Answer:150g of gold
Explanation: There is a lot more gold and gold is significantly significantly more dense than lithium
Answer:
ⁿₐX => ²¹⁸₈₄Po
Explanation:
Let ⁿₐX be the isotope.
Thus, the equation can be written as follow:
²²²₈₆Rn —> ⁴₂α + ⁿₐX
Next, we shall determine the value of 'n' and 'a'. This can be obtained as follow:
222 = 4 + n
Collect like terms
222 – 4 = n
218 = n
Thus,
n = 218
86 = 2 + a
Collect like terms
86 – 2 = a
84 = a
Thus,
a = 84
ⁿₐX => ²¹⁸₈₄Po
²²²₈₆Rn —> ⁴₂α + ⁿₐX
²²²₈₆Rn —> ⁴₂α + ²¹⁸₈₄Po
Hey there!:
Molar mass:
CHCl3 = ( 12.01 * 1 )+ (1.008 * 1 ) + ( 35.45 * 3 ) => 119.37 g/mol
C% = ( atomic mass C / molar mass CHCl3 ) * 100
For C :
C % = (12.01 / 119.37 ) * 100
C% = ( 0.1006 * 100 )
C% = 10.06 %
For H :
H% = ( atomic mass H / molar mass CHCl3 ) * 100
H% = ( 1.008 / 119.37 ) * 100
H% = 0.008444 * 100
H% = 0.8444 %
For Cl :
Cl % ( molar mass Cl3 / molar mass CHCl3 ):
Cl% = ( 3 * 35.45 / 119.37 ) * 100
Cl% = ( 106.35 / 119.37 ) * 100
Cl% = 0.8909 * 100
Cl% = 89.9%
Hope that helps!
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:
Attached in the photo.
Explanation:
Hello,
The answers in the attached photo. Just three things:
- In the second point a parenthesis is missing to properly understand the molecule (after the oxygen), nevertheless, I assumed it was an ether.
- In the sixth point, there's a missing hydrogen for it to be an ether as well.
- In the tenth point the second parenthesis is not clear, it seems there's a missing subscript, nevertheless I draw it assuming complete octates.
Best regards.