Aurous is a cation of gold. Gold takes the name "aurum" (Au) with atomic number of 79. In its purest form, the element is bright, slightly yellow, soft, ductile, and malleable. The charge of aurous is +1. Sulfide, on the other hand, has a charge of -2.
Hence, the chemical formula of the compound is Au₂S and its systematic name is gold (I) sulfide.
The empirical formula is K₂O.
The empirical formula is the <em>simplest whole-number ratio</em> of atoms in a compound.
The <em>ratio of atom</em>s is the same as the <em>ratio of moles</em>.
So, our job is to calculate the <em>molar ratio</em> of K to O.
Step 1. Calculate the <em>moles of each element
</em>
Moles of K = 32.1 g K × (1 mol K/(39.10 g K =) = 0.8210 mol K
Moles of O = 6.57 g O × (1 mol O/16.00 g O) = 0.4106 mol 0
Step 2. Calculate the <em>molar ratio of each elemen</em>t
Divide each number by the smallest number of moles and round off to an integer
K:O = 0.8210:0.4106 = 1.999:1 ≈ 2:1
Step 3: Write the <em>empirical formula
</em>
EF = K₂O
Answer:
1. Option A. Beta decay minus 0 -1e
2. Option B. Alpha 4 2He
3. Option A. Beta decay minus 0 -1e
Explanation:
1. 66 29Cu is undergoing beta decay minus since it produces a daughter nuclei having the same mass number and the atomic number increased by 1 i.e 66 30Zn
2. 238 92U is undergoing alpha decay since the daughter nuclei produced has a decrease of 4 in the mass number and a decrease of 2 in the atomic number ie 234 90Th
3. 14 6C is undergoing beta decay minus since the daughter nuclei produced has the same mass number and the atomic number increased by 1 i.e 14 7N
Please see the attached photo for more details
They are both (polyatomic) ions.
Answer:
Explanation:
use gas law eqation
P1 * V1 / T1 = P2 * V2 /T2
600*V1/227 = 300*800/23
V1 = 300*800*227 / 23*600 = ............ can you solve this and get the answer?
