The answer is A. Plasmodesmata. It's a narrow pathway between cells.
The atomic number increases moving left to right across a period and subsequently so does the effective nuclear charge. Therefore, moving left to right across a period the nucleus has a greater pull on the outer electrons and the atomic radii decreases.
Answer:
The given element is Radon because its atomic weight is 222 amu.
Explanation:
Given data:
Percentage of A-219 = 13.92%
Percentage of B-222 = 72.16%
Percentage of C-225 = 13.92%
Atomic weight of element = ?
Solution:
Average atomic mass = (abundance of A isotope × its atomic mass) +(abundance of B isotope × its atomic mass) + (abundance of C isotope × its atomic mass) / 100
Average atomic mass = (13.92×219)+(72.16×222) + (13.92×225)/100
Average atomic mass = 3048.48 + 16019.52 +3132/ 100
Average atomic mass = 22200 / 100
Average atomic mass = 222 amu.
The given element is Radon because its atomic weight is 222 amu.
Explanation:
We have 2CO + O2 => 2CO2.
Moles of CO = 28.6g / (28g/mol) = 1.021mol.
Moles of CO2 = Moles of CO = 1.021mol.
Answer:
71.7 g of C₆H₅Br are produced at the theoretical yield.
Explanation:
The reaction is:
C₆H₆ + B₂ → C₆H₅Br + HBr
This is an easy excersie, stoichiometry is 1:1
We determine moles of reactants:
42.1 g / 78g/mol = 0.540 moles of benzene
73 g / 159.80 g/mol = 0.457 moles of hydrogen bromide
Certainly, the HBr is the limiting reactant, because I need 0.540 moles of HBr for 0.540 moles of benzene and I only have, 0.457 moles.
0.457 moles of HBr will produce 0.457 of bromobenzene, at the 100 % yield reaction (theoretical yield)
We convert the moles to mass: 0.457 g . 157 g/mol = 71.7 g
