Answer:
6.1×10^8
Explanation:
The reaction is;
Sn^2+(aq) + Cd(s) -----> Sn(s) + Cd^2+(aq)
E°cell = E°cathode - E°anode
E°cathode= -0.14 V
E°anode= -0.40 V
E°cell = -0.14-(-0.40)
E°cell= -0.14+0.40
E°cell= 0.26 V
But
E°cell= 0.0592/n log K
E°cell= 0.0592/2 log K
0.26= 0.0296log K
log K = 0.26/0.0296
log K= 8.7838
K= Antilog (8.7838)
K= 6.1×10^8
Answer:
The mass of
4.6
×
10
24
atoms of silver is approximately 820 g.
Explanation:
In order to determine the mass of a given number of atoms of an element, identify the equalities between moles of the element and atoms of the element, and between moles of the element and its molar mass.
1
mole atoms Ag=6.022xx10
23
atoms Ag
Molar mass of Ag =#"107.87 g/mol"#
Multiply the given atoms of silver by
1
mol Ag
6.022
×
23
atoms Ag
. Then multiply times the molar mass of silver.
4.6
×
10
24
atoms Ag
×
1
mol Ag
6.022
×
10
23
atoms Ag
×
107.87
g Ag
1
mol Ag
=
820 g Ag
Answer:
1) acetylide
2) enol
3) aldehydes
4) tautomers
5) alkynes
6) Hydroboration
7) Keto
8) methyl ketones
Explanation:
Acetylide anions (R-C≡C^-) is a strong nucleophile. Being a strong nucleophile, we can use it to open up an epoxide ring by SN2 mechanism. The attack of the acetylide ion occurs from the backside of the epoxide ring. It must attack at the less substituted side of the epoxide.
Oxomercuration of alkynes and hydroboration of alkynes are similar reactions in that they both yield carbonyl compounds that often exhibit keto-enol tautomerism.
The equilibrium position may lie towards the Keto form of the compound. Usually, if terminal alkynes are used, the product of the reaction is a methyl ketone.
Given :
An atom of a certain element has 12 protons, 14 neutrons, and 2 valence electrons.
To Find :
The name of this element.
Solution :
We know, element with atomic number ( number of protons ) 12 and valance electrons is Magnesium.
Now, isotope of magnesium of proton with 14 neutrons is Mg-16 .
Therefore, name of element is Mg-16 .
Hence, this is the required solution.
Answer:
1 temperature 2 weather 3 the stratosphere 4 decreases 5 increases
Explanation:
