<u>Reactivity</u><u> </u><u>series</u><u>:</u><u>-</u>
#1
Ag is found as too less reactive so
#2
No element in universe can displace potassium (K) from its solution as it's most reactive .
#3
Fe is also a reactive metal.
- Only Cupper is less reactive than it
So
#4
- Only Calcium can displace Mg
#5
- Mg and Ni can do ,Aurum(Gold) is least reactive so can't
#6
#7
#8
#9
#10
#11
As Ca is third most reactive metal
#12
Answer:
A = 674.33mmHg
B = 0.385atm
Explanation:
Both question A and B requires the application of pressure law which states that the pressure of a fixed mass of gas is directly proportional to its temperature provided that volume is kept constant.
Mathematically,
P = kT, k = P / T
P1 / T1 = P2 / T2 = P3 / T3 =.......= Pn/Tn
A)
Data:
P1 = 799mmHg
T1 = 50°C = (50 + 273.15) = 323.15K
P2 = ?
T2 = 273.15K
P1 / T1 = P2 / T2
Solve for P2
P2 = (P1 × T2) / T1
P2 = (799 × 273.15) / 323.15
P2 = 674.37mmHg
The final pressure is 674.37mmHg
B)
P1 = 0.470atm
T1 = 60°C = (60 + 273.15)K = 333.15K
P2 = ?
T2 = 273.15K
P1 / T1 = P2 / T2
Solve for P2,
P2 = (P1 × T2) / T1
P2 = (0.470 × 273.15) / 333.15
P2 = 0.385atm
The final pressure is 0.385atm
Answer:
The density in g/cm3 is 1.9333(put a line above the 3's to show it's repeating).
And no it is not gold because the density of gold in g/cm3 is 19.3.
Explanation:
Divide the mass by the volume in order to get an object's Density.
so, 23.2 divided by 1.20 mm3
=1.93333333 g/cm3
That is the density then you find out the density of gold in grams per cubic centimeter, that is 19.3 g/cm3
For example, gold will always have a density of 19.3 g/cm3; if a mineral has a density other than that, it isn't gold.
Answer:
The chemical formula is NaClO and consists of one atom of sodium (Na), one atom of chlorine (Cl) and one atom of oxygen (O).
Molecular Weight/ Molar Mass: 74.44 g/mol
Density: 1.11 g/cm³
Explanation:
Answer:
The correct option is;
It is always necessary to include a Roman numeral after the symbol of the metal
Explanation:
The transition metals can form ionic compounds with other elements by giving different number of electrons such that the transition metals can combine to form compounds in which they have different oxidation states
Therefore, in a compound formed by a transition metal, the value of the transition metal's valency or oxidation state in the compound is indicated by the inclusion of an equivalent Roman numeral after the transition metal in the name of the chemical formula of the compound
