Which of these is an isoelectronic series? 1) na+, k+, rb+, cs+ 2) k+, ca2+, or, s2– 3) na+, mg2+, s2–, cl– 4) li, be, b, c 5) n
ss7ja [257]
An isoelectronic series is where all of the ions listed have the same number of electrons in their atoms. When an atom has net charge of zero or neutral, it has equal number of protons and electrons. Hence, it means that the atomic number = no. of protons = no. of electrons. If these atoms become ions, they gain a net charge of + or -. Positive ions are cations. This means that they readily GIVE UP electrons, whereas negative ions (anions) readily ACCEPT electrons. So, to know which of these are isoelectronic, let's establish first the number of electron in a neutral atom from the periodic table:
Na=11; K=19; Rb=37; Cs = 55; Ca=20; S=16; Mg=12; Li=3; Be=4; B=5; C=6
A. Na⁺: 11-1 = 10 electrons
K⁺: 19 - 1 = 18 electrons
Rb⁺: 37-1 = 36 electrons
B. K⁺: 19 - 1 = 18 electrons
Ca²⁺: 20 - 2 = 18 electrons
S²⁻: 16 +2 = 18 electrons
C. Na⁺: 11-1 = 10 electrons
Mg²⁺: 12 - 2 = 10 electrons
S²⁻: 16 +2 = 18 electrons
D. Li=3 electrons
Be=4 electrons
B=5 electrons
C=6 electrons
The answer is letter B.
Answer is: the percent by mass of NaHCO₃ is 2,43%.
m(NaHCO₃) = 10 g.
V(H₂O) = 400 ml.
d(H₂O) = 1 g/ml.
m(H₂O) = V(H₂O) · d(H₂O).
m(H₂O) = 400 ml · 1 g/ml.
m(H₂O) = 400 g.
m(solution) = m(H₂O) + m(NaHCO₃).
m(solution) = 400 g + 10 g.
m(solution) = 410 g.
ω(NaHCO₃) = 10 g ÷ 410 g · 100%.
ω(NaHCO₃) = 2,43 %
A mixture called a solution, I think.
Answer:
Precision
Explanation:
It figures out how close all of the data numbers are.
Answer:
The element will be 
Explanation:
Given that,
Number of proton = 80
Number of neutron = 81
Number of electron = 79
We know that,
The atomic number is equal to the number of proton.
So, the atomic number is 80.
According to atomic number,
The element will be mercury.
We need to calculate the atomic mass
Using formula of atomic mass
Put the value into the formula
We need to find the element
Using atomic mass and atomic number
So, the element will be
Put the value of A and Z
Hence, The element will be
