Search

HBr + Sc(OH)3 = H2O + ScBr3

Input interpretation

HBr hydrogen bromide + Sc(OH)_3 scandium hydroxide ⟶ H_2O water + ScBr_3 scandium tribromide
HBr hydrogen bromide + Sc(OH)_3 scandium hydroxide ⟶ H_2O water + ScBr_3 scandium tribromide

Balanced equation

Balance the chemical equation algebraically: HBr + Sc(OH)_3 ⟶ H_2O + ScBr_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HBr + c_2 Sc(OH)_3 ⟶ c_3 H_2O + c_4 ScBr_3 Set the number of atoms in the reactants equal to the number of atoms in the products for Br, H, O and Sc: Br: | c_1 = 3 c_4 H: | c_1 + 3 c_2 = 2 c_3 O: | 3 c_2 = c_3 Sc: | c_2 = c_4 Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 3 c_2 = 1 c_3 = 3 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 HBr + Sc(OH)_3 ⟶ 3 H_2O + ScBr_3
Balance the chemical equation algebraically: HBr + Sc(OH)_3 ⟶ H_2O + ScBr_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HBr + c_2 Sc(OH)_3 ⟶ c_3 H_2O + c_4 ScBr_3 Set the number of atoms in the reactants equal to the number of atoms in the products for Br, H, O and Sc: Br: | c_1 = 3 c_4 H: | c_1 + 3 c_2 = 2 c_3 O: | 3 c_2 = c_3 Sc: | c_2 = c_4 Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 3 c_2 = 1 c_3 = 3 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 HBr + Sc(OH)_3 ⟶ 3 H_2O + ScBr_3

Structures

+ ⟶ +
+ ⟶ +

Names

hydrogen bromide + scandium hydroxide ⟶ water + scandium tribromide
hydrogen bromide + scandium hydroxide ⟶ water + scandium tribromide

Equilibrium constant

Construct the equilibrium constant, K, expression for: HBr + Sc(OH)_3 ⟶ H_2O + ScBr_3 Plan: • Balance the chemical equation. • Determine the stoichiometric numbers. • Assemble the activity expression for each chemical species. • Use the activity expressions to build the equilibrium constant expression. Write the balanced chemical equation: 3 HBr + Sc(OH)_3 ⟶ 3 H_2O + ScBr_3 Assign stoichiometric numbers, ν_i, using the stoichiometric coefficients, c_i, from the balanced chemical equation in the following manner: ν_i = -c_i for reactants and ν_i = c_i for products: chemical species | c_i | ν_i HBr | 3 | -3 Sc(OH)_3 | 1 | -1 H_2O | 3 | 3 ScBr_3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HBr | 3 | -3 | ([HBr])^(-3) Sc(OH)_3 | 1 | -1 | ([Sc(OH)3])^(-1) H_2O | 3 | 3 | ([H2O])^3 ScBr_3 | 1 | 1 | [ScBr3] The equilibrium constant symbol in the concentration basis is: K_c Mulitply the activity expressions to arrive at the K_c expression: Answer: | | K_c = ([HBr])^(-3) ([Sc(OH)3])^(-1) ([H2O])^3 [ScBr3] = (([H2O])^3 [ScBr3])/(([HBr])^3 [Sc(OH)3])
Construct the equilibrium constant, K, expression for: HBr + Sc(OH)_3 ⟶ H_2O + ScBr_3 Plan: • Balance the chemical equation. • Determine the stoichiometric numbers. • Assemble the activity expression for each chemical species. • Use the activity expressions to build the equilibrium constant expression. Write the balanced chemical equation: 3 HBr + Sc(OH)_3 ⟶ 3 H_2O + ScBr_3 Assign stoichiometric numbers, ν_i, using the stoichiometric coefficients, c_i, from the balanced chemical equation in the following manner: ν_i = -c_i for reactants and ν_i = c_i for products: chemical species | c_i | ν_i HBr | 3 | -3 Sc(OH)_3 | 1 | -1 H_2O | 3 | 3 ScBr_3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HBr | 3 | -3 | ([HBr])^(-3) Sc(OH)_3 | 1 | -1 | ([Sc(OH)3])^(-1) H_2O | 3 | 3 | ([H2O])^3 ScBr_3 | 1 | 1 | [ScBr3] The equilibrium constant symbol in the concentration basis is: K_c Mulitply the activity expressions to arrive at the K_c expression: Answer: | | K_c = ([HBr])^(-3) ([Sc(OH)3])^(-1) ([H2O])^3 [ScBr3] = (([H2O])^3 [ScBr3])/(([HBr])^3 [Sc(OH)3])

Rate of reaction

Construct the rate of reaction expression for: HBr + Sc(OH)_3 ⟶ H_2O + ScBr_3 Plan: • Balance the chemical equation. • Determine the stoichiometric numbers. • Assemble the rate term for each chemical species. • Write the rate of reaction expression. Write the balanced chemical equation: 3 HBr + Sc(OH)_3 ⟶ 3 H_2O + ScBr_3 Assign stoichiometric numbers, ν_i, using the stoichiometric coefficients, c_i, from the balanced chemical equation in the following manner: ν_i = -c_i for reactants and ν_i = c_i for products: chemical species | c_i | ν_i HBr | 3 | -3 Sc(OH)_3 | 1 | -1 H_2O | 3 | 3 ScBr_3 | 1 | 1 The rate term for each chemical species, B_i, is 1/ν_i(Δ[B_i])/(Δt) where [B_i] is the amount concentration and t is time: chemical species | c_i | ν_i | rate term HBr | 3 | -3 | -1/3 (Δ[HBr])/(Δt) Sc(OH)_3 | 1 | -1 | -(Δ[Sc(OH)3])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) ScBr_3 | 1 | 1 | (Δ[ScBr3])/(Δt) (for infinitesimal rate of change, replace Δ with d) Set the rate terms equal to each other to arrive at the rate expression: Answer: | | rate = -1/3 (Δ[HBr])/(Δt) = -(Δ[Sc(OH)3])/(Δt) = 1/3 (Δ[H2O])/(Δt) = (Δ[ScBr3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: HBr + Sc(OH)_3 ⟶ H_2O + ScBr_3 Plan: • Balance the chemical equation. • Determine the stoichiometric numbers. • Assemble the rate term for each chemical species. • Write the rate of reaction expression. Write the balanced chemical equation: 3 HBr + Sc(OH)_3 ⟶ 3 H_2O + ScBr_3 Assign stoichiometric numbers, ν_i, using the stoichiometric coefficients, c_i, from the balanced chemical equation in the following manner: ν_i = -c_i for reactants and ν_i = c_i for products: chemical species | c_i | ν_i HBr | 3 | -3 Sc(OH)_3 | 1 | -1 H_2O | 3 | 3 ScBr_3 | 1 | 1 The rate term for each chemical species, B_i, is 1/ν_i(Δ[B_i])/(Δt) where [B_i] is the amount concentration and t is time: chemical species | c_i | ν_i | rate term HBr | 3 | -3 | -1/3 (Δ[HBr])/(Δt) Sc(OH)_3 | 1 | -1 | -(Δ[Sc(OH)3])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) ScBr_3 | 1 | 1 | (Δ[ScBr3])/(Δt) (for infinitesimal rate of change, replace Δ with d) Set the rate terms equal to each other to arrive at the rate expression: Answer: | | rate = -1/3 (Δ[HBr])/(Δt) = -(Δ[Sc(OH)3])/(Δt) = 1/3 (Δ[H2O])/(Δt) = (Δ[ScBr3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| hydrogen bromide | scandium hydroxide | water | scandium tribromide formula | HBr | Sc(OH)_3 | H_2O | ScBr_3 Hill formula | BrH | H_3O_3Sc | H_2O | Br_3Sc_1 name | hydrogen bromide | scandium hydroxide | water | scandium tribromide IUPAC name | hydrogen bromide | scandium(+3) cation trihydroxide | water |
| hydrogen bromide | scandium hydroxide | water | scandium tribromide formula | HBr | Sc(OH)_3 | H_2O | ScBr_3 Hill formula | BrH | H_3O_3Sc | H_2O | Br_3Sc_1 name | hydrogen bromide | scandium hydroxide | water | scandium tribromide IUPAC name | hydrogen bromide | scandium(+3) cation trihydroxide | water |

Substance properties

| hydrogen bromide | scandium hydroxide | water | scandium tribromide molar mass | 80.912 g/mol | 95.977 g/mol | 18.015 g/mol | 284.67 g/mol phase | gas (at STP) | | liquid (at STP) | solid (at STP) melting point | -86.8 °C | | 0 °C | 969 °C boiling point | -66.38 °C | | 99.9839 °C | density | 0.003307 g/cm^3 (at 25 °C) | | 1 g/cm^3 | 3.91 g/cm^3 solubility in water | miscible | | | soluble surface tension | 0.0271 N/m | | 0.0728 N/m | dynamic viscosity | 8.4×10^-4 Pa s (at -75 °C) | | 8.9×10^-4 Pa s (at 25 °C) | odor | | | odorless |
| hydrogen bromide | scandium hydroxide | water | scandium tribromide molar mass | 80.912 g/mol | 95.977 g/mol | 18.015 g/mol | 284.67 g/mol phase | gas (at STP) | | liquid (at STP) | solid (at STP) melting point | -86.8 °C | | 0 °C | 969 °C boiling point | -66.38 °C | | 99.9839 °C | density | 0.003307 g/cm^3 (at 25 °C) | | 1 g/cm^3 | 3.91 g/cm^3 solubility in water | miscible | | | soluble surface tension | 0.0271 N/m | | 0.0728 N/m | dynamic viscosity | 8.4×10^-4 Pa s (at -75 °C) | | 8.9×10^-4 Pa s (at 25 °C) | odor | | | odorless |

Units

Random Queries

total number of protons of palladium(IV) cation
H2SO4 + O2 + Cu = H2O + CuSO4
density of iodic acid
DOT numbers of periodic acid vs methyl chloride
alternate names of phosphorus trioxide vs phosphorus trioxide
alternate names of 2-fluoro-6-iodophenylboronic acid vs 5-bromo-2-ethoxyphenylboronic acid
IUPAC name of lithium benzoate
name of sodium bromide vs rubidium nitrate
HCl + KI + NaNO2 = H2O + I2 + NaCl + KCl + NO
molar heat of vaporization of ammonia