What is the percent zinc in Zn(NO)3
2xZn/189.41]=34.52% am I right
I have no idea what you did, but the percent is correct.
I did Zn(2)=65.39
then I divided that by the amu of Zn(NO3)2 x 100 is that right
To determine the percent zinc in Zn(NO3)2, you need to calculate the molar mass of zinc in the compound and then use it to find the percentage.
First, let's start by calculating the molar mass of Zn(NO3)2.
The molar mass of zinc (Zn) is 65.38 g/mol.
The molar mass of nitrate (NO3) is 62.00 g/mol. There are two nitrate ions in Zn(NO3)2, so the total mass of nitrate is (62.00 g/mol) x 2 = 124.00 g/mol.
Adding the molar masses of zinc and nitrate gives us:
65.38 g/mol + 124.00 g/mol = 189.38 g/mol.
Now, let's calculate the percent zinc.
To do this, divide the molar mass of zinc (65.38 g/mol) by the total molar mass of Zn(NO3)2 (189.38 g/mol), and then multiply the result by 100 to get the percentage:
[(65.38 g/mol) / (189.38 g/mol)] x 100% ≈ 34.51%
So, the calculated value for the percent zinc in Zn(NO3)2 is approximately 34.51%.
Based on your statement, it seems like you have correctly calculated the percentage as 34.52%. However, there might be a slight rounding difference in the calculations. Thus, your answer is very close to the correct value.