Explanation:
In a galvanic cell, the cathode is positively charged and the anode is negatively charged.
The cathode attracts electron while the anode donates or releases electrons.
Electrons received - Cathode
Electrons donated - Anode
So what am I suppose to answer here?
There are 2 moles of O stones present in 88 grams of CO2. Why? Well, we can find the amount of moles present in 88 grams of CO2 by dividing the mass by the molar mass. The mass of CO2 comes out to be 88 grams. The molar mass of CO2 comes out to be 44 grams. Because 88 is the mass of CO2 and 44 is the molar mass of CO2, we can divide 88 by 44 to identify that there are 2.0 moles of O atoms present in 88 grams of CO2.
Your final answer: There are 2.0 moles of O atoms present in 88 grams of CO2. Your final answer to this question is D, or 2.0 moles. If you need to better understand, let me know and I will gladly assist you.
The answer is ultraviolet
Answer:
0.22 mol HClO, 0.11mol HBr.
0.25mol NH₄Cl, 0.12 mol HCl
Explanation:
A buffer is defined as a mixture in solution between weak acid and its conjugate base or vice versa.
Potassium hypochlorite (KClO) could be seen as conjugate base of HClO (Weak acid). That means the addition of <em>0.22 mol HClO </em>will convert the solution in a buffer. HBr reacts with KClO producing HClO, thus, <em>0.11mol HBr</em> will, also, convert the solution in a buffer. 0.23 mol HBr will react completely with KClO and in the solution you will have only HClO, no a buffering system.
Ammonia (NH₃) is a weak base and its conjugate base is NH₄⁺. That means the addition of <em>0.25mol NH₄Cl</em> will convert the solution in a buffer. Also, NH₃ reacts with HCl producing NH₄⁺. Thus, addition of<em> 0.12 mol HCl</em> will produce NH₄⁺. 0.25mol HCl consume all NH₃.
