we have a total of three times the original number (6.923 * 10**-7) moles of all ions, or 2.077 * 10**-6 moles of ions
<h3>What is aragonite-strontianite solid solution dissolution in nonstoichiometric Sr (HCO3)2 solutions?</h3>
Synthetic strontianite-aragonite solid-solution minerals were dissolved in non-stoichiometric CO2-saturated Sr(HCO3)2 and Ca(HCO3)2 solutions at 25°C. The reactions in Sr(HCO3)2 solutions frequently become incongruent, precipitating a Sr-rich phase before attaining stoichiometric saturation. Mechanical mixes of solids approach stoichiometric saturation in terms of the least stable solid in the combination.
This surficial phase has a thickness of 0-10 atomic layers in Sr(HCO3)2 solutions and a thickness of 0-4 layers in Ca(HCO3)2 solutions and dissolves and/or recrystallizes within 6 minutes of reaction.
learn more about Sr (HCO3)2 refer
brainly.com/question/24667072
#SPJ4
Answer:
Si hay algo que sucedió entre ustedes raro o un comportamiento que tuviste con esa persona o que te vio hacer asi como le pudieron decir algo de ti que no le gustara a tu amigo hace que se comporte raro contigo
Explanation:
Answer:
Being flammable means it supports burning,e.g. Oxygen, but being combustible means burning itself too. e.g. Hydrogen.
Explanation:
Hope it helps!!
Answer:
<em>»</em><em> </em><em>λ </em><em>=</em><em> </em><em>0</em><em>.</em><em>0</em><em>2</em><em>m</em>
Explanation:
Given :
Velocity of the wave {v}= 12 m/s
Frequency {f} = 600 Hz
Apply Wavelength formula :
• 
→ λ = 
→ λ = 
→ λ = 0.02m
Answer:

Explanation:
Hello,
In this case, for a concentration of 0.42 M of benzoic acid whose Ka is 6.3x10⁻⁵ in 0.33 M sodium benzoate, we use the Henderson-Hasselbach equation to compute the required pH:
![pH=pKa+log(\frac{[base]}{[acid]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29)
Whereas the concentration of the base is 0.33 M and the concentration of the acid is 0.42 M, thereby, we obtain:
![pH=-log(Ka)+log(\frac{[base]}{[acid]} )\\\\pH=-log(6.3x10^{-5})+log(\frac{0.33M}{0.42M} )\\\\pH=4.1](https://tex.z-dn.net/?f=pH%3D-log%28Ka%29%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29%5C%5C%5C%5CpH%3D-log%286.3x10%5E%7B-5%7D%29%2Blog%28%5Cfrac%7B0.33M%7D%7B0.42M%7D%20%29%5C%5C%5C%5CpH%3D4.1)
Regards.