The reaction is:
CaO(s) + H2O(l)--- --> Ca(OH)2(s)
Ca(OH)2(s) is the product (at the right side of the reaction equation).
To determine the base dissociation constant (Kb) for the reaction involving ch33naq, h2ol, ch33nh, and oh-aq, you need the specific equilibrium expression related to the bases and their respective concentrations in solution. Without additional context or specific reaction details, the Kb value cannot be accurately calculated or provided. Generally, Kb values are determined experimentally or found in literature for specific bases.
2H2,g + O2,g --> H2Ol
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If you added less than the amount needed in part (b) of K2SO3, what would be the limiting reactant? If you added more than the amount needed in part (b) of K2SO3, what would be the limiting reactant? Which situation would create the most SO2 gas?excess HClexcess K2SO3 excess H2Oexcess KClperfectly matched amounts of HCl and K2SO3
To determine the base dissociation constant (Kb) for the reaction involving ch33naq, h2ol, ch33nh, and oh-aq, you need the specific equilibrium expression related to the bases and their respective concentrations in solution. Without additional context or specific reaction details, the Kb value cannot be accurately calculated or provided. Generally, Kb values are determined experimentally or found in literature for specific bases.
2H2,g + O2,g --> H2Ol
It means that it is in a liquid state as opposed to solid (ice) or gas (water vapour or steam)
6.6 x 10-9
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If it's in the form H2O(l) then the l stands for liquid. This is to show that liquid water is different than gaseous water (steam), which would be H2O(g) with a g for gas.
When carbon dioxide in exhaled air mixes with water, it forms carbonic acid. This reaction lowers the pH of the water, making it more acidic. This is the basis for the process of ocean acidification, which can have harmful effects on marine life.
To find the (K_b) for NH2OH, you can use the relationship (K_w = K_a \times K_b). Solving for (K_b), we have (K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.5 \times 10^{-6}} = 6.67 \times 10^{-9}). So, the (K_b) for NH2OH at this temperature is (6.67 \times 10^{-9}).
If you added less than the amount needed in part (b) of K2SO3, what would be the limiting reactant? If you added more than the amount needed in part (b) of K2SO3, what would be the limiting reactant? Which situation would create the most SO2 gas?excess HClexcess K2SO3 excess H2Oexcess KClperfectly matched amounts of HCl and K2SO3
When an acid reacts with a carbonate; a salt, water and carbon dioxide is formed:acid + cabonate --> salt + water + carbon dioxideWhen calcium carbonate (CaCO3) and hydrochloric acid (HCL) react it is called a neutralisation reaction.This is because calcium carbonate is slightly alkaline, meaning when the two react, they cancel each other out; they neutralise each other.Calcium carbonate + hydrochloric acid --> Calcium chloride + water + carbon dioxideCaCO3 + HCL --> CaCl2 + H2O + CO2
Acetic acid + Sodium Bicarbonate ----> Sodium Acetate + Water + Carbon Dioxide CH3COOH + NaHCO3 ----> NaCH3COO + H2O + CO2
Melting