<span>XY4Z2-->Square planar (Electron domain geometry: Octahedral) sp3d2
XY4Z-->Seesaw (Electron domain geometry: Trigonal bipyramidal) sp3d
XY5Z-->Square pyramidal (Electron domain geometry: Octahedral) sp3d2
XY2Z3-->Linear (Electron domain geometry: Trigonal bipyramidal) sp3d
XY2Z-->Bent (Electron domain geometry: Trigonal planar) sp2
XY3Z-->Trigonal pyramidal (Electron domain geometry: Tetrahedral) sp3
XY2Z2-->Linear (Electron domain geometry: Tetrahedral) sp3
XY3Z2-->T shaped (Electron domain geometry: Trigonal bipryamidal) sp3d
XY2-->Linear (Electron domain geometry: Linear) sp
XY3 Trigonal planar (Electron geometry: Trigonal planar) sp2
XY4-->Tetrahedral (Electron domain geometry: tetrahedral) sp3
XY5-->Trigonal bipyramidal (Electron domain geometry: Trigonal bipyramidal) sp3d
XY6-->Octahedral (Electron domain geometry: Octahedral) sp3d2</span>
Answer:
Products will be K2SO4 and Cu(NO3)2. A precipitate will not form.
Explanation:
Double replacement reactions are the switching of cations between two compounds. So in this case, the K+ and Cu 2+ cations will switch places.
KNO3 + CuSO4 -> K2SO4 + Cu(NO3)2
According to solubility rules, any substances containing NO3 or a Group 1 ion (K+ in this case) will be definitely soluble. There are a few exceptions to the Group 1 rule but it does not apply here. Potassium sulfate (K2SO4) contains a Group 1 ion so it will be soluble and not be a precipitate. And copper nitrate (Cu(NO3)2) contains nitrate so it will also be soluble and not be a precipitate.
I would say A because they are two or more different elements bound chemically together and can only be separated chemically.
Answer:
b) dime
Explanation:
a dime is approximately 2.2g
half of this is 1.1g, which can be rounded down to one gram.
hope this helps
1) Balanced chemical equation
H2SO4 + 2NaOH ---> Na2 SO4 + 2H2O
=> 1 mol H2SO4 : 2 moles NaOH
2) Convert 89.3 g of H2SO4 and 96.0 g of NaOH to moles
Molar mass of H2SO4 = 98.1 g/mol
Molar mass of NaOH = 40.0 g/mol
moles = mass in grams / molar mass
moles H2SO4 = 89.3 g / 98.1 g/mol = 0.910 mol
moles NaOH = 96.0 g / 40.0 g/mol = 2.40 mol
3) Theoretical molar ratio = 2 moles NaOH / 1 mol H2SO4
So, all the 0.91 mol of H2SO4 will be consumed along with 1.820 (2*0.91) moles of NaOH, and 0.580 moles (2.40 - 1.82) of NaOH will be left over by the chemical reaction.
4) Convert 0.580 moles NaOH to mass
0.580 moles * 40.0 g/mol = 23.2 g of NaOH will be left over
