How many protons does Thorium have? 90
How many neutrons does Thorium-234 have? 144
Calculate the mass defect for the isotope thorium-234 1.85864 amu
Answer:
Fluoride is commonly used in dentistry to strengthen enamel, which is the outer layer of your teeth. Fluoride helps to prevent cavities. It’s also added in small amounts to public water supplies in the United States and in many other countries. This process is called water fluoridation.
Explanation:
Unburned hydrocarbon on reacting with oxygen undergoes combustion reaction. However, the activation energy of this reaction is significantly high. When a catalyst like Pd is added to the reaction system, it provides active sites for the reaction to occur. It acts are a heterogeneous catalyst. It is pertinent of note that catalyst is refereed as heterogeneous, when it exist in different phase as compared to reactant and products. In present case, reactants and products are in gas phase, while catalyst is in solid phase. Due to availability of larger surface area at active site of Pd, activation energy of reaction decreases and decrease in activation energy favors higher reaction rates.
Answer:
The quantity of ascorbic acid found in sweet lime of 49.6 mg does not meet the daily requirement.
Explanation:
To determine the mass of ascorbic acid knowing the number of moles we use the following formula:
number of moles = mass / molecular weight
mass = number of moles × molecular weight
mass of ascorbic acid = 2.82 × 10⁻⁴ × 176
mass of ascorbic acid = 496 × 10⁻⁴ g = 0.0496 g = 49.6 mg
daily requirement of ascorbic acid = 70 - 90 mg
The quantity of ascorbic acid found in sweet lime of 49.6 mg does not meet the daily requirement.
Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.
When Ni(OH)₂ starts precipitate :
Ksp of Ni(OH)₂ = [ Ni²⁺ ] [ OH²⁻ ]
5.5x10⁻¹⁶ = [ 0.18 ] [ OH²⁻ ]
[ OH²⁻ ] = 5.5x10⁻¹⁶ / 0.18
[ OH⁻ ] = 5.5 × 10⁻⁸ M
pOH = 7.2
therefore , pH = 14 - 7.2
pH = 6.8
Thus, Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.
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