How many bromine atoms are present in 33.8 g of ch2br2?

2.34 x 10^23 Calculate the molar mass of CH2Br2 Atomic weight carbon = 12.0107 Atomic weight hydrogen = 1.00794 Atomic weight Bromine = 79.904 Molar mass of CH2Br2 = 12.0107 + 2 * 1.00794 + 2 * 79.904 = 173.83458 Now divide the given mass of CH2Br2 you have by the molar mass 33.8 g / 173.83458 g/mol = 0.194438 moles Since each molecule of CH2Br2 has 2 bromine atoms, multiply by 2 0.194438 * 2 = 0.388876 moles Finally, multiply by avogadro's number 0.388876 * 6.0221409 x 10^23 = 2.341866 x 10^23 Since you only have 3 significant figures, round the result to 3 significant figures. 2.341866 x 10^23 = 2.34 x 10^23

The answer is b metal denting

For cyclohexane compounds drawn in a chair conformation:   -axial bonds are vertical  -equatorial bonds are parallel to two bonds within the ring.  thus, as i assume the chair conformation is drawn:   -the x group is in an axial position, as are the five other vertical bonds.  -three axial bonds are on the top face of the ring  -the other three axial bonds are on the bottom face  the position of the second substituent on each ring carbon is equatorial.  gauche groups must be on an adjacent carbon such as carbon 6 and carbon 2, and also gauche groups are rotated by 60 degrees relative to each other.  drawing the newman projections for c1–c2 and c1–c6 aids in visualizing the groups that are gauche to the y group on c1.  click this link and watch some academic videos that may you more understand the topic