To find the number of electrons in a neutral gold atom look at it's atomic number. The atomic number is the number of protons in the element. For an atom to be neutral it needs the same number of electrons as protons, therefore the positive and negative charges will be balanced out. Gold's atomic number is 79 therefore it has 79 electrons.
Answer:
8/3 hours
Explanation:
12.5/100=1/8, 1/8*2=2/8=1/4 1/4*2=1/2 1/2*2 = 1 (whole)
we multiplyed 1/8 by two three times to get 1 so 8/3 hours is the answer
by the way i'm in 6th grade
Answer:
(3R,4R)-4-bromohexan-3-ol
Explanation:
In this case, we have reaction called <u>halohydrin formation</u>. This is a <u>markovnikov reaction</u> with <u>anti configuration</u>. Therefore the halogen in this case "Br" and the "OH" must have <u>different configurations</u>. Additionally, in this molecule both carbons have the <u>same substitution</u>, so the "OH" can go in any carbon.
Finally, in the product we will have <u>chiral carbons</u>, so we have to find the absolute configuration for each carbon. On carbon 3 we will have an "R" configuration on carbon 4 we will have also an "R" configuration. (See figure 1)
I hope it helps!
Answer:
atomic mass 112.4 and Cadmium (Cd)
Explanation:
You have 73.35 g of MO.
After the reaction the O is removed and you only have M which the mass is 64.21 g.
With that you can calculate the mass of O removed:
Mass of O = 9.14 g
Mass = AM * moles ; (AM : Atomic Mass)
9.14 = 16 * moles
moles = 9.14 / 16
moles = 0.57125
The formula of the metal oxide is MO, meaning it has 1 mole of M per mole of O. 73.35 had 0.57125 moles of O, then it also had 0.57125 moles of M, and the remaining mass of 64.21 g represents those moles
Mass = AM * moles ; (AM : Atomic Mass)
64.21 = AM * 0.57125
AM = 64.21/0.57125 = 112.4 amu
The metal with an atomic mass of 112.4 is Cadmium (Cd), therefore the metal oxide is CdO
