Answer:
0.17 moles
Explanation:
In the elements of the periodic table, the atomic mass = molar mass. <u>Ex:</u> Atomic mass of Carbon is 12.01 amu which means molar mass of Carbon is also 12.01g/mol.
In order to find the # of moles in a 12 g sample of NiC-12, we will need to multiply the number of each atom by its molar mass and then add the masses of both Nickel and C-12 found in the periodic table:
- Molar Mass of Ni (Nickel): 58.69 g/mol
- Molar Mass of C (Carbon): 12.01 g/mol
Since there's just one atom of both Carbon and Nickel, we just add up the masses to find the molar mass of the whole compound of NiC-12.
- 58.69 g/mol of Nickel + 12.01 g/mol of Carbon = 70.7 g/mol of NiC-12
There's 12g of NiC-12, which is less than the molar mass of NiC-12, so the number of moles should be less than 1. In order to find the # of moles in NiC-12, we need to do some dimensional analysis:
- 12g NiC-12 (1 mol of NiC-12/70.7g NiC-12) = 0.17 mol of NiC-12
- The grams cancel, leaving us with moles of NiC-12, so the answer is 0.17 moles of NiC-12 in a 12 g sample.
<em>P.S. C-12 or C12 just means that the Carbon atom has an atomic mass of 12amu and a molar mass of 12g/mol, or just regular carbon.</em>
Answer:
Genotypes: Homozygous (GG)=50%, Heterozygous (Gg)=50%.
Phenotypes: Homozygous gray (GG)=50%, Heterozygous gray (Gg)=50% or just Gray=100%
Explanation:
Hello,
The Punnett square for this cross turns into:
![\left[\begin{array}{ccc}&G&g\\G&GG&Gg\\G&GG&Gg\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%26G%26g%5C%5CG%26GG%26Gg%5C%5CG%26GG%26Gg%5Cend%7Barray%7D%5Cright%5D)
It means that the genotypes and phenotypes are:
Genotypes: Homozygous (GG)=50%, Heterozygous (Gg)=50%.
Phenotypes: Homozygous gray (GG)=50%, Heterozygous gray (Gg)=50% or just Gray=100%
Best regards.
Answer: 1. 3.914 × ^10-4 | 2. 4.781 × ^10-1
Explanation:
<span><span>S is for soil,</span><span>cl (sometimes c) represents climate,</span><span>o organisms including humans,</span><span>r relief,</span><span>p parent material, or lithology, and</span><span>t time.</span></span>
