (II) oxide will be produced by the decomposition of 2.50 grams of lead (II) carbonate.

Answer:

2.54g

Explanation:

Lead carbonate (PbCO3) can be decomposed to produce lead oxide (PbO) and carbon dioxide (CO2). The equation for this reaction is as follows:

PbCO3(s) -> PbO(s) + CO2(g)

Let's assume that the reaction yield is 95.7%, meaning that 95.7% of the theoretical amount of lead oxide produced in the reaction is actually obtained. To find the actual amount of lead oxide produced, we first need to find the theoretical amount of lead oxide produced by the reaction.

The theoretical amount of lead oxide can be calculated using the stoichiometry of the reaction, where the number of moles of reactants is balanced with the number of moles of products. If we assume that 2.50g of lead carbonate is decomposed, the number of moles of lead carbonate can be calculated as follows:

n = m/M

where n is the number of moles, m is the mass of the substance, and M is the molar mass of the substance. For lead carbonate, the molar mass is:

M = 207.19 g/mol

So, the number of moles of lead carbonate is:

n = 2.50 g / 207.19 g/mol = 0.01202 mol

Since the reaction is balanced, the number of moles of lead oxide produced should be equal to the number of moles of lead carbonate. The mass of lead oxide produced can be calculated using the number of moles and the molar mass of lead oxide:

m = n x M

where M = 223.20 g/mol is the molar mass of lead oxide. So, the mass of lead oxide produced is:

m = 0.01202 mol x 223.20 g/mol = 2.68 g

Since the reaction yield is 95.7%, the actual amount of lead oxide produced is:

actual_mass = 0.957 x 2.68 g = 2.54 g

So, approximately 2.54g of lead oxide will be produced by the decomposition of 2.50g of lead carbonate.

The volume of the given gas is 5.35 L.

Here we use the combined gas law viz. P1V1/T1 = P2V2/T2, which is  derived from the ideal gas equation PV = nRT, where

PV = nRT

where,

P = pressure

V = volume

n = no. of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature.

Here,

\($P_1\) = 1.40 atm, \($P_ 2\)= 0.989 atm  

\($V_1\) = 3.70 L, \($V_2\) = ?

\($T_1\) = 12.20°C = (12.20 + 273) K = 285.2 K  

\($T_2\) = 20.80°C = (20.80 + 273) K = 293.8  K

substituting the values in the equation  P1V1/T1 = P2V2/T2, we get,

(1.40 x 3.70)/285.2 = 0.989 x \($V_2\)/293.8

\($V_2\) = (1.40 x 3.70 x 293.8) / (285.2 x 0.989)

\($V_2\) = 1521.884 / 284.211

\($V_2\) =5.35 L

Thus, the volume of the gas is 5.35 L.

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The reduction potentials of two elements provide information about the redox reaction between those elements, specifically regarding the process of oxidation and reduction. Option C

C. The reduction potentials indicate which element will be oxidized and which element will be reduced. In a redox reaction, one element undergoes oxidation, where it loses electrons, and another element undergoes reduction, where it gains electrons.

The reduction potential values of the elements can determine the relative tendencies for oxidation and reduction. The element with a higher reduction potential is more likely to be reduced (gain electrons), while the element with a lower reduction potential is more likely to be oxidized (lose electrons).

A. The reduction potentials also indirectly provide information about whether electrons will be gained or lost by the oxidized atom. Since oxidation involves the loss of electrons, the element with the lower reduction potential, which is more likely to be oxidized, is associated with electron loss.

Conversely, the element with the higher reduction potential, which is more likely to be reduced, is associated with electron gain.

B. The reduction potentials do not directly indicate which element is more commonly found in nature. The abundance or occurrence of elements in nature is unrelated to their reduction potentials.

D. The reduction potentials are not related to the electronegativity difference between the two elements. Electronegativity is a measure of an atom's tendency to attract electrons in a chemical bond and is distinct from reduction potential, which focuses on the tendency to undergo reduction or oxidation in a redox reaction.

Option C

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Answer:

24.0 g C₃H₈

Explanation:

To find the mass of C₃H₈, you need to (1) convert grams CO/H₂ to moles CO/H₂ (via molar mass), then (2) convert moles CO/H₂ to moles C₃H₈ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles C₃H₈ to grams C₃H₈ (via molar mass). It is important to arrange the conversions/ratios in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 3 sig figs to reflect the sig figs in the given values.

Molar Mass (CO): 12.011 g/mol + 15.998 g/mol

Molar Mass (CO): 28.009 g/mol

Molar Mass (H₂): 2(1.008 g/mol)

Molar Mass (H₂): 2.016 g/mol

Molar Mass (C₃H₈): 3(12.011 g/mol) + 8(1.008 g/mol)

Molar Mass (C₃H₈): 44.097 g/mol

3 CO + 7 H₂ ----> 1 C₃H₈ + 3 H₂O
^           ^              ^

45.8 g CO           1 mole              1 mole C₃H₈          44.097 g
-----------------  x  ------------------  x  --------------------  x  ------------------  =
                           28.009 g           3 moles CO            1 mole

=  24.0 g C₃H₈

87.3 g H₂           1 mole           1 mole C₃H₈          44.097 g
----------------  x  ---------------  x  ---------------------  x  -----------------  =  
                          2.016 g           7 moles H₂             1 mole

=  273 g C₃H₈

It was necessary to find the mass of the products from both of the reactants because you did not know which one was the limiting reagent. The limiting reagent is the reactant which is completely used up first. Because CO produced the smaller amount of product, it must be the limiting reagent. Therefore, the actual amount of C₃H₈ produced is 24.0 grams.

The density of a sample of nitrogen gas (N₂) that exerts a pressure of 5.30 atm in a 3.50-L container at 125°C is 4.54 gm/litre

This law combines the relationships between p, V, T and mass, and gives a number to the constant.

The ideal gas law is:

pV = nRT

where n is the number of moles, and R is universal gas constant.

The value of R depends on the units involved, but is usually stated with S.I. units as: R = 8.314 J/mol·K

The values given in the question are

p= 5.30 atm , V=3.50 L

T (in Kelvin) = 125+273 K

                   = 398 K

R = 0.0821 L·atm /mol·K

The Ideal Gas equation can be re written as

pM=DRT

M is the molar mass , D is the density

M for N₂ is 28 gm/mol

so density can be determined as

\(\rm D= \dfrac{p M}{RT} \\\\\rm D= \dfrac{5.3\; \times 28}{0.082 \;\times398} \\\\\rm D =4.54 gm/litres\)

Therefore the density of a sample of nitrogen gas (N₂) that exerts a pressure of 5.30 atm in a 3.50-L container at 125°C is 4.54 gm/litre

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Answer:

1.904 ppm

Explanation:

Concentration of drug in units of ppm = mass of solute / (mass of solution ) × 1000000

mass of blood = density of blood × volume = 1.05 g / ml × 5000 ml = 5250  g

mass of solution = mass of blood + mass of solute ( statin) = 5250 + 0.01 g  = 5250.01 g

Concentration of drug in units of ppm = (0.01 g / 5250.01 g) × 1000000 = 1.904 ppm

I hope this helps!!

Answer:

Explanation:

1+ 2+ 3+

copper(I), Cu+ cadmium, Cd2+ chromium(III), Cr3+

gold(I), Au+ chromium(II), Cr2+ cobalt(III), Co3+

mercury(I), Hg22+ cobalt(II), Co2+ gold(III), Au3+

silver, Ag+ copper(II), Cu2+ iron(III), Fe3+

Answer:Nothing

Explanation:

The answer is nothing the tempatature isnt matched with the degrees this is false

Answer:5.1L

Explanation:

The pH of the solution made by mixing 15.0cm³ 0.1M NaOH and 35.0 cm³ 0.2 M HCOOH would be 2.39.

The first step in solving this problem is to write the balanced chemical equation for the reaction between NaOH and HCOOH:

NaOH + HCOOH → NaCOOH + H2O

Next, we need to calculate the moles of each reactant:

moles NaOH = 0.1 mol/L x 0.015 L = 0.0015 moles

moles HCOOH = 0.2 mol/L x 0.035 L = 0.007 moles

Since NaOH and HCOOH react in a 1:1 stoichiometry, we know that 0.0015 moles of NaOH reacts with 0.0015 moles of HCOOH. This leaves 0.007 - 0.0015 = 0.0055 moles of HCOOH unreacted.

Now, we need to use the equilibrium constant (Ka) for the reaction between HCOOH and water to determine the concentration of H+ ions in the solution:

HCOOH + H2O ⇌ H3O+ + HCOO-

Ka = [H3O+][HCOO-]/[HCOOH]

We are given the value of Ka, so we can use it to find the concentration of H+ ions:

Ka = 1.82 × 10^-4 M

[H3O+] = [HCOO-] = x (let's assume that the initial concentration of HCOOH is much larger than x, so we can assume that x is negligible compared to the initial concentration of HCOOH)

[HCOOH] = 0.0055 moles / 0.05 L = 0.11 M

Ka = (x^2) / (0.11 - x)

Since x is negligible compared to the initial concentration of HCOOH, we can simplify the equation to:

Ka = (x^2) / 0.11

Solving for x, we get:

x = sqrt(Ka * [HCOOH]) = sqrt(1.82 x 10^-4 * 0.11) = 0.0041 M

Therefore, [H+] = 0.0041 M

Finally, we can use the definition of pH to calculate the pH of the solution:

pH = -log[H+] = -log(0.0041) = 2.39

Therefore, the pH of the solution is 2.39.

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Answer:

1) Ag3N

2)Na2S

3)NaHSO4

4) KNO3

Explanation:

We divide each mass by the element's relative atomic mass

1) 2.90/108-Ag, 0.125/14-N

0.027-Ag, 0.0089-N

Divide by the lowest ratio

0.027/0.0089-Ag, 0.0089/0.0089 N

3-Ag, 1-N

Empirical formula- Ag3N

2)2.22/23-Na, 1.55/32-S

0.097-Na, 0.048-S

Divide by the lowest ratio

0.097/0.048-Na, 0.048/0.048-S

2-Na, 1-S

Empirical formula- Na2S

3) 2.11/23-Na, 0.0900/1-H, 2.94/32-S,5.86/16-O

0.09-Na, 0.09-H, 0.09-S,0.366-O

Divide by the lowest ratio

0.09/0.09-Na, 0.09/0.09-H, 0.09/0.09-S, 0.366/0.09-O

1-Na, 1-H, 1-S, 4-O

Empirical formula- NaHSO4

4)1.84/39, 0.657/14-N, 2.25/16-O

0.047-K, 0.047-N, 0.14-O

Divide through by the lowest ratio

0.047/0.047-K, 0.047/0.047-N, 0.14/0.047-O

1-K, 1-N, O-3

Empirical formula- KNO3

The frequency of a photon of yellow light with a wavelength of 4.63 X 10-⁷m is 6.48 × 10¹⁴ Hz.

The frequency of a wave can be calculated by using the following formula:

v = fλ

Where;

According to this question, a photon of yellow light with a wavelength of 4.63 X 10-⁷m is given. The frequency of this photon can be calculated as follows:

f = 3 × 10⁸m/s ÷ 4.63 X 10-⁷m

f = 6.48 × 10¹⁴ Hz

Therefore, the frequency of a photon of yellow light with a wavelength of 4.63 X 10-⁷m is 6.48 × 10¹⁴ Hz.

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The answer for this question is C.

According to the question for a solution with a pH of 2.68, [H+] = 10-2.68 = 0.00145 M.

pH is a measure of the acidity or alkalinity of a solution. It is measured on a scale of 0 to 14, with 0 being the most acidic, 7 being neutral, and 14 being the most alkaline. The pH of a solution is determined by the amount of hydrogen ions present in the solution. Solutions with a higher concentration of hydrogen ions are more acidic, while solutions with a lower concentration of hydrogen ions are more alkaline. pH is an important factor in determining the solubility of substances, as well as the rate of chemical reactions. It is also important in controlling the growth of microorganisms, as some organisms require a certain pH to survive.

The [H+] concentration can be determined by using the pH equation, which states that pH = -log[H+]. Rearranging this equation to solve for [H+], we get [H+] = 10-pH.
Therefore, for a solution with a pH of 2.68, [H+] = 10-2.68 = 0.00145 M.

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