Magnesium carbonate decomposition on heating to form magnesium oxide and carbon dioxide. how many grams of magnesium oxide will be formed if 20 grams of magnesium

Answer:

d) only one parent and results in offspring that are identical to the parent.

Explanation:

just did the study island

Final answer:

Asexual reproduction requires only one parent and gives rise to genetically identical offspring, seen in processes like binary fission.

Explanation:

Asexual reproduction involves only one parent and results in offspring that are genetically identical to the parent. This method of reproduction is seen in all prokaryotes and some eukaryotes, through various mechanisms such as binary fission, fragmentation, and budding.

In binary fission, a common form of asexual reproduction, the parent cell divides to form two identical daughter cells.

Asexual reproduction requires only one parent and gives rise to genetically identical offspring, seen in processes like binary fission.

An object traveling around another body in space is in orbit. This balance between inertia and gravitational pull defines the path and shape of the orbit, which can be circular or elliptical.

An object that is traveling around another body in space is in orbit around that body. An orbit is a regular, repeating path that one object in space takes around another one. An object in orbit is in a delicate balance between the inertia of its motion in a straight line and the gravitational pull that the other body exerts on it. Familiar examples include the Earth orbiting the Sun or the Moon orbiting Earth. The shape of an orbit can be circular or elliptical, and the mechanics of orbits are governed by the laws of physics, particularly by Newton's Law of Universal Gravitation and his laws of motion.

Answer: Option (a) is the correct answer.

Explanation:

According to the cross multiplication method, charges present on the combining atoms cross multiply with the atoms in order to keep the molecule formed neutral.

For example, oxidation number of lithium is +1 and oxidation number of oxygen is -2.

So, when both of them chemically combine with each other then lithium atom will get multiplied by 2 whereas oxygen atom will get multiplied by 1.

Therefore, we can conclude that formula of this chemical compound will be [tex]Li_{2}O[/tex].

A Characteristic is the unique assets, talents, and qualities of a person. On the other hand, a trait is a feature of a person that is hereditary or inherited from parents or offspring. The characteristics are usually something that a person can learn and built by absorption, a trait, on the other hand, is something you have since birth.

Answer:

the person on top is correct

Explanation:

The amount of Rn produced per day by 9.64 g of Ra is 1.13 [tex]\rm \times\;10^-^8[/tex] L.

Moles of Radon (Ra) in 9.64 g of Ra are;

Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]

Moles of Ra = [tex]\rm \dfrac{9.64}{226}[/tex]

Moles of Ra = 0.0426

Number of atoms of Ra in 0.0426 moles of Ra are:

Number of atoms = moles [tex]\times[/tex] Avagadro number

= 0.426 [tex]\times\;6.203\;\times\;10^2^3[/tex]

number of atoms of Ra produced by 9.64 g of Ra are 2.56 [tex]\rm \times\;10^2^3[/tex]

the proportion of atoms : atom\sec will be:

[tex]\rm \dfrac{atoms\;of\;Ra}{atoms\;of\;Ra/sec}\;=\;\dfrac{atoms\;of\;Rn}{x}[/tex]

[tex]\rm \dfrac{1\;\times\;10^1^5}{1.373\;\times\;10^4}\;=\;\dfrac{2.56\;\times\;10^2^2}{x}[/tex]

Atoms of Rn produced per second are 3.53 [tex]\rn \times\;10^1^0[/tex] atoms/sec.

Moles of Rn produced = [tex]\rm \dfrac{atoms\;per\;sec}{avagadro\;number}[/tex]

Moles of Rn produced = [tex]\rm \dfrac{3.52\;\times\;10^1^0}{6.023\;\times\;10^2^3}[/tex] moles

Moles of Rn produced = 5.85 [tex]\rm \times\;10^-^1^4[/tex]  mol/sec.

From the ideal gas equation,

PV = nRT

The volume of Rn produced = [tex]\rm \dfrac{nRT}{P}[/tex]

= [tex]\rm \dfrac{5.85\;\times\;10^-^1^4\;\times\;0.0821\;\times\;273.15}{1}[/tex]

= 1.31 [tex]\rm \times\;10^-^1^3[/tex] liter/sec.

The amount of Rn produced per day = amount produced per second [tex]\times[/tex] 3600

The amount of Rn produced per day = 1.31 [tex]\rm \times\;10^-^1^3[/tex] [tex]\times[/tex] 24 [tex]\times[/tex] 3600 L

The amount of Rn produced per day by 9.64 g of Ra is 1.13 [tex]\rm \times\;10^-^8[/tex] L.

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The Lewis structure of HClO consists of bonds from Cl to each of H and O, with Cl having 5 pairs of non-bonding electrons and O having 2 pairs.

To create the Lewis structure for HCLO, start by counting valence electrons. Hydrogen (H) has 1, Oxygen (O) has 6, and Chlorine (Cl) has 7. This totals to 14. Next, choose the atom with the least electronegativity (Cl in this case) to be the central atom, and draw bonds to other elements. A bond to each H and O from Cl uses up 4 electrons, leaving 10. Now fill the octets of the H and O atoms.

The Lewis structure should then look like this:

H - Cl - O

With 5 pairs of non-bonding electrons around Cl and two pairs of non bonding electrons around O.

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Answer: The number of moles of sodium phosphate required is 0.33 moles

Explanation:

We are given:

Moles of silver nitrate = 1 mole

The chemical equation for the reaction of sodium phosphate and silver nitrate follows:

[tex]Na_3PO_4+3AgNO_3\rightarrow 3NaNO_3+Ag_3PO_4[/tex]

By Stoichiometry of the reaction:

3 moles of silver nitrate reacts with 1 mole of sodium phosphate.

So, 1.0 moles of silver nitrate will react with = [tex]\frac{1}{3}\times 1.0=0.33mol[/tex] of sodium phosphate.

Hence, the number of moles of sodium phosphate required is 0.33 moles

Answer : The freezing point of a solution is [tex]-0.454^oC[/tex]

Explanation :  Given,

Molal-freezing-point-depression constant [tex](K_f)[/tex] for water = [tex]1.86^oC/m[/tex]

Mass of KCl (solute) = 15 g

Mass of water (solvent) = 1650.0 g  = 1.650 kg

Molar mass of KCl = 74.5 g/mole

Formula used :  

[tex]\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of KCl}}{\text{Molar mass of KCl}\times \text{Mass of water in Kg}}[/tex]

where,

[tex]\Delta T_f[/tex] = change in freezing point

[tex]\Delta T_s[/tex] = freezing point of solution = ?

[tex]\Delta T^o[/tex] = freezing point of water = [tex]0^oC[/tex]

i = Van't Hoff factor = 2  (for KCl electrolyte)

[tex]K_f[/tex] = freezing point constant for water = [tex]1.86^oC/m[/tex]

m = molality

Now put all the given values in this formula, we get

[tex]0^oC-T_s=2\times (1.86^oC/m)\times \frac{15g}{74.5g/mol\times 1.650kg}[/tex]

[tex]T_s=-0.454^oC[/tex]

Therefore, the freezing point of a solution is [tex]-0.454^oC[/tex]

Answer:

The freezing point of the solution containing 15 grams of KCl and 1650 grams of water is [tex]\rm -0.454^\circ\;C[/tex].

Explanation:

The freezing point of the solution can be calculated by:

Molarity = [tex]\rm \frac{mass\;of\;KCl\;}{molar\;mass\;of\;KCl\;\times\;Mass\;of\;water}[/tex]

Molarity = [tex]\rm \frac{15}{74.5;\times\;1650}[/tex]

Molarity = 1.2 [tex]\rm \times\;10^-^4[/tex] M

[tex]\rm \Delta\;T\;=\;i\;\times\;K_f\;\times\;Molarity[/tex]

[tex]\rm T^\circ\;\times\;T_f\;=\;i\;\times\;K_f\;\times\;Molarity[/tex]

[tex]\rm T_f=\;2\;\times\;1.86\;\times\;1.2\;\times\;10^-^4[/tex]

[tex]\rm T_f[/tex] = [tex]\rm -0.454^\circ\;C[/tex]

The freezing point of the solution containing KCl is [tex]\rm -0.454^\circ\;C[/tex].

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Answer: A change in a substance where one or more new substances form

Explanation: Apex

Q> K hence the reaction proceeds in the reverse direction.

First we must obtain the concentration of each of the species;

NO2 - 0.053 mol/2.35-l = 0.023 M

N2O4 - 0.084 mol/2.35-l = 0.036 M

Hence;

Q = [N204]/[NO2]^2

Q = [0.036]/[0.023]^2

Q = 68.05

Since Q > K, it follows that the reaction moves towards the left hand side.

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The reaction quotient Q was found to be 69.86, which is greater than the equilibrium constant Kp of 6.7. Therefore, the reaction will shift towards the formation of NO2 (g) to reach equilibrium.

To determine whether the reaction is at equilibrium, we need to calculate the reaction quotient (Qp) and compare it with the given equilibrium constant (Kp). The reaction we are analyzing is:

2 NO2 (g) ⇌ N2O4 (g), with Kp = 6.7 at 298 K.

The reaction quotient Qp is defined similarly to Kp but for an initial or non-equilibrium state and is calculated using partial pressures. Since we are given the number of moles and the volume of the container, we can use the ideal gas law to find the partial pressures of NO2 and N2O4. However, the student question provides moles instead of partial pressures, so we will assume ideal gas behavior and calculate the reaction quotient (Q) with concentrations, which can be used in place of Qp for our rough estimation, knowing they are related in the circumstance of gasses at same temperature and pressure.

The molar concentrations are:

Now we calculate the reaction quotient Q using the given concentrations:

Q = [N2O4]1 / [NO2]2
  = 0.0357 / (0.0226)2
  = 0.0357 / 0.000511
  = 69.86

Since Q (69.86) is greater than Kp (6.7), the reaction will proceed in the direction that reduces Q to equal Kp, which is towards the formation of NO2 from N2O4.

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The molarity of calcium in maple syrup is 33.4 M.

To find the molarity of calcium in maple syrup, we need to first convert the given amount of calcium from milligrams to grams. Since there are 1000 milligrams in a gram, 67 mg is equal to 0.067 g. Next, we divide the mass of calcium by the density of maple syrup to find the volume of maple syrup.

Volume of maple syrup = mass of calcium / density of maple syrup = 0.067 g / 1.325 g/ml = 0.0506 ml

Now, we can calculate the molarity of calcium (Ca2+) in maple syrup by dividing the moles of calcium by the volume of maple syrup in liters:

Molarity = moles of calcium / volume of maple syrup = (0.067 g / 40.08 g/mol) / (0.0506 ml / 1000 ml/L) = 33.4 M

When the pH of a solution changes from 5 to 3, the hydronium ion concentration increases significantly.

PH of Solution:

The relationship between pH and hydronium ion concentration

[H₃O⁺] is given by the formula:

pH = -log [H₃O⁺]

To find the hydronium ion concentrations at pH 5 and pH 3:

At pH 5:  [H₃O⁺] = 10⁻⁵ M

At pH 3:  [H₃O⁺] = 10⁻³ M

Thus, the hydronium ion concentration increases by a factor of 100 when the pH changes from 5 to 3

By using the equation q = nΔH, where 'q' is the heat evolved or absorbed, 'n' is the number of moles, and 'ΔH' is the enthalpy change per mole, it is calculated that 5.0 kJ of heat is evolved when 0.150 mol of HNO3 is dissolved in 100.0 ml of water.

The amount of heat released or absorbed in a reaction is typically calculated using the equation q = nΔH, where 'q' is the heat evolved or absorbed, 'n' is the number of moles, and 'ΔH' is the enthalpy change per mole. Given that the enthalpy of dissolving HNO3, ΔH°soln, is -33.3 kJ/mol, and we are dissolving 0.150 mol HNO3, we can substitute these values into the equation.

So, q = nΔH = 0.150 mol * -33.3 kJ/mol = -5.0 kJ.

The negative sign indicates that the heat is evolved (released), as per the question. Thus, 5.0 kJ of heat is evolved by dissolving 0.150 mol HNO3 in 100.0 ml of water.

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When 0.150 mol of HNO3 is dissolved in 100.0 ml of water, 5 kJ of heat is released. This is calculated using the equation q = n x δh°soln over the negative valence of the enthalpy change -33.3 KJ/mol.

The amount of heat evolved from dissolving a substance in solution can be found using the equation q = n x δh°soln, where q is the heat evolved, n is the number of moles, and δh°soln is the enthalpy change per mole of solute. In this case, you have been given the enthalpy change per mole of HNO3 (δh°soln) as -33.3 kJ/mol, and the number of moles (n) as 0.150 mol. Plugging these into the equation gives q = 0.150 mol x -33.3 kJ/mol = -5 kj. This means that 5 kJ of heat is released when 0.150 mol of HNO3 is dissolved in 100.0 ml of water.

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The product of the reaction is 3-ethylpentan-3-ol .

The Grignard reagent is a compound that is composed of an alkyl magnesium halide. Hence the general formula of a Grignard reagent is RMgX.

Since we are using two equivalents of ch3ch2mgbr and then is treat with water, the product compound is 3-ethylpentan-3-ol as shown in the image attached to this answer.

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In the presence of AlCl3, excess benzene reacts with dichloromethane to undergo a Friedel-Crafts alkylation, resulting in the product 1,2,4-trichloro-5,5-diphenylpentane.

When dichloromethane is treated with excess benzene in the presence of an aluminum chloride catalyst (AlCl3), a Friedel-Crafts alkylation occurs. This is a type of electrophilic aromatic substitution reaction that introduces an alkyl group into the benzene ring. Here, the final product would be a dichloromethylated benzene molecule.

However, considering biselective alkylation resulting from the use of 2 equivalents of benzene, the product would be 1,2,4-trichloro-5,5-diphenylpentane. The reaction would be as follows: 2 equivalents of Benzene + CH2Cl2 (Dichloromethane) ->[AlCl3 catalyst] 1,2,4-trichloro-5,5-diphenyl pentane

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100mL of  [tex]\rm \bold { Fe_2Cl_3}[/tex] is needed to react with an excess of [tex]\rm \bold {Na_2 S}[/tex] to produce 1.38 g of [tex]\rm \bold { Fe_2S_3}[/tex] .

The reaction is

[tex]\rm \bold{3 Na_2S(aq) + 2 FeCl_2(aq) \rightarrow Fe_2S_3(s) + 6 NaCl(aq)}[/tex]

The mass of [tex]\rm \bold { Fe_2S_3}[/tex] is 1.38 g

The number of moles of [tex]\rm \bold { Fe_2S_3}[/tex] will be 0.00664 moles.

If the if the percent yield for the reaction is 65.0% or 0.65.

[tex]\rm \bold { Theoritical Yield = \frac{0.00664}{0.65} }\\\\\rm \bold { Theoritical Yield = 0.01 Moles}[/tex]

Since 2 mole of [tex]\rm \bold { Fe_2Cl_3}[/tex] is needed to produce 1 mole of [tex]\rm \bold { Fe_2S_3}[/tex].

[tex]\rm \bold { Fe_2Cl_3}[/tex] concentration is 0.020 mol

To find the from molarity formula

[tex]\rm \bold { V = \frac{n}{M} = \frac{0.020 mol}{ 0.2 M} / = 0.1 liter = 100 ml}\\\\\rm \bold { V = \frac{0.020 mol}{ 0.2 M} }\\\\\rm \bold { V = 0.1 liter = 100 ml}[/tex]

Hence we can conclude that  100mL of  [tex]\rm \bold { Fe_2Cl_3}[/tex] is needed to react with an excess of [tex]\rm \bold {Na_2 S}[/tex] to produce 1.38 g of [tex]\rm \bold { Fe_2S_3}[/tex] .

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Answer : The number of moles of water produced would be, 32 moles.

Explanation : Given,

Moles of [tex]O_2[/tex] = 16 mole

The given chemical reaction is:

[tex]2H_2+O_2\rightarrow 2H_2O[/tex]

From the balanced chemical reaction we conclude that,

As, 1 mole of [tex]O_2[/tex] gas react to give 2 moles of [tex]H_2O[/tex]

So, 16 mole of [tex]O_2[/tex] gas react to give [tex]16\times 2=32[/tex] moles of [tex]H_2O[/tex]

Thus, the number of moles of water produced would be, 32 moles.

Answer:

c

Explanation:

Answer is: concentration of the barium ion is 0.245 M.

Chemical reaction: BaF₂ → Ba²⁺ + 2F⁻.

[F⁻] = 1.00·10⁻² M.

Ksp = 2.45·10⁻⁵.

Ksp = [Ba²⁺] · [F⁻]².

[Ba²⁺] = Ksp ÷ [F⁻]².

[Ba²⁺] = 2.45·10⁻⁵ ÷ (1.00·10⁻² M)².

[Ba²⁺] = 0.245 M.

Answer: The concentration of barium ions that must exceed to precipitate the salt is 0.245 M

Explanation:

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio. It is represented as [tex]K_{sp}[/tex]

Barium fluoride is an ionic compound formed by the combination of 1 barium ion and 2 fluoride ions.

The equilibrium reaction for the ionization of barium fluoride follows the equation:

[tex]BaF_2(s)\rightleftharpoons Ba^{2+}(aq.)+2F^-(aq.)[/tex]

The solubility product for the above reaction is:

[tex]K_{sp}=[Ba^{2+}]\times [F^-]^2[/tex]

We are given:

[tex][F^-]=1.00\times 10^{-2}M\\\\K_{sp}=2.45\times 10^{-5}[/tex]

Putting values in above equation, we get:

[tex]2.45\times 10^{-5}=[Ba^{2+}]\times (1.00\times 10^{-2})^2[/tex]

[tex][Ba^{2+}]=\frac{2.45\times 10^{-5}}{(1.00\times 10^{-2})^2}=0.245M[/tex]

Hence, the concentration of barium ions that must exceed to precipitate the salt is 0.245 M