Answer:
2.25×10¯³ mm.
Explanation:
From the question given above, we obtained the following information:
Diameter in micrometer = 2.25 μm
Diameter in millimetre (mm) =?
Next we shall convert 2.25 μm to metre (m). This can be obtained as follow:
1 μm = 1×10¯⁶ m
Therefore,
2.25 μm = 2.25 μm / 1 μm × 1×10¯⁶ m
2.25 μm = 2.25×10¯⁶ m
Finally, we shall convert 2.25×10¯⁶ m to millimetre (mm) as follow:
1 m = 1000 mm
Therefore,
2.25×10¯⁶ m = 2.25×10¯⁶ m /1 m × 1000 mm
2.25×10¯⁶ m = 2.25×10¯³ mm
Therefore, 2.25 μm is equivalent to 2.25×10¯³ mm.
Hey there!:
Sodium cations and phosphate anions .
hope this helps!
Answer:
H2
Explanation:
Critical temperature is the temperature above which gas cannot be liquefied, regardless of the pressure applied.
Critical temperature directly depends on the force of attraction between atoms, it means stronger the force of higher will be the critical temperature. So, from the given options H2 should have the highest critical temperature because of high attractive forces due to H bonding.
Hence, the correct option is H2.
Answer:
The whole molecule is polar because Sulfur has lone pairs but Carbon doesn't. Lone pairs count more toward polarity, shifting dipole toward S.
Explanation:
Even though carbon and sulfur have identical values of electronegativities, the molecule, 
is polar because of the presence of the lone pairs on the sulfur atom.
The C-S bond is not polar because the both the atoms have electronegatiivty. <u>But S has lone pairs which can attract the bond pairs of the bond between the S and H and thus acquires slightly negative charge and H acquires slightly positive charge.</u>
Answer:
B. PROTONS EXHIBIT STRONGER PULL ON OUTER f ORBITALS
Explanation:
Lanthanide contraction is the greater than normal decrease in the ionic radius of the lanthanide series from atomic number 57 to atomic number 71. This decrease is rather not expected of the ionic radii of these elements and they result in the greater decrease in the subsequent series of the lanthanides from the atomic number 72. The cause of which is as a result of the poor shielding effects of the nuclear charge around the electrons of the f orbitals. So therefore, protons are strongly pulled out of the 4f orbital and as a result of the poor shielding effect which causes the electrons of the 6s orbitals to be drawn more closer to the nucleus and hence resulting in a smaller atomic radii. It is worthy to note that the shielding effects of the inner electrons decreasing from s orbital to the f orbital; that is s > p > d > f. So from the decrease in the shielding effects from s to the f orbitals, lanthanide contraction results from the inability of the orbitals far away from s like the 4f orbiatls to shield the outermost shells of the lanthanide elements. So the cause of lanthanide contraction is the action of the protons which strongly pull the electrons of the f orbitals because of the poor shielding effects due to the distance of this orbital from the nucleus.
