Answer:
1610.7 g is the weigh for 4.64×10²⁴ atoms of Bi
Explanation:
Let's do the required conversions:
1 mol of atoms has 6.02×10²³ atoms
Bi → 1 mol of bismuth weighs 208.98 grams
Let's do the rules of three:
6.02×10²³ atoms are the amount of 1 mol of Bi
4.64×10²⁴ atoms are contained in (4.64×10²⁴ . 1) /6.02×10²³ = 7.71 moles
1 mol of Bi weighs 208.98 g
7.71 moles of Bi must weigh (7.71 . 208.98 ) /1 = 1610.7 g
Answer:
Atoms consist of three basic particles: protons, electrons, and neutrons. The nucleus (center) of the atom contains the protons (positively charged) and the neutrons (no charge). The outermost regions of the atom are called electron shells and contain the electrons (negatively charged).
Answer:
There are 5 significant digits in 0.23100.
Explanation:
This is because all non-zero digits are considered significant and zeros after decimal points are considered significant.
The anti periplanar geometry for the E2 reaction of (CH3)2CHCH2Br with base is shown in the image attached as well as the structure of the product formed in the reaction.
In organic chemistry, an antiperiplanar conformation is one in which the groups point up and down at a dihedral angle of 180° away from one another. In the image attached, the antiperiplanar conformation of (CH3)2CHCH2Br is shown.
Recall that an E2 reaction is a synchronous elimination reaction where to atoms leave at the same time. The product of this reaction is also shown in the image attached.
Learn more: brainly.com/question/2510654
Answer:
The percent isotopic abundance of C- 12 is 98.93 %
The percent isotopic abundance of C- 13 is 1.07 %
Explanation:
we know there are two naturally occurring isotopes of carbon, C-12 (12u) and C-13 (13.003355)
First of all we will set the fraction for both isotopes
X for the isotopes having mass 13.003355
1-x for isotopes having mass 12
The average atomic mass of carbon is 12.0107
we will use the following equation,
13.003355x + 12 (1-x) = 12.0107
13.003355x + 12 - 12x = 12.0107
13.003355x- 12x = 12.0107 -12
1.003355x = 0.0107
x= 0.0107/1.003355
x= 0.0107
0.0107 × 100 = 1.07 %
1.07 % is abundance of C-13 because we solve the fraction x.
now we will calculate the abundance of C-12.
(1-x)
1-0.0107 =0.9893
0.9893 × 100= 98.93 %
98.93 % for C-12.
