The equilibrium expression for the reaction (2 \text{NO} + \text{Br}_2 \rightleftharpoons 2 \text{NOBr}) is given by the formula:
[ K_{eq} = \frac{[\text{NOBr}]^2}{[\text{NO}]^2 [\text{Br}_2]} ]
In this expression, (K_{eq}) represents the equilibrium constant, and the square brackets indicate the molar concentrations of the respective species at equilibrium.
In a sealed flask containing both liquid bromine (Br2(l)) and bromine gas (Br2(g)) at 298 K and 1.0 ATM, a dynamic equilibrium exists between the liquid and gas phases. This equilibrium is characterized by the rate of evaporation of the liquid bromine equal to the rate of condensation of the gas bromine. As a result, the concentrations of Br2(l) and Br2(g) remain constant over time, reflecting a balance between the two phases. This type of equilibrium is specifically known as a phase equilibrium.
1 (fully) and 4 (only partial) are decomposition reactions.
The balanced equation for the reaction between zinc (Zn) and bromine (Br2) is: Zn + Br2 -> ZnBr2.
2 Na + Br2 --> 2 NaBr
C8H16 + Br2 -> C8H16Br2
In a sealed flask containing both liquid bromine (Br2(l)) and bromine gas (Br2(g)) at 298 K and 1.0 ATM, a dynamic equilibrium exists between the liquid and gas phases. This equilibrium is characterized by the rate of evaporation of the liquid bromine equal to the rate of condensation of the gas bromine. As a result, the concentrations of Br2(l) and Br2(g) remain constant over time, reflecting a balance between the two phases. This type of equilibrium is specifically known as a phase equilibrium.
1 (fully) and 4 (only partial) are decomposition reactions.
Br2 + 3NaHSO3 = 2NaBr + NaHSO4 + H2O + 2SO2
44.0 grams Br2 ? 44.0 grams Br2 (1 mole Br2/159.8 grams)(6.022 X 10^23/1 mole Br2)(1 mole Br2 atoms/6.022 X 10^23) = 0.275 moles of Br2 atoms
There are two bromine atoms in Br2
The balanced equation for the reaction between zinc (Zn) and bromine (Br2) is: Zn + Br2 -> ZnBr2.
2 Na + Br2 --> 2 NaBr
3.387mL Br2
C8H16 + Br2 -> C8H16Br2
Yes, Br2 is an oxidizing agent.
Ca + Br2 = CaBr2 doesn't need to be balanced.
To calculate the mass of 4.89 x 10^20 molecules of Br2, you need to use the molar mass of Br2, which is 159.808 g/mol. Calculate the number of moles of Br2: 4.89 x 10^20 molecules / 6.022 x 10^23 molecules/mol = 8.13 x 10^-4 moles Use the formula mass = moles x molar mass to find the mass: mass = 8.13 x 10^-4 moles x 159.808 g/mol ≈ 0.130 g