Answer:
Percent error = 2.8
Explanation:
Given data;
Actual value = 17.9 cm
Measured value = 17.4 cm
Percent error = ?
Solution:
Formula:
Percent error = [actual value - measured value / actual value] × 100
Now we will put the values in formula:
Percent error = [17.9 cm - 17.4 cm / 17.9 cm] × 100
Percent error = [0.5 cm /17.9 cm] × 100
Percent error = 0.028 × 100
Percent error = 2.8
Answer is: B₂O₃ + Mg → B + MgO.
B₂O₃ + 3Mg → 2B + 3MgO, balanced.
B₂O₃ - diboron trioxide. White, glassy solid compound.
B - boron. Metalloid, crystalline boron is and amorphous<span> boron is a brown powder.
Mg - magnesium. S</span><span>hiny, gray and solid metal.
MgO - magnesium oxide. W</span><span>hite, </span>hygroscopic<span> solid powder.</span>
Answer:
Electron-pair geometry: tetrahedral
Molecular geometry: trigonal pyramidal
Hybridization: sp³
sp³ - 4 p
Explanation:
There is some info missing. I think this is the original question.
<em>For NBr₃, What are its electron-pair and molecular geometries? What is the hybridization of the nitrogen atom? What orbitals on N and Br overlap to form bonds between these elements?</em>
<em>The N-Br bonds are formed by the overlap of the ___ hybrid orbitals on nitrogen with ___ orbitals on Br.</em>
<em />
Nitrogen is a central atom surrounded by 4 electron domains. According to VESPR, the corresponding electron-pair geometry is tetrahedral.
Of these 4 electron domains, 3 represent covalent bonds with Br and 1 lone pair. According to VESPR, the corresponding molecular geometry is trigonal pyramidal.
In the nitrogen atom, 1 s orbital and 3 p orbitals hybridize to form 4 sp³ orbitals for each of the electron domains.
The N-Br bonds are formed by the overlap of the sp³ hybrid orbitals on nitrogen with 4p orbitals on Br.
