Answer: The ratio of atoms in calcium bicarbonate ; Ca : H : C : O = 1:2:2:6.
The ratio of atoms in lithium sulfide; Li : S = 2 : 1
Explanation:
In calcium bicarbonate:
In a molecular formula of calcium carbonate there are:
Number of Calcium atoms = 1
Number of Hydrogen atom = 1 × 2 = 2
Number of Carbon atoms = 1 × 2 = 2
Number of Oxygen atoms = 3 × 2 = 6
So, Ca : H : C : O = 1 : 2 : 2 : 6
In lithium sulfide :
In a molecular formula of lithium sulfide there are:
Number of Lithium atoms = 1 × 2 = 2
Number of Sulfur atoms = 1
So, the Li : S = 2 : 1
You have to calculate the oxidation estates of the atoms in each compound.
I will start with K2Cr2O7 because I believe that Cr is the best candidate to reduce its oxidation number in 3 units.
In K2Cr2O7:
- K has oxidation state of 1+, then K2 has a charge of 2* (1+) = 2+.
- O has oxidation state of 2*, then O7 has a charge of 7* (2-) = 14-.
That makes that Cr2 has charge of 14 - 2 = +12, so each Cr has +12/2 = +6 oxidation state.
In Cr2O3:
- O has oxidation state of 2-, then O3 has charge 3 * (2-) = - 6
- Then, Cr2 has charge 6+, and each Cr has charge 6+ / 2 = 3+.
So, we have seen that Cr reduced its oxidation state in 3 units, from 6+ to 3+.
Answer: Cr has a change in oxidation number of - 3.
Answer:
71.7 L
Explanation:
Using the ideal gas equation;
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/Kmol)
T = temperature (K)
According to the information provided in this question;
P = 1 atm (STP)
V = ?
n = 3.2mol
T = 273K (STP)
Using PV = nRT
V = nRT/P
V = 3.2 × 0.0821 × 273/1
V = 71.7 L
Answer:
Electrons- 95
Protons- 95
Neutrons-146
Explanation:
An atoms is made up of three fundamental particles; electrons, protons and neutrons,
Americium belongs to the f block in the periodic table. It is an actinide element.
An atom of Am-241 contains 95 protons, 95 electrons and 146 neutrons.
