Total number of atoms in the sample is the sum of number of atoms of all the elements present in the sample.
Number of gold, 
atoms in 
= 
atoms (given)
From the formula of compound that is it is clear that the number of potassium and gold are same whereas those of carbon and nitrogen are 2 times of them.
So, the number of atoms of each element is:
Number of potassium, 
atoms in = atoms
Number of carbon, 
atoms in = 
atoms = 
Number of nitrogen, 
atoms in = atoms =
Total number of atoms in = Number of gold, atoms+Number of potassium, atoms +Number of carbon, atoms + Number of nitrogen, atoms
Total number of atoms in = 
Total number of atoms in = 
atoms
Hence, the total number of atoms in is 
atoms.
Answer:
A. 2.139g of KIO3
B. 26.67mL
Explanation:Please see attachment for explanation
Answer:
will this help ?
Explanation:
(108Hs) is a synthetic element, and thus a standard atomic weight cannot be given. Like all synthetic elements, it has no stable isotopes. The first isotope to be synthesized was 265Hs in 1984. There are 12 known isotopes from 263Hs to 277Hs and 1–4 isomers. The most stable isotope of hassium cannot be determined based on existing data due to uncertainty that arises from the low number of measurements. The confidence interval of half-life of 269Hs corresponding to one standard deviation (the interval is ~68.3% likely to contain the actual value) is 16 ± 6 seconds, whereas that of 270Hs is 9 ± 4 seconds. It is also possible that 277mHs is more stable than both of these, with its half-life likely being 110 ± 70 seconds, but only one event of decay of this isotope has been registered as of 2016.[1][2].
Answer:
b. 11.90 Liters
Explanation:
- The balanced equation for the mentioned reaction is:
<em>3O₂ + 4Al → 2Al₂O₃,</em>
It is clear that 3.0 moles of O₂ react with 4.0 moles of Al to produce 2.0 Al₂O₃.
- Firstly, we need to calculate the no. of moles (n) of 36.12 g of Al₂O₃:
<em>n = mass/molar mass</em> = (44.18 g)/(101.96 g/mol) = <em>0.4333 mol.</em>
<u><em>using cross multiplication:</em></u>
3.0 mol of O₂ produces → 2.0 mol of Al₂O₃.
??? mol of O₂ produces → 0.4333 mol of Al₂O₃.
<em>∴ The no. of moles of O₂ needed to produce 36.12 grams of Al₂O₃</em> = (3.0 mol)(0.4333 mol)/(2.0 mol) = <em>0.65 mol.</em>
- Now, we can find the volume of O₂ used during the experiment:
We can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm (P = 1.3 atm).
V is the volume of the gas in L (V = ??? L).
n is the no. of moles of the gas in mol (n = 0.65 mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (T = 290 K).
<em>∴ V = nRT/P </em>= (0.65 mol)(0.0821 L.atm/mol.K)(290 K)/(1.3 atm) = <em>11.9 L.</em>
<em>So, the right choice is: b. 11.90 Liters.</em>
The molar mass of copper is 63.55 g/mol. So, you convert grams to moles 127.08/63.55 =1.999 moles copper. Now, 1 mole = 6.022e23 atoms, so multiply # of moles by 6.022e23. 1.999 x 6.022e23= # of atoms of copper.
