Answer:
5.96 g/cm^3
Explanation:
Corner atom = 1/8
Atoms in center = 1
Atoms in face of the cube= 1/2
Molar mass of V = 50.94 g/mol <em>(from period table)</em>
1 mole = 6.02x10^23
<em>In BCC unit cell:</em>
(8 x 1/8)+ 1=2 per 1 unit cell
<em>Mass: </em>2(50.94g)/6.02x10^23 = 1.69x10^-22 g/unit cell
305pm=(305x10^-12m÷10^-2m) x (1mL÷1cm^3)
= 2.837 x 10^-23 mL
<em>1pm=10^-12m</em>
<em>1cm=10^-2m</em>
<em>1mL=1cm^3</em>
<em></em>
density=mass/volume
density of V = 1.69x10^-22g÷2.837x10^-23mL
=5.957g/mL
=5.96g/cm^3
Answer:
With regards to the question and equation, an increase in volume will shift the equilibrium position to the left.
Explanation:
As a result of the existence of more moles within the reacting elements or reactants, a corresponding increase in volume will result to a shift of the equilibrium position to the left in order to favor the reactants. on the other hand, a decrease will shift equilibrium position to the right favoring the reaction with fewer molecules
I think Batteries you can connect a batteries to a circuit and reaction between chericals
Answer:
B.
Explanation:
pH = 5.1 has an acid
A. H2O(L) - neutral
B. HC2H3O2(AQ) - acid pH<7
C. Ca(OH)2(AQ) - base pH>7
D. NH3(AQ)- base pH>7
<u>Answer:</u> The standard electrode potential of the cell is 0.62 V.
<u>Explanation:</u>
For the given chemical equation:
The substance having highest positive potential will always get reduced and will undergo reduction reaction.
<u>Oxidation half reaction:</u> ( × 3 )
<u>Reduction half reaction:</u> ( × 2 )
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
To calculate the of the reaction, we use the equation:
Hence, the standard electrode potential of the cell is 0.62 V.