Answer:
Answer in explanation
Explanation:
Argon has 18 electrons. So to get the element in question, we only need to add 18 to the number of the filled electrons.
a. Germanium, atomic number 32
Other group members:
Silicon Si , Carbon C , Tin Sn , Lead Pb and Flerovium Fl
b. Cobalt , atomic number 27
Other group members:
Rhodium Rh , Iridium Ir and Meitnerium Mt
c. Technetium , atomic number 43
Krypton is element 36
Other group members are :
Manganese Mn , Rhenium Re and Bohrium Bh
Explanation:
1 literThe total of water is equal to 1000.0 g of water
we need to find the molality of a solution containing 10.0 g of dissolved in Na₂S0₄1000.0 g of water
1. For that find the molar mass
Na: 2 x 22.99= 45.98
S: 32.07
O: 4 x 16= 64
The total molar mass is 142.05
We have to find the number of moles, y
To find the number of moles divide 10.0g by 142.05 g/mol.
So the number of moles is 0.0704 moles.
For the molarity, you need the number of moles divided by the volume. So, 0.0704 mol/1 L.
The molarity would end up being 0.0704 M
The molality of a solution containing 10.0 g of Na2SO4 dissolved in 1000.0 g of water is 0.0704 Mliter
Answer: After 4710 seconds, 1/8 of the compound will be left
Explanation:
Using the formulae
Nt/No = (1/2)^t/t1/2
Where
N= amount of the compound present at time t
No= amount of compound present at time t=0
t= time taken for N molecules of the compound to remain = 4710 seconds
t1/2 = half-life of compound = 1570 seconds
Plugging in the values, we have
Nt/No = (1/2)^(4710s/1570s)
Nt/No = (1/2)^3
Nt/No= 1/8
Therefore after 4710 seconds, 1/8 molecules of the compound will be left
Answer is: concentratio of H₃O⁺ ions is 4.2·10⁻³ M.<span>
Chemical reaction: HCOOH(aq) + H</span>₂O(l) ⇄ HCOO⁻(aq) + H₃O⁺(aq).<span>
c(HCOOH) = 0,1 M.
[</span>H₃O⁺] = [HCOO⁻] = x.<span>
[HCOOH] = 0,1 M - x.
</span>Ka = [H₃O⁺] · [HCOO⁻] / [HCOOH].
0,00018 = x² / (0,1 M - x).<span>
Solve quadratic equation: x = </span>[H₃O⁺] = 0,0042 M.
From the reduction standard potentials;
The emf of Zinc = -0.76 V
and the emf of Aluminium = -1.66 V
In a galvanic cell the component with lower standard reduction potential gets oxidized and that it is added to the anode compartment.
Therefore. the voltage of a galvanic cell made with zinc and aluminium will be;
Voltage =Ered- Eoxd
= -0.76 - (-1.66)
= 0.9 V
