<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:
0.425M NaOH assuming the volume of KHP was 25.50mL and the volume of the NaOH solution was 30.0mL
Explanation:
The KHP reacts with NaOH as follows:
KHP + NaOH → KNaP + H₂O
<em>Where 1 mole of KHP reacts per mole of KNaP</em>
<em />
That means, the moles of KHP added to the NaOH solution = Moles NaOH at equivalence point. With the moles of NaOH and the volume in liters we can find the molar concentration of NaOH.
<em>Assuming the volume added of KHP was 25.50mL and the solution of NaOH contains 30.0mL (0.0300L), the concentration of the NaOH is:</em>
<em />
<em>Moles KHP = Moles NaOH:</em>
25.50mL = 0.02550L * (0.500mol / L) = 0.01275 moles KHP = Moles NaOH
<em>Molarity NaOH:</em>
0.01275 moles NaOH / 0.0300L =
<h3>0.425M NaOH assuming the volume of KHP was 25.50mL and the volume of the NaOH solution was 30.0mL</h3>
Answer:
Compound D is CH3OPO3 is the best answer
Explanation:
Based on the data provided, there are 25 g of calcium carbonate in 1.505 × 10^23 atoms.
<h3>What is the moles of calcium carbonate in 1.505 × 10^23 atoms of calcium carbonate?</h3>
The mole of a substance can be calculated as follows:
- Moles of substance = number of particles/6.02 × 10^23
Moles of calcium carbonate = 1.505 × 10^23/6.02 × 10^23
Moles of calcium carbonate = 0.25 moles
The mass of calcium carbonate in 0.25 moles is calculated as follows:
- mass = moles × molar mass
molar mass of a calcium carbonate = 100 g/mol
mass of calcium carbonate = 0.25 × 100 = 25 g.
Therefore, there are 25 g of calcium carbonate in 1.505 × 10^23 atoms.
Learn more about molar mass and mass at: brainly.com/question/15476873
