H+ ( aq ) + HCO3- ( aq ) ------> H2O( l ) + CO2 ( g ) The overall reaction when hydroiodic acid interacts with sodium hydrogen carbonate is: HI + NaHCO3 ----> NaI + H2O + CO2. The net reaction shows H2CO3’s instability, leading it to dissociate into water and carbon dioxide found in the reactants. In this reaction, H2CO3 also referred to as carbonic acid, is complemented by iodine and sodium from the reactants, creating NaI in the products. Including the physical states, the balanced equation reads: HI ( aq ) + NaHCO3 ( aq ) ----> NaI ( aq ) + H2O ( l ) + CO2 ( g ). The complete ionic equation represents each compound as ions in the aqueous phase: H+ ( aq ) + I- ( aq ) + Na+ ( aq ) + HCO3- ( aq ) -------> Na+ ( aq ) + I- ( aq ) + H2O( l ) + CO2 ( g ). In this reaction, the spectator ions are I- and Na+, allowing for their cancellation which leads to the net ionic equation: H+ ( aq ) + HCO3- ( aq ) ------> H2O( l ) + CO2 ( g ). Hope that helps!
Answer:
The price will be $1.68
Explanation:
I arrived at this answer by multiplying
.67x or.67lb with 2.5, resulting in $1.68
Result: 40 drops, and the acid percentage in the solution is 0.166%.
Details: Acid is gradually mixed into 1.2 L of water to create the acidic solution. The total volume after mixing is 1.202 L. We can ascertain the amount of acid added to the water by subtracting the initial water volume from the final volume as follows:
volume of acid added = 1.202 L - 1.2 L = 0.002 L
Each drop has a volume of 0.05 mL. Let's convert the added acid volume from liters to milliliters.
= 2 mL
Next, we calculate the total number of drops of acid added as follows:
= 40 drops
With 0.002 L of acid in 1.202 L of solution, we need to determine the percentage concentration of acid. This amounts to how many liters of acid would exist in 100 L of acid solution. We can express this as:
percentage of acid in solution = 
= 0.166%
Thus, 40 drops of acid are needed, indicating the acid solution contains 0.166% acid.
Answer:
[OH⁻] = 3.54 × 10⁻¹⁰ M
Solution:
The relationship between pOH and [OH⁻] is expressed as,
pOH = - log [OH⁻]
Substituting the value of pOH,
9.45 = -log [OH⁻]
Finding [OH⁻],
[OH⁻] = 10⁻⁹·⁴⁵ ∴ 10 = Antilog
[OH⁻] = 3.54 × 10⁻¹⁰ M
The principle of conservation of mass asserts that mass cannot be created or eliminated. Given that element A has a mass of 2 g/mol and element B has a mass of 3 g/mol, the total mass of compound AB equals the combined molar masses: 2 g/mol + 3 g/mol results in 5 g/mol for AB. As for A2B3, the calculation is as follows: A2 has 2 multiplied by 2, yielding 4 g/mol, while B3 equals 3 multiplied by 3, which gives 9 g/mol. Consequently, A2B3 amounts to 13 g/mol.
