Write a balanced complete ionic equation for: hi(aq)+rboh(aq)→

This is a subtraction of expression. You should remember the rules in subtracting or adding expression. You can only add or subtract two terms if they are similar terms which mean they have the same variable with the same exponents.

(7x – 10) – (10x + 9)

You have to distribute the negative to the parenthesis.

7x – 10 – 10x – 9

7x – 10x – 10 – 9

-3x - 19

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: Option (A) is the correct answer.

Explanation:

When Sam placed ice in a pot on a stove then there will be transfer of heat from the stove to the pot and then from the pot to the ice.

As a result, solid state of ice changes into liquid state of water because of melting of ice.

Thus, we can conclude that the heat from the stove is the variable that is causing water to change state.

Final answer:

The balanced net ionic equation for the neutralization reaction involving equal molar amounts of HNO3 and KOH is H+(aq) + OH−(aq) → H2O(l), illustrating the direct interaction between hydrogen and hydroxide ions to form water.

Explanation:

The balanced net ionic equation for the neutralization reaction involving equal molar amounts of HNO3 and KOH is represented by the equation:

H+(aq) + OH−(aq) → H2O(l)

This equation demonstrates that the hydrogen ion (H+) from nitric acid (HNO3) combines with the hydroxide ion (OH−) from potassium hydroxide (KOH) to form water (H2O). The potassium ion (K+) and the nitrate ion (NO3−) are spectator ions and do not participate in the reaction. Therefore, they are not included in the net ionic equation. Neutralization reactions like this involve the combination of an acid and a base to produce water as one of the products.

Answer :

(B) It is likely to form ionic bonds

(D) It is in period seven

(E) It is a metal

Explanation :

Francium is an element whose symbol is 'Fr' and atomic number is 87.

Group 1 alkali metals consists elements Lithium, Sodium, Potassium, Rubidium, Caesium and Francium. Periodic table also shown below.

Francium belongs to the group 1 (alkali metal) and it is in period seven.

It has one valence electron, the electronic configuration of Francium is [Rn} [tex]7s^{1}[/tex].

Alkali metals have tendency to form ionic compounds because they have +1 charge.

The correct statements about Francium (Fr, atomic number 87) are that it is likely to form ionic bonds, is in period seven, and is a metal.

The element Francium (Fr, atomic number 87) has several characteristics based on its position in the periodic table. Firstly, as an element in Group 1, it has one valence electron which makes statement A incorrect. This single valence electron also means Francium is highly probable to form ionic bonds with nonmetals seeking to gain electrons, making statement B true. Francium is not a nonmetal, so statement C is false. Statement D is correct as Francium is located in period seven of the periodic table. Lastly, because Francium is in Group 1, it is indeed an alkali metal, making statement E true. Therefore, the correct statements about Francium are B, D, and E.

Differentiating is a process of loosing or separating. During this stage, variations between the relationship associates are highlighted and what was considered to be connections starts to crumble. The people avoid discussing the relationship because they believe they understand what the other will respond.

The oxidizing agent in the reaction is the NO3- ions.

In the given reaction,

8H+(aq) + 6Cl-(aq) + Sn(s) + 4NO3-(aq) → SnCl62-(aq) + 4NO2(g) + 4H2O(l)

The oxidizing agent is the species that gets reduced. In this reaction, the NO3- ions are reduced from a +5 oxidation state to a +4 oxidation state, so the NO3- ions are the oxidizing agent.

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In the given reaction, the nitrate ion (NO₃⁻) is the oxidizing agent as it gets reduced from an oxidation number of +5 to +4. The tin (Sn) is oxidized from 0 to +4.

The identification is based on changes in oxidation numbers during the reaction.

To identify the oxidizing agent in the given reaction:

Assign oxidation numbers to each element in the reaction:

Determine what is being oxidized and reduced:

Identify the oxidizing and reducing agents:

So, in this reaction, NO₃⁻ (nitrate ion) is the oxidizing agent.

Correct question is: Identify the oxidizing agent in the given reaction:
8H⁺(aq) + 6Cl⁻ (aq) + Sn(s) + 4NO₃⁻(aq) → SnCl₆²⁻(aq) + 4NO₂(g) + 4H₂O(l)

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 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

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: The correct answer is Bohr's atomic model.

Explanation:

For the given options:

Dalton's atomic model: This model states that every matter is made up of smallest unit known as atom.

Thomson's atomic model: He proposed a model known as plum pudding model. He considered atom to be a pudding of positive charge in which negative particles are embedded such as plum.

Rutherford's atomic model: He gave an experiment known as gold foil experiment. In his model, he concluded that in an atom, there exist a small positive charge in the center.

Bohr's atomic model: This model states that electron revolve around the nucleus in discrete orbits in an atom.

Quantum atomic model: This model determines the location of electrons in an atom in a 3-D space.

Hence, the correct answer is Bohr's atomic model.

c.unfavorable geometry

Answer:  C.  Unfavorable geometry

Explanation: When collision occurs between two reactants in order to make a reaction possible there are 3 factors which are responsible.

a) Orientation factor

b) Energy factor

c) rate of collision

Thus out of the given options, unfavorable geometry is the correct one as temperature and concentration as well as surface area will have very little effect on the reaction.

If the geometry of the reactant is not complementary then the reaction would not lead successfully.

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

Explanation: Apex

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

Answer:- [tex][H_3O^+]=4.2*10^-^1^1[/tex]

Solution:- Ammonia is a weak base. So, to calculate the hydroxide ion concentration we make the ice table:

   [tex]NH_3(aq)+H_2O(l)\rightleftharpoons NH_4^+(aq)+OH^-(aq)[/tex]

I            0.0032                                                    0                  0

C             -X                                                          +X               +X

E      (0.0032-X)                                                    X                 X

[tex]K_b=\frac{[NH_4^+][OH^-]}{[NH_3]}[/tex]

Kb value for ammonia is [tex]1.8*10^-^5[/tex] . Let's plug in the values and solve for X.

[tex]1.8*10^-^5=\frac{X^2}{0.0032-X}[/tex]

Kb value is very low so we can neglect the X on the bottom.

[tex]1.8*10^-^5=\frac{X^2}{0.0032}[/tex]

On cross multiply:

[tex]X^2=1.8*10^-^5*0.0032[/tex]

[tex]X^2=5.76*10^-^8[/tex]

On taking square root:

[tex]X=2.4*10^-^4[/tex]

From ice table, [tex][OH^-]]=X[/tex]

So, [tex][OH^-]=2.4*10^-^4[/tex]

hydronium ion and hydroxide ion concentrations are related to each other by the formula:

[tex][H_3O^+][OH^-]=K_w[/tex]

where, Kw is the water dissociation constant and its value is [tex]1.0*10^-^1^4[/tex]

[tex][H_3O^+]=\frac{1.0*10^-^1^4}{2.4*10^-^4}[/tex]

[tex][H_3O^+]=4.2*10^-^1^1[/tex]

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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The molality of the solution is 0.41 molal and the correct option is option 1.

Molality is also known as molal concentration. It is a measure of solute concentration in a solution. The solution is composed of two components; solute and solvent.

The number of moles of solute in a solution corresponding to 1 kg or 1000 g of solvent is known as molality.

Molality = number of moles of solute ÷ mass of solvent in kg

Given,

Mass of AgClO₄ = 75.2g

Mass of benzene = 885g

Molality = number of moles of solute ÷ mass of solvent in kg

moles of AgClO₄ = 75.2 / 207

                              = 0.363 moles

Molality = 0.363 / 0.885

              = 0.41 molal

Therefore, the molality of the solution is 0.41 molal and the correct option is option 1.

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Answer:

c

Explanation:

Answer:

[tex]Keq=\frac{[O_{2}]^{3} }{[H_{2}O]^{2} }[/tex]

Explanation:

The equilibrium constant (Keq) is equal to the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. In Keq we include gases and aqueous species, but not solids nor pure liquids because their concentration remains almost constant over time.

Let's consider the following reaction.

4 KO₂(s) + 2 H₂O(g) ⇄ 4 KOH(s) + 3 O₂(g)

Then, the equilibrium constant is:

[tex]Keq=\frac{[O_{2}]^{3} }{[H_{2}O]^{2} }[/tex]

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