Answer:
i think the first one is gravity and second one is rotation
Explanation:
Answer:
49.2 g/mol
Explanation:
Let's first take account of what we have and convert them into the correct units.
Volume= 236 mL x (
) = .236 L
Pressure= 740 mm Hg x (
)= 0.97 atm
Temperature= 22C + 273= 295 K
mass= 0.443 g
Molar mass is in grams per mole, or MM= 
or MM= 
. They're all the same.
We have mass (0.443 g) we just need moles. We can find moles with the ideal gas constant PV=nRT. We want to solve for n, so we'll rearrange it to be
n=
, where R (constant)= 0.082 L atm mol-1 K-1
Let's plug in what we know.
n=
n= 0.009 mol
Let's look back at MM= and plug in what we know.
MM= 
MM= 49.2 g/mol
Answer:
Your answer has to be B. Camouflage
Explanation:
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.
