Answer:
None are empirical formulas
Explanation:
All are actual compounds. An example of an empirical formula could be CH2O, the empirical formula for carbohydrates like glucose (C6H12O6).
Answer:
bombarding it with an energetic particle
Explanation: nuclear reaction, a change in the identity or characteristics of an atomic nucleus, induced by bombarding it with an energetic particle. The bombarding particle may be an alpha particle, a gamma-ray photon, a neutron, a proton, or a heavy-ion.
5.75 Grams per cm^3
You do mass divided by volume
The average atomic mass of your mixture is 1.03 u
.
The average atomic mass of H is the weighted average of the atomic masses of its isotopes.
We multiply the atomic mass of each isotope by a number representing its relative importance (i.e., its % abundance).
Thus,
0.99 × 1.01 u = 0.998 u
0.002 × 2.01 u = 0.004 u
0.008 × 3.02 u = <u>0.024 u</u>
TOTAL = 1.03 u
<u>61.25 grams</u> of CO can be formed from 35 grams of oxygen.
The molecular mass of oxygen is <u>16 gmol⁻¹</u>
The molecular mass of carbon monoxide is<u> 28 gmol⁻¹</u>
Explanation:
The molar mass of carbon monoxide is molar mass of C added to that of O;
12 + 16 = 28
= 28g/mol
The molar mass of oxygen is 16 g/mol while that of oxygen gas (O₂) is 32 g/mol
Since the ration oxygen to carbon monoxide is 1: 2 moles, we begin to find out how many moles of carbon monoxide are formed by 35 g of oxygen;
35/32 * 2
= 70/32 moles
Then multiply by the molar mass of carbon monoxide;
70/32 * 28
= 61.25 g