This is an acid – base reaction and this always result a salt and water
in a neutralization reaction. <span>
The salt that is formed will be calcium bromide (calcium
is located in group 2 so calcium bromide has a formula of CaBr2)
so essentially we got:
HBr + Ca(OH)2 ------> CaBr2 + H2O </span>
balancing the elements: <span>
<span>2HBr(aq) + Ca(OH)2(aq) --------> CaBr2(aq) +
2H2O(l)</span></span>
Answer:
The reaction will move to the left.
Explanation:
<em>Ba(OH)₂ = Ba²⁺ + 2OH⁻,</em>
<em>Ba(OH)₂ is dissociated to Ba²⁺ and 2OH⁻.</em>
- If H⁺ ions are added to the equilibrium:
H⁺ will combine with OH⁻ to form water.
<em>So, the concentration of OH⁻ will decrease and the equilibrium is disturbed.</em>
<em />
<em>According to Le Châtelier's principle: </em>when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.
- So, the reaction will move to the right to suppress the effect of decreasing OH⁻ concentration.
- The base will dissociate to form more OH⁻ and thus, the quantity of Ba(OH)₂ will decrease.
<em>So, the right choice is: the reaction will move to the left, is the choice that will not happen to the equilibrium.</em>
Answer: It will take 29 years for a 10.0-gram sample of strontium-90 to decay to 5.00 grams
Explanation:
Radioactive decay process is a type of process in which a less stable nuclei decomposes to a stable nuclei by releasing some radiations or particles like alpha, beta particles or gamma-radiations. The radioactive decay follows first order kinetics.
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
Half life is represented by 

= rate constant
Given : Strontium-90 decreases in mass by one-half every 29 years , that is half life of Strontium-90 is 29 years.
As half life is independent of initial concentration, it will take 29 years for a 10.0-gram sample of strontium-90 to decay to 5.00 grams as the amount gets half.
Answer:
The chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.
Explanation:
The chemical potential of 2-propanol in solution relative to that of pure 2-propanol can be calculated using the following equation:
<u>Where:</u>
<em>μ (l): is the chemical potential of 2-propanol in solution </em>
<em>μ° (l): is the chemical potential of pure 2-propanol </em>
<em>R: is the gas constant = 8.314 J K⁻¹ mol⁻¹ </em>
<em>T: is the temperature = 82.3 °C = 355.3 K </em>
<em>x: is the mole fraction of 2-propanol = 0.41 </em>

Therefore, the chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.
I hope it helps you!