Periodic Table & Periodic Trends

Calculate Atomic Mass of The Elements Assuming the Isotopic Composition is Given

Basically, I need help with Group V, this isn't HW or anything I just wanted extra practice for next year but I don't even understand what the question is asking me, I know what an isotope is, what atomic mass is, I understand the A, Z thing (A = mass #, Z = atomic #/protons) and I know the formula for atomic mass. Yet 45 minutes later watching YouTube videos and doing example problems, I still don't know how to solve that one. Help would be much appreciated. Am I looking to find the atomic mass of Group V or the atomic mass of Group VI or what exactly?

Fill in the gaps in the following tables

I have absolutely no idea how to even start answering this question. I need help calculating the protons, neutrons and electrons. I am very lost :( 

I need help with this chemistry question?

Atoms of the element beryllium would most likely behave similar to the way _______ behaves.

Balancing Eqn

Please tell me the step of balancing this eqn.

Creating Periodic Table with Quantum Rules

      For the quantum rules listed, build the periodic table. Use the same element names as our known periodic table. n= 1 to 4, l= 0 to n, ml= 1, ms= 1

Why do beryllium, nitrogen, neon, etc. not follow the electron affinity trend?

Why do beryllium, nitrogen, neon, etc. not follow the  electron affinity trend?

Why did the Element wolfram change its name to Tungsten?

I am taking Chemistry in High School and our teacher told us to find why the Element Wolfram changed its name to Tungsten. I am not able to find information on why the name change. I know where it came from but don't know why it changed. Thank You for your help!!

Could someone please help me understand trends?????

My chem teacher is not the best ever and he didnt do a very good job explaining it. So if anyone could be of any help I would be very thankful...

Explain why Cl2 rather than Br2 would react more vigorously with a solution of I

Explain why Cl2 rather than Br2 would react more vigorously with a solution of I

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