Question

I need each question answered. #6-12

6. Calculate the number of nitrogen atoms are present in each of the following quantities. A) 8.45x10 kg nitric acid B) 1280 dg of calcium nitrate 7. Write the empirical formula of the following compounds. A) H2S2 D) Co:PO B) ZnP203 E) C:H16012 C) C12H1oCl2 F) Au204 8. First balance the equation below the answer the questions that follow. ZrCN(aq) + Al2(S2O3)3(aq) — Zr2S2O3(s) + A) How many moles of ZnSO, of can form from 8.00 moles of Al:(8,0),? Al(CN)3(aq) B) How many molecules of ZrCN are in 1.0023g of ZrCN? C) How many moles of AI(CN), can form from 2.00 g of ZCN2 D) If 5.94 g ZrySO, were formed in the reaction, how many grams of ZrCN must have been used?

Answer 1

Answer 6

(1) 8.45x10² kg Nitric - molar mass of Acid (HNO3) HNO3 = 63.001 g/mol amoan plugging values to of HNO3 as - calculate no. of molecules mass = No. of molecules 6.023 x 1023 Molar mass - - (845x1000) g 63.01 g/mol No. of molecules 6. 023 x 1023 = 27 No. of molecules = 8.08 x 1027 molecules from chemical formula of HNO3, we know that I molecule of HNO3 contains I atom of N. thesefore, 8.08 x 1027 molecules will have 8.08 x 1027 Nitrogen atoms.

2 1280 dg of calcium Nitrate (Ca(NO3)2) - molar mass of Ca(NO3)2 = 164.09 g/mol - mass of Ca(NO3)2 = 1280 dg {1 dg = 0.1g} = 1280 dg x 0.19 = 128.0g lag moles of Ca(NO3)2 = - mass (g) molar mass (g/mol) 128.0g 164.09 g/mol 0.780 moles from chemical formula, Ca(NO3)2 , we know- 1 mole of ca(1103), contains (2X NA) atoms of N : 0.780 moles of Ca(NO3), will have a (2 X NA) atoms x 0.780 mole of N 1 mole = 9.40 x 1023 atoms of N Please Rate.
kindly submit different questions separately.